## Plus One Malayalam Textbook Answers Unit 4 Chapter 6 Shasthrakriya

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 6 Shasthrakriya Text Book Questions and Answers, Summary, Notes.

## Plus One Accountancy Chapter Wise Questions and Answers Kerala

HSE Kerala Board Syllabus HSSLive Plus One Accountancy Chapter Wise Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala Plus One Chapter Wise Questions and Answers. Here HSSLive.Guru has given Higher Secondary Kerala Plus One Accountancy Chapter Wise Questions and Answers based on CBSE NCERT syllabus.

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## Kerala Plus One Accountancy Chapter Wise Questions and Answers

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## Plus One Economics Chapter Wise Questions and Answers Chapter 4 Presentation of Data

Students can Download Chapter 4 Presentation of Data 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 4 Presentation of Data

### Plus One Economics Presentation of Data One Mark Questions and Answers

Plus One Economics Chapter Wise Questions And Answers Pdf Question 1.
Which of the following comes under geometric diagram?
(a) Histogram
(b) Bar diagram
(c) Ogives
(d) Frequency polygon
(b) Bar diagram

Plus One Economics Chapter Wise Questions And Answers Question 2.
Which of the following comes under frequency diagrams?
(a) Bar diagram
(b) Histogram
(c) Pie diagram
(d) All the above
(b) Histogram

Plus One Economics Chapter Wise Questions And Answers Malayalam Question 3.
To draw time-series graph, time is presented on:
(a) X-axis
(b) Y-axis
(c) any of two
(a) X-axis

Plus One Statistics Chapter Wise Questions And Answers Question 4.
Name the types of graphs.

1. One dimensional graph
2. Two-dimensional graph
3. Three-dimensional graph
4. Pictograms

State whether true or false.

1. The width of bars in a bar diagram need not be equal.
2. The width of rectangles in a histogram should essentially be equal.
3. Histograms can only be formed with continuous classification of data.
4. Histogram and column diagram are the same method of presentation of data.
5. Mode of a frequency distribution can be drawn graphically with the help of histogram,
6. The median of a frequency distribution cannot be drawn from the Ogive.

1. true
2. false
3. true
4. true
5. true
6. true

### Plus One Economics Presentation of Data Two Mark Questions and Answers

Hsslive Economics Plus One Chapter Wise Questions And Answers Question 1.
Which of the following is a cumulative frequency curve?
(a) Bar diagram
(b) Histogram
(c) Ogive
(d) Pie diagram
(c) Ogive

Plus One Economics Questions And Answers Question 2.
Distinguish between captions and stubs.
Captions refers to the column headings and stubs refers to the row heading.

Histogram And Frequency Polygon Questions With Answers Question 3.
Match the following.

 A B Source note Row headings Captions Gives origin of data Stubs Explains the specific feature Footnote Column Headings

 A B Source note Gives origin of data Captions Column Headings Stubs Row Headings Footnote Explains the specific feature

### Plus One Economics Presentation of Data Three Mark Questions and Answers

Presentation Of Data Class 11 Question 1.
What kind of diagrams are more effective in representing the following?

1. Monthly rainfall in a year
2. Composition of the population of Delhi by religion
3. Components of cost in a factory

1. Simple bar diagram
2. Sub-divided or component bar diagram
3. Pie diagram

Pie Chart Class 11 Economics Question 2.
Name different types of diagrams.
The different types of diagrams are:
1. Geometric diagram

• Bar diagrams
• Pie diagram

2. Frequency diagram

• Histogram
• Frequency polygon
• Frequency curve -Ogive

3. Arithmetic line graph

Question 3.
“Diagrams and graphs help us visualize the whole meaning of numerical complex data at a single glance”. Comment.
One of the most convincing and appealing ways in which statistical results may be presented is through diagrams and graphs. The special feature of graphs and diagrams is that they do away with figures altogether. Diagrams and graph is a statistical method which can be used for simplifying the complexity of quantitative data and t make them easily intelligible.

It presents dry and uninteresting statistical facts in the shape of attractive and appealing pictures and charts. They are important methods of visual aids and are appealing t the eye and mind of the observer.

Question 4.
“There are generally three forms of diagrammatic presentation of data” explain.
There are various methods to present data. But generally, three forms of presentation of data are there
which are noted below:

1. Geometric diagram
2. Frequency diagram
3. Arithmetic line graph

1. Geometric Diagram:
Bar diagram and pie diagram come in the category of geometric diagram for presentation of data. The bar diagrams are of three types-simple, multiple and component bar diagrams.

2. Frequency Diagram:
Data in the form of grouped frequency distributions are generally represented by frequency diagrams like histogram, frequency polygon, frequency curve, and ogive

3. Arithmetic Line Graph:
An arithmetic line graph is also called time-series graph and is a method of diagrammatic presentation of data. A line graph by joining these plotted points, thus, obtained is called arithmetic line graph or time-series graph.

Question 5.
Explain Ogive?
Cumulative frequency of any class is equal to the sum of the frequencies of all the classes preceding that class and its own frequency e.g., frequencies are 10, 7, 12, 17 and 22. Cumulative frequencies are 10, 10 + 7 = 17, 17 + 12 = 29, 29 + 17 = 46 and 46 + 22 = 68.
Cumulative frequency of the last class = Total frequency.

For drawing an Ogive, cumulative frequency (i.e. number of values) is taken on the Y-axis and limits of class intervals on the X-axis.
Ogive is of two types:

1. less than
2. more than

In a “less than” type Ogive, we plot the upper limit of each class along the X-axis and in a “more than” type Ogive, we plot the lower limit of each class along the X-axis. Along the Y-axis, we plot the cumulative frequencies at the end of each class. Ogive can be drawn even if the class interval are unequal or open end. Ogives are performed over frequency curves for comparative study.

Question 6.
Illustrate how classes can be formed while presenting the data?
Classes can be formed in two ways:

1. Exclusive type
2. Inclusive type

1. Exclusive Type:
When the class intervals are so fixed that the upper limit of one class is the lower limit of the new class, it is known as exclusive method of classification.

 Marks (Percentage) No. of students 0-10 15 10-20 17 20-30 22 30-40 30 40-50 39 50-60 45

In this method, higher value of the variable in the class is not included in that class i.e.,

 Marks (Percentage) No. of students 0 and more but less than 10 15 10 and more but less than 20 17 20 and more but less than 30 22 30 and more but less than 40 30 40 and more but less than 50 39 50 and more but less than 60 45

2. Inclusive Type:
In this method, the students getting say 39% marks will be included in class 30 – 39 itself i.e.,

 Marks (Percentage) No. of students 0-9 5 10-19 8 20-29 7 30-39 13 40-49 25

### Plus One Economics Presentation of Data Four Mark Questions and Answers

Question 1.
a. Bar diagram is a

1. one-dimensional diagram
2. two-dimensional diagram
3. diagram with no dimension
4. none of the above

b. Data represented through a histogram can help in finding graphically the

1. mean
2. mode
3. median
4. all the above

c. Ogives can be helpful in locating graphically the

1. mode
2. mean
3. median
4. none of the above

d. Data represented through arithmetic line graph help in understanding

1. long term trend
2. cyclicity in data
3. seasonality in data
4. all the above

a. 1. one-dimensional diagram
b. 3. mode
c. 3. median
d. 1. long term trend

Question 2.
Point out major parts of a statistical table.

1. Table number
2. Title
4. Stub
6. Body or field
7. Footnote
8. Source note

Question 3.
Give the rules for constructing tables.
The rules of constructing diagrams are:

• Every diagram should be titled.
• It should suit the size of the paper
• It should be neat and attractive
• It should be neatly indexed
• It should contain footnotes
• The details in diagram should be self-explanatory

### Plus One Economics Presentation of Data Five Mark Questions and Answers

Question 1.
Explain the advantages of diagrammatic presentation.
The advantages of diagrammatic presentation are given below.

