Plus Two – Page 32 – HSSLive Guru

Plus Two Computer Application Chapter Wise Questions and Answers Chapter 4 Web Technology

April 13, 2021February 17, 2020 by Prasanna

Students can Download Chapter 4 Web Technology Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 4 Web Technology

Plus Two Computer Application Web Technology One Mark Questions and Answers

Question 1.
Specify an attribute of HTML tag.
Answer:
DIR: Indicates documents direction. It can take values rtl or Itr.
Eg. <HTML DIR= “rtl”> This specify that document is to read from right to left.

Question 2.
Mention the default value of size attribute of <BASEFONT>tag.
Answer:
3.

Question 3.
Name the tag which has ‘Noshade’ attribute.
Answer:
<HR> Tag

Question 4.
Maximum possible value of the size attribute of the <BASEFONT>tag?
Answer:
7.

Question 5.
Salim developed a personal website. In which he has to create an e-mail link. Can you suggest the protocol used to achieve this task?
Answer:
Mailto:
Eg. <A href= “Mailto:[email protected]” >Mail to me</A>.

Question 6.
You want to connect your webpage to the web portal www.yahoo.com. Mention the tag that can be used for this.
Answer:
<A> Anchor Tag
Eg. <A href=”www.yahoo.com”>Yahoo</A>

Question 7.
Specify the main attribute of <IMG> tag used to include an image file in web page.
Answer:
SRC. It specify the name of the image file to be included in the page.
Eg. <IMG SRC=“C:/home,Jpg”>

Question 8.
Select the attribute associated with <IMG> tag from the following: (Name, size, Align, value)
Answer:
Align.

Question 9.
Write HTML code forgiving hyper link in webpage.
Answer:
<A HREF= “page2.htm”>Page2</A>

Question 10.
The default colour of Vlink is______.
(a) Blue
(b) Green
(c) Red
(d) Yellow
Answer:
(c) Red

Question 11.
The default colour of A link is______.
(a) Blue
(b) Green
(c) Red
(d) Yellow
Answer:
(b) Green

Question 12.
Saritha is assigned with a task of writing explana¬tory notes in an HTML code. Which tag she can utilize.
Answer:
Explanatory notes can be given using <comment> tag. Using two mark up elements. as<!—and →

Question 13.
Pick the odd man out.
(a) BODY
(b) HTML
(c) CENTER
(d) ALIGN
Answer:
(d) ALIGN, all others are tag.

Question 14.
Your school has a web site, www.myschool.org. You want to create a link to this site from your website. Write the code for implementing this.
Answer:
<A href=”www. myschool.org”>My School</A>

Question 15.
HTML was developed by_______.
Answer:
Tim Berners-Lee.

Question 16.
What are the two major sections of an HTML document?
Answer:
Head section and Body section.

Question 17.
An HTML file is saved with______extension.
(a) .vbp
(b) .mdb
(c) .htm
(d) .fnm
Answer:
(c) .htm or .html

Question 18.
The software used to view web page is______.
(a) Notepad
(b) Web Browser
(c) Web server
(d) Web Editor
Answer:
(b) Web Browser

Question 19.
The default alignment of image in HTML is______.
(a) Left
(b) Right
(c) Center
(d) Inline with content
Answer:
(a) Left

Question 20.
______is an alternative for centralizing a paragraph other than using tag
(a) <ALIGN>
(b) <C>
(c) <CENTER>
(d) <CENTRE
Answer:
(d) <CENTER>

Question 21.
_______attribute of <A> tag specifies the URL of the hyper linked document.
(a) Name
(b) Target
(c) HREF
(d) SRC
Answer:
(c) HREF

Question 22.
What does HTML stands for?
Answer:
HyperText Markup Language.

Question 23.
Communication on web can be classified into_____and______.
Answer:
Client to Server and Server to Server

Question 24.
The protocol which is responsible for splitting the data into smaller packets is______.
Answer:
TCP

Question 25.
The protocol which is responsible for the routing of data packets through the correct destination is____.
Answer:
IP (Internet Protocol)

Question 26.
TCP / IP stands for_____.
Answer:
Transmission Control Protocol/lntemet Protocal.

Question 27.
Identify the protocol responsible fore-mail commu-nication.
(a) DNS
(b) HTTP
(c) TCP/IP
(d) SMTP
Answer:
(d) SMTP

Question 28.
In server to server communication, authentication is done with help of______.
(a) HTTP
(b) Digital certificate
(c) Client
(d) DNS
Answer:
(b) Digital certificate

Question 29.
Which server acts between merchant server and bank server for transferring data in encrypted format?
Answer:
Payment Gateway.

Question 30.
Identify the name of a place where servers and networking systems are placed with high security,
(a) Head office
(b) DNS
(c) Data centre
(d) IIS
Answer:
(c) Data centre

Question 31.
Identify the port number which request for the service of sending e-mail communication.
(a) 22
(b) 25
(c) 53
(d) 80
Answer:
(b) 25

Question 32.
The IP address corresponding to a domain name is present in____server.
Answer:
DNS

Question 33.
Programs embedded in HTML documents are termed as____.
Answer:
Scripts

Question 34.
Running of ______scripts can be blocked by the user.
(a) Client side
(b) Server side
(c) Both client side and server side
(d) None of these
Answer:
(a) Client side

Question 35.
A platform-independent server-side scripting language is_____.
Answer:
PHP

Question 36.
Which among the following tools is used for easy formatting and defining style of a document written in HTML?
(a) Ajax
(b) CSS
(c) JSP
(d) JavaScript
Answer:
(b) CSS

Question 37.
Pick the Odd one from the following list and give reason. (IMG, FONT, BR, ALIGN, PRE)
Answer:
ALIGN which is an attribute, all others are tags.

Question 38.
Choose the correct HTML statement to display an image with file name “kerala.jpg” as the background of the web page.
(a) <IMG src=”kerala.jpg”>
(b) <BODY src=”kerala.jpg”>
(c) <BODY bgcolor=”kerala.jpg”>
(d) <BODY background=”kerala.jpg”>
Answer:
(d) <BODYbackground=”kerala.jpg”>

Question 39.
Two ofthe following HTMLtags have same attribute Align’. Identify them. (<IMG>, <MARQUEE>, , , <BODY>)
Answer:
<IMG> AND

Question 40.
Identify the correct HTML statement to draw a hori-zontal line with half the width of the screen.
(a) <HR width=“50%” size= “3″>
(b) <HR length=“50%” size= “3″>
(c) <HR size= “50%” width= “3″>
(d) <HR width= “50%” length= “3″>
Answer:
(a) <HR width=“50%” size= “3″>

Question 41.
A student wants to display a poem in a web page just like as he entered in the text editor. Which tag in HTML will help him?
Answer:
<PRE>tag

Question 42.
A student created a webpage about his school. The school name is displayed in the page. He wanted to change the style, colour, and size of the school name. Identify the most appropriate tag in HTML needed for that.
Answer:
 tag

Plus Two Computer Application Web Technology Two Mark Questions and Answers

Question 1.
A student developed a web page about India. He wanted to display a scrolling text moving from right side to left side with a background colour blue. The text is “I Love My Country”.

Identify the tag needed for it.
Write the HTML statement to do the task.

Answer:

<MARQUEE>
<MARQUEE direction-’left” bgcolor=”blue” > I Love My country </MARQUEE>

Question 2.
Write HTML statement for displaying the following text items:

A₂B³
A>B

Answer:

A 2 B 3
A>

Question 3.
Two HTML tags are given. They are <BODY> and . Identify and write the attribute of each from the following list. (Size, Text, Link, Bgcolor, Color).
Answer:
The attributes of <BODY> tag are Text, Link and Bgcolor The attributes of tag are size and color.

Question 4.
Write the use of Border and Aft attribute of <IMG> tag.
Answer:
Border: This attribute is used forgiving border to an image.

Alt: This attribute issued forgiving an alternate text. When there is no image in the specified location or the browser doesn’t support the image them this text will be displayed.

Question 5.
When a client send request to a server, the server must know which service is demanded by the client.

How does the server identify the type of service requested?
Write the name of any one of the services in the web server.

Answer:

Port number
Any service like FTP, SMTP, HTTP, etc.

Question 6.
Following are steps for searching the IP address of a domain name by a browser. Rearrange them in proper order.
(a) Look in the local memory of ISP
(b) Look in the DNS servers starting from the root server
(c) Look in the local memory of brower
(d) Look in the local memory of Operating System
Answer:
Correct order is c, d, a, b.

Question 7.
Categorise the following tags into Containertags and empty tags.
, ,<A> ,<FRAME>, <FRAMESET>,<LI>, <HR>
Answer:
1. Container tags:
,<A> <FRAMESET>

2. Empty tags:
 ,<HR>,<LI>, <FRAME>.

Question 8.
Write the HTML statement to get the following output.

Commerce
Humanities

Answer:

Commerce
<l>Humanities</l>

Question 9.
Write the HTML statement to get the following output.

H₂S0₄
a² + b²
Computer

Answer:

H2SO4
a2 + b 2
Computer

Question 10.
Write HTML code to display as follows. The <IMG> tag is used for placing images.
Answer:
The & lt IMG &gt tag is used for placing images.

Question 11.
Name some browsers.
Answer:

Internet Explorer
Netscape Navigator
Opera
Eudora

Question 12.
What is a website?
Answer:
A website is a collection of webpages. A webpage is created by using HTML tags.

Question 13.
What is the role of attributes in an HTML tag?
Answer:
Attributes are parameters for providing additional information within a tag. Attribute values specify certain characteristics of the tag.
Eg:

Question 14.
What is an HTML?
Answer:
HTML stands for Hyper Text Markup Language. It is used to create webpages. It has two types of tags empty and container. The important thing we have to remember in container tag is first opened tag must be closed last.

Question 15.
Name the main attributes of <HR> tag.
Answer:

Size: It specifies the line thickness.
Width: It specifies the length
Noshade: It specifies no shade is given for the line.
Color: It specifies the color.

Question 16.
Mary wants to display her name in various headings. Name the heading tags available in HTML.
Answer:
Heading can be given in six levels from <H1> to <H6>.The tag <H1 > produces big heading. The tag <H6> produces small heading. The heading size reduce from <H1> to <H6> sequentially.

Question 17.
Differentiate between and<BASEFONT>tags.
Answer:
<BASEFONT>tag sets the normal font for the entire document text. The font specified will be taken as the default font for the entire document. The main attributes are size, face, and color.

The tag defines the font characteristics of the text enclosed. tag change the font property of the text enclosed within and whereas <BaseForrt> tag specifies the default font characteristics.

Question 18.
Explain the different types of hyperlinks.
Answer:
The two types of Hyper Links are

External Hyper Link: This is used two connect the locations of two different web pages
Internal Hyper Link: This is used to connect the different locations of the same web page.

Question 19.
Write the HTML code to display the following list:

Form
TextBox
Label
Command Button

Answer:
Plus Two Computer Application Web Technology Two Mark Questions and Answers 1

Question 20.
The body section forms the content displayed in the browser window. Briefly explain any four attributes in the BODY tag.
Answer:

Bgcolor – It is used to set background colour.
Background – It is used to set a background picture.
Text – it is used to set the foreground colour.
Left margin – It is used to set the left margin.

Question 21.
HTML has the facility to provide External links as well as Internal links.

Which tag is used to include an External link?
How will you construct an Internal link?

Answer:
1. Anchor Tag i.e. <a href=”filename”>

2. Internal link is used to link two places of the same web page
Plus Two Computer Application Web Technology Two Mark Questions and Answers 2

Question 22.
Categorise the following tags in HTML and write the criterian for the categorisation.
 , , <BODY>, , <HR>, <IMG>
Answer:

Empty_tag	Container tag
<BR>	<P>
<HR>	<BODY>
<IMG>	<B>

Question 23.
Differentiate empty tags and container tag with example.
Answer:
1. Empty tags: It has opening tag only, no closing tag
Eg: <hr>, etc.

2. Container tag: It as both opening and closing tag.
Eg: <html> </html>
<body></body>etc.

Question 24.
Write True or False

Text is an attribute of <BODY> tag to insert a text matter in the web page.
tagfunctionssimilarto<l>tag.

Answer:

False
True.

Plus Two Computer Application Web Technology Three Mark Questions and Answers

Question 1.
Write and explain any four text formatting tags in HTML.
Answer:
Text formatting tags are given below.

– This tag is based to make the text Bold Eg: Computer application
<l>: This tag is used to make the text in italics eg: <l> computer aplication</l>
:This tag is used to underline the text eg: computer aplication
<S>: This tag is used for striking out the text eg: <S> computer aplication</S>

Question 2.
Match the following.
Plus Two Computer Application Web Technology Three Mark Questions and Answers 4
Answer:
Correct match as given below
<H2> -Heading-Align
<MARQUEE> – Scrolling text – Bgcolor
<IMG> – Inserting picture – Src.

Question 3.
Briefly explain the use of tags <Q>, <PRE> and <ADDRESS> tags.
Answer:

<Q>: It is used to give text within double-quotes.
<PRE>: This tag is used to display the content as we entered in the text editor.
<ADDRESS>: This tag is used to provide information of the author or owner.

Question 4.
Write a HTML code to develop a web page about Kerala state as shown below:

KERALA STATE

God’s Own Country

Capital: Thiruvananthapuram

The specifications for the page are:

The main heading must be of bigger in size, centralised and bold.
Sub headings must be lesser size than main heading and in italics.
There should be a picture at the center of the page with file name“tree.jpg”.
The background colour of the page must me blue.

Answer:
Plus Two Computer Application Web Technology Three Mark Questions and Answers 5

Question 5.
PHP is a popular scripting language.

Write whether it is client side or server side.
Write a brief note on PHP.

Answer:

Server side
PHP (PHP Hypertext Preprocessor)
- It is an open-source, general-purpose scripting language.
- It is a server side scripting language
- Introduced by Rasmus Lendorf
- A PHP file with extension .php
- It support data base programming the default DBMS is MySQL
- It is platform-independent
- PHP interpreter in Linux is LAMP(Linux, Apache, MySQL, PHP)

Question 6.
How client side scripting differs from server side scripting?
Answer:
Following are the differences

Client Side Scripting	Server Side Scripting
Script is copied to client browser	to the web server
Executed by the client	Executed by the server and result is get back to the browser window
Used for Client level validation	Connect to the database in the server
It is possible to block by the user	Cannot possible
Client side scripts depends the type and version of the browser	It does not depend the type and version of the browser

Question 7.
Briefly explain the two types of communication on the web.
Answer:
The two types of communication on the web are given below.
1. Client to Web server communication:
This communication is carried out between client to the web server (shopping site). The technology used to protect data that are transferred from client to web server is https.

2. Web sever to web server communication:
This communication is usually carried out between web .sever (seller) to another web server (normally bank). For the safe transaction. Digital certificate issued by third party web sites are used.

Question 8.
Compare static and dynamic web pages.
Answer:

Static web pages	Dynamic web pages
Content and layout is fixed	Content and layout is changed frequently
Never use database	Data base is used
Run by browser	It runs on the server and result get back to the client(browser)
Easy to develop	Not at all easy

Question 9.
Differentiate Empty tag and Containertag.
Answer:
There are two types of tags, opening tag, and closing tag

Empty tag: It has only opening tag and no closing tag
Eg: , <hr>,..
Container tag: It has both opening and closing tags. These tag contains some text data Eg:<html>,<head>,<body>,….

Question 10.
Create a webpage using HTML to display the following message. ‘The symbol H₂O represents water’.
Answer:
Plus Two Computer Application Web Technology Three Mark Questions and Answers 6

Question 11.
While designing a webpage Neena wants to explain the meaning of each step. Can you help her?
Answer:
Neena can use Comments while writing the code. It is a good programming practice. Comments improve readability. It is not the part of a program. Comments
<!—and — >
Plus Two Computer Application Web Technology Three Mark Questions and Answers 7

Question 12.
Complete the following table.

RGB code	Colour Name
#0000FF	_________
________	Yellow
#FF0000	_________
#00FF00	_________
________	White
________	Black

Answer:

RGB code	Colour Name
#0000FF	Blue
#FFFF00	Yellow
#FF0000	Red
#00FF00	Green
#FFFFFF	White
#000000	Black

Question 13.
Give HTML tag to display the sentence “WELCOME TO HTML” as centralized heading, having red colour.
Answer:
Plus Two Computer Application Web Technology Three Mark Questions and Answers 8

Question 14.
Explain the Main attributes of font tag.
Answer:
tag defines the font properties of text enclosed
The main attributes are

Face-This specifies the type of font.
Color-This specifies the colour of the text enclosed
Size- This specifies the font size

Eg.

Question 15.
John visited a website, it is found that when clicking on a particular text the browser open a new web page. Name the feature and Identify the tag used for this purpose. Write the HTML code to link to a file name “main.html”.
Answer:

Hyper Linking. By clicking on hypertext we can see or go to other webpages or to other section of same document.
<A> tag is used
<Ahref=”main.html”>Main</A>

Question 16.
Antony visited his school website, he could not see the picture of his school instead of that there is a text message “Your browser could not support images”. Why it is happened and write the html code for this.
Answer:
Because the browser he used cannot load the image. He can use Alt attribute of IMG Tag.
<IMG SRC =” D:\school.jpg” Alt=“Your browser could not support images”>

Question 17.
Suppose you want to display a picture named school.jpg located in the ‘Photos’ sub directory of the directory ‘My documents’ of C drive in your web page.

Name the tag used for this purpose.
Write HTML code for the web page.

Answer:
1. <IMG>tag. It is used to display images in webpage

2. <HEAD><TITLE>
</HEAD>
<Body>
<IMG SRC= ‘C;/My documents/Photos/School.jpg’>
</Body>
<html>

Plus Two Computer Application Web Technology Five Mark Questions and Answers

Question 1.
Explain the main list tags in HTML?
Answer:
HTML provides three basic types of lists-unordered, ordered and definition list.
1. Unordered list:
Unordered list arranges the list items with bullet symbols in front. <UL> and </UL> tag encloses an Unondered list. List items are specified by <LI> tag. The tag <UL> can take values square, circle or disc. The default type is disc.
Eg : <UL>
<LI>COMPUTER
<LI>BIOLOGY
</UL>

2. Ordered list:
In Ordered list, the list items are numbered in sequence. <OL> and </OL> tag encloses an Or-dered list. List items are specified by <LI> tag. The tag <OL> can take values as follows

type = 1 for 1, 2, 3,….
type = i for i, ii, iii,……
type = I for I, II, III,……
type = a for a, b, c,…
type = A for A, B, C,…..

Eg :<OL>
<LI>COMPUTER
<LI>BIOLOGY
</OL>

3. Definition List:
It is formed by a group of definitions and their descriptions. No bullet symbol or number is provided for the list items. The <DL> and </DL> tags enclose the definition list. The <DT> tag contains the definition term and <DD> tag specifies the description.
Eg: <DL>
<DT>Eeheque
<DD>Electronic cheque
</DL>

Question 2.
Explain the use of <BODY> tag and list any four of its attributes
Answer:
Web page contents are given int the body section. Attributes of body tag are:

BGCOLOR-Specifies background color for the document Body
Eg. <BODY BGCOLOR= “RED”>
BACKGROUND -Sets the image as background for the document body
Eg. <BODY BACKGROUNG= “C:\result.jpg”>
TEXT-Specifies the color of the text content of the page
Eg. <BODYTEXT= “Red”>
LINK- Specifies colour of the hyperlinks that are not visited by the user
ALINK-Specifies the colour of hyperlinks
VLINK-Specifies the color of hyperlinks which are already visited by the viewer.
Eg. < BODY ALINK= “Cyan” LINK=” Magenta” VLINK= “Orange’’> *
Left margin and Right margin-Sets margin from left and top of the document window.

Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions

April 13, 2021February 15, 2020 by Prasanna

Students can Download Chapter 3 Functions Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions

Plus Two Computer Application Functions One Mark Questions and Answers

Question 1.
To read a single character for gender i.e. ’rri’ or ’f’.___function is used.
(a) getch()
(b) getchar()
(c) gets()
(d) getline()
Answer:
(b) getchar()

Question 2.
To use getchar(), putchar(), gets() and puts(), which header file is used?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(b) cstdio

Question 3.
To use cin and cout, which header file is needed?
(a) iostream
(b) cstdio
(c) input
(d) output
Answer:
(a) iostrem

Question 4.
Predict the output of the following code snippet
#include<cstdio>
int main()
{
char name[ ] = “ADELINE”;
for(int i=0; name[i]!=’\0′;i++)
putchar(name[i]);
}
Answer:
The output is “ADELINE”.

Question 5.
From the following which is equivalent to the function getc(stdin)
(a) putchar()
(b) gets()
(c) getchar()
(d) puts()
Answer:
(c) getchar()

Question 6.
From the following which is equivalent to the function putc(ch,stdout)
(a) putchar(ch)
(b) ch=gets()
(c) ch=getchar()
(d) puts(ch)
Answer:
(a) putchar(ch)

Question 7.
To print a single character at a time which function is used?
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) putchar()

Question 8.
To read a string____function is used.
(?) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) gets()

Question 9.
To print a string_____function is used.
(a) puts()
(b) putchar()
(c) gets()
(d) getchar()
Answer:
(b) puts()

Question 10.
Consider the following code snippet
main()
{
char str[80];
gets(str);
for(int i=0,len=0;str[i]!-\0′;i++,len++);
cout<<“The length of the string is” <<len;
getch();
}
Select the equivalent for the under lined statement from the following
(a) int len= strlen(str)
(b) int len=strcmp(str)
(c) int len = strcount(str)
(d) None of these
Answer:
(a) int len= strlen(str)

Question 11.
Arjun wants to read a string with spaces from the following which is suitable
(a) cin>>
(b) cin.getline(str,80)
(c) str=getc(stdin)
(d) none of these
Answer:
(b) cin.getline(str,80)

Question 12.
State whether the following statement is true or false. The'<<‘ insertion operator stops reading a string when it encounters a space.
Answer:
True.

Question 13.
The process of dividing big programs into small programs are called____.
Answer:
Modularization.

Question 14.
The big programs are divided into smaller programs. This smaller programs are called_____.
Answer:
Functions.

Question 15.
The execution of the program begins at____function.
Answer:
main function.

Question 16.
One of the following is not involved in the creation and usage of a user defined function
(a) Define a function
(b) Declare a function
(c) invoke a function
(d) None of these
Answer:
(d) None of these

Question 17.
The default data type returned by a function is_____.
(a) float
(b) double
(c) int
(d) char
Answer:
(c) int

Question 18.
After the execution of a function, it is returned back to the main function by executing____keyword.
Answer:
return.

Question 19.
Supplying data to a function from the called function by using______.
Answer:
parameters (arguments).

Question 20.
_____keyword is used to give a value back to the called function.
Answer:
return.

Question 21.
____key word is used to specify a function returns nothing.
Answer:
void

Question 22.
One of the following is not necessary in the function declaration. What is it?
(a) name of the function
(b) return type
(c) number and type of arguments
(d) name of the parameters
Answer:
(d) name of the parameters.

Question 23.
A function declaration is also called_____.
Answer:
prototype.

Question 24.
Considerthe following declaration
int sum(int a , int b)
{
return a+b;
}
From the following which is the valid function call.
(a) n=sum(10)
(b) n=sum(10, 20)
(c) n=sum(10, 20, 30)
(d) n=sum()
Answer:
(b) n=sum(10, 20)

Question 25.
The ability to access a variable or a function from some where in a program is called_____.
Answer:
scope.

Question 26.
A variable ora function declared within a function is have_____scope.
Answer:
local.

Question 27.
A variable or a function declared out side of all the functions is have_____scope.
Answer:
global.

Question 28.
State True/False

A local variable exist till the end of the program
A global variable destroyed when the sub function terminates

Answer:

False
False

Question 29.
consider the following declaration
int x;
int main()
{
–
}
Here x is a____variable.
Answer:
global.

Question 30.
consider the following declaration
int main()
{
int x;
–
}
Here x is a_____variable.
Answer:
local.

Question 31.
______parameter is used when the function call does not supply a value for parameters.
Answer:
default.

Question 32.
Consider the following function declaration with optional (default) arguments and state legal or illegal and give the reasons

int sum(int x=10, int y, int z)
int sum(int x=10, int y=20, int z)
int sum(int x=10, int y=20, int z=30)
int sum(int x, int y=20, int z)
int sum(int x, int y=20, int z=30)
int sum(int x, int y, int z=30)
int sum(int x=10, int y, int z=30)

Answer:
There is a rule to make an argument as default argument,i.e., to set an argument with a value that must be in the order from right to left. All the arguments in the right side of an argument must be set first to make an argument as a default argument.

illegal, because y and z are not have values
illegal, because z has no value
legal
illegal, because z has no value
legal
legal
illegal, because x has a value but y has no value.

Question 33.
The parameter used to call a function is called_____.
Answer:
actual parameter.

Question 34.
The parameters appear in a function definition are_____.
Answer:
formal parameters.

Question 35.
After the distribution of answer scripts, the teacher gives the Photostat copy of the mark list to the students to check the marks. If the students make any change that do not affect the original mark list. There is a similar situation to pass the arguments to a function. What is this method?
(a) call by value
(b) call by reference
(c) call by address
(d) none of these
Answer:
(a) call by value

Question 36.
Your class teacher gives you the original mark list to check the mark. If you make any change it will affect the original mark list. There is a similar situation to pass the arguments to a function. What is this method?
(a) call by value
(b) call by reference
(c) call by function
(d) none of these
Answer:
(b) call by reference

Question 37.
Consider the following function declaration
int sum(int a, int b)
{
Body
}
Here the arguments are passed by______.
Answer:
call by value method.

Question 38.
Consider the following function declaration
int sum(int &a, int &b)
{
Body
}
Here the arguments are passed by______.
Answer:
call by reference method.

Qn. 39
A function calls it self is known as______
Answer:
recursive function.

Question 40.
Varun wants to copy a string by using strcpy() function. From the following which header file is used for this?
(a) cstdio
(b) cmath
(c) cstring
(d) cctype
Answer:
(c) cstring

Question 41.
____is a named group of statements to perform a job /task and returns a value.
Answer:
Function.

Question 42.
To use the function setw(), from the following which header file is used.
Answer:
iomanip.h

Question 43.
In his C++ program Ajith wants to accept a lengthy text of more than one line. Which function in C++ can be used in this situation.
Answer:
gets() function can be used to accept a lengthy text.

Question 44.
Choose the C++ function which can print a string.
(a) getche()
(b) putchar()
(c) getline()
(d) puts()
Answer:
(d) puts()

Question 45.
Which of the following is a console function?
(a) getline()
(b) write()
(c) put()
(d) getchar()
Answer:
(d) getchar()

Question 46.
Pick the odd one out and give reason.
(a) abs()
(b) strlen()
(c) strcmp()
(d) strcpy()
Answer:
(a) abs() – it is a mathematical function. All others are string functions.

Question 47.
Consider the following C++ statement and answer the following question:
char Word[10]=”GOOD DAY”;
Identify the correct output statement to display the string
(a) write (word);
(b) cout.write(word);
(c) cout (word);
(d) cout.write (word, 10);
Answer:
(d) cout.write(wond, 10);

Question 48.
When the number -7 is given as the argument of a predefined function in C++, it returns the value 7. Identify the function.
Answer:
abs(): This function returns the absolute value.

Question 49.
Pick out the correct statement for prototype declaration from the following and also explain the various information it contains.
(a) product (int a, int b);
(b) int product (a,b);
(c) int product (int, int);
(d) product (int, int);
Answer:
(c) int product(int, int);
This prototype specifies the return type, name of function, number, and type of arguments.

Question 50.
One among the following function prototypes is wrongly written. Identify it. Also given reason.
(a) float test (float);
(b) float test (float, int);
(c) test (float);
(d) int test (int);
Answer:
(c) test(float);
Here the prototype contains no return type.

Question 51.
A user defined function definition is given below. Choose the most appropriate function call statement from the options.
float calc(int x, float y)
{
return (x+y) / 2.0;
}
(a) calc (2, 4)
(b) calc (2.5, 4)
(c) calc (2.5, 4.5)
(d) calc (2, 4.5)
Answer:
(d) calc(2, 4.5);

Question 52.
Which of the following statements are FALSE about a local function?
(a) Declared inside a function
(b) Accessible only within the function it is declared
(c) Accessible from anywhere in the program
(d) Declared outside all other functions
Answer:
c and d are false to a local function.

Plus Two Computer Application Functions Two Mark Questions and Answers

Question 1.
In a C++ program, you forgot to include the header file iostream.h. What are the possible errors occur in that Program? Explain ?
Answer:
Proto type error. To use cin and cout the header file iostream is a must.

Question 2.
Pick the odd one out from the following and give reason.

gets()
getline()
getch()
getchar()

Answer:
getline() – It is a stream function where as the others are console functions.

Question 3.
Consider the following code snippet.
using namespace std;
int main()
{
int n;
cout<<“Enter a number”; cin>>n;
cout<<‘The number is “<<n;
}
Write down the names of the header files that must be included in this program
Answer:
Here cin and cout are used so the header file iostream must be included.

Question 4.
How does C++ support modularity in programming
Answer:
The process of converting big programs into smaller programs is known as modularisation. This small programs are called modules or sub programs or functions. C++ supports modularity in programming called functions.

Question 5.
The following assignment statement will generate a compilation error.
char str[20]; str=”Computer”
Write a correct C++ statement to perform the same task
Answer:
char str[20] = “Computer”;

char str[20];
strcpy(str,”Computer”); (The header file should be included).

Question 6.
float area(const float pi=3.1415, const float r)
{
r=10;
return pi*r;
}
Is there any problem? If yes what is it?
Answer:
There is an error. The error is , Y is a constant T must be initialised and cannot be changed during execution.

Question 7.
What are the jobs of a return statement in a program.
Answer:
In the case of a sub function a return statement helps to terminate the sub function and return back to the main function or called function. But in the case of a main function it terminates the program.

Question 8.
Match the following

(a) strcmp()	(1) cctype
(b) tolower()	(2) cstring
(c) sqrt()	(3) cstdlib
(d) abort ()	(4) cmath

Answer:
(a) 2
(b) 1
(c) 4
(d) 3

Question 9.
How to invoke a function in C++ program.
Answer:
A function can be called or invoked by providing the name of the function followed by the arguments in parenthesis Eg. sum(m,n);

Question 10.
Briefly explain constant arguments Constant arguments.
Answer:
By using the key word const we can make argument (parameter) of a function as a constant argument.
The value of the const argument cannot be modified within the function.

Question 11.
void initialise()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 1

Identify the error in the, above code and explain its reasons.
Correct the errors.

Answer:

K is a local variable in the function initialize() – It is not accessible in main()
Making the variable K as global we can correct the error.

Question 12.
List down the advantages of modular programming.
Answer:
Merits of modular programming

It reduces the size of the program
Less chance of error occurrence
Reduces programming complexity
Improves reusability

Question 13.
Some statements are given below. Analyse these statements and predict the output:
char str1 (15], str2[15];
str1[15]=” DATA”;
str2[15]=” STORAGE”;
strcat (str2, str1);
cout<<str2;
Answer:
The output is STORAGE DATA. The strings str2 and str1 are concatenated.

Question 14.
If char name [ ] = “Rajeev Kumar”; then what will be output of the following statement? cout<<strlen(name);
Answer:
The length(number of characters) is 12 including space.

Question 15.
Choose the value of ‘n’ after executing the following statements in C++.
char s1[ ]=”KIRAN”; char s2[ ]=”kiran”;
int n = strcmp (s1,s2);
cout<<n;
(a) 0
(b) >0
(c) <0
(d) None of these
Answer:
(c) <0. Here string 2, i.e. s2 is greaterthan string1 i.e. s1.
strcmp()- It is used to compare two strings and returns an integer.
Syntax: strcmp(string1 ,string2)

if it is 0 both strings are equal.
if it is greater than 0(i.e. +ve) string1 is greater than string2
if it is less than 0(i.e. -ve) string2 is greater than string1

Question 16.
C++ has a built-in function with which we get the result of 4².

Identify the name of the function.
Identify the header file for the above function.

Answer:

pow(4, 2);
The header file used is cmath.

Question 17.
Consider the following C++ statements and predict the output.
int p=isalpha(‘5’);
cout<<p;
Answer:
0.
isalpha() – To check whether a character is an alphabet or not. If the character is an alphabet it returns a value 1 otherwise it returns 0.

Question 18.
Predict the output of the following C++ statements:

cout<<toupper(‘a’);
cout<<(char) toupper(‘a’);

Answer:

It prints 65
It prints A

Question 19.
Differentiate formal arguments and actual arguments.
Answer:
The parameter used to call a function is called actual parameter. The parameters appear in a function definition are formal parameters.