1. Diagram give a clear picture of data
2. Comparison can be made easy
3. Diagrams can be used university at any place
4. It saves time and energy
5. The data can be remembered easily

Question 2.
Show how pie diagram is drawn for the following data?

 Items Production in K.G. Tea 3260 Coffee 1850 Cocoa 900 Total 6010

Question 3.
Give steps in the preparation of pie diagram.
A pie diagram is also a component diagram, but unlike a component bar diagram, a circle whose area is proportionally divided among the components it represents. It is also called a pie chart. The circle is divided into as many parts as there are components by drawing straight lines from the centre to the circumference.

The following steps in the preparation of pie diagram are given below:

• Convert each component as percentage of the total.
• Multiply the percentage by 360/100 = 3.6 to convert into degree.
• Starting with the twelve o’clock position on the circle draw the largest component circle
• Draw other components in clockwise succession in descending order of magnitude except for each all components

Like all others and miscellaneous which are shown last:

• Use different columns or shades to distinguish between different components
• Explain briefly the different components either within the components in the figure or outside by arrow.

### Plus One Economics Presentation of Data Eight Mark Questions and Answers

Question 1.
Write short notes on the following

1. pie diagrams
2. frequency curves
3. frequency polygon
4. ogive
5. arithmetic line graph

1. Pie Diagram:
A pie diagram is also a component diagram, but unlike a component bar diagram, a circle whose area is proportionally divided among the components it represents. It is also called a pie chart. The circle is divided into as many parts as there are components by drawing straight lines from the centre to the circumference. Pie charts usually are not drawn with absolute values of a category.

The values of each category are first expressed as percentage of the total value of all the categories. A circle in a pie chart, irrespective of its value of radius, is thought of having 100 equal parts of 3.6° (3607100) each. To find out the angle, the component shall subtend at the centre of the circle, each percentage figure of every component is multiplied by 3.6°.

2. Frequency Polygon:
A frequency polygon is a plane bounded by straight lines, usually four or more lines. Frequency polygon is an alternative to histogram and is also derived from histogram itself. A frequency polygon can be fitted to a histogram for studying the shape of the curve. The simplest method of drawing a frequency polygon is to join the midpoints of the topside of the consecutive rectangles of the histogram.

3. Frequency Curve:
The frequency curve is obtained by drawing a smooth freehand curve passing through the points of the frequency polygon as closely as possible. It may not necessarily pass through all the points of the frequency polygon but it passes through them as closely as possible

4. Ogive:
Ogive is also called cumulative frequency curve. As there are two types of cumulative frequencies, for example, less than type and more than type, accordingly there are two ogives for any grouped frequency distribution data. Here in place of simple frequencies as in the case of frequency polygon, cumulative frequencies are plotted along y-axis against class limits of the frequency distribution.

For less than give the cumulative frequencies are plotted against the respective upper limits of the class intervals whereas for more than ogives the cumulative frequencies are plotted against the respective lower limits of the class interval. An interesting feature of the two ogives together is that their intersection point gives the median

5. Arithmetic Line Graph:
An arithmetic line graph is also called time-series graph and is a method of diagrammatic presentation of data. Init, time (hour, day/date, week, month, year, etc.) is plotted along x-axis and the value of the variable (time series data) along y-axis. A line graph by joining these plotted points, thus, obtained is called arithmetic line graph (time series graph). It helps in understanding the trend, periodicity, etc. in a long term time series data.

Question 2.
3 Forms of presentation of data

1. Textual
2. Tabular
3. Diagrams & graphs Prepare a flow chart.

## Plus One Malayalam Textbook Answers Unit 4 Chapter 5 Samkramanam

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 5 Samkramanam Text Book Questions and Answers, Summary, Notes.

## Plus One Malayalam Textbook Answers Unit 4 Chapter 4 Vasanavikrithi

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 4 Vasanavikrithi Text Book Questions and Answers, Summary, Notes.

## Plus One Physics Notes Chapter 5 Law of Motion

Students can Download Chapter 5 Law of Motion Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus One Physics Notes Chapter 5 Law of Motion

Summary
Laws Of Motion Class 11 Notes Chapter 5 Introduction
In this chapter we are going to learn about the laws that governs the motion of bodies.
Inertia:
The inability of a body to change by itself it’s state of rest or uniform motion along a straight line is called inertia.
Examples of inertia:
1. When a fast moving bus is suddenly stopped, a standing passenger tends to fall in the forward direction.
Explanation
The passenger has the same velocity as that of the bus. When the bus stops suddenly the lower part of his body is brought to rest suddenly because of the friction between his feet and floor of the bus. But the upper part continues to move because of its inertia.

2. When a bus suddenly takes off, a standing passenger tends to fall in the backward direction. This is because the lower part of the body gets a speed when the bus picks up speed and upper part continues to be at rest because of its inertia.

3. Consider a person sitting inside a stationary train and tossing a coin. The coin falls into his own hand. If he repeats the experiment when the train is moving with uniform speed, then also the coin falls into his own hand.

4. Cleaning a carpet by beating is in accordance with law of inertia.

5. Rabbit chased by a dog runs in zigzag manner. This is to take advantage of the large inertia of the massive dog.

6. A person chased by an elephant runs in a zigzag manner or in a circle. This is to take the advantage of the large inertia of the massive elephant.

Newton’s Laws:
Newton built on Galileo’s ideas and laid the foundation of mechanics in terms of three laws.

• Newtons first law
• Newtons second law
• Newtons third law

Laws Of Motion Class 11 Notes Pdf Chapter 5 Newton’s First Law Of Motion
Everybody continues in its state of rest or of uniform motion along a straight line unless it is compelled by an external unbalanced force to change that state:
Note: Newton’s first law of motion brings the idea of inertia. Inertia of a body is measured by the mass of the body. Heavier the body, greater is the force required to change its state and hence greater is its inertia.

Class 11th Physics Chapter 5 Notes  Newton’s Second Law Of Motion
Linear Momentum ($$\vec{p}$$):
Momentum of a body is defined as the product of its mass m and velocity $$\vec{v}$$

Explanation
Momentum of a body can be produced or destroyed by the application of force on it. Therefore, momentum of a body is measured by the force required to stop the body in unit time.
Force required to stop a moving body depends upon

1. mass of the body
2. velocity of the body.

1. Mass of the body:
When a ball and a big stone are allowed to fall from the same height, we find that a greater force is required to stop the big piece of stone than the ball. Thus larger the mass of a body, greater is its linear momentum.

2. Velocity of the body:
A bullet thrown with the hand can be stopped easily than the same bullet fired from the gun. Therefore, langerthe velocity of a body, greater is its linear momentum.
Note: Momentum is a vector quantity. Its unit is Kgms-1
Newton’s Second Law of motion:
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Mathematically this can be written as

Laws Of Motion Class 11 Notes Pdf Download Chapter 5 Question 1.
Derive F = ma from Newton’s Second law.
Consider a body of mass ‘m’ moving with a momentum $$\vec{p}$$. Let $$\vec{F}$$ be the force acting on it for time internal Dt. Due to this force the momentum is changed from $$\vec{p}$$ to p + Dp. Then according to Newtons second law, we can write

Where K is a constant proportionality. When we take the limit ∆t → 0, we can write

Unit of force:
Unit of force is newton. 1N = 1Kgms-2
Force in terms of the components:
We know force is a vector, Hence we can write as

Impulsive force:
The forces which act on bodies for short time are called impulsive forces.
Example:

• In hitting a ball with a bat
• In firing a gun

Impulse:
An impulse force does not remain constant, but changes from zero to maximum. This impulsive force is not easy to measure, because it changes with time. In such a case, we measure the total effect of the force called impulse.

The impulse of a force is the product of the average force and the time for which it acts.

Relation between impulse and momentum:
We know from Newtons second law
F = $$\frac{\Delta p}{\Delta t}$$

R.H.S. is the impulse and L.H.S. is change of momentum ie; change of momentum = impulse.

Laws Of Motion Notes Class 11 Chapter 5 Question 2.
F = $$\frac{\Delta p}{\Delta t}$$. When we jump to hard soil, Dt is small and F is large. When we jump to loose soil it takes more time for the body to come to rest. Therefore, Dt is large and F will be small.

Plus One Physics Laws Of Motion Chapter 5 Question 3.
A cricketer draws his hand while catching a cricket ball. Why?
When cricketer draws his hand, the Dt will increase. Hence F acting on the hand will decrease.

Newtons Third Law Of Motion
Statement:
To every action, there is always an equal and opposite reaction.
Explanation: When a book is placed on the table, the weight of the book acts on the table downwards. The table exerts an equal force on the book in the upward direction. If the force applied by the book on the table is action, the force applied by the table on the book is reaction.

Newton’s Laws Of Motion Class 11 Notes Chapter 5 Question 4.
If action and reaction are equal and opposite, why they do not cancel?
Though action and reaction are equal and opposite, they do not cancel each other because action is on one body and reaction is on another body.
Consider a pair of bodies A and B. According to the
third law FAB = – FBA
(force on A by B) = – (force on B by A).

Plus One Physics Laws Of Motion Notes Chapter 5 Conservation Of Momentum
Second law and third law lead to conservation of linear momentum.
Statement:
When there is no external force on a body (or system), the total momentum remains constant.

Proof in the case of a single body:
According to Newtons second law, F = $$\frac{d p}{d t}$$. if F = 0, we get p = constant. Which means that momentum of a body remains constant, if there is no external force acting on it.

Conservation of momentum in the case of firing a gun:
Consider a gun of mass M and bullet of mass ‘m’ at rest. On firing the gun exerts a force F on the bullet and bullet exerts an equal force -F in the opposite direction. Because of this action and reaction (due to firing), the gun acquires a momentum Pg and bullet acquires a momentum Pb.
Momentum before firing
The bullet and gun are at rest. Hence momentum before firing = M × 0 + m × 0
Momentum before firing = 0 ________(1)
Momentum after firing
According to Newtons second law, the change in
momentum of bullet. ∆Pb = Pb – 0 = F∆t ______(2)
Since initially both are rest,
Dp = final momentum – initial
momentum Similarly the change in momentum of gun
∆pg = pg – 0 = -F∆t _______(3)
∴ Total momentum after firing = pb + pg
= F∆t + – F∆t.
Total momentum after firing = 0 _______(4)
from eq (1) and eq (4), we get,
Total momentum before firing = Total momentum after firing.

Conservation of momentum in the case of two colliding bodies:

Consider two bodies A and B with initial momenta PA and PB. After collision, they acquire momenta P1A and P1g respectively.
According to Newton’s second law, the change in momentum of A due to the collision with B,

Similarly the change in momentum of B due to the collision with A, FBA∆t = P1B – PB

[Where Dt is time for which the two bodies are in contact].
According Newton’s third law, we can write
FAB = -FBA

Total momentum before collision = Total momentum after collision.
Note: Conservation linear momentum is always satisfied for elastic collision and inelastic collision.

Class 11 Physics Chapter 5 Notes Equilibrium Of A Particle
Equilibrium of a particle in mechanics refers to the situation, when the net external force on the particle is zero.

Common Forces In Mechanics
There are two types of forces commonly used in mechanics,

1. Contact forces
2. Non contact forces

1. Contact forces:
A contact force on an object arises due to contact with some other object. Example : Friction, viscosity, air resistance etc.

2. Non contact forces:
A non contact force on an object arises due to non contact with some other object
Example: Gravitational force

Friction:
Friction is the force that develops at the surfaces of contact of two bodies and impedes (opposes) their relative motion.
There are different types of friction.

• Static friction: The opposing force that comes into play when one body tends to move over the surface of another (but the actual motion has yet not started)
• Limiting friction (fs): The maximum value of static friction is called limiting friction.
• Kinetic friction (fk)(or) dynamic friction: Kinetic friction or dynamic friction is the opposing force that comes into play when one body is actually moving’overthe surface of another body.
• Sliding friction: The opposing force that comes into play when one body is actually sliding over the surface of the other body is called sliding friction.
• Rolling friction: The opposing force that comes into play when one body is actually rolling over the surface of the other body is called rolling friction.

Laws of static Friction:

• The force of maximum static friction is directly proportional to the normal reaction
• The force of static friction is opposite to the direction in which the body tends to move.
• The force of static friction is parallel to the surfaces in contact.
• The force of maximum static friction is independent of the area of contact (as long as the normal reaction remains constant).
• The force of static friction depends only on the nature of surfaces in contact.

a. Laws of Kinetic friction:

1. The force of Kinetic friction is proportional to normal reaction.
2. The force of Kinetic friction is opposite to the dh rection in which the body moves.
3. The force of Kinetic friction is parallel to the surfaces in contact.
4. The force of Kinetic friction is independent of the area contact (as long as the normal reaction remains constant)
5. The force of Kinetic friction depends on the nature of surface.
6. Force of Kinetic friction is almost independent of the speed.
7. Force of Kinetic friction is less than force of static friction.

b. Coefficient of static friction:
The force of static friction (fs)max is directly proportional to the normal reaction N
(fs)max α N

Where ms is called coefficient of static friction.
Definition of ms
Coefficient of static friction is the ratio of the force of the maximum static friction to the nprmal reaction.

c. Coefficient of Kinetic friction:
The force of kinetic friction is directly proportional to the normal reaction N.
i e (fk)max α N

Where µk is called coefficient of Kinetic friction.
Definition of µk
Coefficient of Kinetic friction is the ratio of the force of Kinetic friction to the normal reaction.

d. Angle of friction:
Angle of friction is the angle whose tangent gives the coefficient of friction.

Proof:
Consider a body placed on a surface. Let N be the normal reaction and flimit is the limiting friction. Let ‘θ’ be the angle between Resultant vector and normal reaction. From the triangle OBC,

∴ tanθ = µ
Angle of repose:
The angle of repose is the angle of the inclined plane at which a body placed of it just begins to slide.
Explanation
considers body placed on a inclined plane. Gradually increase the angle of inclination till the body placed on its surface just begins to slide down. If α is the inclination at which the body just begins to slide down, then α is called angle of repose.

The limiting friction F acts in upward direction along the inclined plane. When the body just begins to move, we can write
F = mg sin α ______(1)
from the figure normal reaction,
N = mg cos α ______(2)
dividing eq (1) by eq (2)

Note: Angle of repose is equal to angle of friction.
Rolling friction:
Why rolling friction is less than kinetic friction?
When a body rolls over a plane, there is just one point of contact between the body and plane. The relative motion between point and plane is zero. Hence in this ideal situation, kinetic friction becomes zero.

• Friction helps us to walk on the ground.
• Friction helps us to hold objects.
• Friction helps in striking matches.
• Friction helps in driving automobiles.
• Friction is helpful in stopping a vehicle etc.

• Friction produces wear and tear.
• Friction leads to wastage of energy in the form of heat.
• Friction reduces the efficiency of the engine etc.

Steps to reduce friction

• Polishing the surfaces in contact
• Use of lubricants
• Ball bearing placed between moving parts of machine.