Plus Two Computer Application Functions Three Mark Questions and Answers

Question 1.
Suresh wants to print his name and native place using a C++ program. The program should accept name and native place first
Name is: Suresh Kumar
Address is: Alappuzha
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 2

Question 2.
“Programming is Fun”. Write a C++ program to read a string like this in lowercase and print it in UPPER CASE. Without using toupper() library function.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 3

Question 3.
An assignment, Kumar has written a C++ program which reads a line of text and print the number of vowels in it. What will be his program code?
Answer:
# include<iostream>
# include<cstdio>
# include<cctype>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 5

Question 4.
Write a program to display the following output
A
BB
CCC
Answer:
# include<iostream>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 6

Question 5.
Distinguish getchar and gets.
Answer:
getchar is a character function but gets is a string function. The header file cstdio must be included. It reads a character from the keyboard.
Eg.
char ch;
ch=getchar();
cout<<ch;
gets is used to read a string from the key board. It reads the characters upto enter key. The header file
cstdio must be included.
char str[80];
cout<<” Enter a string”;
gets(str);

Question 6.
Write a program to check whether a string is palindrome or not. (A string is said to be palindrome if it is the same as the string constituted by reversing the characters of the original string. Eg. “MALAYALAM”, “MADAM”, “ARORA”, “DAD”, etc.
Answer:
# include<iostream>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 7

Question 7.
Explain multi character function.
Answer:
getline() and write() functions are multi character functions.
1. getline():
It reads a line of text that ends with a newline character. It reads white spaces also.
Eg.
char line[80];
cin.getline(line,80);

2. write():
It is used to display a string.
Eg.
charline[80];
dn.getline(line, 80);
cout.write(line, 80);

Question 8.
Distinguish between get() and put() functions.
Answer:
1. get() function:
get() is an input function. It is used to read a single character and it does not ignore the white spaces and newline character.
Syntax is cin.get(variable);
Eg. char ch;
cin.get(ch);

2. put() function:
put() is an output function. It is used to print a character.
Syntax is cout.put(variable);
Eg. char ch;
cin.get(ch);
cout.put(ch);

Question 9.
Write a program to read a string and print the number of consonents
Answer:
# include<iostream>
# include<cstdio>
# include<cctype>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 8

Question 10.
Write a program to read a string and print the num¬ber of spaces.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 9

Question 11.
Write a program to count the number of words in a sentence
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 11

Question 12.
Write a program to input a string and display its reversed string using console I / O functions only. For example if the input is “AND” the output should be “DNA”.
Answer:
# include<iostream>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 12

Question 13.
Write a program to input a word(say COMPUTER) and create a triangle as follows.
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 13
Answer:
# include<iostream>
# include<cstring>
using namespace std;
int main()

Question 14.
Write a program to input a line of text and display the first characters of each word. Use only console I /O functions. For example, if the input is “Save Water, Save Nature”, the output should be “SWSN”.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 15

Question 15.
Consider the following code snippet
char ch;
cout<< “Enter an alphabet”; cin>>ch;
cout<<toupper(ch);
What is the output of the above code? Give a sample output. If the above code is used in a computer that has no cctype file, how will you modify the code to get the same output?
Answer:
It reads a character and convert it into uppercase.
Eg:
Enter an alphabet: a
The output is A.
If a computer has no cctype header file the code is as follows.
char ch;
cout<< “Enter an alphabet”; cin>>ch;
if (ch>=97 && ch<<122)
cout<

Question 16.
Read the following program
# include<iostream.h>
int main()
{
cout<<sum(2, 3);
}
int sum(int x, int y)
{return (x + y);}
On compilation on the program, an error will be dis-played. Identify and explain the reason. How can you rectify the problem
Answer:
The compilation of the program starts from the first line and next line and so on( i.e. line by line). While compiling the line cout<<sum(2, 3); The compiler does not understand the word sum(2, 3) because it is not declared yet hence the error prototype required. To rectify this problem there are two methods
First method
Give the function definition just before the main function as follows.
# include<iostream>
using namespace std;
int sum(int x, int y)
{return (x+y);}
int main()
{
cout<<sum(2, 3);
}

Second Method
Give the function declaration(prototype only) in the main function as follows.
# include<iostream>
using namespace std;
int main()
{
int sum(int, int);
cout<<sum(2, 3);
}
int sum(int x, int y)
{return (x+y);}

Question 17.
Considering the following function definition;
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 16
The expected, desired output is 5! = 120
What will be the actual output of the program? It is not the same as above, why? What modification are required in the program to get the desired output.
Answer:
The output is 0! = 120
Because the address of variable ‘a’ is given to the variable ‘n’ of the function fact(call by reference method). So the function changes its value (i.e. n- -) to 0. Hence the result.

To get the desired result call the function as call by value method in this method the copy of the value of the variable ‘a’ is given to the function. So the actual value of ‘a’ will not changed. So instead of int fact(int &n) just write int fact(int n), i.e., no need of & symbol.

Question 18.
A function is defined as follows
int sum (int a, int b=2)
{return (a+b);}
Check whether each of the following function calls is correct or wrong, Justify your answer

cout<<sum(2, 3);
cout<<sum( 2);
cout<<sum();

Answer:
Here the function is declared with one optional argument. So the function call with minimum one argument is compulsory.

0 It is valid. Here a becomes 2 and b becomes 3.
It is also valid . Here a becomes 2 and b takes the default value 2.
It is not a valid call. One argument is compulsory.

Question 19.
How do two functions exchange data between them? Compare the two methods of data transfer from calling function to called function.
Answer:
There are two methods they are call by value and call by reference
1. call by value:
In call by value method, a copy of the actual parameters are passed to the formal parameters. If the function makes any change it will not affect the original value.

2. call by reference:
In call by reference method, the reference of the actual parameters are passed to the formal parameters. If the function makes any change it will affect the original value.

Question 20.
Write down the operation performed by the following statements

int l=strten(“Computer Program”);
charch [] = tolower(“My School”);
cout<<(strcmp(“High”, “Low”)>0 ?

toupper(“High”):tolower(“Low”));
Answer:

The built in function strlen find the length of the string i.e. 16 and assigns it to the variable I.
This is an error because tolower is a character function.
This is also an error because tolower and toupper are character functions.

Question 21.
A line of given length with a particular character is to be displayed. For example, ********** is a line with ten asterisks (*). Define a function to achieve this output
Answer:
#include<iostream>
using namespace std;
void line (char ch, int n)
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 17

Question 22.
Read the following function
int fib(int n)
{
if (n<3)
return 1;
else
return (fib(n-1) + fib(n-2));
}

What is the speciality of this function
How does it work ?
What will be the output of the following code?

for (int i=1; i<5; i++)
cout<<fib(i)<<‘\t’;
Answer:
1. This function is a recursive function. That means the function calls itself.

2. It works as follows
if i = 1, The function fib calls with value 1. i.e. fib(1) returns 1
if i = 2, The function fib calls with value 2. i.e. fib(2) returns 1
if i = 3, The function fib calls with value 3. i.e. fib(3) returns fib(2) + fib(1) i.e. it calls the function again. So the result is 1 + 1 = 2
if i = 4, The function fib calls with value 4. i.e. fib(4) returns fib(3) + fib(2) i.e. it calls the function again. So the result is 2 + 1 = 3

3. The output will be as follows
1 1 2 3.

Question 23.
Explain scope rules of functions and variables in a C++ program
Answer:
1. Local variable or function:
A variable or function declared inside a function is called local variable or function. This cannot be accessed by the outside of the function.
Eg.
main()
{
int k;//local variable ,
cout<<sum(a,b); // local function
}

2. Global variable or function:
A variable or function declared out side of a function is called global variable or function. This can be accessed by any statements.
Eg.
int k; // global variable
int sum(inta, int b); //global function
main()
{
}

Question 24.
Briefly explain default arguments.
Answer:
A default value can be set for a parameter(argument) of a function. When the user does not give a value the function will take the default value. An important thing remember is an argument cannot have a default value unless all arguments on its right side must have default value.

Functions with valid default arguments are given below

float area(int x, int y, int z=30);
float area(int x, int y=20, int z=30);
float area(int x=10, int y=20, int z=30);

Functions with invalid default arguments are given below

float area(int x=10, int y, int z);
float area(intx, inty=20, int z);
float area(int x=10, int y=20, int z);

Question 25.
Write a program to read a character and check whether it is alphabet or not. If it is an alphabet check whether it is upper case or lower case?
Answer:
#include<iostream>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 18

Question 26.
Write a program to read 2 strings and join them
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 19

Question 27.
Write a program to read 2 strings and compare it.
Answer:
# include<iostream>
# include<cstdio>
# include<cstring>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 20

Question 28.
Write a program to read a string and display the number of alphabets and digits and special characters.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 21

Question 29.
A. void change(int&);
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 23
B. void change(int);
int main()

Predict the output of both programs.
Justify your predictions.

Answer:
1. A. Output
value = 40
B. output
value = 0

2. In the first case (A) the argument x is passed by reference method. So the changes made in the function reflects in main()

In the second case (B) the argument x is passed by value method. So the changes made in the function will not reflect in main().

Question 30.
Write a program to read 2 strings and join them using string function
Answer:
# include<iostream>
# include<cstring>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 25

Question 31.
Differentiate the outputs of the folloiwng C++ statements and also give reason

cout< <strcmp(“world”, “WORLD”);
cout<<strcmpi(“world”, “WORLD”);

Answer:

>0 Here first string “world” is greater than “WORLD”.
It prints 0. Because strcmpi is same as strcmp() but it is not case sensitive. That means uppercase and lowercase are treated as same.

Question 32.
Match the following.

1. strcmp()	a. To combine two strings
2. strcpy ()	b. To get 5 from 25
3. strcat ()	c. To get 10 from -10
4. sqrt ()	d. To change a to A
5. abs()	e. To compare two strings
6. toupper()	f. To copy one string another

Answer:
1-e, 2-f, 3-a, 4-b, 5-c, 6-d.

Question 33.
Write a C++ program to find the sum of first ‘N’ natural numbers using a user defined function.
Answer:
# include<iostream>
using namespace std;
int sum(int n)
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 26

Question 34.
Write the need for function prototype in a C++ program.
Answer:
When the function is defined after the main function then there is an error called “function should have a prototype”. This is because of the function is defined after the main function. To resolve this a prototype should be declared inside the main function.

Question 35.
Write suitable function prototype after reading the following cases.

Case I : The function Volume() takes two arguments, one is float the other is int and it returns its volume.
Case II : A function Big() has no arguments and no return type.
Case III: A function PrintO takes two floating point type arguments and nothing is returned.

Answer:

Case I: float Volume(float,int);
Case II: void BigO; or void Big(void);
Case III : void Print(float,float) or void Print(double,double);

Question 36.
Find the error in the following C++ program and rectify it.
#include<iostream>
using namespace std;
int main ()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 27
Answer:
Error 1: Here function prototype is missing.
Error 2: no need for variable z.
Error 3: no need of the statement z = x*y;
#include<iostream>
using namespace std;
int multi(int, int);
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 28

Question 37.
Consider the following function definition.
int add (int a, int b=2, int c=5)
{
int s = a + b + c;
cout<<“Sum is :”<<s;
}
Predict the output of the above code forthe following function calls:

add (5, 8, 10);
add (5, 8);
add (5);

Answer:

add(5, 8, 10). Here a = 5, b = 8 and c = 10. Then It prints 23(5 + 8 + 10)
add(5, 8). Here a = 5, b = 8 and no value for c then c will take the default value 5. Hence it prints 18(5 + 8 + 5(default value for c)).
add(5). Here a = 5 and no values for b and c, then b and c will take the default values 2 and 5 respectively. Hence it prints 12(5 + 2 + 5(default values for b and c)).

Question 38.
Consider the following C++ program, predict the output and justify it.
#include<iostream>
using namespace std;
intsqr(int&);
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 29
Answer:
The output are 25 and 6. Here the function uses call by reference method. The function call sqr(a) passes the original value to the function sqr. The function changes the value of b(here a and b are same) to 6. That means a also becomes 6.

Question 39.
Differentiate local variable and global variable in C++ program.
Answer:
1. Local scope:
A variable declared inside a block can be used only in the block. It cannot be used any other block.
Eg: int sum(int n1,int n2)
{
int s;
s=n1+n2;
return(s);
}
Here the variable s is declared inside the function sum and has local scope;

2. Global scope:
A variable declared outside of all blocks can be used anywhere in the program.
Eg:
int s;
intsum(int n1,int n2)
{
s=n1+n2;
return(s);
}
Here the variable s is declared out side of all functions and we can use variable s any where in the program.

Question 40.
Consider the following C++ code fragment and identify the local function and global function. Also justify your selection.
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 30
Answer:
Here the function print() is declared inside a function hence it is a local function but the function sum() is declared outside of all functions hence it is called global function.

Question 41.
Read the following C++ program and identify the error and give reason.
# include<iostream>
using namespace std;
void disp(int);
int main()
int x=10; disp (x); return 0;
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 31
Answer:
The variable ‘x’ will not be printed because it is declared in the main() function. That is x is a local variable.

Plus Two Computer Application Functions Five Mark Questions and Answers

Question 1.
What will be the output of the following code if the userenterthe value “GOOD MORNING”

char String [80];
gets(string);
cout<<string;
char String [80]; cin>>string;
cout<<string;
char ch;
ch=getchar();
cout<<ch;
char String [80];
cin.getline(string,9);
cout<<string;

Answer:

GOOD MORNING
GOOD
G
GOOD MORN

Question 2.
Read a string and print the number of vowels.
Answer:
# include<iostream>
# include<cstdio>
# include<cctype>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 32

Question 3.
Describe in detail about the unformatted console i/o functions.
Answer:

Single character functions: This function is used to read or print a character at a time.
- getchar(): It reads a character from the keyboard and store it in a character variable.
  Eg. char ch;
  ch=getchar();
- putchar(): This function is used to print a character on the screen.
  Eg. char ch;
  ch=getcharO;
  putchar(ch);
String functions: This function is used to read or print a string.
- gets(): This function is used to read a string from the keyboard and store it in a character variable.
  Eg. char str[80];
  gets(str);
- puts(): This function is used to display a string on the screen.
  Eg. char str[80];
  gets(str);
  puts(str);

Question 4.
Write a program to input a string and find the number of uppercase letters, lowercase letters, digits, special characters, and white spaces.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 33

Question 5.
Write a program to input a string and replace all lowercase vowels by the corresponding uppercase letters.
Answer:
# include<iostream>
# include<cstdio>
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 35

Question 6.
A list of C++ built in functions are given. Classify them based on the usage and prepare a table with proper group names.

strcmp()
sin()
getch()
isalpha()
pow()
puts()
strcat()
tolower()
getchar()
isalnum()
sqrt()
exp()
write()

Answer:
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 37

Question 7.
The factorial of a number, say N is the product of first N natural numbers. Thus, factorial of 5 can be obtained by taking the product of 5 and factorial of 4. Similarly factorial of 4 be found out by taking the product of 4 and factorial of 3. At last the factorial of 1 is 1 itself. Which technique is applicable to find the factorial of a number in this fashion? Write a C++ function to implement this technique. Also explain the working of the function by giving the number 5 as input.
Answer:
A function calls itself is known as recursion.
# include<iostream>
using namespace std;
int fac(int);
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 38
The working of this program is as follows. If the value of n is 5 then it calls the function as fac(5). The function returns value 5*fac(4). That means this function calls the function again and returns 5*4*fac(3). This process continues until the value n=1. So the result is 5*4*3*2*1 = 120.

Question 8.
Read the following program
# include<iostream>
using namespace std;
int a = 0;
int main ()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 39
Write down the value displayed by the output of the above program with suitable explanation. What are the inferences drawn regarding the scope of variables?
Answer:
The output is 061.
1. Global variable: A variable declared out side of all functions it is known as global variable.

2. Local variable: A variable declared inside of a function it is known as local variable.

If a variable declared inside a function(main or other) with the same name of a global variable. The function uses the value of local variable and does not use the value of the global variable.

Here int a=0 is a global variable. In the main function the global variable ‘a’ is used. There is no local vari-able so the value of ‘a’, 0 is displayed. The statement ‘a++’ makes the value of ‘a’ is 1. It calls the function showval with argument ‘a=1’.

The argument ‘x’ will get this value i.e. ‘x=T. But in the function showval there is a local variable ‘a’ its value is 5 is used. So this function returns 6 and it will be displayed. After this the value 1 of the global variable ‘a’ will be displayed. Hence the result 061.

Question 9.
The following are function calling statements. Some of them will be executed, while some other generate compilation error. Write down your opinion on each of them with proper justification
Answer:

char ch=getch();
sqrt(25);
strcat (“Computer”, “Program”);
double num = pow(2, 3, 5)
put char(getchar());

Answer:

getch get a character from the console(key board) but does not echo to the screen. So we can’t read a character from the console.
It returns the square root of 25.
It concatenates Program to computer, i.e. we will get a string “computer program”
The function pow should contains only two arguments. But here it contains 3 arguments so it is an error. We can write this function as follows Double num = pow(pow(2, 3, 5)
It reads a character from the console and display it on the screen.

Question 10.
Short notes about character functions and string functions
Answer:
a. Character functions:
1. isalnum(): It is used to check whether a character is alphabet or digit. It returns a non zero value if it is an alphabet or digit otherwise it returns zero.

2. isalpha(): It is used to check whether a character is alphabet or not. It returns a non zero value if it is an alphabet otherwise it returns zero.

3. isdigit(): It is used to check whether a character is digit or not. It returns a non zero value if it is digit otherwise it returns zero.

4. islower(): It is used to check whether a character is lower case alphabet or not. It returns a non zero value if it is a lower case alphabet otherwise it returns zero.

5. isupper(): It is used to check whether a character is upper case alphabet or not. It returns a non zero value if it is an upper case alphabet otherwise it returns zero.

6. tolower(): It is used to convert the alphabet into lowercase.

7. toupper(): It is used to convert the alphabet into upper case.

b. String functions:
1. strcpy(): This function is used to copy one string into another.

2. strcat(): This function is used to concatenate(join) second string into first string.

3. strlen(): This function is used to find the length of a string.

4. strcmp(): This function is used to compare 2 strings. If the first string is less than second string then it returns a negative value. If the first string is equal to the second string then it returns a value zero and if the first string is greater than the second string then it returns a positive value.

Question 11.
Write a program to perform the following operations on a string

Length of a string
Search a character
Display the string

Answer:
# include<iostream>
# include<cstdio>
using namespace std;
void len()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 40

Question 12.
Write functions to perform the following operations.

sqrt()
power of 2 numbers
sin
cos

Answer:
# include<iostream>
# include<cmath>
using namespace std;
void sqroot()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 44

Question 13.
Read the following C++ programs and answer the questions:
Case I
# include<iostream>
using namespace std;
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 47

Case II
# include<iostream>
using namespace std;
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 3 Functions - 48

Identify the type of function call in each case.
How do they differ?

Answer:

In case I the method used is Call by value and in Case II is Call by reference.
There are two methods they are call by value and call by reference

a. call by value:
In call by value method, a copy of the actual parameters are passed to the formal parameters. If the function makes any change it will not affect the original value.

b. call by reference:
In call by reference method, the reference of the actual parameters are passed to the formal parameters. If the function makes any change it will affect the original value.

Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays

April 13, 2021February 14, 2020 by Prasanna

Students can Download Chapter 2 Arrays Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays

Plus Two Computer Application Arrays One Mark Questions and Answers

Question 1.
From the following which is not true for an array
(a) It is easy to represent and manipulate array variable
(b) Array uses a compact memory structure
(c) Readability of program will be increased
(d) Array elements are dissimilar elements
Answer:
(d) Array elements are dissimilar elements.

Question 2.
Consider the following declaration. int mark (50).
Is it valid? If no give the correct declaration.
Answer:
It is not valid. The correct declaration is as follows, int mark[50]. Use square brackets instead of parenthesis.

Question 3.
Consider the following declaration. int mark[200].
The index of the last element is____.
Answer:
199.

Question 4.
Consider the following declaration int mark[200]
The index of the first element is_____
Answer:
0.

Question 5.
Consider the following int age[4]={15, 16, 17, 18};
From the following which type of initialisation is this.
(a) direct assignment
(b) along with variable declaration
(c) multiple assignment
(d) None of these
Answer:
(b) along with variable declaration

Question 6.
From the following which is used to read and display array elements
(а) loops
(b) if
(c) switch
(d) if else ladder
Answer:
(a) loops

Question 7.
Write down the corresponding memory consumption in bytes

int age[10]=_____
charname[10] =_____
intage[10][10]=____

Answer:

4*10 = 40 bytes (4 bytes for one integer)
1*10=10 (one byte for each character)
4*10*10 = 400 (4 * 100 elements)

Question 8.
Consider the following intage[4] = {12, 13, 14};
cout<<age[3]; What will be the output?
(а) 14
(b) 12
(c) 13
(d) 0
Answer:
(d) 0

Question 9.
The elements of 2-dimensional array can be read using_____loop
Answer:
nested loop.

Question 10.
_____is the process of reading / visiting elements of an array
Answer:
traversal.

Question 11.
Anjaly wants to read the 10 marks that already stored in an array and find the total. This process is known as_____
(a) insertion
(b) deletion
(c) traversal
(d) linear search
Answer:
(c) traversal

Question 12.
The elements of an array of size ten are numbered from____to___.
Answer:
0 to 9.

Question 13.
Element mark[6] is which element of the array?
(a) The sixth
(b) the seventh
(c) the eighth
(d) impossible to tell
Answer:
(b) the seventh

Question 14.
When a multidimensional array is accessed, each array index is
(a) Separated by column.
(b) Surrounded by brackets and separated by commas.
(c) Separated by commas and surrounded by brackets.
(d) Surrounded by brackets.
Answer:
(d) surrounded by brackets

Question 15.
Write a C++ statement that defines a string variable called ‘name’ that can hold a string of upto 20 characters.
Answer:
char name[21];

Question 16.
_____is a collection of elements with same data type
Answer:
Array

Question 17.
Following are some of the statements regarding array. Identify the correct statement.
(a) Array is a collection of elements of same data type.
(b) Array cannot be initialised during the time of declaration.
(c) Array allocates continuous memory.
(d) An array element can be accessed using index or subscript.
Answer:
(a) Array is a collection of elements of same data type.

Question 18.
Which of the following is the correct declaration of an array?
(a) int a(10);
(b) int 10[a];
(c) a[1] int;
(d) inta[10];
Answer:
(d) int a[10];

Question 19.
Which is the last subscript of the array int m[25]?
(a) 24
(b) 25
(c) 0
(d) 26
Answer:
(a) 24

Question 20.
The memory size of the data type float is 4 bytes. What is the total bytes required for the array declaration float salary[10];?
(a) 10
(b) 4
(c) 40
(d) 400
Answer:
(c) 4*10 = 40.

Question 21.
int num[100]; The above statement declares an array named num that can store maximum_____integer numbers.
(a) 99
(b) 100
(c) 101
(d) Any number
Answer:
(b) 100

Question 22.
If int a[10]; is array, then which element of the array will be referenced as a[4].
Answer:
Fifth element.

Question 23.
Consider the following array declaration int A[ ] = {4, 5, 8}; int B[ ]={2, 10};
Write a valid C++ statement for finding the difference between the last element of the array ‘B’ and the first element of the array ‘A’.
Answer:
B[1] – A[0]; ORA[0] – B[1];

Question 24.
Consider the following code and predict the output.
int sum=0;
int a[5] = {1, 2, 3, 4, 5};
for(i=0;i<4;++i)
{
sum=sum+a[i];
}
cout<<sum;
Answer:
10.

Question 25.
Which data type is used to declare a variable to hold string data?
Answer:
The data type char is used for this.

Question 26.
The terminating character of string array is______
Answer:
\0 or NULL character.

Question 27.
Write a statement for storing the string “NO SMOKING” using a character array with name ‘ARR’
of minimum size.
Answer:
char APR[11] = ”NO SMOKING”;

Plus Two Computer Application Arrays Two Mark Questions and Answers

Question 1.
Given some array declaration. Pick the odd man out.
Float a[+40], int num[0-10], double [50]. char name[50], amount[20] of float.
Answer:
char name[50]. It is a valid array decalaration the remaining are not valid.

Question 2.
Whether the statement char text[] = “COMPUTER”; is True / False ? Justify.
Answer:
It is a single-dimensional array. If the user doesn’t specify the size the operating system allocates the number of characters + one (for null character for text) bytes of memory. So here OS allocates 9 bytes of memory.

Question 3.
Suppose you are given Total mark of 50 students in a class

How will you store these values using ordinary variable?
Is there any other efficient way to store these values? Give reason

Answer:
We have to declare 50 variables individually to store total marks of 50 students. It is a laborious work. Hence we use array, it is an efficient way to declare 50 variables.

With a single variable name we can store multiple elements. Eg: int mark[50]. Here we can store 50 marks. The first mark is in mark[0], second is in mark[1], …etc the fiftieth mark is in mark[49].

Question 4.
Consider the statement charstr[ ] = “PROGRAM” What will be stored in last location of this array. Justify
Answer:
The last location is the null character(\0) because each string must be appedend by a null character.

Question 5.
Explain the needs for arrays
Answer:
Array is collection of same type of elements. With the same name we can store more elements. The elements are distinguished by using its index or subscript. To store 50 marks of 50 students we have to declare 50 variables, it is a laborious work.

Hence the need for arrays arise. By using array this is very easy as follows int mark[50]. Here the index of first element is 0, then 1, 2, 3, etc upto 49.

Question 6.
Consider the following code and predict the output.
int a[5]= {6, 8, 10,20,40};
cout<<“\n”<<a[3];
cout<<“\n”<<a[1]+1[4];
Answer:
20(The fourth element).
48(Sum of second element 8 and fifth element 40, i.e 8 + 40 =48).

Question 7.
Suppose you need to store the value 10, 20, 30, 40 and 50 into an array. Write different methods to do this problem. Answer:
Method 1:
int a[5]={10, 20, 30, 40, 50};

int a[ ]={10, 20, 30, 40, 50};

Method 2:
int a[5];
a[0]=10;
a[1]=20;
a[2]=30;
a[3]=40;
a[4]=50;

Question 8.
What would be the appropriate array declaration to store the following?

Name of a student
Age of 20 students
Mark of 6 subject
Average mark of 10 students in 5 subjects

Answer:

charname[20];
intage[20];
intmark[6];
float mark[10];

Question 9.
Consider the following code and predict the output.
int A[5] = {11, 12, 13, 14, 15};
int i;
for (i=4;i>=0;–i)
{
cout<<“\n”<<A[i];
}
Answer:
It prints the array in reverse order as follows.
15
14
13
12
11.

Question 10.
Predict the output of the following code segment.
int K[ ] = {1, 2, 3, 4};
for (int i=0; i<4; i++)
cout<<K[i] * K[i]<<“\t”;
Answer:
The output is as follows
1 4 9 16
Hint: 1(1*1) 4(2*2) 9(3*3) 16(4*4).

Question 11.
Consider the following code and predict the output.
Justify your answer.
int A[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int sum=0, i;
for(i=0;i<10;++i)
{
if (A[i]%2==0)
{
sum=sum+A[i];
}
}
cout<<“\nSum=”<<sum;
Answer:
The output is 30. That is sum of all even numbers in the array.

Question 12.
Predict the output of the following C++ statement.
char str[8] = “WELCOME”;
cout<<“\n’’<,str[3];
cout<<“\n”<<str;
Answer:
The output is as follows
C(The fourth character)
WELCOME.

Question 13.
How many bytes will be allocated in the memory for storing the string “MY SCHOOL”? Justify your answer.
Answer:
A total of 10 bytes. 9 bytes is used to store 9 characters in the string MY SCHOOL(including 1 byte for space) and 1 byte is used to store the NULL character.

Plus Two Computer Application Arrays Three Mark Questions and Answers

Question 1.
Total mark of 50 students in a class are given in an array. A bonus of 10 marks is awarded to all of them. Write the program code for making such a modification.
Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 1

Question 2.
Write the program code for counting the number of vowels from your school name
Answer:
# include
# include
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 2

Question 3.
Write the program code for counting the number of words from the given string.” Directorate of Higher Secondary Examination”
Answer:
# include
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 4

(Program for more than one space between words)
# include
# include
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 5

Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 6

Question 4.
Given a word like “ECNALUBMA” Write the program code for arranging it in into a meaningful word Answer:
# include
# include
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 8

Question 5.
Explain different types of arrays.
Answer:
1. Single dimensional:
It contains only one index or subscript. The index starts from 0 and ends with size-1.
Eg. int n[50]; charname[10];

2. Multidimensional:
It contains more than one index or subscript. The two dimensional array contains two indices, one for rows and another for columns. The row index starts from 0 and end at row size-1 and column index starts at 0 and ends at colunn size-1.
Eg. int n[10][10] can store 10 * 10 =100 elements. The index of the first element is n[0][0] and index of the 100^th element is n[9][9].

Question 6.
Write a program to read the 5 marks of a students and display the marks and total. Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 9

Question 7.
Explain different array operations in detail.
Answer:

Traversal:- All the elements of an array is visited and processed is called traversal
Search:- Check whether the given element is present or not
Sorting:-Arranging elements in an order is called sorting.

Question 8.
Given a word “COMPUTER”, write a C++ program to reverse the word without using any string functions.
Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 10

Question 9.
Write statements to declare an array and initialize it with the numbers 1, 2, 3, 4, 5 and print 5, 4, 3, 2, 1.
Answer:
# include
using namespace std;
int main()
int a[5]={1, 2, 3, 4, 5},i;
for(i=4;i>=0;i-)
cout<<a[i]<<“,”;
}

Question 10.
Consider the following array declaration. Write statements to count how many numbers are greater than zero.
int p[ ] = {-5, 6, -7, 0, 8, 9};
Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 11

Question 11.
Write a C++ program to read 10 integer values and find the largest number among them using array.
Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 12

Question 12.
Write a C++ program to accept a string from the keyboard and find its length without using function. For example if “WELCOME” is accepted, the output will be 7.
Answer:
# include
# include
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 13

Question 13.
Considerthe following C++ statements

charword[20];
cin>>word;
cout<<word;
char word[20];
gets(word);
puts(word);

If the string entered is “HAPPY NEW YEAR”. Predict the output in both cases and justify your answers.
Answer:

HAPPY. When we use cin to accept string then space is the delimiter. The string after space is truncated.
HAPPY NEW YEAR. gets() reads all the characters (including space) upto the user presses the enter key.

Question 14.
Consider the following C++ statements
char str[] = “NO/nSMOKING”;
cout<<str;

What is the output of the above code?
How many bytes will be allocated in the memory for the variable str?

Answer:

NO
SMOKING. The output is in 2 lines.
A total of 11 bytes is used to store this string.

1 byte for \n. 1 byte for \0(The null character that is automatically appended) and 9 bytes for the remaining characters (N, 0, S, M, 0, K, l, N AND G).

Question 15.
Write a C++ program to store the given string in an array and display it in reverse order without using string function. For example if ABCD is given, the output should be DCBA.
Answer:
# include
#include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 14

Plus Two Computer Application Arrays Five Mark Questions and Answers

Question 1.
Collect the heights of 12 students from your class in which 7 students are male and others are female students. Suppose these male and female students be seated in two separate benches and you are given a place which is used for sitting these 12 students in linear form. How will you combine and make them sit without mixing male/female students? Write a program for the same.
Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 15

Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 16

Question 2.
Write a program to read 3 marks of 5 students and find the total and display it
Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 17

Question 3.
Write a program to read a string and a character and find the character by using linear search.
Answer:
# include
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 18

Question 4.
Write a program to read a string and find the no. of vowels consonents and special characters.
Answer:
# include
#include
#include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 19

Question 5.
Write a program to accept marks of 10 students and find out the largest and smallest mark from the list.

Write a C++ program to store the scores of 10 batsmen of a school cricket team and find the largest and smallest score.
Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 21

Question 6.
Write a C++ program to read 6 marks of a student and find total and average mark.
Answer:
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 22

Question 7.
Write a C++ proram to accept a sentence and count the number of times the letter ‘s’ occurs in it. For example if the sentence is This is my school’, the output should be 3.
Answer:
# include
# include
using namespace std;
int main()
Plus Two Computer Application Chapter Wise Questions and Answers Chapter 2 Arrays - 23

Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting

April 13, 2021February 14, 2020 by Prasanna

Students can Download Chapter 6 Data Base Management System for Accounting Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting

Plus Two Accountancy Data Base Management System for Accounting One Mark Questions and Answers

Question 1.
DBMS stands for __________
Answer:
Data Base Management system.

Question 2.
Name of database object to hold data
(a) Tables
(b) Forms
(c) Queries
(d) Reports
Answer:
(a) Tables

Question 3.
LibreOffice Base is a
a) Word Processing Software
b) Presentation Software
c) Spread sheet Software
d) Data Base Management Software
Answer:
d) Data Base Management Software

Question 4.
The common fields used in a relationship between tables are called
(a) Joint field
(b) Main field
(c) Key field
(d) Table field
Answer:
(c) Key field

Question 5.
The result of Query can be displayed by clicking on ___________
Answer:
Run Button

Question 6.
SQL stands for _______
Answer:
Structural Query Language

Question 7.
______ denotes the number of rows in the table
(a) Tuple
(b) Cardinality
(c) DBMS
(d) Domain
Answer:
(b) Cardinality

Question 8.
A single row in the table is called
(a) Attribute
(b) Tuple
(c) Column
(d) Cardinality
Answer:
(b) Tuple

Question 9.
Database in LibreOffice Base is called ________
(a) Data group
(b) Mass Data
(c) Rows and columns
(d) Data Source
Answer:
(d) Data Source

Question 10.
A data base consists of a number of that ______ contains the individual pieces of data
Answer:
fields

Question 11.
A _________ of the database is a group of fields
(a) Table
(b) Forms
(c) Query
(d) None of these
Answer:
(a) Table

Question 12.
The data type suitable to the name of a person
Answer:
Text: [VARCHAR]

Question 13.
The default extension of LibreOffice Base file is
(a) .bmp
(b) .xls
(c) .lob
(d) .odb
Answer:
(d) .odb

Question 14.
Which among the format searches for all values ending with R?
(a) LIKE ‘*R*’
(b) LIKE ‘*R’
(c) LIKE ‘R*’
(d) LIKE ‘END\‘R’
Answer:
(b) LIKE ‘*R’

Question 15.
ODBC stands for _______
Answer:
Open Database Connectivity

Question 16.
The data type suitable to the age of an employee is;
Answer:
Number (Numeric)

Question 17.
__________are used to store the data in the database
(a) Reports
(b) Forms
(c) Tables
(d) Queries
Answer:
(c) Tables

Question 18.
________ is a tool to connect tables in a database
(a) Forms
(b) Queries
(c) Relationships
(d) Fields
Answer:
(c) Relationships

Question 19.
Choose the right path to start up LibreOffice Base
(a) Applications → Office → LibreOffice Base
(b) Applications → Create → LibreOffice Base
(c) Applications → Login → Office → LibreOffice Base
(d) Applications → Create → Office → LibreOffice Base
Answer:
(a) Applications → Office → LibreOffice Base

Question 20.
LibreOffice Base runs on _____ and _____ operating system
Answer:
Windows and Linux

Question 21.
Which among the following is not an advantage of LibreOffice Base
(a) The information is portable
(b) Ensure Data security
(c) Initial training is required for all users
(d) Many people can access the same database at the same time
Answer:
(c) Initial training is required for all users

Question 22.
Which among the following is not a component of Database system?
(a) Data
(b) Hardware
(c) Software
(d) None of these
Answer:
(d) None of these

Question 23.
RDBMs stands for ____________
Answer:
Relational Data Base Management System

Question 24.
The data type suitable to basic pay of employee:
Answer:
Number (Numeric)

Question 25.
What field type is used to store picture in a table?
Answer:
OLE object.