Circular Motion
When a body moves along circumstances of a circle, there is an acceleration towards it’s centre. This acceleration is called centripetal acceleration. The force providing this acceleration is called centripetal force.
Centripetal force f = $$\frac{\mathrm{mv}^{2}}{\mathrm{R}}$$

1. For a stone rotated in a circle by a string, the centripetal force is provided by the tension in the string.
2. The centripetal force for motion of a planet around the sun is the gravitational force on the planet due to sun.
3. For a car on circular road, the centripetal force is provided by the friction between tire and road.

1. Motion of a car on a level road:

Consider a vehicle moving overa level curved road. The two forces acting on it are

• Weight (mg) vertically down
• The reaction (N)

The normal reaction can’t produce sufficient centripetal force required for circular motion. The centripetal force for circular motion is provided by friction. This friction opposes the motion of the car moving away from the circular road. Hence condition for circular motion can be written as Centripetal force ≤ force of friction

The maximum speed of circular motion of the car
vmax = $$\sqrt{\mu_{s} \mathrm{rg}}$$

Question 5.
Why surface of the road is kept inclined to the horizontal?
Consider a vehicle moving along a level curved road. The vehicle will have a tendency to slip outward. This outward slip is prevented by frictional force. But friction causes unnecessary wear and tear. More over, for typical value of µ and R the maximum speed v = $$\sqrt{\mu_{s} \mathrm{rg}}$$ rg will be very small.

These defects can be avoided if we raise the outer edge of the road slightly above the inner edge. This process is called banking of curve. The angle made by the surface of the road with the horizontal is called the angle of banking.

2. Motion of a car on a banked road:

Consider a vehicle along a curved road with angle of banking q. Then the normal reaction on the ground will be inclined at an angle q with the vertical.

The vertical component can be divided into N Cosq (vertical component) and N sinq (horizontal component). Suppose the vehicle has a tendency to slip outward. Then the frictional force will be developed along the plane of road as shown in the figure. The frictional force can be divided into two components. Fcosq (horizontal component) and F sinq (vertical component).
From the figure are get
N cos q = F sinq + mg
N cosq – F sinq = mg ______(1)
The component Nsinq and Fsinq provide centripetal force. Hence

Dividing both numerator and denominator of L.H.S by N cosq. We get

This is the maximum speed at which vehicle can move over a banked curved road.

Optimum speed:
Optimum speed is the speed at which a vehicle can move over a curved banked road without using unnecessary friction.
When a car is moved with optimum speed Vo, m can be taken as zero.
putting m = 0 in the above equation we get

## Plus One Malayalam Textbook Answers Unit 4 Chapter 3 Muhyadheen Mala

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 3 Muhyadheen Mala Text Book Questions and Answers, Summary, Notes.

## Plus One Malayalam Textbook Answers Unit 4 Chapter 2 Anukampa

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 2 Anukampa Text Book Questions and Answers, Summary, Notes.

## Plus One Maths Chapter Wise Questions and Answers Chapter 5 Complex Numbers and Quadratic Equations

Students can Download Chapter 5 Complex Numbers and Quadratic Equations Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 5 Complex Numbers and Quadratic Equations

### Plus One Maths Complex Numbers and Quadratic Equations Three Mark Questions and Answers

Plus One Maths Chapter Wise Questions And Answers Pdf Chapter 5 Question 1.
If z1 = 2 – i, z2 = 1 + i

1. Find | z1 + z2 + 1| and |z1 – z2 + i| (2)
2. Hence find $$\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\right|$$ (1)

1. |z1 + z2 + 1| = |2 – i + 1 + i + 1| = 4
|z1 – z2 + i| = |2 – i – 1 – i + i| = |1 – i|
$$=\sqrt{1+1}=\sqrt{2}$$

2.

Hsslive Maths Textbook Answers Plus One Chapter 5 Question 2.
Find the square root of -15 – 8i.
Let x + iy = $$\sqrt{-15-8 i}$$
Then (x + iy)2 = -15 – 8i
⇒ x2 – y2 + 2xyi = – 15 – 8i
Equating real and imaginary parts, we have
x2 – y2 = -15 ______(1)
2xy = – 8
We know the identity
(x2 + y2)2 = (x2 – y2)2 + (2xy)2
= 225 + 64
= 289
Thus, x2 + y2 = 17 _______(2)
From (1) and (2), x2 = 1 and y2 = 16 or x = ±1 and y = ±4
Since the product xy is negative, we have
x = 1, y = -4 or, x = -1, y = 4
Thus, the square roots of -15 – 8i are 1 – 4i and -1 + 4i.

### Plus One Maths Complex Numbers and Quadratic Equations Four Mark Questions and Answers

Plus One Maths Questions And Answers Chapter 5 Question 1.
Consider the complex number $$\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$$

1. Express in a + ib form. (2)
2. Convert into polar form. (2)

1.

2.

The complex number lies in the first quadrant;
⇒ θ = α = $$\frac{5 \pi}{12}$$

Plus One Maths Chapter Wise Questions And Answers Chapter 5 Question 2.

1. Express the complex number $$\frac{2-i}{(1-i)(1+2 i)}$$ in the form a + ib (2)
2. Solve the equation 27x2 – 10x + 1 = 0 (2)

1.

2. 27x2 – 10x + 1 = 0

Complex Numbers Class 11 Extra Questions Chapter 5  Question 3.

1. For what value of x and y 4x + i(3x – y) = 3 – 6i (2)
2. Solve the equation 21x2 – 28x + 10 = 0 (2)

1. Given; 4x + i(3x – y) = 3 – 6i
⇒ 4x = 3; 3x – y = -6

2. 21x2 – 28x + 10 = 0

Complex Numbers And Quadratic Equations Chapter 5 Question 4.
Consider the complex number z = $$\frac{1+i}{1-i}$$
1. Write z in a + ib form.
2.

In the figure radius of the circle is 1. Write the polar form of the complex number represent by the points P and Q. (2)
3. Find the square root of i. (2)
1.

2. Polar form of the point P is $$1\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$$
Polar form of the point Q is $$1\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$$

3. i = 0 + i ⇒ $$\sqrt{i}$$ = x + iy ⇒ i = x2 + y2 + 2xyi x2 + y2 = 0; 2xy = 1
(x2 + y2)2 = (x2 – y2)2 + 4x2y2
(x2 + y2)2 = 0 + (1)2 = 1
x2 + y2 = 1; x2 + y2 = 0

### Plus One Maths Complex Numbers and Quadratic Equations Six Mark Questions and Answers

Complex Numbers And Quadratic Equations Class 11 Pdf Question 1.

1. Express the complex number $$\frac{3-\sqrt{-16}}{1-\sqrt{-9}}$$ in the form a + ib (2)
2. Represent the complex number $$\frac{5+i \sqrt{3}}{-4+2 \sqrt{3 i}}$$ in the polar form. (2)
3. Solve the equation ix2 – x + 12i = 0 (2)

1.

2.

The complex number lies in the third quadrant;

3. ix2 – x + 12i = 0

### Plus One Maths Complex Numbers and Quadratic Equations Practice Problems Questions and Answers

Complex Numbers And Quadratic Equations Class 11 Solutions Question 1.
Express each of the following in a + ib form. (1 score each)

1. (2 – 4i) + (5 + 3i)
2. (1 – i) – (-1 + 6i)
3. 3(7 + 7i) + i(7 + 7i)
4. $$\left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+\frac{5}{2} i\right)$$

1. (2 – 4i) + (5 + 3i) = (2 + 5) + (-4 + 3)i = 7 – i

2. (1 – i) – (-1 + 6i) = 1 – i + 1 – 6i = 2 – 7i

3. 3(7 + 7i) + i(7 + 7i) = 21 + 21i + 7i – 7 = 14 + 28i

4.

Question 2.
Express each of the following in a + ib form. (1 score each)

1. (-5i)($$\frac{1}{8}$$i)
2. (-i)(2i)(-$$\frac{1}{8}$$i)3
3. i99
4. i111 + i222 + i333
5. (7 – i)(2 + 7i)
6. (-1 – i)(4 + 2i)
7. (5 – 3i)2
8. ($$\frac{1}{3}$$ + 3i)3

1.

2.

3. i99 = i96 + 3 = i96i3 = -i

4. i111 + i222 + i333 + i108 + i220 + 2 + i332 + 1
= i3 + i2 + i1 = -i – 1 + i = -1

5. (7 – i)(2 + 7i) = 7 × 2 – 2i + 7 × 7i – i × 7i
= 14 – 2i + 49i + 7 = 21 + 47i

6. (-1 – i)(4 + 2i) = -4 – 4i – 2i + 2 = – 2 – 6i

7. (5 – 3i)2 = 52 – 2 × 5 × 3i + (3i)2
= 25 – 30i – 9 = 16 – 30i

8.

Question 3.
Find the multiplicative inverse of the following; (1 score each)

1. 3 – 4i
2. 2 – 3i
3. $$\sqrt{5}$$ + 3i

1. Multiplicative inverse = $$\frac{1}{3-4 i}$$

2. Multiplicative inverse = $$\frac{1}{2-3 i}$$

3. Multiplicative inverse = $$\frac{1}{\sqrt{5}+3 i}$$

Question 4.
Express each of the following in a + ib form. (2 score each)

Question 5.
Convert the following into polar form. (2 score each)

1. 1 + i
2. -1 + i
3. $$\sqrt{3}$$ – i
4. $$\frac{5-\sqrt{3} i}{4+2 \sqrt{3} i}$$

1. Given; 1 + i = r(cosθ + isinθ)
r = $$\sqrt{1+1}=\sqrt{2}$$
tanα = $$\left|\frac{1}{1}\right|$$ = 1 ⇒ α = $$\frac{\pi}{4}$$
The complex number lies in the first quadrant;

2. -1 + i = r(cosθ + isinθ)
r = $$\sqrt{1+1}=\sqrt{2}$$
tanα = $$\left|\frac{1}{-1}\right|$$ = 1 ⇒ α = $$\frac{\pi}{4}$$
The complex number lies in the second quadrant;

3.

The complex number lies in the fourth quadrant;

4.

The complex number lies in the fourth quadrant;

## Plus One Economics Notes Chapter 1 Indian Economy on the Eve of Independence

Students can Download Chapter 1 Indian Economy on the Eve of Independence Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus One Economics Notes Chapter 1 Indian Economy on the Eve of Independence

Low level of economic development under the colonial rule:
The British rule started in India in 1757 and came to an end in 1947. The Indian economy underwent rapid changes under British rule. The economic policies pursued by the colonial government in India were concerned more with the protection and promotion of the economic interests of their home country than the development of the Indian economy. The twin objectives of British rule in India were

1. To use India as a supplier of raw materials for British Industries.
2. To convert India into a market for the finished products produced in Britain.

Plus One Economics Notes Chapter 1 Agricultural Sector:
Agricultural Sector was the backbone of the Indian economy.
During the British colonial rule India remained fundamentally an agrarian economy. Around eighty percent of India’s population lived in villages. Agriculture was stagnant and it was the main source of livelihood of the population. People depended directly or indirectly on agriculture and its productivity was very slow. The agricultural sector stagnated during British rule.
Major reasons for agricultural stagnation were:

1. The exploitative land settlement system followed by British rulers
2. Use of low level of technology
3. Rural indebtedness
4. Low agricultural productivity
5. Use of limited chemical fertilizer

Economics Plus One Notes Chapter 1 Industrial Sector:
India’s industrial sector could not make progress during British rule. Their aim was to collect raw materials from India and sell their final products in India.

By the second half of the nineteenth century, modem industry began to take root in India. Initially, cotton industries in Maharashtra and Gujarat (Bombay presidency) and the jute industry in Bengal were established. Then industries of fertilizers, rayon, rubber, cement, sugar, pepper, etc., were established in some regions of the country. The setting up of Tata Iron and Steel Company (TISCO) in 1907 was a landmark in the industrialization of India. Jemshedji Tata established TISCO in Jamshedpur in Bihar. During the British rule hardly any capital goods industries were established in the country.

Plus One Economics Chapter 1 Foreign Trade:
Though India exported value-added products before the British period, we started exporting primary products during their rule. Consequently, India became an exporter of primary products such as raw silk, cotton, wool, sugar, indigo, jute, etc. and an importer of finished consumer goods like cotton, silk and woollen clothes and capital goods like light machinery produced in the factories of Britain.
The most important characteristic of India’s foreign trade, throughout the colonial period was the generation of a large export surplus.

Plus One Economics Chapter 1 Notes Demographic Condition:
Various details about the population of British India were first collected through a census in 1881. Through Suffering from certain limitations, it revealed the unevenness in India’s population growth. Subsequently, every ten years such census operations were carried out. Before 1921, India was in the first stage of demographic transition. The second stage of transition began after 1921.

Economics Notes Plus One Chapter 1 Occupational Structure:
Occupational structure refers to the distribution of working persons across different industries and sectors. Broadly we divide occupations into three types. Agriculture, animal husbandry, forestry, fisheries, etc., are collectively known as ‘primary’ activities. Manufacturing industries, both small and large scale, are known as ‘secondary’ activities. Transport, communication, banking, financial services, etc., are ‘tertiary’ activities.

Hsslive Economics Plus One Chapter 1 Infrastructure:
Infrastructural facilities developed in India during the British period. Infrastructure means some kind of permanent installation, which are used over a long period of time for the supply of basic inputs like railway lines, roads, dams, canal systems, power stations, pipelines, hospitals, educational institutions like schools, colleges, etc. Basic infrastructure facilities such as railways, ports, water transport, and telegraph did develop during the British rule. The real intention behind such a development was to serve the various colonial interests of Britain.

## Plus One Accountancy Notes Chapter 3 Recording of Transactions – I & II

Students can Download Chapter 3 Recording of Transactions – I & II Notes, Plus One Accountancy Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus One Accountancy Notes Chapter 3 Recording of Transactions – I & II

Summary:
Plus One Accountancy Chapter 3 Meaning of source documents:
Various business documents such as invoice, bills, cash memos, vouchers, which form the basis and evidence of a business transaction recorded in the books of account are called source documents.

Hsslive Plus One Accountancy Notes Meaning of accounting equation:
A statement of equality between debits and credits signifying that the assets of a business are always equal to the total liabilities and capital.

Plus One Accountancy Notes Rules of debit and credit:
An account is divided into two sides. The left side of an account is known as debit and the credit. The rules of debit and credit depend on the nature of an account. Debit and Credit both represent either increase or decrease, depending on the nature of an account.

These rules are summarised as follows:

Accountancy Class 11 Chapter 3 Notes Books of original entry:
The transactions are first recorded in these books in a chronological order. Journal is one of the books of original entry. The process of recording entries in the journal is called journalising.

Format of Journal

Chapter 3 Accounts Class 11 Notes Ledger:
A book containing all accounts to which entries are transferred from the books of original entry. Posting is process of transferring entries from books of original entry to the ledger.

Accountancy Class 11 Chapter 3 Solutions Journalising Format of a Ledger

Plus One Accountancy Chapter 3 Notes Special Journals:
Special journals are also called day books or subsidiary books. Transactions that cannot be recorded in any special journal are recorded in journal is called the “Journal Proper.”