Question 26.
‘Join Line’ in the context of LibreOffice Base Tables means:
Answer:
Graphical representation of relationship between tables.

Question 27.
Reports are created from
(a) Tables
(b) Forms
(c) Relationships
(d) Tabs
Answer:
(a) Tables

Question 28.
Database is referred to as _______________
(a) Front-end program
(b) Back – end program
(c) User-end program
(d) None of these
Answer:
(b) Back – end program

Question 29.
Which among the following is not a DBMS?
(a) Base
(b) Access
(C) Oracle
(d) None of these
Answer:
(d) None of these

Question 30.
Choose the right pairs.
(a) Desktop data base – Single user applications
(b) Desktop data base – Multi user applications
(c) Server data base – Multi user applications
(d) Server data base – Single user applications.
Answer:
(a) & (c)

Question 31.
_______ are the fundamental building blocks fo the database.
(a) Tables
(b) Forms
(c) Queries
(d) Reports
Answer:
(a) Tables

Question 32.
Each column of the table is called …..(a)…… and characteristics of which is called …….(b)…….
Answer:
(a) field
(b) attributes

Question 33.
The format for getting the employees whose names begins with ‘K’ is ______
Answer:
LIKE ‘K*’

Question 34.
To expect a well formatted printable data from LibreOffice Base database, we may use
Answer:
Report

Question 35.
__________ are used for connecting tables in database to get the advantage of data redundancy
(a) Relationships
(b) Primary key
(c) Forms
(d) Tables
Answer:
(a) Relationships

Question 36.
In a database context is a window or screen that contain numerous fields, or spaces to enter data
(a) Tables
(b) Forms
(c) Query
(d) Reports
Answer:
(b) Forms

Question 37.
________ section contains the tittle of the report
(a) Field name
(b) Report header
(c) Form Header
(d) Report name
Answer:
(b) Report header

Question 38.
A ________ is a reference to a field in another relation or table
(a) Primary key
(b) Candidate key
(c) Foreign key
(d) Super key
Answer:
(c) Foreign key

Question 39.
The name which indicate the number of columns in the table
(a) Domain
(b) Tuple
(c) Degree
(d) Attribute
Answer:
(c) Degree

Question 40.
The end result of normalisation is known as ___________
Answer:
Refinement

Question 41.
Data type ‘Text’ can store up to characters
(a) 65,535
(b) 255
(c) 35,423
(d) 555
Answer:
(b) 255

Question 42.
Relationship between primary key of one table to primary key of another table is called
(a) One to one
(b) One to many
(c) Many to many
(d) Many to one
Answer:
(a) One to one

Question 43.
A query criteria LIKE ‘RAJU*’ returns all names that: LIKE ‘RAJU*’
(a) Contains RAJU
(b) Starts with RAJU
(c) Ends with RAJU
(d) None of these
Answer:
(b) Starts with RAJU

Question 44.
The characteristics of an entity is ________________
Answer:
Attributes

Question 45.
The data type _______ can store upto 65, 535 characters
(a) Text
(b) Number
(c) Memo
(d) Date
Answer:
(c) Memo

Question 46.
Anything which has a real life existence is called ___________
Answer:
Entities

Question 47.
What criteria is used to get return a text starts with ‘A’
Answer:
LIKe ‘A*’

Question 48.
A text data field is used to hold ______ values
(a) Alpha numeric
(b) Numbers only
(c) Alphabets only
(d) Any data
Answer:
(a) Alpha numeric

Question 49.
DBMS is an aggregate of data, hardware, software and ______
Answer:
Users

Question 50.
A _______ is a two dimensional array containing rows and columns
Answer:
Table

Plus Two Accountancy Data Base Management System for Accounting Two Mark Questions and Answers

Question 1.
What do you mean by Data base?
Answer:
Database/ Data source – Introduction
A database is a collection of related data. It is organised in such a way that its contents can easily be accessed, managed and updated. In LibreOffice, database is also called data source.

Database consists of interrelated data tables that are structured in a manner that ensures-data consistency and integrity. LibreOffice base, MS Access, Oracle, SQL server, etc. Are the commonly used softwares for data base management.

Question 2.
Define Primary Key.
Answer:
Primary key is a unique key which identify a row in a table. A primary key comprises a single column or set of columns.

Question 3.
What are the different ‘views’ of a form?
Answer:
Form view, Layout view and Design view

Question 4.
What are the components of Database system?
Answer:

Data
Hardware
Software
Users

Question 5.
What do you mean by Normalisation?
Answer:
Normalisation is the Alteration of tables that reduces data redundancy. Data redundancy means data duplication.

Question 6.
What are the different objects in LibreOffice Base?
Answer:
Tables, Relations, Forms, Queries and Reports.

Question 7.
Write the step for starting up Libre Office Base.
Answer:

Step 1 – Click on Applications
Step 2 – Select Office
Step 3 – Select LibreOffice Base

Path → Application → office → LibreOffice Base

Question 8.
What is an attribute?
Answer:
Attributes: These define the characteristics of an entity. Eg: Name, Age, Caste, Salary etc.

Question 9.
What do you mean by Query?
Answer:
Query:
Query is a question. Queries are used to view, change and analyse data in different ways. It creates a new table from the existing tables based upon the question/ request asked to the data base.

Question 10.
What is Join line?
Answer:
It is the graphical representation of relationship between tables.

Question 11.
What is relationships?
Answer:
These are links between tables.

Question 12.
How to run saved Query?
Answer:
Select Queries option from the left panel of the LibreOffice Base window. The saved query can be seen in the right side. Double click on the query name to run the query.

Question 13.
What are the different methods to create Forms in LibreOffice Base?
Answer:

Create Form in Design view
Use wizard to create Form

The second method is the easy way to create Forms in LibreOffice Base

Question 14.
How to close a Database file and Exit from it?
Answer:

Step 1 – Click on the File menu
Step 2 – Select Close

Then the file will be closed. To exit from the Base file, follow these steps:

Step 1 – Click on the File menu
Step 2 – Select exit LibreOffice

Question 15.
Give short note on Tables in LibreOffice Base
Answer:
A table is a database object used to store data about a particular subject. A table consists of records and fields, the columns are called fields the and rows are called records.

Question 16.
Write down the character length of the ‘Text’ data type and ‘memo’ data type.
Answer:

Text – 225 Characters
Memo – 65535 Characters

Question 17.
List down the disadvantages of LibreOffice Base
Answer:
Disadvantages of database / Data source:

Designing of database is a complex and time consuming process.
Initial training is required for all the users.
Installation cost is high.

Question 18.
Distinguish between database and database management system
Answer:
Database is a collection of data. It consists of inter-related data tables.
Database management system is a collection of programs that enables users to work on database. DBMS enables the user to create and maintain a database.

Question 19.
Match the following

One to One Relationship – Non Primary Key to Non Primary Key
One to Many Relationship – Primary key to Primary Key
Many to Many Relationship – Non Primary key to Primary Key
Many to One Relationship – Primary key to Non Primary Key

Answer:

One to One Relationship – Primary key to Primary Key
One to Many Relationship – Primary key to Non Primary Key
Many to Many Relationship – Non Primary key to Non Primary Key
Many to One Relationship – Non Primary key to Primary Key

Question 20.
What are the different methods for creating Queries?
Answer:

Using the Query Wizard
Using Design view

Plus Two Accountancy Data Base Management System for Accounting Four Mark Questions and Answers

Question 1.
List down the advantages of using Database.
Answer:
Advantage of database/ data source:

All of the information is together
The information is portable
Information can be accessed at any time
Many users can access the same database at the same time.
Reduced data entry, storage and retrieval cost.

Question 2.
Explain the importance of Database Management System.
Answer:

It helps to maintain records for ongoing use.
It helps to generate reports based on the database.
Mass volume of data can be managed easily.
It is very useful; if the information stored in the system is subject to many changes.

Question 3.
Three options are available in database wizard to create a new database. What are they?
Answer:

Create a new Database.
Open an existing database.
Connect to an existing database.

1.Create a new Database:
This option helps to create a new database with new tables, forms, queries, reports, etc.

2. Open an existing database:
This option helps to open an existing database file which had been created earlier.

3. Connect to an existing database:
This option helps to connect a database which is created on a server.

Question 4.
Distinguish between Desktop Database and Server Database.
Answer:

Desktop Database	Server Database
(a) It is single user database.	(a) It is multi user database.
(b) It helps to do simple analysis and calculations.	(b) It helps to do complex analysis and calculations.
(c) It is less expensive.	(c) It is expensive.
(d) It is residing on personal computers.	(d) It is residing on server computer.

Question 5.
List down the steps to create a Database in LibreOffice base
Answer:

Step 1 – Creation of Blank Database
Step 2 – Creation of Tables
Step 3 – Creation of Relationships
Step 4 – Creation of Forms
Step 5 – Creation of Queries
Step 6 – Creation of Reports

Question 6.
What are the different Data types commonly used in LibreOffice Base?
Answer:

Text – Used to hold alphanumeric values
Memo- Used to enter long pieces of text
Number – Used to enter Numeric data
Date – Used to enter data information
Time – Used to enter time information

Question 7.
Some query criteria are given below. Write the query format.

Search for all values which contain salary
Search for all values beginning with K
Search for all values ending with P
Searches for all values beginning with Dr
Searches for all values ending with Cr.

Answer:

LIKE ‘*salary*’
LIKE ‘K*’
LIKE ‘*P’
LIKE ‘Dr*’
LIKE ‘*Cr’

Plus Two Accountancy Data Base Management System for Accounting Practical Lab Work Questions and Answers

Question 1.
Create a Table named ‘Tblstudents’ in LibreOffice Base with the following fields. Set the Admn No. as Primary Key
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 1
procedure:
Step 1 – Open Libre Office Base
Application → Office → LibreOffice Base.

Step 2 – Create a new database
Database wizard → Create a New Database → Next → Yes register the database for me → Open the database for editing Finish.

Step 3 – In Save Dialogue box, give the name STUDENTFEE and select a location to save the database and click Save button.
The new database file STUDENTFEE.odb. is created

Step 4 – Create Data Tables
From the left Database Pane, click on the icon Tables and below the Tasks section, click on ‘Create Table in Design view’
Database Pane → Tables → Create Table in Design view

Step 5 – Field Name Entry
In Table creation Screen, Enter the Field names and select appropriate Data Types as given below.

Field Name	Data Type
Admn No.	Number (Numeric)
Name of student	Text (VARCHAR)
Class	Text (VARCHAR)
Admission date	Date(DATE)
Fee Paid	Number (Numeric)

Step 6 – Setting Primary Key
To set the Admn No. as the Primary key, Right click on the row selector of Admn No. and select the Primary key from the drop down menu

Step 7 – Save Table
Click on Save button (or press Ctrl+S) to save the table. In save as dialogue box, Enter Tblstuderrts as table name and click on OK button Close the table creation screen

Step 8 – Data Entry
Database Pane → Tables → Created Tables → Select Tbl students → Double click to open it

Enter all the data given one by one

Output:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 2

Question 2.
Create Two Data Tables named TBLSTUDENT and TBLMARKS in LibreOffice Base with the following fields and show the relationship assuming that the first fields in both tables are set as primary keys

Table – 1
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 3
Table – 2

procedure:
Step 1 – Open LibreOffice Base
Application → Office LibreOffice Base

Step 2 – Create new Database
Datebase wizard → Create a new database Next → Yes, register database for me → Open the data base for editing → click on Finish button

Step 3 – In save Dialogue box, give the name STUDENTFILE and select a location to save the database and click Save button.
The new database file STUDETFILE.odb is created

Step 4 – Create Data Tables
Create Table – 1
From the left Data base pane, click on the Icon Tables and below the Tasks section, click on ‘Create Table in Design view’,
Data base Pane → Tables → Create Table in Design view

Step 5 – Field name Entry:
In Table creation screen, Enter the field names and select appropriate Data Types as given below

Field Name	Data Type
Admn No.	Number (Numeric)
Name	Text (VARCHAR)
Class	Text(VARCHAR)

Step 6 – Setting Primary key
To set Admn No as the primary key Right click → Primary key from the drop-down menu

Step 7 – Save Table
Press Ctrl+S on click on save button to save the table. Enter the name TBLSTUDENT → Click OK button.

Step 8 – Repeat the same steps for creating the second Table
Create Table -2

Field name	Data Type
Class No.	Number (Numeric)
Mark	Number (Numeric)

Set ‘Class No’ as the ‘Primary key’ save the table by naming TBLMARKS. Then close the table creation screen.

Step 9 – Data Entry
Data base Pane → Tables → Created Tables → Select → (Two tables TBLSTUDENT and TBLMARKS one by one) → Double click to open it
Enter all the details one by one.

Step 10 – Create Relationship
(a) Go to the Tools menu, select Relationships. Now Add Table dialogue box will appear. Add both the tables to the relationship window and close the Add Table dialogue box

(b) Create relationship between TBLSTUDENT and TBLMARK. Position the mouse pointer over the primary key of TBLSTUDENT table, hold down the left mouse button, drag the pointer right to the primary key in the TBLMARK table and then release the mouse button.

(c) Click on Save button in relation design window to save the relationship File → Save Click Close button to close the relation design window
Output:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 5

Question 3.
Create a Form in LibreOffice Base to Manage data in a table named customers with the following details.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 6
Step 1 – Open LibreOffice base
Application → Office LibreOffice Base

Step 2 – Create New Database
Database wizard → Create a new database Next → Yes, register database for me Open the database for edting → Click on Finish button

Step 3 – In save dialogue box, give the name CUSTOMERFILE . Select the location to save the database and cfick Save button.
The new database file CUSTOMERFILE.odb is created.

Step 4 – Create Data Tables
From the left database pane, click on the Icon Tables, and below the Tasks section, click on ‘Create Table in Design View’
Database Pane → Tables Create Tables in Design View

Step 5 – Enter necessary field names and select appropriate Data Type. Save the Table in the name TBLCUSTOMER and close the window

Step 6 – Create Form
From the left database pane, click on the Forms button and in right side, under Tasks section, click on Use Wizard to Create a Form.
Now Form wizard window opens
Data base Pane → Forms → Use wizard to create a Form

Step 7 – Selection of fields

In Form wizard, select TBLCUSTOMER from the combo box under the head Tables/Queries. → Click on Next button.
Add all Fields to Fields in the Form section in the right side by clicking >> button → Next button
Setup subform → Next → Arrange controls → Next
Set Data entry → Select The form is to display all data → Next → Apply styles → Next
Enter the name of the form inset Name Option as FORM CUSTOMER → select work with the Form → Finish

Now the data entry form will be opened

Step 8 – Data Entry in the form
Enter data in each field and Press Enter Key or Tab Key
Output:
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 7

Question 4.
Enter the following data in a database table
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 8

Display the names of students who scored greater than or equal to 500.
Display the name of student whose name begin with ‘S’

Procedure:
Step 1 – Open LibreOffice Base
Application → Office → LibreOffice Base

Step 2 – Create New Database
Datebase wizard → Create New Data base Next Yes, register the database form → Open the database for editing → Finish

Step 3 – In Save dialogue box, give the name STUDENTMARK and select the location to Save the data base and click Save button, the new data base file STUDENTMARK.odb. is created.

Step 4 – Create Data Table
From the left Database Pane, Click on the icon Tables, and below the Tasks section, Click on ‘Create Table in Design view’
Database Pane → Tables → Create Tables in Design view

Step 5 – Filed Name Entry
In Table creation Screen, Enter the Field names and select appropriate Data Types as given below

Step 6 – Setting Primary Key
To set STUDROLL_NO as primary key Right click → Select Primary key from the drop down menu

Step 7 – Save Table
Click on Save button or Press Ctrl+S to save the table. In save as dialogue box, Enter TBLSTUDENTDETAILS as the name and dick on OK Button
Close the Table creation screen.

Step 8 – Data Entry
Data base pane → Tables → Created Tables → Select TBLSTUDENTDETAILS → Double click to open it Enter all the data on by one

Step 9 – Create Query

From the Data base Pane, Click on Queries button and in the right side under Task section, Click on Create Query in Design view. Now Add Table or Query Window opens.
Select the table TBLSTUDENTDETAILS and click Add button, then close the window.
In the Query Design window, double click on each Field in Table window to add all the fields to query design grid.

Step 10 – Enter Query Criteria (Question 1)
In Query design Grid, set the Criteria for the query.

Enter >=500, in the Criterion row in STUDMARK column.
Click on the RUN Query Button or Press F5 to display the result.
Save and close the query. File → Save. ‘Save as’ window opens. Give the name Query 1 and click on OK button.

Step 11 – Enter the Query Criteria (Question 2)
In the Query design grid, set the criteria for the query

a) Enter LIKE ‘S*’, in the criterion row in STUDNAME column.
b) Click on the Run Query Button or Press F5 to display the result.
c) Save and close the query.
File → Save. ‘Save as’ window opens. Give the name ‘Query 2’ and click on OK button.
Output – 1
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 9

Output – 2
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 10

Question 5.
Enter the following list of employees in Tblemployee and make a report from this table sorted in the order of Designation.
Plus Two Accountancy Chapter Wise Questions and Answers Chapter 6 Data Base Management System for Accounting - 11
Procedure:
Step 1 – Open Libre Office base

Step 2 – Create new database
Database wizard → Create a new database → Next → Yes, register database for me → Open the database for editing → Click on Finish Button.

Step 3 – In save dialogue box, give the name Employee file. Select the location to save the database and click Save button.
Now, the new database file Employeefile.odb is created.

Step 4 – Create data tables
From the left database pane, click on the icon Tables, and below the Tasks section, Click on Create Table in Design View. Database Pane → Tables → Create Tables in design view.

Step 5 – Enter the necessary field names and select appropriate Data Type. Save the Table in the name Tbl employee and close the window.

Step 6 – Create Report
In the left Pane of the data base window, click on the Reports button and in right side, Under Tasks section click on Use Wizard to create a Report. This will open a Report wizard.

Step 7 – In Field selection step, under Tables Queries, Select the table Tblemployee. Then Press Add All buttons [>>] to all fields to report. Then click Next button.

Step 8 – Labelling fields (Skip this Step) → Next → Grouping Level (Skip this Step) → Next → In sort Option, Select Designation against sort by, and click Next button.

Step 9 – Choose Layout section, Select Tabular in the layout. → Click Next → In create Report Section → Give the tittle of Report as Report-Designation and select modify I Report Layout → Click Finish button.

Step 10 – Report Design window, click on Page Header area → Insert → Report controls → Label field to insert a Header for the report

Step 11 – Double click on the Label Field inserted and in the property window edit the Label as “Designation List of Employees” and Press Enter.

Step 12 – Click on Execute Report button to view the Report.

Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency

April 13, 2021February 13, 2020 by Prasanna

Students can Download Chapter 5 Measures of Central Tendency Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency

Plus One Economics Measures of Central Tendency One Mark Questions and Answers

Question 1.
The midpoint of the class ‘5-10’ is:
(i) 5
(ii) 7.5
(iii) 10
(iv) 15
Answer:
(ii) 7.5

Question 2.
Mode is equal to:
(i) 3 median – 2 mean
(ii) 2 median – 3 mea
(iii) 3 median – 3 mean
(iv) 3 median -1 mean
Answer:
(i) 3 median – 2 mean

Question 3.
Which of the following is a positional average?
(i) Mean
(ii) Median
(iii) Mode
(iv) All the above
Answer:
(ii) Median

Question 4.
Which of the following divides the data into four equal parts?
(i) decile
(ii) percentile
(iii) quartiles
(iv) none of the above
Answer:
(iii) quartiles

Question 5.
Which of the following is the most commonly used average?
(i) Arithmetic Mean
(ii) Median
(iii) Mode
(iv) Percentile
Answer:
(i) Arithmetic Mean

Question 6.
Mode can be graphically located by means of
(i) bio diagram
(ii) pie diagram
(iii) histogram
(iv) ogive
Answer:
(iii) histogram

Question 7.
The most suitable average for qualitative measurement is
(i) Arithmetic mean
(ii) Median
(iii) Mode
(iv) Geometric mean
(v) None of the above
Answer:
(ii) Median

Question 8.
Which average is affected most by the presence of extreme items?
(i) median
(ii) Mode
(iii) Arithmetic mean
(iv) Geometric mean
(v) Harmonic mean
Answer:
(iii) Arithmetic mean

Question 9.
The algebraic sum of deviation of a set of n values from A.M. is
(i) n
(ii) 0
(iii) 1
(iv) None of the above
Answer:
(ii) 0

Question 10.
The average value of a given variable is known as ____
Answer:
A.M

Question 11.
Total of given variables can be depicted by _____
Answer:
Σx

Question 12.
Common factor is depicted by ____
Answer:
‘c’

Question 13.
A.M should be
(i) Simple
(ii) Based on all items
(iii) Rigidly defined
(iv) All the above
Answer:
(iv) All the above

Question 14.
Median is the _____ value in a series.
Answer:
Middle.

Question 15.
Q₃ represents ____ Quartile.
Answer:
Middle.

Question 16.
_____ is the division of the series into 100 equal parts.
Answer:
Percentiles.

Question 17.
\(\left(\frac{N+1}{10}\right)^{t h}\) is used to calculate _____.
Answer:
Deciles

Question 18.
Value of median is equal to _____Answer:
Answer:
II Quartile – 50^th percentile, 5^th Decile

Question 19.
Pick out the odd one out and Justify.
Arithmetic mean, Median, Standard deviation, Mode
Answer:
Standard deviation. Others are measures of central tendency.

Question 20.
Which average would be suitable in the following cases?

Average size of readymade garments.
Average intelligence of students in a class.
Average production in a factory per shift.
Average wages in an industrial concern.
When the sum of absolute deviations from average is least.
When quantities of the variable are in ratios.
In case of open-ended frequency distribution.

Answer:

Mode
Median
Mode or median
Mode or median
Mean
Mode or mean
Median

Plus One Economics Measures of Central Tendency Two Mark Questions and Answers

Question 1.
Name the types of positional averages.
Answer:
Positional averages are median and mode.

Question 2.
Explain weighted Arithmetic mean
Answer:
When calculating Arithmetic means it is important to assign weights to various items according to their importance. The arithmetic mean calculated with the relative importance to different items is known as weighted arithmetic mean.

Question 3.
Give the special features of arithmetic mean.
Answer:
It is interesting to know and useful for checking your calculation that the sum of deviations of items about arithmetic mean is always equal to zero. Symbolically, S (X-X) = 0.However, arithmetic mean is affected by extreme values. Any large value, on either end, can push it up or down.

Question 4.
If median and mean of distribution are respectively 18.8 and 20.2. What would be its made?
Answer:
Mode = 3 Median – 2 Mean
= 3 × 18.8 – 2 × 20.2
= 56.4 – 40.4 = 16

Question 5.
Mention two demerits of median.
Answer:

Median is not based on all observations
It cannot be given for further mathematical treatment

Question 6.
Give the formulae of median in all series
Answer:
Formulae of Median
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 1

Question 7.
If median is 15 and mean is 17, calculate mode?
Answer:
Mode = 3 median – 2 mean
= 3 × 15 – 2 × 17
= 45 – 34 = 11

Question 8.
Can there be a situation where mean, median and mode are equal?
Answer:
Yes. Mean, median and mode will be equal when all given variables are the same.

Question 9.
Mark the missing value of the following data. The mean marks are 10.05.
5, 6, 7, 8, 12, ?, 15, 17, 18, 3, 10 5, 10, 12, 15, 11, 13, 1
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 2

Plus One Economics Measures of Central Tendency Three Mark Questions and Answers

Question 1.
What is the relative position of arithmetic mean, median and mode?
Answer:
Relative position of arithmetic mean, median and mode can be understood from the following narration. Suppose we express,
Arithmetic Mean = Me
Median = Mi
Mode = Mo
The relative magnitude of the three is
Me>Mi>Moor
Me<Mi<Mo
That is the median is always between the arithmetic mean and the mode.

Question 2.
Write down the advantages of median.
Answer:
Merits of median:

It is easy to understand
It is not affected by extreme values It can be graphically determined
It is suitable in case of open-end classes
It is suitable for qualitative measurement

Question 3.
Write the merits of mode.
Answer:
Merits of mode

It is easy to understand and simple to calculate
It is not affected by extreme values
It can be graphically determined
It is suitable in case of open-end classes.

Question 4.
Complete the following

………………. divides the series into two equal parts
The central tendency based on all values is …………..
The average which can be determined through ogive is ………

Answer:

median
mean
median

Question 5.
The mean mark of 60 students in section A are 40 and mean mark if 40 student in section B is 35. Calculate the combined mean of all the students.
Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 3

Plus One Economics Measures of Central Tendency Four Mark Questions and Answers

Question 1.
Point out important features of a good average.
Answer:
The important features of a good average are given below.

It should be easy to understand
It should be simple to calculate
It should be rigidly defined
It should be based on all observations
It should not be affected by extreme values.
It should be capable for further statistical calculations.

Question 2.
Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 4
Answer:

Question 3.
Calculate the median from the following data.

X	F
0-10	5
10-20	8
20-30	10
30-40	14
40-50	3

Answer:
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 6

Plus One Economics Measures of Central Tendency Five Mark Questions and Answers

Question 1.
Comment whether the following statements are true or false.

The sum of deviation of items from median is zero.
An average alone is not enough to compare series.
Arithmetic mean is a positional value.
The upper quartile is the lowest value of top 25% of items.
Median is unduly affected by extreme observations.

Answer:

False
True
False
True
False

Question 2.
There are three types of averages. Name them. Also, give appropriate definitions.
Answer:
There are several statistical measures of central tendency or “averages”. The three most commonly used averages are:

Arithmetic Mean
Median
Mode

1. Arithmetic mean:
Arithmetic mean is the most commonly used measure of central tendency. It is defined as the sum of the values of all observations divided by the number of observations

2. Median:
Median is that positional value of the variable which divides the distribution into two equal parts, one part comprises all values greater than or equal to the median value and the other comprises all values less than or equal to it. The Median is the “middle” element when the data set is arranged in order of the magnitude.

3. Mode:
The word mode has been derived from the French word “la Mode” which signifies the most fashionable values of distribution because it is repeated the highest number of times in the series. Mode is the most frequently observed data value. It is denoted by Mo.

Question 3.
What are the merits and demerits of arithmetic mean?
Answer:
1. Merits

It is simple to calculate
It is regidly defined
It is easy to understand
It is based on all observations

2. Demerits

It is affected by extreme values.
It cannot be calculated in open-end series.
It cannot be determined graphically.
It may sometimes give misleading results.

Plus One Economics Measures of Central Tendency Eight Mark Questions and Answers

Question 1.
Prepare a list of peculiarities of median, quartiles, and percentiles.
Answer:
1. Median

The arithmetic mean is affected by the presence of extreme values in the data.
If you take a measure of central tendency which is based on middle position of the data, it is not affected by extreme items.
Median is that positional value of the variable which divides the distribution into two equal parts, one part comprises all values greater than or equal to the median value and the other comprises all values less than or equal to it.
The Median is the middle element when the data set is arranged in order of the magnitude.

2. Quartiles

Quartiles are the measures that divide the data into four equal parts; each portion contains equal number of observations. Thus, there are three quartiles.
The first Quartile (denoted by Q₁) or lower quartile has 25% of the items of the distribution below it and 75% of the items are greater than it.
The second Quartile (denoted by Q₂) or median has 50% of items below it and 50% of the observations above it.
The third Quartile (denoted by Q₃) or upper Quartile has75% of the items of the distribution below it and 25% of the items above it.
Thus, Q₁ and Q₃ denote the two limits within which central 50% of the data lies.

3. Percentiles

Percentiles divide the distribution into hundred equal parts, so you can get 99 dividing positions denoted by P₁ P₂, P_{3, ……….}, P₉₉.
P50 is the median value.
If you have secured 82 percentile in a management entrance examination, it means that your position is below 18 percent of total candidates appeared in the examination.

Question 2.
The daily sales of car of 20 distributing companies is given below. Calculate:

Median, upper quartile and lower quartile.
Interpret the result obtained.

Daily Sales	No. of Companies
0-20	1
20-40	3
40-60	9
60-80	5
80-100	2
	20

Answer:
1.
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Measures of Central Tendency img 7

2. The median divides the values into two equal parts. Lower quartile (Q₁) divides the values into 1/4 and upper quartile (Q₃) divides the values into 3/4.

Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

April 13, 2021February 13, 2020 by Prasanna

Students can Download Chapter 13 Probability Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 13 Probability

Plus Two Maths Probability Three Mark Questions and Answers

Question 1.
Determine P(E/F). A die is thrown three times, E: ‘4 appears on the third toss’, F: ‘6 and 5 appears respectively on the two tosses’.
Answer:
n(S) = 6³ = 216
E = {( 1, 1, 4), (1, 2, 4), (1, 3, 4)……….(1, 6, 4),
(2, 1, 4), (2, 2, 4), (2, 3, 4)……..(2, 6, 4),
(3, 1, 4), (3, 2, 4), (3, 3, 4)……..(3, 6, 4),
(4, 1, 4), (4, 2, 4), (4, 3, 4)…….(4, 6, 4),
(5, 1, 4), (5, 2, 4), (5, 3, 4)……..(5, 6, 4),
(6, 1, 4), (6, 2, 4), (6, 3, 4)……..(6, 6, 4)}
F = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
⇒ E ∩ F = {(6, 5, 4)}
P(F) = \(\frac{6}{216}\) and P(E ∩ F) = \(\frac{1}{216}\)
Then, P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}\).

Question 2.
Determine P(E/F). Mother, Father and son lineup at random for a photograph.
E: ‘Son on one end’, F: ‘ Father in middle.
Answer:
Let Mother-M, Father-F and Son-S.
n(S) = 3! = 6
E = {SMF, SFM, MFS, FMS},
F = {MFS, SFM}
⇒ E ∩ F = {SFM, MFS}
P(F) = \(\frac{2}{6}\) = \(\frac{1}{3}\) and P(E ∩ F) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Then, P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{3}}{\frac{1}{3}}=1\).

Question 3.
A black and a red dice are rolled

Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Answer:
We have, n(S) = 36
1. E = Event of 5 on black die.
E = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
P(E) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
F = Getting a sum greater than 9.
F = {(4, 6), (5, 5), (6, 4)(5, 6), (6, 5), (6, 6)}
⇒ E ∩ F = {(5,5), (5,6)}
P(E ∩ F) = \(\frac{2}{36}\)= \(\frac{1}{18}\)
Therefore the required probability
P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{18}}{\frac{1}{6}}=\frac{1}{3}\).

2. E = Event of a number less than 4 on red die.
E = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3),
(3, 1), (3, 2), (3, 3), (3, 1), (3, 2), (3, 3),
(5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
P(E) = \(\frac{18}{36}\) = \(\frac{1}{2}\)
F = Getting a sum 8.
F = {(4, 4), (5, 3), (3, 5)(2, 6), (6, 2), (6, 6)}
⇒ E ∩ F = {(5, 3),(6, 2)}
P(E ∩ F) = \(\frac{2}{36}\)= \(\frac{1}{18}\)
Therefore the required probability
P(E/F) = \(\frac{P(E \cap F)}{P(E)}=\frac{\frac{1}{18}}{\frac{1}{2}}=\frac{1}{9}\).

Question 4.
An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult true/False questions, 500 easy multiple choice questions, and 400 difficult multiple choice questions. If a question is selected from the test question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Answer:
Describe the events as follows
E: ‘getting an easy question.’
F: ‘getting a multiple choice question.’
Total Questions = 300 + 200 + 500 + 400 = 1400
n(F) = 500 + 400 = 900, n(E ∩ F) = 500
P(F) = \(\frac{900}{1400}=\frac{9}{14}\), P(E ∩ F) = \(\frac{500}{1400}=\frac{5}{14}\)
Therefore the required probability
= P(E/F) = \(\frac{P(E \cap F)}{P(F)}=\frac{\frac{5}{14}}{\frac{9}{14}}=\frac{5}{9}\).

Question 5.
Two cards are drawn at random without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Answer:
Describe the events as follows
B1: ‘getting a black card in the first draw.’
B2: ‘getting a black card in the second draw.’
P(B1) = \(\frac{26}{52}\) = \(\frac{1}{2}\)
When the first event is executed and since no replacement is allowed, the remaining total number of cards become 51 and black cards become 25.
P(B2/B1) = \(\frac{25}{51}\)
Therefore the required probability
P(B1 ∩ B2) = P(Bl) × P(B2/B1) = \(\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}\).

Question 6.
If P(A) = \(\frac{6}{11}\), P(B) = \(\frac{5}{11}\) and P(A ∪ B) = \(\frac{9}{11}\) Find

P(A ∩ B)
P(A/B)
P(B/A)

Answer:
1. P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

2. P(A/B)
Plus Two Maths Probability 3 Mark Questions and Answers 2

3. P(B/A)
Plus Two Maths Probability 3 Mark Questions and Answers 3

Question 7.
Events A and B are such that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\) and P(not A or not B) = \(\frac{1}{4}\) State whether A and B are independent.
Answer:
P(not A or not B) = \(\frac{1}{4}\) ⇒ \(P(\bar{A} \cup \bar{B})=\frac{1}{4}\)
Plus Two Maths Probability 3 Mark Questions and Answers 4
We have, P(A) × P(B) = \(\frac{1}{2} \times \frac{7}{12}=\frac{7}{24}\)
Therefore, P(A ∩ B) ≠ P(A) × P(B)
Hence A and B are not independent.

Question 8.
Consider two events such that P(A) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{3}{5}\) and P(B) = p. Find p, if A and B are independent events.
Answer:
If A and B are independent then
P(A ∩ B) = P(A) × P(B)
We have,
P(A ∪ B) = P(A) + P(B) – P(A) × P(B)
Plus Two Maths Probability 3 Mark Questions and Answers 5

Question 9.
One card is drawn at random from a well shuffled pack of 52 cards. In which of the following cases are the events E and F independent? (3 scores each)

E: ‘the card drawn is a spades.’
F: ‘the card drawn is an ace.’
E: ‘the card drawn is a black.’
F: ‘the card drawn is a king.’
E: ‘the card drawn is a king or a queen.’
F: ‘the card drawn is queen or a jack.’

Answer:
1. P(E) = \(\frac{13}{52}=\frac{1}{4}\), P(F) = \(\frac{4}{52}=\frac{1}{13}\)
There is only one card which is an ace of spade.
P(E ∩ F) = \(\frac{1}{52}\)
We have,
P(E) × P(F) = \(\frac{1}{4} \times \frac{1}{13}=\frac{1}{52}\) = P(E ∩ F)
Hence E and F are independent events.

2. P(E) = \(\frac{26}{52}=\frac{1}{2}\), P(F) = \(\frac{4}{52}=\frac{1}{13}\)
There are two king of black.
P(E ∩ F) = \(\frac{2}{52}\) = \(\frac{1}{26}\)
We have,
P(E) × P(F) = \(\frac{1}{2} \times \frac{1}{13}=\frac{1}{26}\) = P(E ∩ F)
Hence E and F are independent events.

3. There are 4 king and 4 queen cards
P(E) = \(\frac{8}{52}\) = \(\frac{2}{13}\),
There are 4 queen and 4 jack cards.
P(F) = \(\frac{8}{52}\) = \(\frac{2}{13}\)
There 4 queen common for both.
P(E ∩ F) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
We have,
P(E) × P(F) = \(\frac{2}{13} \times \frac{2}{13}=\frac{4}{169}\) ≠ P(E ∩ F)
Hence E and F are not independent events.

Question 10.
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.
Answer:
P(A) = \(\frac{1}{2}\) and P(B) = \(\frac{1}{6}\)
When a coin and die are tossed the sample space will be as follows.
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Here head and 3 come only once.
⇒ P(A ∩ B) = \(\frac{1}{12}\)
P(A) × P(B) = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\) = P(A ∩ B)
Hence A and B are independent.

Question 11.
Rani and Joy appear in an interview for two vacancies in the same post. The probability of Rani’s selection is \(\frac{1}{7}\) and that of Joy’s selection is \(\frac{1}{5}\) .What is the probability that

Rani will not be selected? (1)
Both of them will be selected? (1)
None of them will be selected? (1)

Answer:
1. Let Rani’s selection be the event A and Joy’s selection be the event B.P(Rani will not be selected)

2. P(Both of them will be selected)
P(A ∩ B) = P(A).P(B) = \(\frac{1}{7} \cdot \frac{1}{5}=\frac{1}{35}\).

3. P(None selected)

Question 12.
Find the probability distribution of number heads in two tosses of a coin.
Answer:
S = {HH, HT, TH, TT}
Let X denotes the random variable of getting a head. Then X can take values 0, 1, 2.
P(X = 0) = P(no heads) = P({TT})
= P(T) × P(T) = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
P(X = 1) = P(one heads)
= P({HT, TH})
= P(H) × P(T) + P(T) × P(H)
= \(\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}=\frac{1}{2}\)
P(X = 2) = P(two heads) = P({HH})
= P(H) × P(H) = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
The required Probability Distribution is
Plus Two Maths Probability 3 Mark Questions and Answers 8

Question 13.
Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.
Answer:
Let X denotes the random variable of number of defective eggs in the 10 eggs drawn.
Clearly X has a Binomial Distribution with n = 10 and p = 10% = \(\frac{1}{10}\), q = 1 – p = \(\frac{9}{10}\)
⇒ P(X = x) = 10C_xq^10-xp^x = 10C_x\(\left[\frac{9}{10}\right]^{10-x}\left[\frac{1}{10}\right]^{x}\)
P(at least 1 defective egg) = P(X ≥ 1)
1 – P(X = 0) = 1 – 10C₀\(\left[\frac{9}{10}\right]^{10}=1-\frac{9^{10}}{10^{10}}\).

Plus Two Maths Probability Four Mark Questions and Answers

Question 1.
In a hostel 50 % of the girls like tea, 40 % like coffee and 20% like both tea and coffee. A girl is selected and random.

Find the probability that she likes neither tea nor coffee. (2)
If the girl likes tea, then find the probability that she likes coffee. (1)
If she likes coffee then find the probability she likes tea. (1)

Answer:
Let T denotes the set of girls who like tea and C denotes who like coffee.
1. P(T) = 50% = \(\frac{1}{2}\); P(C) = 40% = \(\frac{2}{5}\);
P(T ∩ C) = 20% = \(\frac{1}{5}\)
P(T ∪ C)^‘ = 1 – P(T ∪ C)
= 1 – {P(T) + P(C)-P(T ∩ C)}
\(=1-\left\{\frac{1}{2}+\frac{2}{5}-\frac{1}{5}\right\}=\frac{3}{10}\).

2.
Plus Two Maths Probability 3 Mark Questions and Answers 9

3.
Plus Two Maths Probability 3 Mark Questions and Answers 10

Question 2.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Answer:
For the box to be approved all the three oranges should be selected from the 12 good ones. Since the events are executed without replacement the number for good oranges and total oranges reduce by one on each draw.
O1: ‘getting a good orange in the first draw.’
O2: ‘getting a good orange in the second draw.’
O3: ‘getting a good orange in the third draw.’
P(good orange in the first draw) =P(O1)= \(\frac{12}{15}\) = \(\frac{4}{5}\),
P(good orange in the second) = p(O2/O1) = \(\frac{11}{14}\),
P(good orange in the third)
= P(O3/(O1 ∩ O2)) = \(\frac{10}{13}\)
Therefore the required probability
= P(O1 ∩ O2 ∩ O3)
= P(O1)P(O2/O1)P(O3/(O1 ∩ O2))
\(=\frac{4}{5} \times \frac{11}{14} \times \frac{10}{13}=\frac{44}{91}\).

Question 3.
Let two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find

P(A and B)
P(A and not B)
P(A or B)
P (neither A nor B)

Answer:
1. P(A and B) = P(A ∩ B) = P(A) × P(B)
= 0.3 × 0.6 = 0.18.

2. P(A and not B) = P(A ∩ \(\bar{B}\)) = P(A) × P(\(\bar{B}\))
= 0.3 × 0.4 = 0.12.

3. P(A or B) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) × P(B)
= 0.3 + 0.6 – 0.3 × 0.6 = 0.72.

4. P(neither A nor B) = \(P(\bar{A} \cap \bar{B})\)
= P(\(\bar{A}\)) × P(\(\bar{B}\)) = 0.7 × 0.4 = 0.28.

Question 4.
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

both balls are red.
the first ball is a black and the second is red
one of them is black and the other red.

Answer:
Describe the events as follows.
Black-B and Red-R.n(S) = 18,
P(B) = \(\frac{10}{18}\) = \(\frac{5}{9}\) and P(R) = \(\frac{8}{18}\) = \(\frac{4}{9}\)
Since the event is executed with replacement, is independent.
1. P( both ball is red) = P(R) × P(R)
= \(\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}\).

2. P( first black and second red) = P(B) × P(R)
= \(\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}\).

3. P( one of them is a black and the other red)
= P(B) × P(R) + P(R) × P(B)
\(=\frac{5}{9} \times \frac{4}{9}+\frac{4}{9} \times \frac{5}{9}=\frac{40}{81}\).

Question 5.
Bag 1 contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the Bags and it is found to be red. Find the probability that it was from Bag II.
Answer:
Describe the events as follows.
A: ‘getting a defective ball’.
E₁: ‘choosing Bag I.’
E₂: ‘choosing Bag II.’
P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = P (drawing a red ball from Bag I) = \(\frac{3}{7}\)
P(A/E₂) = P (drawing a red ball from Bag II) = \(\frac{5}{11}\)
P (a ball from Bag II, being given that it is red)
Plus Two Maths Probability 4 Mark Questions and Answers 11

Question 6.
A Bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One ball of the two Bag is drawn at random and the ball is drawn the Bag is found to be red. Find the probability that the ball is drawn from first Bag.
Answer:
Describe the events as follows.
A: ‘getting a red ball’.
E₁: ‘choosing Bag I.’
E₂: ‘choosing Bag II.’
P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = P (a red ball from Bag I) = \(\frac{4}{8}\) = \(\frac{1}{2}\)
P(A/E₂) = P (a red ball from Bag II) = \(\frac{2}{8}\) = \(\frac{1}{4}\)
P (a ball from Bag II, being given that it is red)
Plus Two Maths Probability 4 Mark Questions and Answers 12

Question 7.
In a factory which manufactures blots, machines A, B, and C manufacture respectively 25%, 35%and 40% of the bolts. Of their outputs 5%, 4% and 2%are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?
Answer:
Describe the events as follows.
A: ‘getting a defective bolt’.
E₁: ‘choosing machine A.’
E₂: ‘choosing machine B.’
E₃: ‘choosing machine C.’
P(E₁) = 0.25, P(E₂) = 0.35, P(E₃) = 0.4
P(A/E₁) = P
(a defective bolt from machine A)
= 5% = 0.05
P(A/E₂) = P
(a defective bolt from machine B)
= 4% = 0.04
P(A/E₃) = P
(a defective bolt from machine C)
= 2% = 0.02
P (a bolt from machine B, being given that it is defective)
= P(E₂/A)
Plus Two Maths Probability 4 Mark Questions and Answers 13

Question 8.
Suppose 5% of men and 0.25% of women have hair grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume there are equal number males and females.
Answer:
Describe the events as follows.
A: ‘person is grey haired’.
E₁: ‘choosing man.’
E₂: ‘choosing woman.’
P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = P (a grey haired person from men)
= 5% = 0.05
P(A/E₂) = P (a grey haired person from women) = 0.25% = 0.0025
P(selecting a male, being given that it is grey haired)
Plus Two Maths Probability 4 Mark Questions and Answers 14

Question 9.
An Insurance company insured 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. The probabilities of an accident are .01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer:
Describe the events as follows.
A: ‘accident happens’.
E₁: ‘choosing Scooter driver.’
E₂: ‘choosing Car driver.’
E₃: ‘choosing Truck driver.’
Total drivers = 2000 + 4000 + 6000 = 12000
Plus Two Maths Probability 4 Mark Questions and Answers 15
P(A/E₁) = P
(accident of a Scooter driver) = 0.01
P(A/E₂) = P
(accident of a Car driver) = 0.03
P(A/E₃) = P
(accident of a Truck driver) = 0.15
P (accident happens, given that it is a Scooter driver).
= P(E₁/A)

Question 10.
A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Answer:
Describe the events as follows.
A: ‘getting a defective item.’
E₁: ‘choosing machine A.’
E₂: ‘choosing machine B.’
P(E₁) = 60% = 0.6, P(E₂) = 40% = 0.4
P(A/E₁) = P (a defective from machine A) = 2% = 0.02
P(A/E₂) = P (a defective from machine B) = 1% = 0.01
P (a defective from machine B)
Plus Two Maths Probability 4 Mark Questions and Answers 17

Question 11.
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Answer:
Describe the events as follows.
A: ‘getting exactly one head.’
E₁: ‘she getting 5 or 6.’
E₂: ‘she getting 1, 2, 3 or 4.’
When a die is thrown the sample space is{1, 2, 3, 4, 5, 6}
P(E₁) = \(\frac{2}{6}\) = \(\frac{1}{3}\), P(E₂) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
When she gets 5 or 6, throws a coin 3 times. Then sample space is {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
P(A/E₁) = P (one head given that 5 or 6 happened)
\(\frac{3}{8}\)
When she gets 1, 2, 3 or 4, throws a coin once. Then sample space is {H, T}
P(A/E₂) = P
(one head given that 1, 2, 3 or 4 happened)
\(\frac{1}{2}\)
P (She gets exactly one head threw 1, 2, 3 or 4)
Plus Two Maths Probability 4 Mark Questions and Answers 18

Question 12.
Vineetha and Reshma are competing for the post of school leader. The probability Vineetha to be elected is 0.6 and that of Reshma is 0.4 Further if Vineetha is elected the probability of introducing a new pattern of election is 0.7 and the corresponding probability is 0.3 if Resma is elected. Find the probability that the new pattern of election is introduced by Reshma.
Answer:
Let E₁ and E₂ be the respectively probability that Vineetha and Reshma will be elected. Let Abe the probability that a new pattern of election is introduced.
P(E₁) = 0.6; P(E₂) = 0.4
P(A|E₁) = 0.7; P(A|E₂) = 0.3
Plus Two Maths Probability 4 Mark Questions and Answers 19

Question 13.
Find the probability of number of doublets in three throws of a pair of dice.
Answer:
Let X denotes the random variable of getting a Doublet. Possible doublets are (1, 1),(2, 2),(3, 3),(4, 4),(5, 5),(6, 6).
Then X can take values 0, 1, 2, 3.
P(getting a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
P(not getting a doublet) = \(\frac{30}{36}\) = \(\frac{5}{6}\)
P(X = 0) = P(no doublet) = \(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{125}{216}\)
P(X = 1) = P(one doublet and 2 non-doublets)

P(X = 2) = P(2 doublet and 1 non-doublets)
Plus Two Maths Probability 4 Mark Questions and Answers 21
P(X = 3) = P(3 doublet) = \(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{216}\)
The required Probability Distribution is

Question 14.
Find the probability distribution of the number of white balls drawn when three balls are drawn one by one without replacement from a bag containing 4 white and 6 red balls.
Answer:
Let X denotes the random variable of number of white balls. Clearly X can take values 0, 1, 2, 3. Describe the events as follows.
W: ‘getting white ball.’
R: ‘getting red ball.’
P(X=0) = P(no white balls)
= P(RRR)= \(\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}=\frac{5}{30}\)
P(X=1) = P(1white, 2red balls)
= P(WRR) + P(RWR) + P(RRW)
Plus Two Maths Probability 4 Mark Questions and Answers 23
P(X=2) = P(2white, 1 red balls)
= P(WWR) + P(RWW) + P(WRW)

P(X=3) = P(3white)
P(WWW) = \(\frac{4}{10} \times \frac{3}{9} \times \frac{2}{8}=\frac{1}{30}\)
The required Probability Distribution is

Question 15.
Two dice are thrown simultaneously. If X denotes the number of sixes, find expectation of X. Also find the variance.
Answer:
Let X denotes the random variable of getting a 6. Clearly X can take values 0, 1, 2.
P(X = 0) = P(non-six, non-six) = \(\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}\)
P(X = 1) = P((six, non-six),(non-six, six))
Plus Two Maths Probability 4 Mark Questions and Answers 26
P(X = 2) = P(six, six) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)
The required Probability Distribution is

Variance = σ² = E(X²) – [E(X)]²

Question 16.
A random variable X has the following probability distribution
Plus Two Maths Probability 4 Mark Questions and Answers 29
Determine

k
P(X < 3)
P(X > 6)
P(0 < X < 3)

Answer:
1. We know that sum of the probabilities is = 1
0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
10k² + 9k – 1 = 0
(k + 1)(10k – 1) = 0; k= -1 or k = \(\frac{1}{10}\)
(negative value cannot be accepted).

2. P(X < 3) = P(0) + P( 1) + P(2)
0 + k + 2k + 2k = 3k = \(\frac{3}{10}\).

3. P(X < 3) = P(7) = 7k² + k
Plus Two Maths Probability 4 Mark Questions and Answers 30

4. P(0 < X < 3) = P(1) + P(2) = k + 2k = 3k = \(\frac{3}{10}\). Question

Question 17.
(i) P(A) = \(\frac{7}{13}\); P(B) = \(\frac{9}{13}\); (A ∩ B) = \(\frac{4}{13}\), then P(A/B) is
(a) \(\frac{9}{4}\)
(b) \(\frac{16}{13}\)
(c) \(\frac{4}{9}\)
(d) \(\frac{11}{13}\)
(ii) Probability of solving a specific problem independently by A and B are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. If both try to solve the problem independently, then
(a) Find the probability that the problem is solved. (2)
(b) Find the probability that exactly one of them solve the problem. (1)
Answer:
(i) (c) \(\frac{4}{9}\)

(ii)
Plus Two Maths Probability 4 Mark Questions and Answers 31
P(Problem is solved)
= P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

(iii) P(Exactly one of them solve)
Plus Two Maths Probability 4 Mark Questions and Answers 33

Plus Two Maths Probability Six Mark Questions and Answers Question

Question 1.
(i) A and B are two events such that P(A) = \(\frac{1}{5}\) and P(A ∪ B) = \(\frac{2}{5}\) Find P(B) if they are mutually exclusive
(a) \(\frac{1}{5}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{5}\)
(d) \(\frac{4}{5}\)
(ii) A box contains 3 red and 4 blue balls. Two balls are drawn one by one without replacement. Find the probability of getting both balls red.
(iii) Three cards are drawn successively without replacement from a pack of 52 cards. What is the probability that first two cards are queen and the third is king.
Answer:
(i) (a) \(\frac{1}{5}\).

(ii) Let A be the event that the first ball drawn is red and B be the event of drawing red ball in the second draw
P(A) = \(\frac{3}{7}\)
Probability of getting one red ball in the second draw = P(B/A) = \(\frac{2}{6}\) = \(\frac{1}{3}\). P(A ∩ B) = P(A).P(B/A)
\(=\frac{3}{7} \times \frac{1}{3}=\frac{1}{7}\).

(iii) Let Q denote the event that the card drawn is Queen and K denote the event of drawing a King
P(Q) = \(\frac{4}{52}\), P(Q/Q) = \(\frac{3}{51}\)
P(K/QQ) is the probability of drawing the third card is a king
P(K/QQ) = \(\frac{4}{50}\)
P(QQk) = P(Q)P(Q/Q)P(K/QQ)
\(=\frac{4}{52} \cdot \frac{3}{51} \cdot \frac{4}{50}=\frac{2}{5525}\).

Question 2.
60 shirts of different colours are on sale. If one shirt is chosen at random.
Plus Two Maths Probability 4 Mark Questions and Answers 34

What is the probability that it is red? (1)
What is the probability that it is plain and extra-large? (1)
What is the probability that it is small, given that it is blue? (2)
If A is the event ‘the shirt is medium’ and B is the event ‘the shirt is blue’. Are the events A and B independent? (2)

Answer:
1. P(Red) = \(\frac{8+8+2+4}{60}=\frac{11}{30}\).

2. P (Plain and extra-large) = \(\frac{4+5}{60}=\frac{3}{20}\).

3. P(small/blue)
Plus Two Maths Probability 4 Mark Questions and Answers 35

4.
Plus Two Maths Probability 4 Mark Questions and Answers 36
∴ Not independent.

Question 3.
From a box containing balls numbered from, 1 to 100, one ball is drawn at random. The events X and Y are as follows. X: A perfect square is drawn. Y: An even number is drawn.

Find P(X) and P(Y). (2)
Compute P (X/Y). (2)
Are X and V independent? Justify. (2)

Answer:
Perfect square numbered balls are
1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Therefore, there are 10 perfect square numbered balls, 50 even numbered balls and 5 perfect square even numbered balls.
1. P(X) = \(\frac{10}{100}=\frac{1}{10}\), P(Y) = \(\frac{50}{100}=\frac{1}{2}\)
P (Even perfect square number) = P (X ∩ Y) \(\frac{50}{100}=\frac{1}{20}\).

2. P(X/Y)
= P(Drawing a perfect square numbers from even numbers) = \(\frac{5}{50}=\frac{1}{10}\).

3. We have, P (X ∩ Y) = \(\frac{1}{20}=\frac{1}{10} \times \frac{1}{2}\)
= P(X).P(Y).
Therefore X and Y are independent events.

Question 4.
(i) The probability of three mutually exclusive events A, B, and C are given by 2/3, 1/4, 1/6 respectively. Is this statement ________ (1)
(a) true?
(b) false?
(c) cannot be said?
(d) data not sufficient?
A husband and wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is 1/7 and that of wife’s selection is 1/5. What is the probability that
(ii) Only one of them will be selected? (3)
(iii) None will be selected? (2)
Answer:
(i) (b) Probability should be less than or equal to 1.
Here A, B, C are mutually exclusive.
Then,
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = \(\frac{2}{3}+\frac{1}{4}+\frac{1}{6}=\frac{26}{24}>1\)
∴ statement is false.

(ii) H – Event of husband selected,
W – Event of wife selected
Plus Two Maths Probability 4 Mark Questions and Answers 37

(iii) P (None of them will be selected)
Plus Two Maths Probability 4 Mark Questions and Answers 38

Question 5.
(i) Find P(A∩B) if A and B are independent events with P(A) = \(\frac{1}{5}\) and P(B) = \(\frac{5}{8}\)
Plus Two Maths Probability 4 Mark Questions and Answers 39
(ii) An unbiased die is thrown twice. Let the event A be getting prime number in the first throw and B be the event of getting an even number in the second throw. Check the independence of the events A and B. (3)
(iii) The probability of solving a problem independently by A and B are \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively. Find the probability that exactly one of them solves the problem. (2)
Answer:
(i) (c) \(\frac{1}{8}\).

(ii) P(A) = \(\frac{18}{36}\) = \(\frac{1}{2}\)
P(B) = \(\frac{18}{36}\) = \(\frac{1}{2}\)
P(A∩B) = P( prime number in first throw and even number in the second throw)
Plus Two Maths Probability 4 Mark Questions and Answers 40
∴ A and B are independent events.

(iii)
Plus Two Maths Probability 4 Mark Questions and Answers 41
Probability of exactly one of them solves the problem = P(A)P(B’) + P(B)P(A’)

Question 6.
(i) A set of events E₁ + E₂,…….E_n are said to be a partition of the Sample Space, then which of the following conditions is always not true (1)
(a) E₁∪ E₂∪……..∪ E_n = S,
(b) E₁∩ E_n = Φ,
(c) P(E₁) > 0,
(d) P(E₁) ≥ P(E_n)
(ii) A person has undertaken a business. The probabilities are 0.80 that there will be a crisis, 0.85 that the business will be completed on time if there is no crisis and 0.35 that the business will be completed on time if there is a crisis. Determine the probability that the business will be completed on time. (2)
(iii) A box contains 5 red and 10 black balls. A ball is drawn at random, its colour is noted and is returned to the box. More over 2 additional balls of the colour drawn are put in the box and then a ball is drawn. What is the probability that the second ball is red? (3)
Answer:
(i) P(E₁) ≥ P(E_n).

(ii) Let A be the event that the business will be completed on time and B be the event that there will be a crisis
P(B) = 0.80
P(no crisis) = P(B’) = 1 – P(B) = 0.20
P(A/B) = 0.35 P(A/B’) = 0.85
By theorem on total probability
P(A) = P(B)P(A/B) + P(B’)P(A/B’)
= 0.8 × 0.35 + 0.2 × 0.85 = 0.45.

(iii) Let a red ball be drawn in the first attempt P(drawing a red ball) = \(\frac{5}{15}=\frac{1}{3}\)
If two red balls are added to the box, then the box contains 7 red balls and 10 black balls
P(drawing a red ball) = \(\frac{7}{17}\)
Let a black ball be drawn in the first attempt P(drawing a black ball) = \(\frac{10}{15}=\frac{2}{3}\)
If two black balls are added to the box, then the box contains 5 red and 12 black balls
P(drawing a red ball) = \(\frac{5}{17}\)
Probability of drawing the second ball red is
Plus Two Maths Probability 4 Mark Questions and Answers 43

Question 7.

Bag I contains 5 red and 6 black balls. Bag II contains 7 red and 5 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag I. (3)
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being diamond. (3)

Answer:
1. Let E₁ be the event of choosing Bag I and E₂ be the event of choosing Bag II
A be the event of drawing a red ball
Plus Two Maths Probability 4 Mark Questions and Answers 44

2. Let E₁ be the event of choosing a diamond and E₂ be the event of choosing a non diamond card
A be the event that a card is lost

When a diamond card is lost, there are 12 diamond cards in 52 cards. Then
Plus Two Maths Probability 4 Mark Questions and Answers 46
When a non diamond card is lost, there are 13 diamond cards in 51 cards. Then

Question 8.
(i) If X denotes number of heads obtained in tossing two coins. Then which of the following is false (1)
(a) X(HH) = 2
(b) X(HT) = 1
(c) X(TH)= 0
(d) X(TT) = 0
(ii) Find the probability distribution of the number of tails in the simultaneous toss of two coins. (2)
(iii) A coin is tossed so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails. (3)
Answer:
(i) (c) X(TH)= 0.

(ii) Sample space is S = {HH, HT, TH, TT}
Let X denote the number of tails, then
X(HH) = 2, X(HT) = 1, X(TH) = 1, X(TT) = 0
Therefore X can take the values 0, 1 or 2
P(HH) = P(HT) = P(TH) = P(TT) = \(\frac{1}{4}\)
P(X = 0) = \(\frac{1}{4}\) P(X = 1) = \(\frac{1}{2}\) P(X = 2) = \(\frac{1}{4}\)
Then the Probability distribution is
Plus Two Maths Probability 4 Mark Questions and Answers 48

(iii) Let the probability of getting a tail in the biased coin be x.
P(T) = x P(H) = 3x
P(T) + P(H) = 1 ⇒ x + 3x = 1 x = \(\frac{1}{4}\)
P(T) = \(\frac{1}{4}\) P(H) = \(\frac{3}{4}\)
Let X denote the random variable representing the number of tails
P(X = 0) = P(HH) = P(H).P(H) = \(\frac{9}{16}\)
P(X = 1) = P(HT) + P(TH)
Plus Two Maths Probability 4 Mark Questions and Answers 49
P(X = 2) = P(TT) = P(T).P(T) = \(\frac{1}{16}\)
Then the Probability distribution is

Question 9.
If a fair coin is tossed 10 times, find the probability of

Exactly 6 heads. (2)
At least 6 heads. (2)
At most 6 heads. (2)

Answer:
Let X denotes the random variable of number of heads in an experiment of 10 trials.
Clearly X has a Binomial Distribution with
n = 10 and p = \(\frac{1}{2}\) ⇒ P(x) = nC_xq^n-x P^x
Plus Two Maths Probability 4 Mark Questions and Answers 51
1. P(x = 6)

2. P(at least 6 heads) = P(X ≥ 6)
= P(X = 6) + P(X = 1)+P(X = 8)
Plus Two Maths Probability 4 Mark Questions and Answers 53

P(at most 6 heads) = P(X ≤ 6)
= P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)
Plus Two Maths Probability 4 Mark Questions and Answers 54

Question 10.
Five cards are drawn successively with a replacement from a pack of 52 cards. What is the probability that

All the 5 cards are spades? (2)
only 3 cards are spade? (2)
none is a spade? (2)

Answer:
Let X denotes the random variable of number of spades cards in an experiment of 5 trials. Clearly X has a Binomial Distribution with n = 5
Plus Two Maths Probability 4 Mark Questions and Answers 56
1. P(all 5 are spades) = P(X = 5)

2. P(3 are spades) = P(X = 3)
Plus Two Maths Probability 4 Mark Questions and Answers 58

3. P(non-spade) = P(X = 0) =
Plus Two Maths Probability 4 Mark Questions and Answers 59

Question 11.
Find the probability distribution, Mean and Variance of the number of success in two tosses of a die, where a success is defined as

number greater than 4. (3)
6 appears on at least on die. (3)

Answer:
1. Let X denotes the random variable of getting a 5, 6. Clearly X can take values 0, 1, 2.
When number 1, 2, 3, 4 appears in both die. Number of such cases = 4 × 4 = 16
P(X=0) = P(no success) = \(\frac{16}{36}=\frac{4}{9}\)
When 5, 6 in one die and other with 1, 2, 3, 4 and visa versa.
Number of such cases is 2 × 4 + 4 × 2 = 16.
P(X= 1) = P(1 success and 1 no success) = \(\frac{16}{36}=\frac{4}{9}\)
When number 5, 6 appears in Jjoth die. Number Hof such cases = 2 × 2 = 4.
P(success) = \(\frac{4}{36}=\frac{1}{9}\)
The required Probability Distribution is
Plus Two Maths Probability 4 Mark Questions and Answers 60

2. Let X denotes the random variable of getting at least 6 on one die. Clearly X can take values 0, 1.
No success means 1, 2, 3, 4, 5 appears on both die. Number of such cases is 5 × 5 = 25.
P(X=0) = P(no success) = \(\frac{25}{36}\)
When 6 in one die and other with 1, 2, 3, 4, 5 and visa versa. Number of such cases is 1 × 5 + 5 × 1 = 10
When both the die is 6. Number of such case is 1. Therefore total cases is 1 + 10 = 11
P(X=1) = P(1 success) = P(at least 1 six) = \(\frac{11}{36}\)
The required Probability Distribution is
Plus Two Maths Probability 4 Mark Questions and Answers 61

Question 12.
(i) If A and B are two events such that A ⊂ B and P(A) ≠ 0 then P(A/B) is (1)
Plus Two Maths Probability 4 Mark Questions and Answers 62
(ii) There are two identical bags. Bag I contains 3 red and 4 black balls while Bag II contains 5 red and 4 black balls. One ball is drawn at random from one of the bags.
(a) Find the probability that all the ball drawn are red. (3)
(b) If the balls drawn is red what is the probability that it was drawn from Bag I? (2)
Answer:
(i) Describe the events as follows.
A: ‘getting a red ball’.
E₁: ‘choosing Bag I.’
E₂: ‘choosing Bag II.’
P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = P
(drawing a red ball from Bag I) = \(\frac{3}{7}\)
P(A/E₂) = P
(drawing a red ball from Bag II) = \(\frac{5}{9}\)
P (All the balls drawn is red)
= P(A) = P(E₁)P(A/E₁) + P(E₂)P(A/E₂)
Plus Two Maths Probability 4 Mark Questions and Answers 63

Question 13.
Consider the following probability distribution of a random, variable X.
Plus Two Maths Probability 4 Mark Questions and Answers 65

Find the value of k. (2)
Determine the Mean and Variance of X. (4)

Answer:
(i) We have; Σp_i = 1
Plus Two Maths Probability 4 Mark Questions and Answers 66

(ii)
Plus Two Maths Probability 4 Mark Questions and Answers 67

Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear Programming

April 13, 2021February 11, 2020 by Prasanna

Students can Download Chapter 12 Linear Programming Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 12 Linear Programming

Plus Two Maths Linear Programming Four Mark Questions and Answers

Question 1.
Solve the following LPP Graphically;
Maximise; Z = 60x + 15y
Subject to constraints;
x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0.
Answer:
1. In the figure the shaded region OABC is the fesible region. Here the region is bounded. The corner points are O(0, 0), A(30, 0), B(20, 30), C(0, 50).
Plus Two Maths Linear Programming 4 Mark Questions and Answers 1
Given; Z = 60x + 15y

Corner points	Value of Z
O	Z = 0
A	Z = 60(30) + 15(0) = 1800
B	Z = 60(20) + 15(30) = 1650
C	Z = 60(0) + 15(50) = 750

Since maximum value of Z occurs at A, the soluion is Z = 1800, (30, 0).

Question 2.
Solve the following LPP Graphically;
Minimise; Z = -3x + 4y
Subject to constraints;
x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Answer:
In the figure the shaded region OABC is the fesible region. Here the region is bounded. The corner points are O(0, 0), A(4, 0) B(2, 3), C(0, 4).
Plus Two Maths Linear Programming 4 Mark Questions and Answers 2
Given; Z = -3x + 4y

Corner points	Value of Z
O	Z = 0
A	Z = -3(4) + 4(0) = -12
B	Z = -3(2) + 4(3) = 6
C	Z = -3(0) + 4(4) = 16

Since minimum value of Z occurs at A, the soluion is Z = -12, (4, 0).

Question 3.
Solve the following LPP Graphically;
Maximise; Z = 3x + 5y
Subject to constraints;
x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0
Answer:
In the figure the shaded region ABC is the fesible region. Here the region is unbouded.
Plus Two Maths Linear Programming 4 Mark Questions and Answers 3
The corner points are A(3, 0), B\(\left(\frac{3}{2}, \frac{1}{2}\right)\), C(0, 2)
Given; Z = 3x + 5y

Corner points	Value of Z
A	Z = 3(3) + 5(0) = 9
B	Z = 3\(\left(\frac{3}{2}\right)\) + 5\(\left(\frac{1}{2}\right)\) = 7
C	Z = 3(0) + 5(2) = 10

Form the table, minumum value of Z is 7 at B\(\left(\frac{3}{2}, \frac{1}{2}\right)\). The feasible region is unbounded, so consider the inequality 3x + 5y < 7. Clearly the feasible region has no common points with 3x + 5y < 7, Thus minimum value of Z occurs at B, the soluion is Z = 7.