The special-purpose journals are:

• Cash Book
• Petty Cash Book
• Purchase Book
• Purchase Return Book
• Sales Book
• Sales Return Book
• Journal Proper

(a) Cash Book
A book used to record all cash receipts and payments. Cash book may be single column cash book, doulbe column cash book and three column cash book.

Single Column Cash book
This is cash book containing only one column for cash and prepared as cash account in ledger.

Format of Single Column Cash Book

Double Column Cash book:
This is cash book containing one more column for bank along with the cash column, it serves the purpose of cash and bank account.

Format of Double Column Cash Book

(b) Petty Cash Book:
A book used to record small cash payments

(c) Purchase Book / Purchase Journal:
A special journal in which only credit purchases are recorded.

Accounts Class 11 Chapter 3 Notes Format of Purchase Day Book

(d) Purchase Return Book:
A book in which return of purchased goods on credit is recorded.

Accounts Chapter 3 Class 11 Notes Format of Purchase Return Book

(e) Sales Book / Sales Journal:
A special journal in which only credit sales are recorded.

Format of Sales Day Book

(f) Sales Return Book:
A special book in which return of goods sold on credit is recorded.

Format of Sale Return Book

Balancing the Accounts:
Accounts in the ledger are periodically balanced, generally at the end of the accounting period with the object of ascertaining the net position of each amount.

Balancing of an account means that the two sides are totaled and the difference between them is shown on the side which is shorter in order to make their totals equal. The words ‘balance carried down (c/d)’ are written against the amount of the difference between the two sides.

The amount of balance is brought down (b/d) in the next accounting period indicating that it is a continuing account until finally settled or closed. In case the debit side exceeds the credit side.

The difference is written on the side, if the credit side exceeds the debit side, the difference between the two appears on the debit side and is called debit and credit balance respectively. The accounts of expenses losses, gains and revenues are not balanced but are closed by transferring to trading and profit and loss account.

## Plus One Malayalam Textbook Answers Unit 4 Chapter 1 Peeli Kannukal

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 1 Peeli Kannukal Text Book Questions and Answers, Summary, Notes.

## Plus One Chemistry Notes Chapter 2 Structure of Atom

Students can Download Chapter 2 Structure of Atom Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus One Chemistry Notes Chapter 2 Structure of Atom

Plus One Chemistry Chapter 2 Notes Pdf Introduction
The atomic theory of matter was first proposed by John Dalton. His theory, called Dalton’s atomic theory, regarded the atom as the ultimate particle of matter.

Sub-Atomic Particles
Discovery Of Electron
The experiments of Michael Faraday in discharge tubes showed that when a high potential is applied to a gas taken in the discharge tube at very low pres-sures, certain rays are emitted from the cathode. These rays were called cathode rays.

The results of these experiments are summarised below:
1. The cathode rays start from cathode and move towards the anode.

2. In the absence of electrical or magnetic field, these rays travel in straight lines.ln the presence of electrical or magnetic field, they behave as negatively charged particles, i.e.,they consist of negatively charged particles, called electrons.

3. The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.Thus, we can conclude that electrons are the basic constituent of all the atoms.

Charge To Mass Ratio Of Electron
In 1897, the British physicist J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (m<sub>e</sub>) by using cathode ray tube and applying electrical and magnetic field perpendicular to each other as well as to the path of electrons.

From the amount of deviation of the particles from their path in the presence of electrical or magnetic field, the value of e/m was found to be 1.75882 × 1011 coulomb per kg or approximately 1.75288 × 10<sup>8</sup> cou-lomb per gram. The ratio e/m was found to be same irrespective of the nature of the gas taken in the dis-charge tube and the material used as the cathode.

Structure Of Atom Class 11 Notes Charge Of The Electron
Millikan (1868-1953) devised a method known as Oil drop experiment (1906-14), to determine the charge on the electrons. He found the charge on the electron to be – 1.6 × 10-19C.

Mass of the electron (m)

Discovery Of Protons And Neutrons
Electrical discharge earned out in the modified cathode ray tube led to the discovery of canal rays. The characteristics of these positively charged particles are listed below:

• unlike cathode rays, the e/m ratio of the particles depend upon the nature of gas present in the cathode ray tube.
• Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
• The behaviour of these particles in the magnetic or electrical field is opposite to that observed for cathode rays.

The smallest and lightest positive ion was obtained from hydrogen and was called proton. Later, electrically neutral particles were discovered by Chadwick (1932) by bombarding a thin sheet of beryllium by α – particles when electrically neutral particles having a mass slightly greater than that of the protons was emitted. He named these particles as neutrons.

Atomic Models

Structure Of Atom Class 11 Notes Hsslive Thomson Model Of Atom
J.J. Thomson was the first to propose a model of the atom. According to him, the atom is a sphere in which positive charge is spread uniformly and the electrons are embedded in it so as to make the atom electrically neutral. This model is also known as “plumpudding model’. But this model was soon discarded as it could not explain many of the experimental observations.

Hsslive Structure Of Atom Notes Rutherford’s Nuclear Model of Atom
Rutherford and his students (Hans Geiger and Ernest Marsden) bombarded very thin gold foil with α – particles. The experiment is known as α -particle scattering experiment. On the basis of the observations, Rutherford drew the following conclusions regarding the structure of atom :

1. Most of the space in the atom is empty as most of the α -particles passed through the foil undeflected.

2. A few α – particles were deflected. Since the α – particles are positively charged, the deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson had presumed. The positive charge has to be concentrated in a very small volume that repelled and deflected the positively charged α – particles.

3. Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly small as compared to the total volume of the atom.

On the basis of above observations and conclusions, Rutherford proposed the nuclear model of atom (after the discovery of protons). According to this model:
1.The positive charge and most of the mass of the atom was densely concentrated in extremely small region. This very small portion of the atom was called nucleus by Rutherford.

2. The electrons move around the nucleus with a very high speed in circular paths called orbits. Thus, Rutherford’s model of atom resembles the solar system in which the nucleus plays the role of sun and the electrons that of revolving planets.

3. Electrons and the nucleus are held together by electrostatic forces of attraction.

Chemistry Notes For Class 11 Chapter 2 Atomic Numberand Mass Number
’ Knowing the atomic number Z and mass number A of an element, we can calculate the number of protons, electrons and neutrons present in the atom of the element.
Atomic Number (Z) = Number of protons = Number of electrons
Mass Number (A) – Atomic number (Z) = Number of neutrons

Isotopes, Isobars And Isotones
Isotopes are atoms of the same element having the same atomic number but different mass numbers. They contain different number of neutrons. For ex-ample, there are three isotopes of hydrogen having mass numbers 1,2 and 3 respectively. All the three isotopes have atomic number 1. They are represented as $$_{ 1 }^{ 1 }{ H }$$, $$_{ 1 }^{ 2 }{ H }$$ and $$_{ 1 }^{ 3 }{ H }$$ and named as hydrogen or protium, deuterium (D) and tritium (T) respectively. Isobars are atoms of different elements which have the same mass number. For example, $$_{ 6 }^{ 14 }{ C }$$ and $$_{ 7 }^{ 14 }{ N }$$ are isobars.
Isotones may be defined as atoms of different elements containing same number of neutrons. For example $$_{ 6 }^{ 13 }{ C }$$ and $$_{ 7 }^{ 14 }{ N }$$ are isotones.

Developments Leading To The Bohr’S Model Of Atom
Neils Bohr improved the model proposed by Rutherford. Two developments played a major role in the formulation of Bohr’s model of atom. These were:

1. electromagnetic radiation possess both wave like and particle like properties(Dual character)
2. Experimental results regarding atomic spectra which can be explained only by assuming quantized electronic energy levels in atoms.