Plus Two Maths Linear Programming Six Mark Questions and Answers

Question 1.
One kind of a cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1kg of fat assuming that there is no shortage of the other ingredients, used in making the cake.
Answer:
Let the number of cakes made of type I are x and that of type II are y. Then the total number of cakes will be Z = x + y
Flour constraint 200x + 100y ≤ 5000
Fat constraint 25x + 50y ≤ 1000
Therefore;
Maximise; Z = x + y
2x + y ≤ 50; x + 2y ≤ 40; x ≥ 0, y ≥ 0
Plus Two Maths Linear Programming 6 Mark Questions and Answers 4
In the figure the shaded region OABC is the feasible region. Here the region is bounded. The corner points are O(0, 0), A(25, 0), B(20, 10), C(0, 20)
Given; Z = x + y

Corner points	Value of Z
O	Z = 0
A	Z = 25 + 0 = 25
B	Z = 20 + 10 = 30
C	Z = 0 + 20 = 20

Since maximum value of Z occurs at B, the soluion is Z = 30, (20, 10).

Question 2.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine and 3hours of craftman’s time in its making, while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has availability of not more than 42 hours of machine time and 24 hours of craftman’s time.

What no. of rackets and bats must be produced if the factory is to work at full capacity?
If the profit on a racket and a bat is 10 find maximum profit.

Answer:
Let the number of rackets made = x and that of bats = y.
Maximise; Z = x + y
Machine constraints 1.5x + 3y ≤ 42
Craftsman’s constraint 3x + y ≤ 24
Therefore; Maximise; Z = x + y
x + 2y ≤ 14, 3x + y ≤ 24, x ≥ 0, y ≥ 0
In the figure the shaded region OABC is the fesible region. Here the region is bounded. The corner points are O(0, 0), A(8, 0), B(4, 10), C(0, 14).
Plus Two Maths Linear Programming 6 Mark Questions and Answers 5
Given; Z = x + y

Corner points	Value of Z
O	Z = 0
A	Z = 8 + 0 = 8
B	Z = 4 + 12 = 16
C	Z = 0 + 14 = 14

Since maximum value of Z occurs at B, the soluion is Z = 16, (4, 12).

Question 3.
Two godowns A and B have grains capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E, and F whose requirement are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops is given in the following table; Transportation cost per quintal(in Rs.)
Plus Two Maths Linear Programming 6 Mark Questions and Answers 6
Hence should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
Answer:

Express the problem diagrammatically as shown above. The total transportation cost is given by
Z = 6x + 3y + 2.5{100 – (x + y)} + 4(60 – x) + 2(50 – y) + 3(-60 + (x + y))
⇒ Z = 2.5x + 1.5y + 410
100 – (x + y) ≥ 0 ⇒ x + y ≤ 100
60 – x ≥ 0 ⇒ x ≤ 60
50 – y ≥ 0 ⇒ y ≤ 50 – 60 + x + y ≥ 0 ⇒ x + y ≥ 60
Then the given LPP is
Minimise; Z = 2.5x + 1.5y + 410
x + y ≤ 100, x + y ≥ 60
0 ≤ x ≤ 60, 0 ≤ y ≤ 50
Plus Two Maths Linear Programming 6 Mark Questions and Answers 8
In the figure the shaded region ABCD is the feasible region. Here the region is bounded. The corner points are
A(60, 0), B(60, 40), C(50, 50), D(10, 50).
Given; Z = 2.5x + 1.5y + 410

Corner points	Value of Z
A	Z = 2.5(60) + 1.5(0) + 410= 560
B	Z = 2.5(60) + 1.5(40) + 410 = 620
C	Z = 2.5(50) + 1.5(50) + 410 = 610
D	Z = 2.5(10) + 1.5(50) + 410 = 510

Since minimum value of Z occurs at D, the soluion is Z = 510.

Question 4.
(i) Choose the correct answer from the bracket. If an LPP is consistent, then its feasible region is always
(a) Bounded
(b) Unbounded
(c) Convex region
(d) Concave region
(ii) Maximize Z = 2x + 3y subject to the constraints x + y ≤ 4, x ≥ 0, y ≥ 0.
Answer:
(i) (c) Convex region.

(ii)
Plus Two Maths Linear Programming 6 Mark Questions and Answers 9
Corner points of the feasible region are as follows

Corner points	Z = 2x + 3y
0(0, 0)	0
A(0, 4)	12 → Maximum
B(4, 0)	8

∴ the maximum value of Z is 12 attained at (0, 4).

Question 5.
The graph of a linear programming problem is given below. The shaded region is the feasible region. The objective function is Z = px + qy
Plus Two Maths Linear Programming 6 Mark Questions and Answers 10

What are the co-ordinates of the comers of the feasible region.
Write the constraints
If the Max. Z occurs at A and B, what is the relation between p and q?
If q = 1, write the objective function
Find the Max Z

Answer:
1. Corner points are O(0, 0), A(5, 0), B(3, 4), C(0, 5).

2. Constraints are 2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0, y ≥ 0.

3. At (3, 4), Z = 3p + 4q
At (5, 0), Z = 5p
⇒ 3p + 4q = 5p ⇒ p = 2q.

4. If q = 1, p=2
Then the objective function is,
Maximize Z = 2x + y.

5. At (3, 4) Z = 2 × 3 + 4 = 10 is the maximum value.

Question 6.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs Rs 4 per unit food and F₂ costs Rs 6 per unit. One unit of food F₁ contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum costs for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Answer:
Let x units of food F₁ and y units of food F₂ be in the diet
Total cost Z = 4x + 6y
Then the LPP is
Minimize Z = 4x + 6
Subject to the constraints
3x + 6y ≥ 80
4x + 3y ≥ 100
x, y ≥ 0
Plus Two Maths Linear Programming 6 Mark Questions and Answers 11
The feasible region is unbounded

As the feasible region is unbounded, 104 may or may not be the minimum value of Z. For this we draw a graph of the inequality 4x + 6y < 104 or 2x + 3y < 52 and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common points with 2x + 3y < 52 Therefore minimum cost of the mixture will be 104.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

April 13, 2021February 11, 2020 by Prasanna

Students can Download Chapter 11 Three Dimensional Geometry Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Plus Two Maths Three Dimensional Geometry Three Mark Questions and Answers

Question 1.

Write the Cartesian equation. (1)
Find the angle between the line. (2)

Answer:
1.

2. cosθ
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 3

Question 2.
Find the vector equation of the plane passing through the intersection of the planes \(\bar{r}\).(i + j + k) = 6 and \(\bar{r}\).(2i + 3 j + 4k) = -5 at the point (1,1,1).
Answer:
The Cartesian equation of the planes are x + y + z = 6 and 2x + 3y + 4z = – 5. Therefore the equation of the plane passing through the intersection of these planes is
x + y + z – 6 + k(2x + 3y + 4z + 5) = 0
Since it pass through (1, 1, 1) we get,
1 + 1 + 1 – 6 + k(2 + 3 + 4 + 5) = 0 ⇒ -3 + k14 ⇒ k = \(\frac{3}{14}\)
∴ the equation is
x + y + z + -6 + \(\frac{3}{14}\) (2x + 3 y + 4z + 5) = 0
14x + 14y + 14z – 84 + 6x + 9y + 12z + 15 = 0
20x + 23y + 26z = 69
Vector equation is \(\bar{r}\). (20i + 23j + 26k) = 69.

Question 3.
Find the equation of the plane passing through the intersection of the planes x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 and contains the point (1, 0, 0).
Answer:
The equation of the planes passing through the intersection of the planes
x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 is
x + y + 4z + 5 + k(2x – y + 3z + 6) = 0 ____(1)
Since (1) pass through (1, 0, 0)
⇒ 1 + 0 + 0 + 5 + k(2 – 0 + 0 + 6) = 0
⇒ 6 + 8k = 0 ⇒ k = –\(\frac{3}{4}\); Then (1)
⇒ x + y + z + 5 + \(\frac{3}{4}\)(2x – y + 3z + 6) = 0
⇒ 4x + 4y + 16z + 20 – 6x + 3y – 9z – 18 = 0
⇒ 2x – 7y – 7z = 2.

Plus Two Maths Three Dimensional Geometry Four Mark Questions and Answers

Question 1.
Consider the point (-1, -2, -3).

In which octant, the above point lies.(1)
Find the direction cosines of the line joining (-1, -2, -3) and (3, 4, 5). (1)
If P is any point such that OP = \(\sqrt{50}\) and direction cosines of OP are \(\frac{3}{\sqrt{50}}\), \(\frac{4}{\sqrt{50}}\) and \(\frac{5}{\sqrt{50}}\), then find the co-ordinate of P. (2)

Answer:
1. The point lies in the octant X’OY’Z’.

2. Direction ratios of the line joining (-1, -2, -3) and (3, 4, 5) are (3 + 1), (4 + 2), (5 + 3) ⇒ 4, 6, 8 ⇒ 2, 3, 4.
Therefore direction cosines are

3. Given, OP = \(\sqrt{50}\)
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 5
Therefore the point is (3, 4, 5).

Question 2.
Consider a cube of side ‘a’ unit has one vertex at the origin O.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 6

Write down the co-ordinate of 0, 0′, A and A’ (1)
Find the direction ratios of OO’ and AA’. (2)
Show that the angle between the main diagonals of the above cube is cos^-1\(\left(\frac{1}{3}\right)\) (1)

Answer:
1. O(0, 0, 0), O'(a, a, a), A(a, 0, 0) and A'(0, a, a).

2. Direction ratios along OO’ is a – 0, a – 0, a – 0
⇒ a, a ,a ⇒ 1, 1, 1

3. cosθ
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 7

Question 3.
Consider two points A and B and a line L as shown in the figure.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 8

Find \(\overline{A B}\) (1)
Find the Cartesian equation of the line L. (1)
Find the foot of the perpendicular drawn from ( 2, 3, 4 ) to the line L. (2)

Answer:
1. \(\overline{A B}\) = (3 – 1)i + (- 3 – 3)j + (3 – 0)k = 2i – 6j + 3k.

2. The Cartesian equation of a line passing through the point (4, 0, -1 ) and parallel to the vector \(\overline{A B}\) is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 9

3. We take, \(\frac{x-4}{2}=\frac{y}{-6}=\frac{z+1}{3}\) = r then any point of the line can be taken as (2r + 4, -6r, 3r – 1). Assume that this point be the foot of the perpendicular drawn from (2, 3, 4 ). The dr’s of the line is 2 : – 6 : 3 and dr’s of the perpendicular line L is
2r + 4 – 2 : -6r – 3 : 3r – 1 – 4 ⇒ 2r + 2: -6r – 3 : 3r – 5
Since perpendicular,
2(2 r + 2) – 6(-6 r -3) + 3(3r – 5) = 0
49r = -7 ⇒ r = \(\frac{-7}{49}\) = –\(\frac{1}{7}\). Therefore the foot of the
perpendicular is

Question 4.
Cartesian equation of two lines are

(i) Write the vector equation of the lines. (2)
(ii) Shortest distance between the lines. (2)
Answer:
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 12

Question 5.
Consider the points (1, 3, 4) & (-3, 5, 2)

Find the equation of the line through P and Q. (1)
At which point that the above line cuts the plane 2x + y + z + 3 = 0. (3)

Answer:
1. Equation of a line passing through( 1, 3, 4) and (-3, 5, 2) is given by,

2. Let \(\frac{x-1}{-2}=\frac{y-3}{1}=\frac{z-4}{-1}\) = λ
Then any point on the line is (-2λ + 1, λ + 3, -λ + 4)
Since the plane 2x + y + z + 3 = 0 cuts the aboveline. We have,
⇒ 2(-λ + 1) + λ + 3 – λ + 4 + 3 = 0
⇒ -2λ + 2 + λ + 3 – λ + 4 + 3 = 0
⇒ -2λ = -12 ⇒ λ = 6
∴ point of intersection is (-2 × 6 + 1, 6 + 3, -6 + 4)
⇒ (-11, 9, -2).

Question 6.
Let the equation of a plane be \(\bar{r}\). (2i – 3j + 5k) = 7, then

Find the Cartesian equation of the plane. (1)
Find the equation of a plane passing through the point (3, 4, -1) and parallel to the given plane. (2)
Find the distance between the parallel planes. (1)

Answer:
1. Given, \(\bar{r}\).(2i – 3j + 5k) = 7 and if we substitute \(\bar{r}\) = xi + yj + zk Then we get the Cartesian equation as 2x – 3y + 5z – 7 = 0.

2. The equation of a plane parallel to the above plane differ only by a constant, therefore let the equation be 2x – 3y + 5z + k = 0.
⇒ 6 – 12 – 5 + k = 0 ⇒ k = 11
Therefore the equation is 2x – 3y + 5z + 11 = 0

3. The distance between the parallel planes

Question 7.

State the condition for the line \(\bar{r}\) = \(\bar{a}\) + λ \(\bar{b}\) is parallel to the plane \(\bar{r}\).\(\bar{n}\) = d. (2)
Show that the line \(\bar{r}\) = i + j + λ(2i + j + 4k) is parallel to the plane \(\bar{r}\). (-2i + k) = 5. (1)
Find the distance between the line and The Plane in (ii). (1)

Answer:
1. The line \(\bar{r}\) = \(\bar{a}\) + λ \(\bar{b}\) is parallel to the plane \(\bar{r}\).\(\bar{n}\) = d, if the normal of the plane is perpendicularto the line.
∴ \(\bar{b}\).\(\bar{n}\) = 0.

2. Given,
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 17
The line \(\bar{r}\) = i + j + λ(2i + j + 4k) is parallel to the plane \(\bar{r}\). (-2i + 4k) = 5 ⇒ -2x + 4y = 5.

3. Distance = Distance between – 2x + 4y = 5 and point (1, 1, 0) on the line
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 19

Question 8.
Choose the correct answer from the bracket,
(i) If a line in the space makes angle α, β and γ with the coordinates axes, then cos²α + cos²β + cos²γ is equal to (1)
(a) 1
(b) 2
(c) 0
(d) 3
(ii) The direction ratios of the line are \(\frac{x-6}{1}=\frac{2-y}{2}=\frac{z-2}{2}\) (1)
(a) 6, -2, -2
(b) 1, 2, 2
(c) 6, 1, -2
(d) 0, 0, 0
(iii) If the vector equation of a line is \(\bar{r}\) = i + j + k + µ(2i – 3j – 4k), then the Cartesian equation of the line
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 20
(iv) If the Cartesian equation of a plane is x + y + z =12, then the vector equation of the line is (1)
(a) \(\bar{r}\).(2i + j + k) = 12
(b) \(\bar{r}\).(i + j + k) = 12
(C) \(\bar{r}\).(i + y + 2k) = 12
(d) \(\bar{r}\).(i + 3j + k) = 12
Answer:
(i) (a) 1

(ii) (b) 1, 2, 2

(iii) (b) \(\frac{x-I}{2}=\frac{y-1}{-3}=\frac{z-1}{-4}\)

(iv) (b) \(\bar{r}\).(i + j + k) = 12.

Question 9.
Consider the lines \(\bar{r}\) = (i + 2j – 2k) + λ(i + 2 j) and \(\bar{r}\) = (i + 2j – 2k) + µ(2j – k)

Find the angle between the lines.
Find a vector perpendicular to both the lines.
Find the equation of the line passing through the point of intersection of lines and perpendicular to both the lines.

Answer:
1. \(\bar{b}_{1}\) = i + 2j; \(\bar{b}_{2}\) = 2j – k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 21

2. Perpendicular vector = \(\bar{b}_{1}\) × \(\bar{b}_{2}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 22
= i(-2 – 0) -j(-1 – 0) + k(2 – 0)
= -2i + j + k.

3. Equation of line is \(\bar{r}\) = (i + 2j – k) + µ(-2i + j + 2k).

Question 10.
Consider the line \(\bar{r}\) = (2i – j + k) + λ(i + 2j + 3k)

Find the Cartesian equation of the line.
Find the vector equation of the line passing through A (1, 0, 2) and parallel to the above line.
Write two points on the line obtained in (ii) which are equidistant from A.

Answer:
1. \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-1}{3}\).

2. \(\bar{r}\) = (i + 2k) + λ(i + 2j + 3k).

3. Put λ = a and λ = -a for any real value ‘a’.
Let us put λ = 1 and λ = -1
\(\bar{r}\) = (i + 2k) + 1(i + 2j + 3k) = 2i + 2j + 5k
⇒ (2, 2, 5)
\(\bar{r}\) = (i + 2k) – 1(i + 2j + 3k) = 0i – 2j – k
⇒ (0, -2, -1)
The equidistant points are(2, 2, 5) and (0, -2, -1).

Question 11.

Find the equation of the plane through the point(1, 2, 3) and perpendicular to the plane x – y + z = 2 and 2x + y – 3z = 5 (2)
Find the distance between the planes x – 2y + 2z – 8 = 0 and 6y – 3x – 6z = 57 (2)

Answer:
1. Required equation is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 23
(x – 1)(3 – 1) – (y – 2)(-3 – 2) + (z – 3)(1 + 2) = 0
2(x – 1) + 5(y – 2) + 3(z – 3) = 0
2x + 5y + 3z – 2 – 10 – 9 = 0
2x + 5y + 3z – 21 = 0

2. The planes are
x – 2y + 2z – 8 = 0 and 3x – 6y + 6z + 57 = 0
ie, 3x – 6y + 6z – 24 = 0 and 3x – 6y + 6z + 57 = 0
Distance
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 24

Question 12.
Consider the Cartesian equation of a line \(\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-5}{-2}\)

Find the vector equation of the line. (1)
Find its intersecting point with the plane 5x + 2y – 6z – 7 = 0 (2)
Find the angle made by the line with the plane 5x + 2y – 6z – 7 = 0 (1)

Answer:
1. The vector equation is \(\bar{r}\) = (3i – j + 5k) + λ(2i + 3 j – 2k).

2. Any point on the line is
\(\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-5}{-2}\) = λ
x = 2λ + 3, y = 3, λ – 1, z = -2λ + 5
Since this lies on the plane ,it satisfies the plane
5(2λ + 3) + 2(3λ – 1) -6(-2λ + 5) – 7 = 0
10λ + 6λ + 12λ + 15 – 2 – 30 – 7 = 0
28λ = 24
λ = 6/7
The point of intersection is \(\left[\frac{33}{7}, \frac{11}{7}, \frac{23}{7}\right]\).

3. Let θ be the angle between the line and the plane. The direction of the line and the plane
\(\bar{b}\) = 2i + 3j + k; \(\bar{m}\) = 5i + 2j – 6k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 25

Question 13.
From the following figure
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 26

Find \(\overline{A B}\). (1)
Find the vector equation of line L. (1)
Find a point on line L other than C. (2)

Answer:
1. P.v of A = i – j + 4k,
P.v. of B = 2i + j + 2k
\(\overline{A B}\) = p. v. of B – p. v. of A
= 2i + j + 2k -(i – j + 4k) = i + 2j – 2k.

2. The line L passes through (1, -2, -3) and parallel to \(\overline{A B}\)

∴ Vector equation of line L is \(\bar{r}=\bar{a}+\lambda \bar{m}\)

3. From (1) of part (ii), we have
xi + yj + zk = (l + λ)i + (-2 + 2λ)j + (-3 – 2λ)k
Put λ = 1
⇒ xi +yj + zk = (1 +1)i + (-2 + 2)j + (-3 – 2 )k
⇒ xi + yj + zk = 2i + 0j – 5k
Therefore a point on line L is (2, 0, -5).

Question 14.
Find the vector equation of the plane which is at a distance of \(\frac{6}{\sqrt{29}}\) from the origin with perpendicular vector 2i – 3j + 4k. Convert into Cartesian form. Also, find the foot of the perpendicular drawn from the origin to the Plane.
Answer:
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 30
Perpendicular distance from origin = d = \(\frac{6}{\sqrt{29}}\)
The equation of the Plane is \(\bar{r}\).\(\hat{n}\) = d

Cartesian equation is 2x – 3y + 4z = 6
The direction cosines perpendicular to the Plane is \(\frac{2}{\sqrt{29}},-\frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}\).
Perpendicular distance to the Plane is as \(\frac{6}{\sqrt{29}}\)
Hence the foot of the perpendicular is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 32

Question 15.
Consider the Plane \(\bar{r}\).(-6i -3j – 2k) + 1 = 0, find the direction cosines perpendicular to the Plane and perpendicular distance from the origin.
Answer:
Convert the equation of the plane into normal form
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 33
Direction cosines perpendicular to the Plane is \(\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\)
Perpendicular distance from the origin is \(\frac{1}{7}\).

Question 16.
Consider three points (6, -1, 1), (5, 1, 2) and (1, – 5, -4) on space.

Find the Cartesian equation of the plane passing through these points. (2)
Find direction ratios normal to the Plane.(1)
Find a unit vector normal to the Plane. (1)

Answer:
1. Equation of a plane passing through the points (6, -1, 1),(5, 1, 2) and (1, -5, 4)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 34
⇒ (x – 6)(-10 + 4) – (y + 1)(5 + 5) + (z – 1)(4 + 10) = 0
⇒ (x – 6)(-6) – (y + 1)(10) + (z – 1)(14) = 0
⇒ -6x + 36 – 10y – 10 + 14z – 14 = 0
⇒ 6x +10y – 14z -12 = 0.

2. Dr’s normal to the plane are 6 : 10 : -14 ⇒ 3 : 5 : -7.

3. Since the dr’s normal to the plane are 3 : 5 : -7, a unit vector in this direction is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 35

Question 17.
Consider a straight line through a fixed point with position vector 2i – 2j + 3k and parallel to i – j + 4k.

Write down the vector equation of the straight line. (1)
Show that the straight line is parallel to the plane \(\bar{r}\).(i + 5y + k) = 5 (1)
Find the distance between the line and plane. (2)

Answer:
1. Vector equation of a straight line is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where a is \(\bar{a}\) fixed point and \(\bar{b}\) is a vector parallel to the line. Here \(\bar{a}\) = 2i – 2y + 3 it and \(\bar{b}\) = i – j + 4k. Therefore vector equation of the line \(\bar{r}\) = 2i – 2j + 3k + λ(i – j + 4k).

2. The vector parallel to the line is i – j + 4k and vector normal to the plane is i + 5j + k.
Then, (i – j + 4k). (i + 5j + k) = 1 – 5 + 4 = 0
implies that straight line and plane are parallel.

3. A point on the line is 2i – 2j + 3k. Then the distance of 2i – 2j + 3k to the given plane
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 36

Question 18.
Consider the vector equation of two planes \(\bar{r}\).(2i + j + k) = 3, \(\bar{r}\).(i – j – k) = 4

Find the vector equation of any plane through the intersection of the above two planes. (2)
Find the vector equation of the plane through the intersection of the above planes and the point (1, 2, -1 ) (2)

Answer:
1. The cartesian equation are 2x + y + z – 3 = 0 and x – y – z – 4 = 0 Required equation of the plane is
(2x + y + z – 3) + λ(x – y – z – 4) = 0
(2+ λ)x + (1 – λ)y + (1 – λ)z + (-3 – 4λ) = 0.

2. The above plane passes through (1, 2, -1)
(2+ λ)1 + (1 – λ)2 + (1 – λ)(-1) + (-3 – 4λ) = 0
3 – 3 + 4λ = 0
λ = 0
Equation of the plane is 2x + y + z – 3 = 0
\(\bar{r}\).(2i + j + k) = 3.

Question 19.
(i) Distance of the point(0, 0, 1) from the plane x + y + z = 3
(a) \(\frac{1}{\sqrt{3}}\) units
(b) \(\frac{2}{\sqrt{3}}\) units
(c) \(\sqrt{3}\) units
(d) \(\frac{\sqrt{3}}{2}\) units
(ii) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to x – y + z = 0 (3)
Answer:
(i) (b) \(\frac{2}{\sqrt{3}}\) units.

(ii) Equation of the plane passing through the intersection is of the form
x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0 _____(1)
(1 + 2λ)x + (1 + 3λ)j + (1 + 4λ)z – 1 – 5λ = 0
Thr Dr’s of the required plane is
(1 + 2λ), (1 + 3λ), (1 + 4λ)
Thr Dr’s of the Perpendicular plane is 1, -1, 1
⇒ (1 + 2λ)(1) + (1 + 3λ)(-1) + (1 + 4λ)(1) = 0
⇒ 1 + 2λ – 1 – 3λ + 1 + 4λ = 0
⇒ 3λ + 1 = 0 ⇒ λ = \(\frac{-1}{3}\)
(1) ⇒ x + y + z – \(\frac{1}{3}\)(2x + 3y + 4z – 5) = 0
⇒ 3x + 3y + 3z – 2x – 3y – 4z + 5 = 0
⇒ x – z + 2 = 0.

Question 20.
Consider a plane \(\bar{r}\).(6i – 3j – 2k) + 1 = 0

Find dc’s perpendicular to the plane. (2)
Find a vector of magnitude 14 units perpendicular to given plane. (1)
Find the equation of a line parallel to the above vector and passing through the point (1, 2, 1 ). (1)

Answer:
1. Given, \(\bar{r}\).(6i – 3j – 2k) + 1 = 0 ____(1)
Now, |6i – 3j – 2k| = \(\sqrt{36+9+4}\) = 7
∴ \(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\) is a unit perpendicular to the plane (1)
⇒ the dc’s perpendicular to the plane (1) are \(\frac{6}{7},-\frac{3}{7},-\frac{2}{7}\).

2. We have, \(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\) is a unit perpendicular to the Plane (1). Therefore, a vector of magnitude 14 units perpendicular to the Plane (1) is 14(\(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\))
⇒ 12i – 6j – 4k.

3. Equation of a line parallel to the vector 12i – 6j – 4k and passing through the point (1, 2, 1 )is given by
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 37

Plus Two Maths Three Dimensional Geometry Six Mark Questions and Answers

Question 1.
Consider the pair of lines whose equations are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 38

Write the direction ratios of the lines. (1)
Find the shortest distance between the above skew lines. (4)
Find the angle between these two lines. (1)

Answer:
1. The direction ratios are 2, 5, – 3 and – 1, 8, 4.

2. The given lines are \(\bar{r}\) = (2i + j – 3k) + λ(2i + 5j – 3k)
i.e. \(\bar{r}\) = \(\overline{a_{1}}+\lambda \overline{b_{1}}\),
where \(\overline{a_{1}}\) = 2i + j – 3k) + λ(2i + 5j – 3k)
and \(\bar{r}\) =(-i + 4j + 5k) + µ(-i + 8j + 4k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 39

3. cosθ
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 40

Question 2.
Consider the pair of lines \(\bar{r}\) = 3i + 4j – 2k + λ(-i + 2j + k) ——L_1,\(\bar{r}\) = i – 7j – 2k + µ(i + 3j + 2k) ——L₂

Find one point each on lines L₁ and L₂. (1)
Find the distance between those points. (2)
Find the shortest distance between L₁ and L₂. (3)

Answer:
1. By putting λ = 0 in line L₁ and µ = 0 in L₂ we get the required points. L₁ ⇒ \(\bar{r}\) = 3i + 4j – 2k
∴ Co-ordinate is (3, 4, -2)
L₂ ⇒ \(\bar{r}\) = i – 7j – 2k
∴ Co-ordinate is (1, -7, -2).

2. Distance between (3, 4, -2) and (1, -7, -2)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 41

3. Let L₁ ⇒ \(\bar{r}\) = 3i + 4j – 2k + λ(-i + 2j + k) is of the form \(\bar{r}=\overline{a_{1}}+\lambda \overline{b_{1}}\) where
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 42

Question 3.
Consider the points A (2, 2, -1), B (3, 4, 2) and C (7, 0, 6).

Are A, B, and C collinear? Explain.
Find the vector and Cartesian equation of the plane passing these three points. (2)
Find the angle between the above plane and the line \(\bar{r}\) = (i + 2j – k) + λ(i – j + k) (2)

Answer:
1. Direction ratios along A and B is 3 -2, 4 -2, 2 + 1 ⇒ 1, 2, 3
Direction ratios along B and C is
7 -3, 0 -4, 6 -2 ⇒ 4, -4, 4
Since the direction ratios are not proportional they are not collinear.

2. Cartesian equation of the Plane is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 43
⇒ (x – 2)(14 + 6) -(y – 2)(7 – 15) + (z + 1)(-2 -10) = 0
⇒ 20(x – 2) + 8(y – 2) – 12(z + 1) = 0
⇒ 20x – 40 + 8y – 16 – 12z – 12 = 0
⇒ 20x + 8y – 12z = 68
⇒ 5x + 2y – 3z = 17
Vector Equation is \(\bar{r}\).(5i + 2j – 3k) = 17.

3. Angle between the Plane and the Line
\(\bar{r}\) = (i + 2j – k) + λ(i – j + k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 44

Question 4.
Consider three points on space ( 2, 1, 0 ), (3, -2, -2)and(3, 1, 7)

Find the Cartesian equation of the plane passing through the above points. (2)
Convert the above equation into vector form.
Hence, find a unit vector perpendicular to the above plane and also find the perpendicular distance of the plane from the origin. (2)

Answer:
1. Equation of the plane is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 45

⇒ (x – 2)(-21) – (y – 1)(7 + 2) + z(0 + 3) = 0
⇒ 21x + 42 – 9y + 9 + 3z = 0 ⇒ -21x – 9y + 3z + 51 = 0
⇒ 7x + 3y – z = 17.

2. Vector form is \(\bar{r}\).(7i + 3j – k) = 17 _____(1)

3. Now, |7i + 3j – k| = \(\sqrt{49+9+1}=\sqrt{59}\)
Dividing equation (1) by \(\sqrt{59}\), we get
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 47
Therefore the above equation is the normal form of the plane. Then \(\frac{7 i+3 j-k}{\sqrt{59}}\) is the unit vector perpendicular to the plane and \(\frac{17}{\sqrt{59}}\) is the perpendicular distance from the origin.

Question 5.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 48
\(\overline{O A}\) = i + 2j + 3k
\(\overline{O B}\) = i – 2j + 4k
\(\overline{O C}\) = 2i + 3j + k
are adjacent sides of the parallelopiped.

Find the base area of the parallelopiped. (2)
(Base determined by \(\overline{O A}\) and \(\overline{O B}\))
Find the volume of the parallelopiped. (2)
Find the height of the parallelopiped. (2)

Answer:
1. \(\overline{O A}\) × \(\overline{O B}\) =
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 49
= 14i – j – 4k
Base area = |l4i – j – 4k|
\(=\sqrt{196+1+16}=\sqrt{213}\)

2. Volume of the parallelopiped is

= (14i – j – 4k).(2i + 3 j + k)
= 28 – 3 – 4 = 21.

3. Height = \(\frac{\text {volume}}{\text {base area}}=\frac{21}{\sqrt{213}}\).

Question 6.

Find the equation of the line passing through the point (2, 1, 0) and (3, 2, -1) (3)
Find the shortest distance of the above line from the line \(\bar{r}\) = (i – j + 2k) + λ(2i + j – 3k) (3)

Answer:
1.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 51

2.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 52

= i(-3 + 1) – j(-3 + 2) + k(1 – 2)
= -2i + j – k

Question 7.
The equation of two lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 55

Find the dr’s of the given lines. (2)
Find the angle between the given lines. (2)
Find the equation of the line passing through (2, 1, 3) and perpendicular to the given lines. (2)

Answer:
1. The given lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 56
The dr’s of (1) are 2, 2, 3 and dr’s of (2) are -3, 2, 5.

2. The angle between (1) and (2) is given by
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 57

3. Let a, b, c be the dr’s of the line perpendicular to lines (1) and (2).
∴ 2a + 2b + 3c = 0, -3a + 2b + 5c = 0
Solving by the rule of cross-multiplication, we get
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 58
∴ dr’s of the required line are 4, -19, 10 and the line passes through (2, 1, 3).
∴ Equation of the required line is
\(\frac{x-2}{4}=\frac{y-1}{-19}=\frac{z-3}{10}\).

Question 8.

Find the direction cosines of the vector 2i + 2j – k. (1)
Find the distance of the point (2, 3, 4) from the plane \(\bar{r}\).(3i – 6j + 2 k) = -11. (2)
Find the shortest distance between the lines \(\bar{r}\) = (2i – j – k)+ λ(3i – 5 j + 2k) an \(\bar{r}\) = (i+ 2 j + k)+ µ(i – j + k) (3)

Answer:
1. Direction ratios of the vector 2i + 2j – k is 2, 2, -1
Direction cosines of the vector 2i + 2j – k is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 59

2. The equation of the plane in the Cartesian form is 3x – 6y + 2z + 11 = 0 . Then distance from the point (2, 3, 4) is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 60

3. The given lines are \(\bar{r}\) = (2i – j – k) + λ(3i – 5j + 2k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 61

Plus Two Maths Chapter Wise Questions and Answers Chapter 10 Vector Algebra

April 13, 2021February 10, 2020 by Prasanna

Students can Download Chapter 10 Vector Algebra Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 10 Vector Algebra

Plus Two Maths Vector Algebra Three Mark Questions and Answers

Question 1.
Find \(\bar{a}+\bar{b}, \bar{a}-\bar{b}\) and \(\bar{b}+\bar{c}\) using the vectors.
\(\bar{a}\) = 3i + 4j + k, \(\bar{b}\) = 2i – 7 j – 3k and \(\bar{c}\) = 2i + 3j – 9k.
Answer:
\(\bar{a}+\bar{b}\) = 3i + 4j + k + 2i – 7j – 3k = 5i – 3j – 2k
\(\bar{a}-\bar{b}\) = 3i + 4j + k – (2i – 7j -3k) = i + 11j + 4k
\(\bar{b}+\bar{c}\) = 2i – 7j -3k + 2i +3j – 9k
= 4i – 4j – 12k.

Question 2.

Find the vector passing through the point A( 1, 2, -3) and B(-1, -2, 1).
Find the direction cosines along AB.

Answer:
1. \(\overline{A B}\) = \(\overline{O B}\) – \(\overline{O A}\) = -i – 2j + k – (i + 2j – 3k) = -2i – 4j + 4k.

2. Unit Vector
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 1
Direction cosines are \(\frac{-2}{6}\), \(\frac{-4}{6}\), \(\frac{4}{6}\).

Question 3.
Show that the points A, B and C with position vectors \(\bar{a}\) = 3i – 4j – 4k, \(\bar{b}\) = 2i – j + k and \(\bar{c}\) = i – 3j – 5k respectively from the vertices of a right angled triangle.
Answer:
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 2
41 = 35 + 6 ⇒ BC² = AB² + CA²
Hence right angled triangle.

Question 4.
Prove that \([\bar{a}+\bar{b} \bar{b}+\bar{c} \bar{c}+\bar{a}]=2[\bar{a} \bar{b} \bar{c}]\).
Answer:
LHS
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 3
Note: If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are coplanar then so is \([\bar{a}+\bar{b} \bar{b}+\bar{c} \bar{c}+\bar{a}]\).

Question 5.
Consider the vector \(\bar{p}\) = 2i – j + k. Find two vectors \(\bar{q}\) and \(\bar{r}\) such that \(\bar{p}\), \(\bar{q}\) and \(\bar{r}\) are mutually perpendicular.
Answer:
Find a vector \(\bar{q}\) such that \(\bar{p} \cdot \bar{q}\) = 0, for this use any \(\bar{q}\) whose two components are randomly selected. Let \(\bar{q}\) = 2i + 2j + xk
\(\bar{p} \cdot \bar{q}\) = (2i – j + k) . (2i + 2 j + xk) = 0
⇒ 4 – 2 + x = 0 ⇒ x = -2
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 4
= 6j + 6k.

Question 6.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 5
Answer:

= i(-12 + 7) – j(-9 – 2) + k(-21 – 8)
= -5i + 11j – 29k

= i(63 + 9) – j(-18 + 6) + k(6 – 14)
= 72i + 12 j – 8k.

Question 7.
If \(\bar{a}\) = 3i + j + 2k,
(i) Find the magnitude of \(\bar{a}\). (1)
(ii) If the projection of \(\bar{a}\) on another vector \(\bar{b}\) is \(\sqrt{14}\), which among the following could be \(\bar{b}\) ? (1)
(a) i + j + k
(b) 6i + 2j + 4k
(c) 3i – j + 2k
(d) 2i + 3j + k
(iii) If \(\bar{a}\) makes an angle 60° with a vector \(\bar{c}\), find the projection of \(\bar{a}\) on \(\bar{c}\) (1)
Answer:

(ii) Since projection of \(\bar{a}\) on another vector \(\bar{b}\) and magnitude of \(\bar{a}\) is \(\sqrt{14}\), then \(\bar{a}\) and \(\bar{b}\) are parallel, (b) 6i + 2j + 4k.

(iii) Projection of \(\bar{a}\) on \(\bar{c}\)
= |\(\bar{a}\)|cos60° = \(\sqrt{14}\) × \(\frac{1}{2}\) = \(\frac{\sqrt{14}}{2}\).

Question 8.
(i) The projection of the vector 2i + 3j + 2k on the vector i + j + k is (1)
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 9
(ii) Find the area of a parallelogram whose adjacent sides are the vectors 2i + j + k and 6i – j (2)
Answer:

(ii) Let \(\bar{a}\) = 2i + j + k, \(\bar{b}\) = i – j
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 11
= i(0 + 1) – j(0 – 1) + k(-2 – 1 ) = i + j -3k
Area =

Question 9.
(i) The angle between the vectors i + j and j + k is (1)
(a) 60°
(b) 30°
(c) 45°
(d) 90°
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 13
Answer:
(i) (a) 60°

Plus Two Maths Vector Algebra 3 Mark Questions and Answers 14

Question 10.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 15
Answer:

(ii) Given, \(\bar{a}\) + \(\bar{b}\) + \(\bar{a}\) = \(\bar{0}\), squaring both sides we get
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 17

Plus Two Maths Vector Algebra Four Mark Questions and Answers

Question 1.
Let A (2, 3), B (1, 4), C (0, -2) and D (x, y) are vertices of a parallelogram ABCD.

Write the position vectors A, B, C and D. (2)
Find the value of x and y. (2)

Answer:
1. Position vector of A = 2i + 3 j
Position vector of B = i + 4j
Position vector of C = 0i – 2j
Position vector of D = xi + yj.

2. Since ABCD is a parallelogram, then
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 18
(1) ⇒ -i + j = -xi – (y + 2 )j
x = 1, -2 – y = 1 ⇒ y = -3
∴ D is (1, -3).

Question 2.
Find the position vector of a point R which divides the line joining the two points P and Q whose vectors i + 2j – k and -i + j + k in the ratio 2:1

internally and
externally.

Answer:
\(\overline{O P}\) = i + 2j – k, \(\overline{O Q}\) = -i + j + k
Let R be the position vector of the dividing point,
1.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 19

2.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 20

Question 3.
(i) Choose the correct answer from the bracket. If a unit vector \(\widehat{a}\) makes angles \(\frac{\pi}{4}\) with i and \(\frac{\pi}{3}\) with j and acute angle θ with k. then θ is
(a) \(\frac{\pi}{6}\),
(b) \(\frac{\pi}{4}\),
(c) \(\frac{\pi}{3}\),
(d) \(\frac{\pi}{2}\) (1)
(ii) Find a unit vector \(\widehat{a}\) (1)
(iii) Write down a unit vector in XY plane, making an angle 60°of with the positive direction of x – axis. (2)
Answer:
(i) (c), Since I = cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\), m = cos\(\frac{\pi}{3}\) = 1/2;
n = cos θ
l² + m² + n² = 1
n² = 1 – (\(\frac{1}{2}\))² – (\(\frac{1}{\sqrt{2}}\))² = 1/4
n = \(\frac{1}{2}\), cosθ = 1/2 , θ = \(\frac{\pi}{3}\).

(ii)
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 21

Question 4.
Let the vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) denoted the sides of a triangle ABC.
(i) Prove that (2)

(ii) Find the projection of the vector i + 3j + 7k on the vector 7i – j + 8k (2)
Answer:
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 24

(ii) Projection of the vector i + 3j + 7k on the vector 7i – j + 8k
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 25

Question 5.
(i) If \(\bar{a}\) and \(\bar{b}\) are any two vectors, then axb is (1)
(a) a vector on the same plane where \(\bar{a}\) and \(\bar{b}\) lie.
(b) ab cosθ, if θ is the angle between them.
(c) a vector parallel to both \(\bar{a}\) and \(\bar{b}\).
(d) a vector perpendicular to both \(\bar{a}\) and \(\bar{b}\).
(ii) Let \(\bar{a}\) = 2i + 4j – 5k, \(\bar{b}\) = i + 2j + 3k. Then find a unit vector perpendicular to both \(\bar{a}\) and \(\bar{b}\). (2)
(iii) Find a vector of magnitude 5 in the direction perpendicular to both \(\bar{a}\) and \(\bar{b}\) (1)
Answer:
(i) (d) a vector perpendicular to both \(\bar{a}\) and \(\bar{b}\).

(ii) \(\bar{a}\) = 2i + 4j-5k, \(\bar{b}\) = i + 2j+3k
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 26
= i(12 + 10) – j(6 + 5) + k(4 – 4) = 22i – 11j

Therefore unit vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is

(iii) 5 × unit vector perpendicular to both \(\bar{a}\) and \(\bar{b}\)
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 29

Question 6.
Consider a vector that is inclined at an angle 45° to x-axis and 60° to y-axis

Find the dc’s of the vector. (2)
Find a unit vector in the direction of the above vector. (1)
Find a vector which is of magnitude 10 units in the direction of the above vector. (1)

Answer:
1. Let l, m, n are the direction ratios.
Given that, l = cos 45° = \(\frac{1}{\sqrt{2}}\), m = cos 60° = \(\frac{1}{2}\)
⇒ l² + m² + n² = 1
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 30
∴ the dc’s of the vector are \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), \(\frac{1}{2}\)

2. A unit vector in the direction of the above vector is given by li + mj + nk ⇒ \(\frac{1}{\sqrt{2}}\)i + \(\frac{1}{2}\)j + \(\frac{1}{2}\)k.

3. A vector, which is of magnitude 10 units in the direction of the above vector is given by
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 31

Question 7.
Consider the point A(2, 1, 1) and B(4, 2, 3)

Find the vector \(\overline{A B}\) (1)
Find the direction cosines of \(\overline{A B}\) (2)
Find the angle made by \(\overline{A B}\) with the positive direction of x-axis. (1)

Answer:
1. \(\overline{A B}\) = 2i + j + 2k

2. |\(\overline{A B}\)| = \(\sqrt{4+1+4}\) = 3
The direction cosines are \(\frac{2}{3}\), \(\frac{1}{3}\), \(\frac{2}{3}\).

3. cos α = \(\frac{2}{3}\) ⇒ α = cos^-1(\(\frac{2}{3}\)).

Question 8.
If i + j + k, 2i + 5j, 3i + 2 j – 3k, i – 6j – k respectively are the position vector of points A, B,C and D. Then

Find \(\overline{A B}\) and \(\overline{C D}\). (1)
Find the angle between the vectors \(\overline{A B}\) and \(\overline{C D}\). (2)
Deduce that \(\overline{A B}\) parallel to \(\overline{C D}\). (1)

Answer:
1.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 32

2.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 33

3. Since the angle between \(\overline{A B}\) and \(\overline{C D}\) is π they are parallel.

Question 9.
Let ABCD be a parallelogram with sides as given in the figure.

Find area of the parallelogram. (2)
Find the distance between the sides AB and DC. (2)

Plus Two Maths Vector Algebra 3 Mark Questions and Answers 34
Answer:
1. Given;
\(\overline{A B}\) = i – 3j + k and \(\overline{A D}\) = i + j + k

2. Let h be the distance between the parallelsides AB and DC. Then ; Area = Base × h _____(2)
Here, Base = |\(\overline{A B}\)|
|i – 3j + k| = \(\sqrt{1+9+1}=\sqrt{11}\)
From (1) and (2)
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 36

Question 10.
Consider \(\bar{a}\) = i + 2j – 3k, \(\bar{b}\) = 3i – j + 2k, \(\bar{c}\) = 11i + 2j

Find \(\bar{a}\) + \(\bar{b}\) and \(\bar{a}\).\(\bar{b}\) (2)
Find the unit vector in the direction of \(\bar{a}\) + \(\bar{b}\). (1)
Show that \(\bar{a}\) + \(\bar{b}\) and \(\bar{a}\) – \(\bar{b}\) are orthogonal. (1)

Answer:
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 37

(ii) Unit vector in the direction of
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 38

(iii) We have,
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 39
Therefore, they are orthogonal.

Question 11.
Let A (1, -1, 4), B ( 2, 1, 2 ) and C (1, -2, -3 )

Find \(\overline{A B}\). (1)
Find the angle between \(\overline{A B}\) and \(\overline{A C}\).(2)
Find the area of the parallelogram formed by \(\overline{A B}\) and \(\overline{A C}\) as adjacent sides. (1)

Answer:
1. \(\overline{A B}\) = P.v of B – P.v of A
= 2 i + j + 2 k – (i – j + 4k) = i + 2 j – 2k

2. \(\overline{A C}\) = P.v of C – P.v of A
= i – 2 j – 3 k -(i – j + 4k) = – j – 7k
Let A be the angle between \(\overline{A B}\) and \(\overline{A C}\)
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 40

3.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 41
Area of the parallelogram

Plus Two Maths Vector Algebra Six Mark Questions and Answers

Question 1.
Using this figure answer the following questions.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 43

Find \(\overline{O A}\), \(\overline{O B}\), \(\overline{O C}\) (2)
Find \(\overline{O D}\) (2)
Find the coordinate of the vertex D. (2)

Answer:
1. \(\overline{O A}\) = (3 – 1)i + (-1 – 2)j + (7 – 3)k = 2i – 3j + 4k
\(\overline{O B}\) = (2 – 1)i + (4 – 2)j +(2 – 3)k = i + 2j – k
\(\overline{O C}\) = (4 – 1)i + (1 – 2 )j + (5 – 3 )k = 3i – j + 2 k.

2. From the figure,
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 44

3. Let the vertex of D be (x , y , z),
Then, \(\overline{O D}\) = (x – 1)i + (y – 2)j + (z – 3)k.
But we have,
\(\overline{O D}\) = 6i – 2j + 5k = (x – 1)i + (y – 2)j +(z – 3)k
x – 1 = 6 ⇒ x = 7, y – 2 = -2 ⇒ y = 0, z – 3 = 5 ⇒ z = 8.

Question 2.
OABCDEFG is a cube with edges of length 8 units and axes as shown. L, M, N are midpoints of the edges FG, GD, GB respectively.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 45

Find p.v’s of F, B,D and G. (1)
Show that the angle between the main diagonals is θ = cos^-1\(\left(\frac{1}{3}\right)\). (2)
Find the p.v’s of L, M, N. (1)
Show that \(\overline{L M}+\overline{M N}+\overline{N L}=0\). (1)

Answer:
1. \(\overline{O F}\) = 8 j + 8k, \(\overline{O B}\) = 8i + 8k, \(\overline{O D}\) = 8i + 8k, \(\overline{O G}\) = 8i + 8j + 8k.

2. Consider the main diagonals \(\overline{O G}\) and \(\overline{E B}\)
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 46

3. P.V of L = \(\overline{O L}\) = 4i + 8j + 8k
P.V of M = \(\overline{O M}\) = 8i + 8j + 4k
P.V of N = \(\overline{O N}\) = 8i + 4j + 8k

4.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 47

Question 3.
Using the figure answer the following questions
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 48

Evaluate \(\overline{A B}\).\(\overline{A C}\) (2)
Find \(\overline{A D}\) . (2)
Find the coordinates of D.

Answer:
1. \(\overline{A B}\) = p.v of B – p.v of A= -4i + 0j + 3k
\(\overline{A C}\) = p.v of C – p.v of A = 0i – 4 j + 4k
\(\overline{A B}\).\(\overline{A C}\) = -4 × 0 + 0 × -4 + 3 × 4 = 12

2.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 49

3. Let the coordinate of D be (x, y ,z)
⇒ \(\overline{A D}\) = (x – 3)i + (y – 2)j + (z – 1)k,
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 50

Question 4.
Consider the Parallelogram ABCD
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 51

Find \(\overline{A B}\) and \(\overline{A D}\) (1)
Find the area of the parallelogram ABCD. (1)
Find \(\overline{A C}\). (2)
Find co-ordinate of C. (2)

Answer:
1. \(\overline{A B}\) = p.v of B – p. v of A
= 3i + 5j + 8k – (i + 2j + k) = 2i + 3j + 7k
\(\overline{A D}\) = p.v of D – p. v of A
= i + 3j + 2k – (i + 2j + k)= 0i + j + k.

2.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 52

3. By triangle inequality;
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 53

4. Let the co-ordinate of C be (x, y, z)
Then, \(\overline{A C}\) = (x – 1)i + (y – 2)j + (z – 1)k = 2i + 4j + 8k
x – 1 = 2 ⇒ x = 3, y – 2 = 4 ⇒ y = 6,
z – 1 = 8 ⇒ z = 9
Co-ordinate of C is (3, 6, 9).

Question 5.
Consider the following quadrilateral ABCD in which P, Q, R, S are the midpoints of the sides.
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 54

Find \(\overline{P Q}\) and \(\overline{S R}\) in terms of \(\overline{A C}\) (2)
Show that PQRS is a parallelogram. (2)
If \(\bar{a}\) is any vector, prove that (2)

Answer:
1. Using triangle law of addition, we get
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 56

2. From the above explanation we have,

and parallel. Similarly, |\(\overline{S P}\)| = |\(\overline{R Q}\)|
Therefore, PQRS is a parallelogram.

3. Let \(\bar{a}\) = a₁ i + a₂ j + a₃ k
Plus Two Maths Vector Algebra 3 Mark Questions and Answers 58

Plus Two Maths Chapter Wise Questions and Answers Chapter 9 Differential Equations

April 13, 2021February 10, 2020 by Prasanna

Students can Download Chapter 9 Differential Equations Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 9 Differential Equations

Plus Two Maths Differential Equations Three Mark Questions and Answers

Question 1.
y = e^2x(a + bx), a and b are arbitrary constants.
Answer:
y = e^2x(a + bx) ____(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = e^2xb + (a + bx)2e^2x
\(\frac{d y}{d x}\) = 2y + be^2x ⇒ \(\frac{d y}{d x}\) – 2y = be^2x ____(2)
Differentiating (2) with respect to x,
Plus Two Maths Differential Equations 3 Mark Questions and Answers 1

Question 2.
y = e^x(acosx + 6sinx), a and b are arbitrary constants.
Answer:
y = e^x(acosx + 6sinx) ___(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = e^x(-asinx + bcosx) + e^x(acosx + bsinx) \(\frac{d y}{d x}\) = e^x(-asin x + b cos x) + y
\(\frac{d y}{d x}\) – y = e^x(-a sin x + b cos x) ____(2)
Differentiating (2) with respec to x,
Plus Two Maths Differential Equations 3 Mark Questions and Answers 2

Question 3.
y = c₁e^x + c₂ e^-x , c₁ and c₁ are arbitrary constants.
Answer:
y = c₁e^x + c₂ e^-x ___(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = c₁e^x + c₂ e^-x __(2)
Differentiating (2) with respect to x,
Plus Two Maths Differential Equations 3 Mark Questions and Answers 3

Question 4.
(x – a)² + 2y² = a², a is a arbitrary constants.
Answer:
(x – a)² + 2y² = a² ___(1)
Differentiating with respect to x,
Plus Two Maths Differential Equations 3 Mark Questions and Answers 4

Question 5.
Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
Answer:
y\(\frac{d y}{d x}\) = x ⇒ ydy = xdx
Integrating on both sides,
∫ydy = ∫xdx + c
⇒ \(\frac{y^{2}}{2}=\frac{x^{2}}{2}\) + c ____(1)
Since it passes through (0, -2),
Plus Two Maths Differential Equations 3 Mark Questions and Answers 5

Question 6.
Form the DE representing the family of parabolas having a vertex at origin and axis along positive direction of x-axis.
Answer:
Let (a, 0) be focus of the given family of parabolas.
y² = 4ax ____(1)
Differentiating with respect to x,
Plus Two Maths Differential Equations 3 Mark Questions and Answers 6

Question 7.
For the DE xy \(\frac{d y}{d x}\) = (x + 2)(y + 2), find the solution curve passing through the point(1,- 1).
Answer:
xy \(\frac{d y}{d x}\) = (x + 2)(y + 2)
⇒ \(\frac{y}{y+2} d x=\frac{x+2}{x} d x\)
Integrating on both sides,
Plus Two Maths Differential Equations 3 Mark Questions and Answers 8
⇒ y – 2 log|y + 2| = x + 2log|x| + c ____(1)
Since it passes through (1, -1),
⇒ -1 – 2log|-1 + 2| = 1 + 2log|l| + c
⇒ -2 = c
(1) ⇒ y – 2log|y + 2| = x + 2log|x| – 2.

Question 8.
Solve the initial value problem: \(\frac{d y}{d x}\) = y tan 2x; y(0) = 2.
Answer:
\(\frac{d y}{d x}\) = y tan 2x
⇒ \(\frac{d y}{y}\) tan 2xdx,
This is a variable type
∴∫\(\frac{d y}{y}\) = ∫tan 2xdx ⇒ log y = \(\frac{1}{2}\) log|sec 2x| + c
Given y(0) = 2 ⇒ log 2 = \(\frac{1}{2}\) log|sec 0| + c ⇒ c = log 2
log y = \(\frac{1}{2}\) log|sec 2x| + log 2 ⇒
Plus Two Maths Differential Equations 3 Mark Questions and Answers 9

Plus Two Maths Differential Equations Four Mark Questions and Answers

Question 1.
(i) Consider the differential equation given below. (1)
\(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0\). Write the order and degree of the DE (if defined)
(ii) Find the Differential equation satisfying the family of curves y² = a(b² – x²), a and b are arbitrary constants. (3)
Answer:
(i) 4; degree is not defined

(ii) y² = a(b² – x²) ____(1)
Differentiating with respect to x,
2y \(\frac{d y}{d x}\) = -a2x ⇒ y\(\frac{d y}{d x}\) = -ax ____(2)
Differentiating (2) with respect to x,
Plus Two Maths Differential Equations 3 Mark Questions and Answers 10
Which is the differential equation.

Question 2.

Find the Differential equation satisfying the family of curves y = ae^3x + be^-2x, a and b are arbitrary constants. (3)
Hence write the degree and order of the DE. (1)

Answer:
1. y = ae^3x + be^-2x ____(1)
Differentiating with respect to x,
\(\frac{d y}{d x}\) = ae^3x × 3 + be^-2x × -2 ____(2)
Differentiating (2) with respect to x,
⇒ \(\frac{d^{2} y}{d x^{2}}\) = 9ae^3x + 4be^-2x ____(3)
Now, (3) + 2 × (2) ⇒ \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=15 a e^{3 x}\)
Plus Two Maths Differential Equations 3 Mark Questions and Answers 11
Using (4), (5) in (1), we have,

2. Order: 2; degree: 1.

Question 3.
Consider the equation of all circles which pass through the origin and whose centres are on the x-axis.

Define the general equation of the circle.(1)
Find the DE corresponding to the above equation. (3)

Answer:
1. The general equation of the circle, passing through the origin and whose centers lies on x-axis can be taken as (x – h)² + y² = h² where h being an arbitrary constant.

2. Simplifying (x – h)² + y² = h² we get,
x² – 2hx + h² + y² = h² ⇒ x² – 2hx + h² = 0 _____(1)
Differentiating we get,
2x + 2y \(\frac{d y}{d x}\) – 2h = 0 ⇒ h = x + y \(\frac{d y}{d x}\)
Substituting in (1) we can eliminate h
Plus Two Maths Differential Equations 3 Mark Questions and Answers 13

Question 4.
Find a particular solution satisfying the given condition. (x³ + x² + x +1)\(\frac{d y}{d x}\) = 2x² + x, y = 1, when x = 0.
Answer:
Plus Two Maths Differential Equations 3 Mark Questions and Answers 14
Integrating on both sides,
∫dy = ∫\(\frac{2 x^{2}+x}{\left(x^{2}+1\right)(x+1)} d x\)
Splitting into partial fractions we have, (see Unit:7)

Question 5.

Write the degree of the DE y’ = 2xy. [0, 1, 2, 3] (1)
Express y’ = 2xy in the form Mdx = Ndy. Where M is a function of x and N is the function of y. (2)
Solve y’ = 2xy, y(0) = 1 (1)

Answer:

1. Degree = 1

2. We have, \(\frac{d y}{d x}\) = 2xy ⇒ \(\frac{d y}{y}\) = 2xdx, which is of the form Mdx = Ndy.

3. Solution is ∫\(\frac{d y}{y}\) = 2∫xdx ⇒ log|y| = x² + c
Given y(0) = 1 ⇒ log|1| = 0 + c ⇒ c = 0
⇒ log|y| = x² ⇒ y = e^x².

Question 6.
Solve the following DE \(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\).
Answer:
\(\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}\), this is a Homogeneous DE.
Therefore, put y = vx and \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) to convert it into variable separable form.
The DE becomes,

Plus Two Maths Differential Equations 3 Mark Questions and Answers 18
Therefore integrating we get,

Question 7.
Solve the linear differential equation \(x \frac{d y}{d x}-y=(x-1) e^{x}\).
Answer:
Given, x\(\frac{d y}{d x}\) – y = (x – 1)e^x, dividing both sides by x ,we get
Plus Two Maths Differential Equations 3 Mark Questions and Answers 21
Solution is
y × IF = ∫Q(IF)dx + c

Question 8.
(i) Choose the correct answer from the bracket. The solution of the differential equation xdy + ydx = 0 represents (1)
(a) a rectangular hyperbola
(b) a parabola whose centre is origin
(c) a straight line whose centre is origin
(d) a circle whose centre is origin.
(ii) From the DE of the family of circles touching the x-axis at origin. (3)
Answer:
(i) (c) a straight line whose centre is origin.

(ii) Let (0, a) be the centre of the circle. Therefore the equation of the circle is
x² + (y – a)² = a²
⇒ x² + y² = 2ay
⇒ \(\frac{x^{2}+y^{2}}{y}\) = 2a ____(1)
Plus Two Maths Differential Equations 3 Mark Questions and Answers 23
Differentiating with respect to x,

Question 9.
Solve the DE x²\(\frac{d y}{d x}\) = x² – 2y² + xy.
Answer:
x²\(\frac{d y}{d x}\) = x² – 2y² + xy
Plus Two Maths Differential Equations 3 Mark Questions and Answers 25
this is a Homogeneous DE.
Put y = vx and \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

Integrating on both sides,

Question 10.
Choose the correct answer from the bracket

The DE \(\frac{d y}{d x}+\frac{y}{x}\) = e^x, x > 0 is of order …..[0,1,2,3] (1)
The integrating factor \(\frac{d y}{d x}+\frac{y}{x}\) = e^x, is……..[x, e^x, -x, e^-x] (1)
Solve \(\frac{d y}{d x}+\frac{y}{x}\) = e^x (2)

Answer:
1. Order = 1

2. \(\frac{d y}{d x}+\frac{y}{x}\) = e^x is of the form \(\frac{d y}{d x}\) + Py = Q,
where P = \(\frac{1}{x}\), Q = e^x
IF = e^∫Pdx = e^{∫\(\frac{1}{x}\)dx} = e^logx = x

3. Solution is y.IF = ∫e^x. IFdx
⇒ yx = ∫x.e^xdx ⇒ yx = x.e^x – ∫e^xdx
⇒ yx = x.e^x – e^x + c ⇒ yx = e^x(x – 1) + c.

Question 11.
Solve the DE \(\frac{d y}{d x}=\frac{x+y}{x-y}\).
Answer:
\(\frac{d y}{d x}=\frac{x+y}{x-y}\), this is a Homogeneous DE.
Put y = vx and \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\)
Plus Two Maths Differential Equations 3 Mark Questions and Answers 28
Integrating on both sides,

Plus Two Maths Differential Equations Six Mark Questions and Answers

Question 1.
Consider the DE \(\frac{d y}{d x}=\frac{y^{3}+3 x^{2} y}{x^{3}+3 x y^{2}}\)

Identify the DE ? Give reason. (1)
Explain the method of solving the DE. (1)
Solve the DE. (4)

Answer:
1. Given DE is a Homogeneous DE. Since y³ + 3x²y and x³ + 3xy² are Homogeneous functions of same degree (deg = 3).

2. By giving a substitution y = v x and \(\frac{d y}{d x}\) = v + x\(\frac{d y}{d x}\)
we can convert the DE into variable separable.

3. Now we have, \(\frac{d y}{d x}=\frac{y^{3}+3 x^{2} y}{x^{3}+3 x y^{2}}\)
Plus Two Maths Differential Equations 3 Mark Questions and Answers 30
∴ Integrating we get,
\(\int \frac{1+3 v^{2}}{2 v\left(1-v^{2}\right)} d v=\int \frac{d x}{x}\)

\(\frac{1}{2}\) log v – log(1 – v²) = log x + log c

Question 2.
Consider the D.E \(\frac{d y}{d x}+\frac{y}{x}=x^{2}\)

Find degree and order of DE . (1)
Solve the D.E. (4)
Find the particular solution when x = 1, y = 1. (1)

Answer:
1. Degree: 1, Order: 1.

2. The given D.E is first order linear DE of the form
\(\frac{d y}{d x}\) + Py = Q. Comparing we get, P = \(\frac{1}{x}\), Q = x²
∴ ∫Pdx = ∫\(\frac{1}{x}\)dx = logx
Integrating factor (I.F) = e^∫Pdx = e^logx = x
y.x = ∫x².xdx + c = ∫x³ dx + c
⇒ y.x = \(\frac{x^{4}}{4}\) + c ___(1)

3. Given, y = 1 when x = 1, then (1)
⇒ 1 × 1 = \(\frac{1}{4}\) + c ⇒ c = \(\frac{3}{4}\)
Therefore particular solution is
y.x = \(\frac{x^{4}}{4}\) + \(\frac{3}{4}\) ⇒ 4xy = x³ + 3.

Question 3.
Consider the equation.\(\frac{d y}{d x}\) + y = sin x

What is the order and degree of this equation? (1)
Find the integrating factor. (2)
Solve this equation. (3)

Answer:
1. Order = 1, Degree = 1

2. Given, \(\frac{d y}{d x}\) + y = sin x is of the form
\(\frac{d y}{d x}\) + Py = Q ⇒ P = 1, Q = sinx
Integrating factor = e^∫Pdx = e^∫1dx = e^x

3. Therefore solution is
y.IF = ∫Q.IFdx + c ⇒ ye^x = ∫e^x sinxdx + c ____(1)
∫sinx.e^xdx = e^x sinx – ∫cosx.e^xdx
= e^x sin x – cosx.e^x – ∫sinx.e^x dx
⇒ 2∫e^x sin xdx = e^x(sin x – cos x)
⇒ ∫e^x sinxdx = \(\frac{e^{x}}{2}\)(sinx – cosx)
(1) ⇒ ye^x = \(\frac{e^{x}}{2}\)(sinx – cosx) + c.

Question 4.
Considerthe D.E (x² – 1)\(\frac{d y}{d x}\) + 2(x + 2)y = 2(x + 1)

Find \(\frac{d y}{d x}\), degree and order of the above differential equation. (1)
Find the integrating factor of the above differential equation. (2)
Solve the differential equation. (3)

Answer:
1. Given, (x² – 1)\(\frac{d y}{d x}\) + 2(x + 2)y = 2(x + 1)
Plus Two Maths Differential Equations 3 Mark Questions and Answers 33
Here, Degree = 1, Order = 1.

2. The given DE is of the form \(\frac{d y}{d x}\) + Py = Q
Where,
Plus Two Maths Differential Equations 3 Mark Questions and Answers 34
Splitting it into partial fractions we get,

Put x = 1, ⇒ 6 = 2B ⇒ B = 3,
put x = -1, ⇒ 2 = -2A ⇒ A = -1

3. Solution is y × IF = ∫Q × IFdx + c
Plus Two Maths Differential Equations 3 Mark Questions and Answers 37

Question 5.
(i) The degree of the differential Equation \(\frac{d^{2} y}{d x^{2}}+\cos \left(\frac{d y}{d x}\right)=0\) is
(a) 2
(b) 1
(c) 0
(d) Not defined
(ii) Solve \(\frac{d y}{d x}\) + 2y tanx = sinx; y = 0, x = \(\frac{\pi}{3}\) (5)
Answer:
(i) (d) Not defined.

(ii) \(\frac{d y}{d x}\) + 2y tanx = sinx
Then, P = 2tanx, Q = sinx
IF = e^∫Pdx = e^∫2tanxdx = e^{2log sec x} = sec² x Solution is; y × IF = ∫Q(IF)dx + c
⇒ ysec² x = ∫sinx sec² xdx + c
⇒ ysec² x = ∫tanx secx dx + c
⇒ ysec²x = secx + c
Here; y = 0, x = \(\frac{\pi}{3}\)
⇒ 0 × sec² \(\frac{\pi}{3}\) = sec\(\frac{\pi}{3}\) + c ⇒ c = -2
⇒ ysec² x = secx – 2.

Question 6.
(i) The order of the differential equation \(x^{4} \frac{d^{2} y}{d x^{2}}=1+\left(\frac{d y}{d x}\right)^{3}\) is
(a) 1
(b) 3
(c) 4
(d) 2
(ii) Find the particular solution of the (1 + x²) \(\frac{d y}{d x}\) + 2 xy = \(\frac{1}{1+x^{2}}\); y = 0, when x = 1 (5)
Answer:
(i) (d) 2

(ii) (1 + x²) \(\frac{d y}{d x}\) + 2 xy = \(\frac{1}{1+x^{2}}\); y = 0, when x = 1
Plus Two Maths Differential Equations 3 Mark Questions and Answers 38
⇒ 0(1 + 1²) = tan^-11 + c ⇒ c = \(-\frac{\pi}{4}\)
⇒ y(1 + x²) = an^-1x – \(\frac{\pi}{4}\).

Plus One Economics Chapter Wise Questions and Answers Chapter 6 Rural Development

April 13, 2021February 7, 2020 by Prasanna

Students can Download Chapter 6 Rural Development Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 6 Rural Development

Plus One Economics Rural Development One Mark Questions and Answers

Question 1.
Rural developments include:
(i) education and skill developments
(ii) land reforms
(iii) measures for alleviation of poverty
(iv) all of the above
Answer:
(iv) all of the above

Question 2.
Shift of workforce from agriculture to other allied activities like livestock, poultry, fisheries, etc, is known as.
Answer:
Diversification.