Wave Nature Of Electromagnetic Ra-Diation
Light is the form of radiation and it was supposed to be made of particles known as corpuscules.
As we know, waves are characterised by wavelength (λ), frequency (υ) and velocity of propagation (c) and these are related by the equation
c = vλ or v = $$\frac { c }{ \lambda }$$

The wavelengths of various electro magnetic radia-tions increase in the order.
γ rays < X-rays< uv rays < visible < IR < Microwaves < Radio waves

Particle Nature Of Electro Magnetic Radiation: Planck’S Quantum Theory
Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete quantities and not in a continuous manner, a belief popular at that time. Planck gave the name quantum to the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency (υ) and is expressed by the equation E = hυ

Class 11 Chemistry Chapter 2 Notes Photoelectric Effect
When a metal was exposed to a beam of light, electrons were emitted. This phenomenon is called photoelectric effect. Obseravations of the photoelectric effect experiment are the following:

• There is no time lag belween the striking of light beam and the ejection of electrons from the metal surface.
• The number of electrons ejected is proportional to the intensity or brightness of light.
• For each metal, there is a characteristic minimum frequency, u0 (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency u>u0, the ejected electrons come out with certain kinetic energy.

The kinetic energies of these electrons increase with the increase of frequency of the light used.

Using Plank’s quantum theory Einstein explained photoelectric effect. When a light particle, photon with sufficient energy strikes an electron instantaneously to the electron during the collision and the electron is ejected without any time lag. Greater the energy of photon greater will be the kinetic energy of ejected electron and greater will be the frequency of radiation.

If minimum energy to eject an electron is hv0 and the photon has an energy equal to hv. Then kinetic en-ergy of photoelectron is given by, hv=hv0 + 1/2 mev2 where me is the mass of electron and hv0 is called the work function.

Duel Behaviour Of Electromagnetic Ra-Diation
Light has dual behaviour that is it behaves either as a wave or as a particle. Due to this wave nature, it shows the phenomena of interference and diffraction.

Evidence For The Quantized Electronic Energy Levels : Atomic Spectra
It is observed that when a ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength. Since ordinary white light consists of waves with ail the wave-lengths in the visible range, a ray of white light is spread out into a series of coloured bands called spectrum. In a continuous spectrum light of different colours merges together. For example violet merges into blue, blue into green and soon.

Chapter 2 Class 11 Chemistry Notes Emission and absorption spectra
The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. Atoms, molecules or ions that have absorbed radiation are said to be “excited”.

A continuum of radiation is passed through a sample which absorbs radiation of certain wavelengths. The missing wavelength which corresponds to the radiation absorbed by the matter, leave dark spaces in the bright continuous spectrum. The study of emission or absorption spectra is referred to as spectroscopy Line spectra or atomic spectra is the spectra where emitted radiation is identified by the appearance of bright lines in the spectra.

Line spectrum of Hydrogen
The hydrogen spectrum consists of several series of lines named after their discoverers. Balmershowed in 1885 on the basis of experimental observations that if spectral lines are expressed in terms of wavenumber ($$\overline { v }$$), then the visible lines of the hydrogen spectrum obey the following formula :
$$\overline { v }$$ = 109,677 $$\left[\frac{1}{2^{2}}-\frac{1}{n^{2}}\right] \mathrm{cm}^{-1}$$
where n = 3, 4, 5, ………….
The series of lines described by this formula are called the Balmer series.

The value 109,677cm-1 is called the Rydberg constant for hydrogen. The first 5 series of lines correspond to n1 = 1, 2, 3, 4, 5 are known as Lyman, Balmer, Paschen, Bracket and Pfund series respectively. Line specrum becomes more complex for heavier atoms.

Chapter 2 Chemistry Class 11 Notes Bhor’S Model For Hydrogen Atom
Bhors model for hydrogen atom says that
1. the energy of an electron does not change with time.
The diagram shows the Lyman, Balmer and Paschen series of transitions for hydrogen atom.

2. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ∆E, is given by :
$$v=\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}$$
E1 and E2 are the energies of the lower and higher allowed energy states respectively.
The angular momentum of an electron in a given stationary state can be expressed as in equation,

Chemistry Chapter 2 Class 11 Notes Bohr’s theory for hydrogen atom:
1. The stationary states for electron are numbered n = 1,2,3. These integral numbers are known as Principal quantum numbers.
2. The radii of the stationary states are expressed as:
rn = n² a0
where a0 = 52.9 pm

3. The most important property associated with the electron, is the energy of its stationary state. It is
given by the expression, $$E_{n}=-R_{H}\left(\frac{1}{n^{2}}\right)$$
where RH is called Rydberg constant and its value is 2.18 × 10-18 J. The energy of the lowest state, also called as the ground state, is
E1 = -2.18 × 10-18 $$\left(\frac{1}{1^{2}}\right)$$ = -2.18 × 10-18 J. The energy of the stationary state for n = ∝, will be :
E2 = -2.18 × 10-18 J$$\left(\frac{1}{2^{2}}\right)$$ = -0.545 × 10-18 J.

When the electron is free from the influence of nucleus(n = ∞), the energy is taken as zero. When the electron is attracted by the nucleus and is present in orbit n, the energy is emitted and its energy is lowered. That is the reason for the presence of negative sign and depicts its stability relative to the reference state of zero energy and n = ∞

4. Bohr’s theory can also be applied to the ions containing only one electron, similar to that present in hydrogen atom. For example, He<sup>+</sup> Li<sup>2+</sup>, Be<sup>3+</sup> and so on. The energies of the stationary states associated with these hydrogen-like species are given by the expression,

Structure Of Atom Class 11 Notes Pdf Explanation of Line Spectrum of Hydrogen
The frequency (v) associated with the absorption and emission of the photon can be evaluated by using equation,

Class 11 Chapter 2 Chemistry Notes Limitations of Bohr’s Model
Bohr’s model was too simple to account for the following points:
1. It fails to account for the finer details (doublet, that is two closely spaced lines) of the hydrogen atom spectrum. This model is also unable to explain the spectrum of atoms other than hydrogen Further, Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
2. It could not explain the ability of atoms to form molecules by chemical bonds.

Towards Quantum Mechanical Model Of The Atom
Two important developments which contributed significantly in the formulation of a more suitable and general model for atoms were:

1. Dual behaviour of matter
2. Heisenberg uncertainty principle

Structure Of Atom Class 10 Notes Pdf Dual Behaviour of Matter
The French physicist, de Broglie proposed that matter, like radiation, should also exhibit dual behaviour i. e., both particle and wavelike properties. This means that just as the photon, electrons should. also have momentum as well as wavelength. de Broglie, from this analogy, gave the following relation between wavelength (λ) and momentum (p) of a material particle.
$$\lambda=\frac{h}{m v}=\frac{h}{p}$$

Heisenberg’s Uncertainty Principle
Werner Heisenberg a German physicist in 1927, stated uncertainty principle which is the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as in equation,

∆x is the uncertainty in position and ∆p<sub>x</sub> (or ∆v<sub>x</sub>) is the uncertainty in momentum (or velocity) of the particle. If the position of the electron is known with high degree of accuracy (∆x is small), then the velocity of the electron will be uncertain ∆v<sub>x</sub> is large]. On the other hand, if the velocity of the electron is known precisely ( ∆v<sub>x</sub> is small), then the position of the electron will be uncertain (∆x will be large). Thus, if we carry out some physical measurements on the electron’s position or velocity, the outcome will always depict a fuzzy or blur picture.