Question 3.
Golden revolution is associated with?
(i) Horticulture
(ii) Livestock
(iii) Fisheries
(iv) Pulses
Answer:
(i) Horticulture

Question 4.
How many banks were nationalised in 1969?
(i) 6
(ii) 9
(iii) 14
(iv) 20
Answer:
(iii) 14

Question 5.
White revolution is related to
(i) Bamboo
(ii) Cash crop
(iii) Milk
(iv) Foodgrain
Answer:
(iii) Milk

Question 6.
Blue revolution is related to
(i) Milk
(ii) Fish
(iii) Egg
(iv) None of these
Answer:
(ii) Fish

Question 7.
Which among the following is an advantage of organic fanning
(i) It is labour intensive
(ii) It is environment-friendly
(iii) It uses locally produced organic inputs
(iv) All the above
Answer:
(iv) All the above

Question 8.
Complete the following statement.
The credit provisions of SHGs are generally referred to as ………………..
Answer:
Microcredit programs.

Question 9.
“Regulated Market yards benefit farmers as well as consumers”. Do you agree with this statement?
Answer:
Yes, I agree with this statement that the Regulated Market yields benefit farmers as well as consumers.

Plus One Economics Rural Development Two Mark Questions and Answers

Question 1.
What is TANWA?
Answer:
TANWA stands for Tamil Nadu Women in Agriculture. It is a project initiated in Tamil Naud to train women in latest agricultural techniques. It includes women to actively participate in raising agricultural productivity and family income. They promote small scale household activities like soap manufacturing, doll making etc.

Question 2.
Name few horticultural crops.
Answer:
Names of horticultural crops are:

Fruits
Vegetables
Medicinal plants

Plus One Economics Rural Development Three Mark Questions and Answers

Question 1.
Complete the following.
a. Percentage of population living in the rural area is
(i) 25%
(ii) 50%
(iii) 75%
(iv) 90%
b. NABARD is the open bank providing credit to sector
(i) ombined sector
(ii) industrial sector
(iii) agricultural sector
(iv) tertiary sector
c. The Golden revolution was a period of very high productivity in
(i) food grain production
(ii) horticulture
(iii) organic farming
(iv) pisciculture
Answer:
a. (iii) 75%
b. (iii) agricultural sector
c. (ii) horticulture

Question 2.
Why the co-operative marketing could not gain momentum in India?
Answer:
Co-operative marketing has failed to gain adequate success in India because of inadequate coverage of farmer members and lack of appropriate link between marketing and processing management.

Question 3.
What do you mean by organic farming?
Answer:
Eco-friendly technology of cultivation is known as organic farming, it is the farming without using chemical fertilizers or pesticides. It uses natural mannure.

Question 4.
Give reasons for the declining share of agriculture in GDP of India.
Answer:
After the initiation of reforms, the growth rate of agriculture sector has declined to 2.3% during the 1990s which was lower than earlier years. The reasons for this decline are inadequate infrastructure, alternate employment opportunities in the industry or service sector, increase in casualisation of employment in agriculture, etc.

Question 5.
Explain the term ‘golden revolution’.
Answer:
The growth of horticulture sector at a spectacular speed is called ‘Gloden Revolution’. The period between 1991 – 2003 is known as the period of golden revolution. During this period, the investment in horticulture became highly productive and the sector emerged as a sustainable livelihood option. Thus India could become leading producer of mangoes, bananas, coconuts etc.

Question 6.
Name a few traditional industries in India.
Answer:
The traditional industries of India are:

Pottery
Crafts
Handloom
Textiles
Bamboo

Question 7.
Prepare a note on ‘Kudumbashree’ in Kerala.
Answer:
Kudumbasree is a women-oriented community-based poverty reduction programme being implemented in kerala. The objective of this group is to encourage savings. The small savings in women’s group creates more employment and productive atmosphere in women’s groups.

Question 8.
Prepare a note on NABARD?
Answer:
NABARD stands for National Bank for Agriculture and Rural Development. It was established in 1982, July. It has been assigned the role performed by agricultural refinance and development corporation and also the role performed by RBI for rural credit. NABARD provides short term loans for agricultural operation.

Plus One Economics Rural Development Four Mark Questions and Answers

Question 1.
What are the alternative channels available for agricultural marketing? Give examples:
Answer:
Following are the alternative channels available for agricultural marketing.

Co-operative Credit Societies
Reserve Bank of India
Regional Rural Banks
Commercial Banks
NABARD
Self Help Groups (SHGs)

Question 2.
Critically appraise Rural banking operations in the light of recent farmers suicides in India.
Answer:
Operation of rural banking had a positive effect on rural farm and non-farm output, income, and employment. Farmers were given a variety of loans for meeting production needs. This helped India to achieve food security and to gain a huge buffer stock of grains.

However our banking system in the rural area could not meet all loan needs of the farmers. Some of the farmers still depend on local moneylenders. This system failed to develop a culture of deposit mobilization. Agriculture loan default was very high.

Question 3.
Mention some obstacles that hinder the mechanism of agricultural marketing.
Answer:
The farmers suffer from many problems relating to agricultural marketing. Some obstacles that hinder the mechanism of agricultural marketing are mentioned below.

faulty weighing
manipulation of accounts
forced to sell at lower prices
lack of proper storage facilities
lack of infrastructure facilities
inadequate coverage of farmers by cooperatives
inefficient financial management

Question 4.
Match the following.
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Rural Development img1
Answer:

Question 5.
‘Rainbow revolution is meant for overall agricultural development’, comment.
Answer:
The rainbow revolution is meant for overall development of Indian agriculture. Under rainbow revolution, the following revolutions are included.
Plus One Economics Chapter Wise Questions and Answers Chapter 6 Rural Development img3

Plus One Economics Rural Development Five Mark Questions and Answers

Question 1.
State whether true or false.

Organic food has higher nutritional value compared to non-organic food
Ten commercial banks were nationalized in 1969
Golden revolution happened in fisheries
SHG stands for Social Health Groups
Use of chemical fertilizers lead to pollution.

Answer:

True
False. Fourteen commercial banks were nationalized in 1969
False. Golden revolution happened in horticulture
False. SHG stands for Self Help Groups
True

Question 2.
“Organic farming and sustainable development are closely related to each other”. Substantiate.
Answer:
In recent years, awareness of the harmful effect of chemical-based fertilizers and pesticides on our health is on a rise. Conventional agriculture relies heavily on chemical fertilizers and toxic pesticides etc., which enter the food supply, penetrate the water sources, harm the livestock, deplete the soil and devastate natural eco-systems.

Efforts in evolving technologies which are ecofriendly are essential for sustainable development and one such technology which is eco-friendly is organic farming. In short, organic agriculture is a whole system of farming that restores, maintains and enhances the ecological balance.

There is an increasing demand for organically grown food to enhance food safety throughout the world. Thus it can be stated that organic farming and sustainable development are closely related to each other.

Question 3.
Identify the benefits and limitations of organic farming.
Answer:
A. Benefits of Organic Farming
1. Organic agriculture offers a means to substitute costlier agricultural inputs with locally produced organic inputs that are cheaper and thereby generate good returns on investment.

2. Organic agriculture also generates incomes through international exports as the demand for organically grown crops is on a rise. Studies across countries have shown that organically grown food has more nutritional value than chemical farming thus providing us with healthy foods.

3. Since organic farming requires more labour input than conventional farming, India will find organic farming an attractive proposition

4. the organic produce is pesticide-free and produced in an environmentally sustainable way

5. Organic fanning helps in sustainable development of agriculture and India has a clear advantage in producing organic products for both domestic and international markets.

B. Limitations of organic farming
1. Popularizing organic farming requires awareness and willingness on the part of farmers to adapt to new technology.

2. Inadequate infrastructure and the problem of marketing the products are major concerns which need to be addressed apart from appropriate agriculture policy to promote organic farming.

3. It has been observed that the yields from organic farming are less than modem agricultural farming in the initial years. Therefore, small and marginal farmers may find it difficult to adapt to large scale production.

4. Organic produce may also have more blemishes . and a shorter shelf life than sprayed produce. Moreover choice in production of off-season crops is quite limited inorganic farming.

Question 4.
Narrate the importance of agricultural diversification in promoting rural sector?
Answer:
Agricultural diversification is a system of fanning that encourages production of a variety of plant and animals and their products. Non-farm employment provides greater income stability, economies of size also reduces uncertainty of business. It involves maximum number of persons as it is labour intensive in nature.

The majority of small and marginal farmers cultivate mainly low-value subsistence crops. In the absence of farm and non-farm employment opportunities, they are forced to live below poverty line. So this problem is solved with the help of commercialisation and diversification of small farmers within and outside agriculture and their proper integration with local and global markets.

This is helpful not only in liberating the small and marginal farmers from the poverty trap, but also to meet the country’s growing demands for fruits, vegetables, milk and milk products, meat, fishes, eggs, etc. which generally show rising trends with increasing levels of per capita income in the economy.

Question 5.
“Rural development focuses on actions for the development of the areas that are lagging behind in the overall development of the development of the village economy” point out the areas of rural development.
Answer:
Rural development is quite a comprehensive term but it essentially means a plan of action for the development of areas which are lagging behind in socio-economic development. Some of the areas which are challenging and need fresh initiatives for development in India include

a. Development of human resources including:

Literacy, more specifically, female literacy, education and skill development
Health, addressing both sanitation and public health

b. Land reforms

c. Development of infrastructure development like electricity, irrigation, credit, marketing, transport facilities including construction of village roads and feeder roads to nearby highways, facilities for agriculture research and extension, and information dissemination

d. Introduction of special measures for alleviation of poverty and bringing about significant improvement in the living conditions of the weaker sections of the population emphasizing access to productive employment opportunities.

Question 6.
Explain some measures taken by the government to improve agricultural marketing.
Answer:
Agricultural marketing encompasses all activities in moving farm products from the producers to the final consumers such as storage, transport, processing, etc. In India, agricultural marketing was regarded as operationally inefficient and exploitative in character. Hence to improve the position of agricultural marketing following things are required :

Facilities for storing goods.
Adequate and economical transport.
Eliminination of intermediaries.
Basic information of market conditions to get better prices for the produce.

The above mentioned requirements for efficient agricultural marketing can be also classified into the following.

Promotion of cooperative agricultural marketing societies.
Developing regulated markets where marketing practices have been standardized so as to provide encouragement to farmers to come to mandis to dispose off their produce and prevent them from exploitation in the hand of Dallas.
Granding and standardisation of produce to help the farmer to fetch better prices.
Developing warehousing facilities, road transport, etc.
Declaration of support price that offers a minimum price to the farmers for their produce.

Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals

April 13, 2021February 7, 2020 by Prasanna

Students can Download Chapter 8 Application of Integrals Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 8 Application of Integrals

Plus Two Maths Application of Integrals Four Mark Questions and Answers

Question 1.
Consider the following figure.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 1

Find the point of intersection (P) of the given parabola and the line. (2)
Find the area of the shaded region. (2)

Answer:
1. We have, y = x² and y = x ⇒ x = x² ⇒
⇒ x² – x = 0 ⇒ x(x – 1) = 0 ⇒ x = 0, 1
When x = 0, y =0 and x = 1, y = 1.
Therefore the points of intersections are (0, 0) and(1, 1).

2. Required area
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 2

Question 2.
1. Find the point of intersection ‘p’ of the given parabola and the line. (2)
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 3
2. Find the area of the shaded region. (2)
Answer:
1. Given, y = x², y = 2x
⇒ 2x = x² ⇒ x² – 2x = 0 ⇒ x(x – 2) = 0 ⇒ x = 0, 2
We have, y = 2x
⇒ when x = 0 ⇒ y = 0, when x = 2 ⇒ y = 4
‘P’ has co-ordinate (2, 4)

2. Area = \(\int_{0}^{2} 2 x d x-\int_{0}^{2} x^{2} d x=\left(x^{2}\right)^{2}-\left(\frac{x^{3}}{3}\right)_{0}^{2}=4-\frac{8}{3}=\frac{12-8}{3}=\frac{4}{3}\).

Question 3.
Consider the following figure.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 4

Find the point of Intersection P of the circle x² + y² = 32 and the line y = x. (1)
Find the area of the shaded region. (3)

Answer:
1. x² + x² = 32 ⇒ 2x² = 32 ⇒ x² = 16 = 4
Therefore the point of intersection P is (4, 4).

2. We have x² + y² = 32 ⇒ y = \(\sqrt{32-x^{2}}\).
The required area =
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 5
= 8 + [8π – 8 – 4π] = 4π.

Question 4.

Shade the area enclosed by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant. (2)
Find the area of the region bound by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant. (2)

Answer:
1.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 6

2. Area
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 7

Question 5.

Draw a rough sketch of the graph of the function y² = 4x. (2)
Find the area by the curve and the line x= 2. (2)

Answer:
1.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 8

2. Area
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 9

Plus Two Maths Application of Integrals Six Mark Questions and Answers

Question 1.

Draw the graph of y² = 4x and y = x. (2)
Find the points of intersection of y² = 4x and y = x. (2)
Find the area bounded by the graphs.(2)

Answer:
1. y² = 4x is a parabola and y = x is a straight line passing through the origin.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 10

2. x² = 4x ⇒ x² – 4x = 0 ⇒ x(x – 4) = 0 ⇒ x = 0, 4
When x = 0, y = 0 and when x = 4, y =4. Therefore the points of intersection are (0, 0) and (4, 4).

3. Area bounded by the graphs = Area under the parabola in the first quadrant – Area under the line.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 11

Question 2.

Draw the graph of the function y = x² and x = y² in a coordinate axes. (2)
Find the point of intersection of the above graphs. (2)
Find the area of the region bounded by the above two curves. (2)

Answer:
1. The two function are parabolas as shown in the figure.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 12

2. We have, y = x² and x = y²
x = (x²)² ⇒ x – x⁴ = 0 ⇒ x(1 – x³) = 0 ⇒ x = 0, 1
When x= 1, y= 1 and x = 0, y = 0.
Therefore the point is (0, 0) and (1, 1).

3. The required area = \(\int_{0}^{1} \sqrt{x} d x-\int_{0}^{1} x^{2} d x\)
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 13

Question 3.
Using the figure answer the following questions
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 14

Find the area of the shaded region as the sum of the area of two triangles. (2)
Define the function of the given graph. (2)
Verify the area of the shaded region using integration. (2)

Answer:
1. Area = Area of ∆OAD + Area of ∆ ABC
\(\frac{1}{2}\) × 3 × 3 + \(\frac{1}{2}\) × 2 × 2 = \(\frac{9}{2}\) + 2 = \(\frac{13}{2}\).

2. Area = \(\int_{0}^{5}\)f(x) dx = \(\int_{0}^{3}\)(-x + 3) dx + \(\int_{3}^{5}\)(x – 5) dx
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 15
Therefore verified.

Question 4.
The figure given below contains a straight line L with slope \(\sqrt{8}\) and a circle.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 16

Find the equation of the line L and circle. (1)
Find the point of intersection P. (2)
Find the area of the shaded region. (3)

Answer:
1. The line L passes through origin and have slope 3, therefore its equation is y = \(\sqrt{8}\) x. The circle passes through origin and have radius 3, therefore its equation is x² + y² = 9.

2. We have, y =3x and x² + y² = 9
⇒ x² + (\(\sqrt{8}\)x)² = 9 ⇒ 9x² = 9
⇒ x = 1
∴ y = \(\sqrt{8}\) × 1 = \(\sqrt{8}\).
Therefore, coordinate of ‘P’ is (1, \(\sqrt{8}\)).

3. Area of the shaded region
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 17

Question 5.
Using the given figure answer the following
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 18

Define the equation of the circle and ellipse in the figure. (1)
Find the area of the ellipse using integration. (4)
Find the area of the shaded region. (Use formula to find the area of the circle.) (1)

Answer:
1. From the figure equation of the circle is x² + y² = 4 and that of the ellipse is \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\).

2. We have, \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)
⇒ y² = 1 ⇒ y = \(\frac{1}{2} \sqrt{4-x^{2}}\)
Area of the ellipse = 4 \(\int_{0}^{2}\)y dx
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 19

3. Area of the circle of radius 2 = π (2)² = 4π
∴ Area of the shaded region = Area of the circle – Area of the ellipse
= 4π – 2π = 2π.

Question 6.

Find the area bounded by the curve y = sin x with X – axis, between x = 0 and x = 2π. (2)
Find the area of the region bounded by the curve y = x² and y = |x| (4)

Answer:
1. Area of y = sin x in each loop is same. Therefore;
2\(\int_{0}^{\pi}\)sin xdx = \(-2(\cos x)_{0}^{\pi}\) = -2 (cos π – cos0)
= -2(-1 – 1) = 4

2.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 20
Area

Question 7.
Using integration, find the area of the region bounded by the triangle whose vertices are(-1, 1), (0, 5) and (3, 2). (6)
Answer:
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 22
Equation of BC is \(\frac{y-5}{1-5}=\frac{x-0}{-1-0}\)
⇒ y – 5 = 4x ⇒ 4x – y + 5 = 0 ⇒ y = 4x + 5
Equation of AB is x + y – 5 = 0 ⇒ y = 5 – x
Equation of AC is x – 4y + 5 = 0 ⇒ y = \(\frac{x}{4}+\frac{5}{2}\)
The required area = Area of the region PABCQP – Area of the region PACQP
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 23

Question 8.
Consider the functions f(x) = sin x and g(x) = cosx in the interval [0, 2π]

Find the x coordinates of the meeting points of the functions. (1)
Draw the rough sketch of the above functions. (2)
Find the area enclosed by these curves in the given interval. (3)

Answer:
1. f(x) = sin x and g(x) = cos x meet at multiples of \(\frac{\pi}{4}\)
x = \(\frac{\pi}{4}\), \(\frac{5 \pi}{4}\).

2.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 24

3. Area = 2{Area under f(x) = sinx from \(\frac{\pi}{4}\) to π – Area under g(x) = cosx from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\)}
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 25

Question 9.

Evaluate \(\int_{0}^{r} \sqrt{r^{2}-x^{2}} d x\), where r is a fixed positive number. Hence prove the area of the circle of radius r is π r². (2)
Find the area of the circle, x² + y² = 16 which is exterior to parabola y² = 6x. (4)

Answer:
1.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 26

2. Given, x² + y² = 16 and y² = 6x ⇒ x² + 6x = 16 ⇒ x² + 6x – 16 = 0 ⇒ (x + 8)(x – 2) = 0 ⇒ x = -8, 2.
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 28
Area = Area of the circle – Interior area of the parabola.

Question 10.
Using the figure answer the following questions
Plus Two Maths Application of Integrals 4 Mark Questions and Answers 30

Define the equation of the ellipse and circle in the given figure. (1)
Find the area of the ellipse using integration. (4)
Find the area of the shaded region. (Area of the circle can be found using direct formula). (1)

Answer:
1. Equation of the ellipse is \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\) and circle is x² + y² = 1.

3. Area of the circle = πr² = π × 1 = π
Required area = Area of ellipse – area of the circle = 2π – π = π.

Plus One Economics Chapter Wise Questions and Answers Chapter 5 Human Capital Formation in India

April 13, 2021February 7, 2020 by Prasanna

Students can Download Chapter 5 Human Capital Formation in India Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 5 Human Capital Formation in India

Plus One Economics Human Capital Formation in India One Mark Questions and Answers

Question 1.
As per 2011 census, the literacy rate in India was.
(i) 74%
ii) 78%
(iii) 80%
(iv) 82%
Answer:
(i) 74%

Question 2.
Which among the following is related to health sector?
(i) UGC
(ii) NCERT
(iii) ICMR
(iv) AICTE
Answer:
(iii) ICMR

Question 3.
The state in which the literacy rate is the highest in the country is ……………..
(i) Delhi
(ii) Bihar
(iii) Mizoram
(iv) Kerala
Answer:
(iv) Kerala

Question 4.
The Adult literacy rate among females in India in the year2000 was ………………..
(i) 68.4%
(ii) 61.9%
(iii) 45.4%
(iv) 37.9%
Answer:
(iii) 45.4%

Question 5.
State true or false:
‘New Educational Policy was announced in 1987-88’.
Answer:
True.

Question 6.
Which state is having highest literacy rate in India according to 2011 census?
(i) Kerala
(ii) Punjab
(iii) Manipur
(iv) Hariyana
Answer:
(i) Kerala

Plus One Economics Human Capital Formation in India Two Mark Questions and Answers

Question 1.
Mention two government organizations that regulate the health and education sector.
Answer:
1. Organizations that regulate the education sector:

The ministries of education at the union and state level
National Council of Educational Research and Training (NCERT)
University Grants Commission (UGC)
All India Council of Technical Education (AICTE)

2. Organizations that regulate the health sector:

The ministries of health at the union and state level
Indian Council for Medical Research (ICMR)

Question 2.
Classify the following into preventive medicines, curative medicines, and social medicines.

Provision of clean drinking water.
Good sanitation
Medical intervention during illness
Vaccination

Answer:

Preventive Medicine – Vaccination
Curative Medicine – Medical intervention during illness.
Social Medicine – Provision of clean drinking water, good sanitation

Question 3.
Classify the following into the features of human capital and human development

Education and health are considered as a means to increase human productivity.
Education and health are integral to human well-being.
Human beings are treated as a means to an end.
Human beings are treated as ends in themselves.

Answer:
a. Human Capital:
1. Education and health are considered as a means to increase human productively.
3. Human beings are treated as a means to an end.
b. Human development:
2. Education and health are integral to human well being.
4. Human beings are treated as ends in themselves.

Question 4.
Name the two major sources of human capital in a country.
Answer:
Two major sources of human capital in a country are:

Investment in education and
Investment in health.

Question 5.
Bring out the need for on-the-job training for a person.
Answer:
Firms give on-the-job-training to enhance the productive skills of the workers so as to enable them to absorb new technologies and modern ideas. It can be given in two ways:

Workers may be trained in the firm itself under the assistance of an experienced worker.
Workers may be sent off the firm campus for the training.

Plus One Economics Human Capital Formation in India Three Mark Questions and Answers

Question 1.
Give full form of the following

Answer:

GER – Gross Enrolment Ratio
SSA – Sarva Shiksha Abhiyan

Question 2.
Find the odd one out

Roads, factory, teachers, power plants
Teachers, doctors, ports, scientists
UGC, NCERT, SCERT, NREGP

Answer:

Teachers. Others are physical capital
Ports. Others are human capital
NREGP. Others are educational regulatory institutions

Question 3.
“Adequate education and training to farmers can raise productivity in farms.”
Do you agree or not? Substantiate your answer.
Answer:
Yes, I do agree to the statement that adequate education and training to farmers can raise productivity in farms.
This is because the attainment of more information and training will help the farmers to attain more knowledge about new methods of farming and plant protection. The application of the new knowledge in the farms will enable them to raise productivity in the farms so that the overall productivity in the agricultural sector can be improved.

Question 4.
What do you mean by Human Development Index? How it is calculated?
Answer:
The quality of life index prepared and published by United Nations Development Programme is termed as Human Development Index. HDI studies the following three basic human capabilities:

Living a Long Life (Longevity)
Being Knowledgable (Educational Attainment)
Enjoying decent standard of living (Real Per Capita GDP)

In orderto calculate HDI we are required to construct the following three indices.
1. Life expectancy index: This index measures the degree of the achievement of the country regarding the expectancy of the people in a country. The value of index varies between 0-1.

2. Educational attainment index: This index measures the level of educational attainment of the people. The value of this index also varies between 0-1. Higher index shows higher level of educational attainment.

3. Real GDP per capita index: Real GDP is calculated at constant price. It shows the changes in physical production in real terms. Per capita, real GDP is GDP apanstant price divided by population. It is better measure of the quality of life of the people as compared to per capita income.
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Human Capital Formation in India img1

Plus One Economics Human Capital Formation in India Four Mark Questions and Answers

Question 1.
Point out the major problems of educational development in India.
Answer:
The major problems of educational development in India are pointed out below.

Wastage of resources.
Illiteracy
Disparities in standard
Lack of funds
Problems of brain drain.

Question 2.
Discuss the following as a resource of human capital formation

Health infrastructure
Expenditure on migration

Answer:
1. Health infrastructure:
Expenditure on health is an important source of human capital formation. Health expenditure directly increases the supply of healthy labour force and is a source of human capital formation.

Preventive medicine, social medicine, curative medicine, and provision of clean drinking water and good sanitation are the various forms of health expenditure. In short provision of health facilities directly improve the efficiency of human capital.

2. Expenditure on migration:
Existence of unemployment is the main reason for the rural-urban migration in India. People migrate in search of jobs. Technically qualified persons migrate to other countries because of higher salaries that they may get in other countries.

Migration in any of these cases involves cost of transport, higher cost of living in the migrated places. The enhanced earnings in the new place outweigh the cost of migration. Hence expenditure on migration is a source of human capital formation.

Question 3.
Bring out the differences between human capital and human development.
Answer:

Human capital	Human development
1. Considers education and health to increase productivity.	1. It is based on the idea that education and health are integral to human well being.
2. Treats human beings as a means to achieve higher productivity or the end.	2. It is a broader concept which considers human beings as ends in themselves.

Question 4.
The following table shows the indicators of development in Education and health sectors. Analyze the table and make inference.
Plus One Economics Chapter Wise Questions and Answers Chapter 5 Human Capital Formation in India img2
Answer:
1. The per capita income of the country was Rs.3687 in the year 1951. This value gradually increased over the years and in 2001, the real per capita income became Rs. 10306. this increase is a good sign of development as far a developing country like India is concerned.

2. Crude Death rate (Per 1000 population) of the country Was 25.1 in1951 which has fallen to 8.1 in 2001. Similarly, the infant mortality rate in India also declined significantly from 146 in 1951 to 63 in 2001. This shows the developments undergone in the health sector of the country.

3. Another important achievement in the field of education is rising literacy rate. It was 16.67% in 1951 whereas, the rate increased to 65.2% in 2001. The above table shows India’s achievements in the field of education, health and per capita income since independence. In all these three fields, country made very significant development during the last 50 years.

Plus One Economics Human Capital Formation in India Five Mark Questions and Answers

Question 1.
Trace the relationship between human capital and economic growth.
Answer:
Economic growth means the increase in real national income of a country. The contribution of the educated person to economic growth is more than that of an illiterate person. Education and health are important factors of economic growth.

Human capital contributes not only towards increasing labour productivity but also stimulates innovations and creates the ability to absorb new technologies. Education provides knowledge to understand changes in society and scientific advancements. Higher-income causes building of high level of human capital and also the high level of human capital causes growth of income.

Question 2.
“The percentage of ‘education expenditure of total government expenditure’ indicates the importance of education in the scheme of things before the government”. Critically analyze the changes in government spending on education.
Answer:
The government is supposed to make expenditure on education so as to improve the efficiency of human capital in the country. The percentage of ‘education expenditure of GDP’ expresses how much of our income is being committed to the development of education in the country.

During 1952-2002, education expenditure as percentage of total government expenditure increased from 7.92 to 13.17 and as percentage of GDP increased from0.64 to 4.02. Throughout this period the increase in education expenditure has not been uniform and there has been irregular rise and fall.

In addition to lowering expenditure on education, there exists a regional disparity in case of spending on education in various parts of the country. For example, the per capita education expenditure differs considerably across states from as high as Rs 3,440 in Lakshadweep to as low as Rs 386 in Bihar. This leads to differences in educational opportunities and attainments across states.

Thus it can be concluded that the government is not giving proper attention on the education sector. The relevance of education in the development of the country is to be recognized and therefore, given top priority to the expenditure in this sector.

Question 3.
“Physical capital is well distinguished from human capital”. Narrate them distinguishing points of physical capital and human capital.
Answer:
Physical capital is well distinguished from human capital. The major points of differences are given below.

1. Physical capital is tangible and can be easily sold in the market like any other commodity. Human capital is intangible; it is endogenously built in the body and mind of its owner.

2. Human capital is not sold in the market; only the services of human capital are sold and hence the necessity of the owner of the human capital to be present in the place of production. The physical capital is separable from its owner, whereas, human capital is inseparable from its owner.

3. The two forms of capital differ in terms of mobility across space. Physical capital is completely mobile between countries except for some artificial trade restrictions. Human capital is not perfectly mobile between countries as movement is restricted by nationality and culture.

4. In the case of human capital, depreciation takes place with ageing but can be reduced, to a large extent, through continuous investment in education, health, etc. This investment also facilitates the human capital to cope with change in technology which is not the case with physical capital.

5. Nature of benefits flowing from human capital are different from that of physical capital. Human capital benefits not only the owner but also society in general. This is called external benefit. An educated person can effectively take part in a democratic process and contribute to the socio-economic progress of a nation. A healthy person, by maintaining personal hygiene and sanitation, stops the spread of contagious diseases and epidemics.

6. Human capital creates both private and social benefits, whereas physical capital creates only private benefit. That is, benefits from a capital good flow to those who pay the price for the product and services produced by it.

Question 4.
Make some argument in favour of the need for different forms of government intervention in education and health sectors.
Answer:
Both education and health have been considered as important for human capital formation. The provision of basic education and health facilities has been accepted as a goal in All societies.

It is important that these facilities be made available free for the poor classes who can not afford them in order to ensure better skills and health of manpower. However, the private sector will not be forthcoming in making large investments in nonprofitable or free distribution of such services. Hence, the government has to play an important role.

Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals

April 13, 2021February 7, 2020 by Prasanna

Students can Download Chapter 7 Integrals Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals

Plus Two Maths Integrals Three Mark Questions and Answers

Question 1.
Integrate the following. (3 Score each)

∫sin x sin 2x sin 3 xdx
∫sec²x cos²2x dx

Answer:
1. We have sinxsin2xsin3x
= 1/2 (2sinxsin3x) sin2x
= 1/2 (cos2x – cos4x) sin2x
= 1/4 (2sin2xcos2x – 2cos4xsi n2x)
= 1/4 [sin4x – (sin6x – sin2x)]
= 1/4(sin4x + sin2x – sin6x)
∫sin x sin 2x sin 3 xdx
= \(\frac{1}{4}\) ∫(sin 4x + sin 2x – sin 6x)dx
= –\(\frac{1}{16}\) cos4x – \(\frac{1}{8}\) cos2x + \(\frac{1}{24}\) cos6x + c.

2. sec²x cos²2x = \(\frac{\left(2 \cos ^{2} x-1\right)^{2}}{\cos ^{2} x}\)
= \(\left(\frac{2 \cos ^{2} x}{\cos x}-\frac{1}{\cos x}\right)^{2}\) = (2cosx – secx)²
= 4cos²x + sec²x – 4
= 2(1 + cos2x) + sec²x – 4
= 2cos2x + sec²x – 2
∫sec² x cos² 2x dx = ∫(2 cos 2x + sec² x – 2)dx
= sin 2x + tan x – 2x + c.

Question 2.
Find \(\int \frac{2+\sin 2 x}{1+\cos 2 x} e^{x} d x\)?
Answer:
Plus Two Maths Integrals 3 Mark Questions and Answers 1
= ∫e^x [sec² x + tan x]dx
= ∫e^x[tanx + sec²x]dx = e^x tanx + c.

Question 3.
Evaluate \(\int \frac{\sec ^{2} x d x}{\sqrt{\tan ^{2} x+4}}\)?
Answer:
Put tanx = u, sec²xdx = dy
Plus Two Maths Integrals 3 Mark Questions and Answers 2

Question 4.
Find the following integrals.
Plus Two Maths Integrals 3 Mark Questions and Answers 3
Answer:
(i) I = \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)
Put cosx = t ⇒ -sin xdx = dt
When x = 0 ⇒ t = cos0 = 1,

(ii) I = \(\int_{0}^{1} x e^{x^{2}} d x\)
Put x² = t ⇒ 2xdx = dt
When x = 0 ⇒ t = 0,
x = 1 ⇒ t = 1
I = \(\frac{1}{2} \int_{0}^{1} e^{t} d t\) =

= [e¹ – e⁰] = e – 1.
Plus Two Maths Integrals 3 Mark Questions and Answers 6
Put sin x = t ⇒ cos xdx = dt
When x = 0 ⇒ t = sin0 = 0,

(iv) I = \(\int_{0}^{2} x \sqrt{x+2} d x\)
Put x + 2 = t² ⇒ dx = 2tdt
When x = 0 ⇒ t = \(\sqrt{2}\), x = 2 ⇒ t = 2
Plus Two Maths Integrals 3 Mark Questions and Answers 8

(v) I = \(\int_{0}^{\frac{\pi}{2}} \sqrt{\sin x} \cos x d x\)
Put sin x = t ⇒ cos xdx = dt
When x = 0 ⇒ t = sin0 = 0,
Plus Two Maths Integrals 3 Mark Questions and Answers 9

Put tan x = t ⇒ sec² xdx = dt
When x = 0 ⇒ t = tan 0 = 0,

Question 5.
(i) If f (x) is an odd function, then \(\int_{-a}^{a} f(x)\) = ?
(a) 0
(b) 1
(c) 2\(\int_{0}^{a} f(x)\) dx
(d) 2a
Evaluate
(ii) \(\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x\)
(iii) \(\int_{-1}^{1} e^{|x|} d x\)
Answer:
(i) (a) 0.

(ii) Here, f(x) = sin⁹⁹x.cos¹⁰⁰x .then,
f(-x) = sin⁹⁹(- x).cos¹⁰⁰(- x) = – sin⁹⁹ x. cos¹⁰⁰ x = -f(x)
∴ odd function ⇒ \(\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x=0\).

(iii) Here, f(x) = e^|x|, f(-x) = e^|-x| = e^|x| = f(x)
∴ even function.
Plus Two Maths Integrals 3 Mark Questions and Answers 12
we have |x| = x, 0 ≤ x ≤ 1

Question 6.

Show that cos² x is an even function. (1)
Evaluate \(\int_{-\pi / 4}^{\pi / 4} \cos ^{2} x d x\) (2)

Answer:
1. Let f(x) = cos²x ⇒ f(-x) = cos² (-x) = cos² x = f(x) even.

2.
Plus Two Maths Integrals 3 Mark Questions and Answers 14

Question 7.
Find the following integrals.
Plus Two Maths Integrals 3 Mark Questions and Answers 15
Answer:

Question 8.
Find the following integrals.
Plus Two Maths Integrals 3 Mark Questions and Answers 18
Answer:

Add (1) and (2)
Plus Two Maths Integrals 3 Mark Questions and Answers 21

Plus Two Maths Integrals 3 Mark Questions and Answers 23

Question 9.
Find the following integrals.

\(\int \frac{1}{3+\cos x} d x\)
\(\int \frac{2 x}{x^{2}+3 x+2} d x\)

Answer:
1. \(\int \frac{1}{3+\cos x} d x\)
Put t = tanx/2 ⇒ dt = 1/2 sec² x/2 dx
Plus Two Maths Integrals 3 Mark Questions and Answers 25

2. \(\int \frac{2 x}{x^{2}+3 x+2} d x\) = \(\int \frac{2 x}{(x+2)(x+1)} d x\)
Plus Two Maths Integrals 3 Mark Questions and Answers 26
2x = A(x + 1) + B (x + 2)
when x = -1, -2 = B ; B = -2
when x = -2, -4 = -A ; A = 4

= 4log(x + 2) – 2log (x + 1) + C.

Plus Two Maths Integrals Four Mark Questions and Answers

Question 1.
Find the following integrals.
Plus Two Maths Integrals 3 Mark Questions and Answers 28
Answer:

x² + x +1 = A(x² + 1) + (Bx + C)(x + 2)
Put x = -2 ⇒ 4 – 2 + 1 = 5A ⇒ A = \(\frac{3}{5}\)
Equating the coefficients of x²
⇒ 1 = A + B ⇒ B = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)
Equating the constants
⇒ 1 = A + 2C ⇒ 2C = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\) ⇒ C = \(\frac{1}{5}\)
Plus Two Maths Integrals 3 Mark Questions and Answers 30

⇒ 1 = A(x – 1) + B(x + 3)
Put x = 1 ⇒ 1 = 2A ⇒ A = \(\frac{1}{2}\)
Put x = -3 ⇒ 1 = -4B ⇒ B = – \(\frac{1}{4}\)

Equating the constants; ⇒ 1 = A
Equating the coefficients if t;
⇒ 0 = A + B ⇒ B = -1

Question 2.
Find the following integrals.

∫ e^2x sin3xdx
∫ x sin^-1xdx

Answer:
1. I = ∫e^2x sin3xdx = ∫ sin 3x × e^2xdx
Plus Two Maths Integrals 3 Mark Questions and Answers 35

2. ∫ x sin^-1xdx = ∫ sin^-1x × xdx
Plus Two Maths Integrals 3 Mark Questions and Answers 37

Question 3.
(i) Which of the following is the value of \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}\)? (1)
Plus Two Maths Integrals 3 Mark Questions and Answers 38
(ii) Evaluate \(\int \frac{2 x}{x^{2}+3 x+2} d x\) (3)
Answer:
(i) [sin^-1\(\frac{x}{a}\) + c]

(ii)
Plus Two Maths Integrals 3 Mark Questions and Answers 39
⇒ 2x = A(x + 1) + B(x + 2) ⇒
Put x = -2 and x = -1, we get A = 4, B = -2

Question 4.

Choose the correct answer from the bracket.
∫e^x dx = — (e^2x + c, e^-x + c, e^2x + c) (1)
Evaluate: ∫ e^x sinxdx

Answer:
1. e^x + c

2. I = ∫e^x sinxdx = sinx.e^x – ∫cos x.e^xdx
= sin x.e^x – (cos x.e^x – ∫(- sin x).e^x dx)
= sinx.e^x – cosxe^x – ∫sinx.e^xdx
= sin x.e^x – cos xe^x – I
2I = sin x.e^x – cos xe^x
I = \(\frac{1}{2}\)e^x(sinx – cosx) + c.

Question 5.
(i) f(x)∫g(x) dx – ∫(f'(x)∫g(x) dx)dx (1)
(a) ∫f'(x)g{x)dx
(b) ∫f(x)g'(x)dx
(c) ∫\(\frac{f(x)}{g(x)}\)dx
(d) ∫f(x)g(x)dx
(ii) Integrate sin^-1\(\sqrt{\frac{x}{a+x}}\)dx w.r.to x. (3)
Answer:
(i) (d) ∫f(x)g(x)dx

(ii) ∫sin^-1\(\sqrt{\frac{x}{a+x}}\)dx,
Put x = a tan²θ, θ = tan^-1\(\sqrt{\frac{x}{a}}\)
⇒ dx = 2a tanθ sec²θ dθ
I = ∫sin^-1\(\left(\frac{\tan \theta}{\sec \theta}\right)\) 2a tanθ sec²θ dθ
= ∫sin^-1(sinθ)2a tanθ sec²θ dθ
= 2a∫θ tanθ sec²θ dθ
Put tanθ = t, θ = tan^-1 t ⇒ sec²θ dθ = dt
= 2a ∫ tan^-1 t (t) dθ
Plus Two Maths Integrals 3 Mark Questions and Answers 41
= a[tan²θ.θ – tanθ + θ] + c
= a[θ(1 + tan²θ) – tanθ] + c

Question 6.
Match the following. (4)
Plus Two Maths Integrals 3 Mark Questions and Answers 43
Answer:

Question 7.
Evaluate \(\int \frac{x}{\sqrt{x+a}+\sqrt{x+b}} d x\)?
Answer:
Plus Two Maths Integrals 3 Mark Questions and Answers 45

Question 8.
Match the following.
Plus Two Maths Integrals 3 Mark Questions and Answers 46
Answer:
1.

2. ∫sec x(sec x + tan x)dx = ∫(sec² x + sec x. tan x)dx
= tanx + secx + c.

3. ∫e^3xdx = \(\frac{e^{3 x}}{3}\) + c.

4. ∫(sin x + cos x)dx = sin x – cosx + c.

Question 9.
Consider the integral I = \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x\)?

What substitution can be given for simplifying the above integral? (1)
Express I in terms of the above substitution. (1)
Evaluate I. (2)

Answer:
1. Substitute sin^-1 x = t.

2. We have, sin^-1 x = t ⇒ x = sint
Differentiating w.r.t. x; we get,
\(\frac{1}{\sqrt{1-x^{2}}}\)dx = dt
∴ I = ∫t sin t dt.

3. I = ∫t sin t dt = t.(-cost) -∫(-cost)dt = -t cost + sint + c
= -sin^-1 x. cos (sin^-1 x) + sin(sin^-1 x) + c
x – sin^-1 x.cos(sin^-1 x) + c.

Question 10.
Evaluate \(\int_{0}^{\pi / 4} \log (\tan x) d x\).
Answer:
Plus Two Maths Integrals 3 Mark Questions and Answers 48

Question 11.
Find the following integrals.

\(\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x\) (2)
\(\int \frac{1}{x^{2}-6 x+13} d x\) (2)

Answer:
1. \(\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x\) = \(\int \frac{\sin ^{2} x}{\cos ^{2} x} d x\) = ∫tan² xdx
= ∫(sec²x – 1)dx = tanx – x + c.

2. \(\int \frac{1}{x^{2}-6 x+13} d x\)
Plus Two Maths Integrals 3 Mark Questions and Answers 49

Question 12.
Match the following. Justify your answer.
Plus Two Maths Integrals 3 Mark Questions and Answers 50
Answer:

Question 13.
(i) ∫sin2x dx = ? (1)
(a) 2 cos x + c
(b) -2 sin x + c
(c) \(\frac{\cos 2 x}{2}\) + c
(d) \(-\frac{\cos 2 x}{2}\) + c
(ii) Evaluate ∫e^x sin 2x dx (3)
Answer:
(i) (d) \(-\frac{\cos 2 x}{2}\) + c.

(ii) Consider I = ∫e^x sin 2x dx
= ∫sin 2x. e^xdx = sinx.e^x – 2∫cos 2x. e^xdx
= sin 2x.e^x – 2 (cos 2x.e^x + 2∫sin 2x. e^xdx)
= sin 2x. e^x – 2 cos 2x e^x – 4 ∫sin 2x. e^xdx
= sin 2x. e^x – 2 cos 2x e^x – 4I
5 I = sin 2x. e^x – 2 cos 2x e^x
I = \(\frac{e^{x}}{5}\) (sin 2x – 2 cos 2x).

Question 14.

Resolve \(\frac{x^{2}+1}{x^{2}-5 x+6}\) into partial fractions. (2)
Hence evaluate ∫\(\frac{x^{2}+1}{x^{2}-5 x+6}\). (2)

Answer:
1.
Plus Two Maths Integrals 3 Mark Questions and Answers 52

2.

5x – 5 = A(x – 2) + B(x – 3)
x = 2, 5 = -B, B = -5
x = 3, 10 = A, A = 10
(1) ⇒ I = ∫ 1dx + ∫\(\frac{10}{x-3}\) dx – ∫\(\frac{5}{x-2}\) dx
= x + 10log(x – 3) – 5log(x – 2) + c.

Question 15.
Evaluate \(\int_{0}^{4}\) xdx as a limit of sum.
Answer:
By definition,
\(\int_{a}^{b}\) f(x) dx =
(b – a)\(\lim _{n \rightarrow \infty} \frac{1}{n}\){f(a) + f(a + h) +…….+f(a + {n – 1)h)}
Here, a = 0, b = 4, f(x) = x, h = \(\frac{4-0}{n}=\frac{4}{n}\) ⇒ nh = 4
Plus Two Maths Integrals 3 Mark Questions and Answers 54

Question 16.

Define the real valued function f(x) = |x² + 2x – 3| (2)
Evaluate \(\int_{0}^{2}\)|x² + 2x – 3|dx. (2)

Answer:
1. f(x) = |x² + 2x – 3| = |(x – 1) (x + 3)|
We have;
Plus Two Maths Integrals 3 Mark Questions and Answers 55

2. I = \(\int_{0}^{2}\)|x² + 2x – 3|dx
Plus Two Maths Integrals 3 Mark Questions and Answers 56

Question 17.
Consider the function f(x) = |x|+|x + 1|

Define the function f (x) in the interval [-2, 1]. (2)
Find the integral \(\int_{-2}^{1}\) f(x) dx (2)

Answer:
1. Given, f(x) = |x|+|x + 1|.
We have,
Plus Two Maths Integrals 3 Mark Questions and Answers 57
Combining these two functions, we get the function f(x).

2.
Plus Two Maths Integrals 3 Mark Questions and Answers 59

Question 18.
Evaluate \(\int_{\sqrt{6}}^{\sqrt{3}} \frac{d x}{1+\sqrt{\tan x}} d x\). (4)
Answer:
Plus Two Maths Integrals 3 Mark Questions and Answers 60

Plus Two Maths Integrals Six Mark Questions and Answers

Question 1.
(i) Fill in the blanks. (3)
(a) ∫ tan xdx = —
(b) ∫ cos xdx = —
(c) ∫\(\frac{1}{x}\)dx = —
(ii) Evaluate ∫sin³ xcos² xdx (3)
Answer:
(i) (a) log|secx| + c
(b) sinx + c
(c) log|x| + c.

(ii) ∫sin³ xcos² xdx = ∫sin² xcos² x sin xdx
= ∫(1 – cos² x)cos² x sin xdx
Put cos x = t ⇒ – sin xdx = dt
∴ ∫(1 – cos² x)cos² xsin xdx = -∫(1 – t² )t²dt
= ∫(t⁴ – t²)dt = \(\frac{t^{5}}{5}-\frac{t^{3}}{3}\) + c
= \(\frac{\cos ^{5} x}{5}-\frac{\cos ^{3} x}{3}\) + c.

Question 2.
Find the following integrals.
Plus Two Maths Integrals 3 Mark Questions and Answers 61
Answer:
(i) I = ∫(3x – 2)\(\sqrt{x^{2}+x+1} d x\)
Let 3x – 2 = A(2x + 1) + B
⇒ 3 = 2 A ⇒ A = \(\frac{3}{2}\)
⇒ -2 = A + B ⇒ -2 = \(\frac{3}{2}\) + B
⇒ B = -2 – \(\frac{3}{2}\) = – \(\frac{7}{2}\)

Using (2) and (3) in (1) we have;
Plus Two Maths Integrals 3 Mark Questions and Answers 64

(ii) I = \(\int \frac{2 x-3}{x^{2}+3 x-18} d x\)
Let 2x – 3 = A(2x + 3) + B
⇒ 2 = 2A ⇒ A = 1
⇒ -3 = 3A + B ⇒ -3 = 3 + B ⇒ B = -6
Plus Two Maths Integrals 3 Mark Questions and Answers 65

(iii) I = \(\int \frac{5 x+2}{1+2 x+3 x^{2}} d x\)
Let 5x + 2 = A{6x + 2) + B
⇒ 5 = 6 A ⇒ A = \(\frac{5}{6}\)
⇒ 2 = 2A + B ⇒ 2 = \(\frac{5}{3}\) + B ⇒ 2 – \(\frac{5}{3}\) = \(\frac{1}{3}\)
Plus Two Maths Integrals 3 Mark Questions and Answers 67

(iv) I = \(\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x\)
Let 5x + 3 = A(2x + 4) + B
⇒ 5 = 2A ⇒ A = \(\frac{5}{2}\)
⇒ 3 = 4A + B ⇒ 3 = 10 + B ⇒ B = -7
Plus Two Maths Integrals 3 Mark Questions and Answers 69

Using (2) and (3) in (1) we have;

Question 3.
Consider the expression \(\frac{1}{x^{3}-1}\)

Split it into partial fraction. (2)
Evaluate ∫ \(\frac{1}{x^{3}-1}\) dx (4)

Answer:
1.
Plus Two Maths Integrals 3 Mark Questions and Answers 72
1 = A (x² + x + 1) + (Bx + c)(x + 1),
Put x = -1 ⇒ 1 = A(1 + 1 + 1) ⇒ A= \(\frac{1}{3}\)
Equating like terms.
0 = A + B ⇒ B = – \(\frac{1}{3}\), 1 = A + C ⇒ C = \(\frac{2}{3}\)

2.
Plus Two Maths Integrals 3 Mark Questions and Answers 74
Put, x – 2 = D (2x – 1) + E ,
1 = 2 D ⇒ D = \(\frac{1}{2}\),
-2 = -D + E ⇒ E = –\(\frac{3}{2}\)

Question 4.
(i) Match the following (4)
Plus Two Maths Integrals 3 Mark Questions and Answers 77
(ii) Consider the function f(x) = \(\frac{x^{4}}{x+1}\) Evaluate ∫f(x)dx (2)
Answer:
(i)

(ii) Here the numerator is of degree 4 and denominator of degree 1. So to make it a proper fraction we have to divide Nr by Dr.
Plus Two Maths Integrals 3 Mark Questions and Answers 79

Question 5.

Evaluate the as \(\int_{0}^{2}\)x²dx the limit of a sum. (3)
Hence evaluate \(\int_{-2}^{2}\)x²dx (1)
If \(\int_{0}^{2}\) f(x)dx = 5 and \(\int_{-2}^{2}\) f(x)dx = 0, then \(\int_{-2}^{0}\) f(x)dx = …….. (2)

Answer:
1. Here the function is f(x) = x², a = 0, b = 2 and h = \(\frac{b-a}{n}=\frac{2}{n}\)
\(\int_{0}^{2}\)x²dx =
Plus Two Maths Integrals 3 Mark Questions and Answers 80

2. \(\int_{-2}^{2}\) x²dx = 2 \(\int_{0}^{2}\)x²dx = \(\frac{16}{3}\)

3.
Plus Two Maths Integrals 3 Mark Questions and Answers 81

Question 6.
Find ∫\(\sqrt{\tan x}\)xdx.
Answer:
Given;
I = ∫\(\sqrt{\tan x}\)xdx,
Put tanx = t² ⇒ sec²xdx = 2tdt ⇒ dx = \(\frac{2 t d t}{1+t^{4}}\)
Plus Two Maths Integrals 3 Mark Questions and Answers 82

Question 7.
(i) Match the following. (2)
Plus Two Maths Integrals 3 Mark Questions and Answers 85
(ii) Integrate \(\frac{\sec ^{2} x}{5 \tan ^{2} x-12 \tan x+14}\) w.r.to x. (4)
Answer:
(i)

Question 8.

Evaluate \(\int_{0}^{1} \sqrt{x} d x\) (1)
If \(\int_{0}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x\), find the value of a. (3)
Hence find \(\int_{a}^{a+1}\)x dx. (2)

Answer:
1.
Plus Two Maths Integrals 3 Mark Questions and Answers 89

2. Given;
Plus Two Maths Integrals 3 Mark Questions and Answers 90

3. When a = 0
Plus Two Maths Integrals 3 Mark Questions and Answers 91
When, a = 4

Question 9.
(i) Let f (x) be a function, then \(\int_{0}^{a}\) f(x) dx = ? (1)
(a) 2 \(\int_{0}^{a}\) f(x – a) dx
(b) \(\int_{0}^{a}\) f(a – x) dx
(c) f(a)
(d) 2\(\int_{0}^{a}\) f(a – x) dx
Evaluate
Plus Two Maths Integrals 3 Mark Questions and Answers 93
Answer:
(i) (b) \(\int_{0}^{a}\) f(a – x) dx

(ii)
Plus Two Maths Integrals 3 Mark Questions and Answers 94
(1) + (2)

⇒ I = 1.

(iii)
Plus Two Maths Integrals 3 Mark Questions and Answers 96

Question 10.
Find the following integrals.

∫\(\frac{2 e^{x}}{e^{3 x}-6 e^{2 x}+11 e^{x}-6} d x\)
∫\(\frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x\)

Answer:
1.
Plus Two Maths Integrals 3 Mark Questions and Answers 97
⇒ 1 = A(t – 2)(t – 3) + B(t – 1)(t – 3) + C(t – 1)(t – 2)
Put t = 1 ⇒ 1 = A(-1)(-2) ⇒ A = \(\frac{1}{2}\)
Put t = 2 ⇒ 1 = B(1)(-1) ⇒ B = -1
Put t = 3 ⇒ 1 = B(2)(1) ⇒ B = \(\frac{1}{2}\)

2. I = ∫\(\frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x\)dx
Put sin x = t ⇒ cosxdx = dt
Plus Two Maths Integrals 3 Mark Questions and Answers 100
⇒ 3t – 2 = A(t – 2) + B
Equating the coefficients if t; ⇒ 3 = A
Equating the constants
⇒ -2 = -2A + B ⇒ -2 = -6 + B ⇒ B = 4

Question 11.

Find ∫\(\frac{1}{x^{2}+a^{2}}\)dx (1)
Show that 3x + 1 = \(\frac{3}{4}\)(4x – 2) + \(\frac{5}{2}\) (2)
Evaluate \(\int \frac{3 x+1}{2 x^{2}-2 x+3} d x\) (3)

Answer:
1. ∫\(\frac{1}{x^{2}+a^{2}}\)dx = 1/a tan^-1 x/a + c.

2. 3x + 1 = A \(\frac{d}{d x}\)(2x² – 2x + 3) + B
= A(4x – 2) + B
3 = 4A; A = 3/4
1 = -2A + B
1 = -3/2 + B, B = 1 + 3/2 = 5/2
∴ 3x + 1 = 3/4(4x – 2) + 5/2

3.
Plus Two Maths Integrals 3 Mark Questions and Answers 102

Plus Two Computer Science Chapter Wise Questions and Answers Chapter 12 ICT and Society

April 13, 2021February 6, 2020 by Prasanna

Students can Download Chapter 12 ICT and Society Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 12 ICT and Society

Plus Two Computer Science ICT and Society One Mark Questions and Answers

Question 1.
IPR stands for ______.
Answer:
Intellectual Property Right.

Question 2.
WIPO stands for _____.
Answer:
World Intellectual Property Organisation

Question 3.
______ is the exclusive rights to prevent unauthorized copying of inventions by a Creator from the Unauthorised person or company.
Answer:
Patent

Question 4.
_____ is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company.
Answer:
Trademark

Question 5.
A product or article is designed so beautifully to attract customers. This type of design is called
Answer:
Industrial Design.

Question 6.
Aranmula Kannadi, Palakkadan Matta, Marayoor Sarkkara, etc are example of _______.
Answer:
Geographical indications.

Question 7.
_____ is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator.
Answer:
Copyright

Question 11.
Unauthorized copying or use of Intellectual property rights such as Patents, Copyrights and Trademarks are called ____.
Answer:
Intellectual Property Infringement.

Question 12.
_____ prevents others from the unauthorized or intentional copying or use of Patent without the permission of the creator.
Answer:
Patent Infringement.

Question 13.
______ is the illegal copying, distribution, or use of software.
Answer:
Piracy.

Question 14.
______ prevents others from the unauthorized or intentional copying or use of Trademark without the permission of the creator.
Answer:
Trademark Infringement

Question 15.
_____ prevents others from the unauthorized or intentional copying or use of Copy right without the permission of the creator.
Answer:
Copy right Infringement

Question 16.
______ is a virtual environment created by computer systems connected to the internet
Answer:
Cyberspace

Question 17.
A person committing crimes and illegal activities with the use of computers over Internet. This crime is included as _____ crime.
Answer:
Cybercrime

Question 18.
State True or False.
Cybercrimes can be classified into three categories such as against individual, property, and Government.
Answer:
True

Question 19.
Phishing, hacking, denial of service attacks, etc are ____ crimes.
Answer:
Cyber

Question 20.
Odd one out
(а) Identity theft
(b) Harassment
(c) violation of privacy
(d) credit card fraud
Answer:
(d) credit card fraud, it is a cybercrime against individual others are cyber crimes against property.

Question 21.
Odd one out
(a) Credit card theft
(b) Intellectual property theft
(c) Internet time theft
(d) Dissemination of obscene material
Answer:
(d) Dissemination of obscene material, It is cyber , crime against individual, the others are cyber against property.

Question 22.
Odd one out
(a) cyberterrorism
(b) Attacks against e-Governance websites
(c) Impersonation and cheating
(d) Website defacement
Answer:
(c) Impersonation and cheating, it is cybercrime against individual others are cyber crimes against Government.

Question 23.
IT Act amended in _____.
(a) 2015
(b) 2008
(c) 1900
(d) 1998
Answer:
(b) 2008

Question 24.
IT Act passed in Indian parliament is ____.
Answer:
2000.

Question 25.
The laws to prevent cyber crimes is termed as ____.
Answer:
Cyberlaw

Question 26.
_____ excessive enthusiasm for acquiring knowledge.
Answer:
Infomania

Question 27.
Phishing is an example of ______.
Answer:
Cybercrime.

Question 28.
ICT stands for _______.
(a) Internet and Communication Technology
(b) Information and Computer Technology
(c) Information and Communication Technology
(d) Integrated Communication Technology
Answer:
(c) Information and Communication Technology

Question 29.
Which of the following e-Governance helps citizens for interacting with the Government?
(a) G2E
(b) G2B
(c) G2C
(d) G2G
Answer:
(c) G2C

Question 30.
What are the different types of interactions in e-Governance?
Answer:
G2G, G2E.G2B, G2C.

Question 31.
The unauthorized use of intellectual property rights is termed as
Answer:
Infringement

Question 32.
Expand the term WIPO in connection with IPR.
Answer:
World Intellectual Property Organization.

Question 33.
The exclusive right granted to an invention is called
(a) Trademark
(b) Copy right
(c) Patent
(d) Design
Answer:
(c) Patent

Question 34.
The exclusive right given to a person over the creation of his/her mind for a period of time is called
Answer:
Patent / Intellectual Property Right

Question 35.
What is the name given to the process of using scientific knowledge for analyzing and presenting evidence of cyber related crimes before court?
Answer:
Cyber forensics

Question 36.
Which among the following are considered as violation to privacy?
1. Keeping hidden cameras in private places
2. Publishing private photos of individual in social media without their permission
3. Use of unauthorized software
4. Using simple password
(A) All the above are correct
(B) 1,2 and 3 only
(C) 1 and 4 only
(D) 1 and 2
Answer:
(D) 1 and 2

Plus Two Computer Science ICT and Society Two Mark Questions and Answers

Question 1.
“IPR (Intellectual Property Right) encourages innovation” Justify.
Some people spend lots of money,time body and mental power to create some products such as a classical movie, album, artistic work, discoveries, invention, software, etc. These type of Intellectual properties must be protected from unauthorized access by law. This is called Intellectual Property right(IPR). It enables to earn recognition, financial benefit, can sell the innovation, etc. It motivates further innovation.

Question 2.
Define the following terms.

Cyber space
Cyber crime

Answer:
1. CyberSpace:
Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too.

Later telephones were used for voice communication. Nowadays telephone system and computer system are integrated and create a virtual(un real) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

2. Cyber crime:
Just like normal crimes (theft, trespassing private area, destroy, etc,) Cyber crimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc) also increased significantly. Due to cyber crime, the victims lose money, reputation, etc and some of them commit suicide.

Question 3.
Write a short note on

Trademark
Industrial design

Answer:
1. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

2. Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 4.
Compare patent and Trademark.
1. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

2. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

Question 5.
Write any one website for the following services.

e-Governance
e-Business
e-Banking
e-Learning

Answer:

e-Governance(any One) www.dhsekerala.gov.in, www.incometaxindia.gov.in, www.spark.gov.in,www.ceo.kerala.gov. in
e-Business www.indane.co.in, www.amazon.com,www.ebay.in
e-Banking www.onlinesbi.co.in
e-Learning www.ignouonline.ac.in,www.nptel.iitm.ac.in

Question 6.
Write a short note about EPS.
Answer:
Electronic Payment System(EPS): It is also called plastic money that is electronically exchange money between two individuals or firms(buyers and sellers) in an online environment.

Question 7.
What is cyberspace?
Answer:
Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too.

Later telephones were used for voice communication. Nowadays telephone system and computer system are integrated and create a virtual(unreal) environment. This is called cyberspace. The result for this integration is that tremendous speed and it is very cheap.

Question 8.
Why is cyberspace called a virtual world?
Answer:
The telephone system and computer system are integrated and create a virtual(un real) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap. This is an imaginary world. We can see persons with different behaviour. Because of good and bad people we can’t believe blindly. If we search a solution for a problem thousands of answers will get instantly and may confused us.

Question 9.
What is copyright? How does it differ from patent?
Answer:
1. Copyright:
The trademark is ©, copyright is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator for 60 years after the death of the author.

2. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

Question 10.
Explain the exclusive right given to the owner by IPR?
Answer:
The exclusive right given to the owner by I PR is owner can disclose their creations for money.

Question 11.
it is the unauthorized copying, distribution, and use of a creation without the permission of the creator. It is against the copyright act and hence the person committed deserve the punishment.

Question 12.
Match the following
Plus Two Computer Science Chapter Wise Questions and Answers Chapter 12 ICT and Society img1
Answer:
a – 2
b – 3
c – 4
d – 1

Question 13.
What do you meant by infringement?
Answer:
Unauthorized copying or use of Intellectual property rights such as Patents, Copy rights and Trademarks are called intellectual property lnfringement(violation). It is a punishable offence.

Plus Two Computer Science ICT and Society Three Mark Questions and Answers

Question 1.
Write a short note on the importance of IT Act 2000.
Answer:
Information Technology Act 2000(amended in 2008):
IT Act 2000 controls the use of Computer(client), Server, Computer Networks, data and Information in Electronic format and provide legal infrastructure for E-commerce, in India. This is developed to promote IT industry, control e-commerce also ensures the smooth functioning of E-Governance and it prevents cyber crimes.

The person those who violate this will be prosecuted. In India, IT bill introduced in the May 2000 Parliament Session and it is known as Information Technology Act 2000. Some exclusions and inclusions are introduced in December 2008.

Question 2.
“Infomania affects peoples’ lives and their loved ones.”
Comment on this statement.
Answer:
Info mania is excessive desire(infatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media. Instant Message Application(WhatsApp) and Smart Phones. Due to this the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired.

They give first preference to Internet than others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Question 3.
Define the term e-Business. What are the advantages and challenges of e-Business? Write any two e-Business websites.
Answer:
E-business(electronic Business): Providing ser¬vices or running business through internet is called E-business.
Advantages of e-business:

It overcomes geographical limitations
It reduces the operational cost
It minimizes the time and cost
It remains open all the time
We can locate the product faster from a wider range of choices
Challenges to E business
Peoples are unaware of IT applications and its uses
Most peoples don’t have plastic money(credit / debit card) and net banking
It requires high security measurements otherwise you may lose money
We can’t touch or smell products through online
Some companies may not have proper Goods delivery service
Useful e-Business websites

Plus Two Computer Science Chapter Wise Questions and Answers Chapter 12 ICT and Society img2

Question 4.
How do trademark and industrial design differ?
Answer:
Trademark:
This is a unique, sirhple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 5.
Why is Cyberlaw important?
Answer:
Just like normal crimes (theft, trespassing private area, destroy, etc.) Cybercrimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc.) also increased significantly. Due to cybercrime, the victims lose money, reputation, etc. and some of them commit suicide.

Cyberlaw ensures the use of computers and Internet by the people safely and legally. It consists of rules and regulations like Indian Penal Code (IPC) to stop crimes and for the smooth functions of Cyberworld. Two Acts are IT Act 2000 and IT Act Amended in 2008.

Question 6.
“Infomania has became a psychological problem”. Write your opinion.
Answer:
Info mania is the excessive desire(lnfatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media, Instant Message Application(WhatsApp) and Smart Phones. Due to this the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired.

They give first preference to Internet others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Plus Two Computer Science ICT and Society Five Mark Questions and Answers

Question 1.
“Due to anonymous nature of Internet it is possible for the people to engage in variety of criminal activities.” Justify the statement with special reference to cyber crimes taking place against individual.
Answer:
Cyber crimes against individuals
i. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number etc) , Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

ii. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friend ship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

iii. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original ICT and Society one for the purpose of cheating or misleading others. Eg: Fake accounts in Social Medias (Facebook, Twitter, etc), fake SMS, fake emails etc.

iv. Violation of privacy:
Trespassing into another person’s life and try to spoil life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

v. Dissemination of obscene material: With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

Question 2.
Explain different categories of cyber crimes in detail.
Answer:
Just like normal crimes( theft, trespassing private area, destroy, etc,) Cyber crimes(Virus, Trojan Horse, Phishing, Denial of Service, Pornography etc) also increased significantly . Due to cyber crime, the victims lose money, reputation,etc and some of them commit suicide.
Answer:
A. Cyber crimes against individuals:
1. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number etc.), Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

2. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friend ship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

3. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original one for the purpose of cheating or misleading others. Eg: F.ake accounts in Social Medias (Facebook, Twitter,etc), fake sms, fake emails, etc.

4. Violation of privacy:
Trespassing into another person’s life and try to spoil the life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

5. Dissemination of obscene material:
With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

B. Cyber crimes against property:
Stealing credit card details, hacking passwords of social media accounts or mail account or Net banking, uploading latest movies etc, are considered as cyber crimes against property.
1. Credit card fraud:
Stealing the details such as credit card number, company name, expiry date, cw number,password etc. and use these details to make payment for purchasing goods or transfer funds also.

2. Intellectual property theft:
The violation of Intellectual Property Right of Copy right, Trademark, Patent, etc. In film industry crores of investment is needed to create a movie. Intellectual Property thieves upload the movies on the Releasing day itself. Hence the revenue from the theatres are less significantly and undergoes huge loss.(Eg: Premam, Bahubali, etc) Copying a person’s creation and present as a new creation is called plagiarism. This can be identified some tools(programs) available in the Internet

3. Internet time theft:
This is deals with the misuse of WiFi Internet facility. If it is not protected by good password there is a chance of misuse our devices(Modem/Router) to access Internet without our consent by unauthorized persons. Hence our money and volume of data(Package) will lose and we may face the consequences if others make any crimes.

C. Cyber crimes against government:
The cyber crimes against Govt, websites is increased significantly. For example in 2015 the website of Registration Department of Kerala is hacked and destroys data from 2012 onwards.

1. Cyber terrorism:
It is deals with the attacks against very sensitive computer networks like computer controlled atomic energy power plants, air traffic controls, Gas line controls, telecom, Metro rail controls, Satellites, etc. This is a very serious matter and may lead to huge loss (money and life of citizens). So Govt, is very conscious and give tight security mechanism for their services.

2. Website defacement:
It means spoil or hacking websites and posting bad comments about the Govt.

3. Attacks against e-governance websites :
Its main target is a Web server. Due to this attack the Web server/ computer forced to restart and this results refusal of service to the genuine users. If we want to access a website first you have to type the web site address in the URL and press Enter key, the browser requests that page from the web server. Dos attacks send huge number of requests to the web server until it collapses due to the load and stops functioning.

Question 3.
“For the implementation of e-Learning different tools.
Answer:
e Learning tools
1. Electronic books reader(eBooks): With the help of a tablet or portable computer or any other device we can read digital files by using an s/w is called electronic books reader.

2. e-text: The electronic format of textual data is called e-Text.

3. Online chat: Realtime exchange of text or audio or video messages between two or more person over the Internet.

4. e-Content: The data or information such as text, audio, video, presentations, images, animations, etc, are stored in electronic format.

5. Educational TV channels: TV channels dedicated only for the e-Learning purpose.
Eg. VICTERS (Virtual Classroom Technology on Edusat for Rural Schools OR Versatile ICT Enabled Resources for Students)