Significance of Uncertainty Principle
Heisenberg Uncertainty Principle rules out existence of definite paths or trajectories of electrons and other similar particles. The trajectory of an object is determined by its location and velocity at various moments. If we know where a body is at a particular instant and if we also know its velocity and the forces acting on it at that instant, we can tell where the body would be sometime later. We, therefore, conclude that the position of an object and its velocity fix its trajectory. The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects.

Reasons for the Failure of the Bohr Model
In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. The wave character of the electron is not considered in Bohr model. Further, an orbit is a clearly defined path and this path can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Bohr.model of the hydrogen atom, therefore, not only ignores dual behaviour of matter but also contradicts Heisenberg uncertainty principle. There was no point in extending Bohr model to other atoms. In fact, an insight into the structure of the atom was needed which could account for wave-particle duality of matter and be consistent with Heisenberg uncertainty principle. This came with the advent of quantum mechanics.

Quantum Mechanical Model Of Atom
Quantum mechanics is a theoretical science that deals with the study of motions of microscopic objects such as electrons.

In quantum mechanical model of atom, the behaviour of an electron in an atom is described by an equation known as Schrodinger wave equation. Fora system, such as an atom or molecule whose energy does not change with time, the Schrodinger equation written as Hψ = Eψ where H is a mathematical operator, called Hamiltonian operator, E is the total energy and ψ is the amplitude of the electron wave called wave function.

Hydrogen Atom And The Schrodinger Equation
The wave function ψ as such has no physical significance. It only represents the amplitude of the electron wave. However ψ² may be considered as the probability density of the electron cloud. Thus, by determining ψ² at different distances from the nucleus, it is possible to trace out or identify a region of space around the nucleus where there is high probability of locating an electron with a specific energy.

According to the uncertainty principle, it is not possible to determine simultaneously the position and momentum of an electron in an atom precisely. So Bohr’s concept of well defined orbits for electron in an atom cannot hold good. Thus, in quantum mechanical mode, we speak of probability of finding an electron with a particular energy around the nucleus. There are certain regions around the nucleus where probability of finding the electron is high. Such regions are called orbitals. Thus an orbital may be defined as the region in space around the nucleus where there is maximum probability of finding an electron having a specific energy.

Orbitals and Quantum Numbers
Orbitals in an atom can be distinguished by their size, shape and orientation. An orbital of smaller size means there is more chance of finding the electron near the nucleus. Similarly, shape and orientation mean that there is more probability of finding the electron along certain directions than along others. Atomic orbitals are precisely distinguished by what are known as quantum numbers. Each orbital is designated by three quantum numbers labelled as n, l and m<sub>l</sub>

The principal quantum number n’ is a positive integer with value of n= 1, 2, 3 ……………

The principal quantum number determines the size and to large extent the energy of the orbital.

The principal quantum number also identifies the shell. With the increase in the value of ‘n’, the number of allowed orbital increases and are given by ‘n²’ Ait the orbitals of a given value of ‘n’ constitute a single shell of atom and are represented by the following letters
n= 1 2 3 4 ………………
Shell = K LM N ………………

Size of an orbital increases with increase of principal quantum number ‘n’. Since energy of the orbital will increase with increase of n.

Azimuthal quantum number, ‘F is also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. For a given value of n, l can have n values ranging from 0 to (n – 1), that is, for a given value of n, the possible value of l are: l = 0, 1, 2, ……….. (n – 1)

Each shell consists of one or more subshells or sub-levels. The number of subshells in a principal shell is equal to the value of n. For example h the first shell (n = 1), there is only one sub-shell which corresponds to l = 0. There are two sub-shells (l= 0, 1) in the second shell (n = 2), three l= 0, 1, 2) and so on. Each sub-shell is assigned an azimuths! quantum number (l). Sub-shells corresponding to different values of l are represented by the following symbols.
l : 0 1 2 3 4 5 …………….
Notation for sub-shell : s p d f g h …………….

Magnetic orbital quantum number. ‘m<sub>l</sub>’ gives information about the spatial orientation of the or bital with respect to standard set of co-ordinate axis. For any sub-shell (defined by T value) 21+ 1 values of m,are possible and these values are given by:
m, = -l, -(l-1), (l-2)… 0, 1… (l-2), (l-1), l Thus for l = 0, the only permitted value of m,= 0, [2(0) + 1 = 1, one s orbital].

Electron spin ‘s’:
George Uhlenbeck and Samuel Goudsmit proposed the presence of the fourth quantum number known as the electron spin quantum number (m<sub>s</sub>). Spin angular momentum of the electron — a vector quantity, can have two orientations relative to the chosen axis. These two orientations are distinguished by the spin quantum numbers ms which can take the values of +½ or -½. These are called the two spin states of the electron and are. normally represented by two arrows, ↑ (spin up) and ↓ (spin down). Two electrons that have different m<sub>s</sub> values (one +½ and the other -½) are said to have opposite spins. An orbital cannot hold more than two electrons and these two electrons should have opposite spins.

Shapes of Atomic Orbitals
The orbital wave function or V for an electron in an atom has no physical meaning. It is simply a mathematical function of the coordinates of the electron.

According to the German physicist, Max Bom, the square of the wave function (i.e., ψ²) at a point gives the probability density of the electron at that point.

For 1 s orbital the probability density is maximum at the nucleus and it decreases sharply as we move away from it. The region where this probability I density function reduces to zero is called nodal surfaces or simply nodes. In general, it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n.

These probability density variation can be visualised . in terms of charge cloud diagrams.

Boundary surface diagrams of constant probability density for different orbitals give a fairly good representation of the shapes of the orbitals. In this representation, a boundary surface or contour surface is drawn in space for an orbital on which the value of probability density |ψ|² is constant. Boundary ‘ surface diagram for a s orbital is actually a sphere centred on the nucleus. In two dimensions, this sphere looks like a circle. It encloses a region in which probability of finding the electron is about 90%. The s-orbitals are spherically symmetric, that is, the probability of finding the electron at a given distance is equal in all the directions.

unlike s-orbitals, the boundary surface diagrams of p orbitals are not spherical. Instead, each p orbital consists of two sections called lobes that are on either side of the plane that passes through the nucleus. The probability density function is zero on the plane where the two lobes touch each other. The size, shape and energy of the three orbitals are identical. They differ, however, in the way the lobes are oriented. Since the lobes may be considered to lie along the x, y or z-axis, they are given the designations 2px, 2py, and 2pz. It should be understood, however, that there is no simple relation between the values of m, (-1, 0 and+1) and the x, y and z directions. For our purpose, it is sufficient to remember that, because there are three possible values of m, there are, therefore, three p orbitals whose axes are mutually perpendicular. Like s orbitals, p orbitals increase in size and energy with increase in the principal quantum number

The number of nodes are given by (n -2), that is number of radial node is 1 for 3p orbital, two for 4p orbital and so on.

For l = 2, the orbital is known as d-orbital and the minimum value of principal quantum number (n) has to be 3 as the value of l cannot be greater than n-1. There are five m; values (-2, -1, 0, +1 and +2) for l = 2 and thus there are five d orbitals. The five d-orbitals are designated as dxy, dyz, dxz, dx²-y² and d. The shapes of the first fourd-orbitals are similarto each other, where as that of the fifth one, d, is different from others, but all five 3d orbitals are equivalent in energy. The d orbitals for which n is greater than 3 (4d, 5d…) also have shapes similar to 3d orbital, but differ in energy and size.

Besides the radial nodes (i.e., probability density function is zero), the probability density functions for the np and nd orbitals are zero at the plane (s), passing through the nucleus (origin). For example, in case of pz orbital, xy-plane is a nodal plane, in case of dxy orbital, there are two nodal planes passing through the origin and bisecting the xy plane containing z-axis. These are called angular nodes and number of angular nodes are given by T, i.e., one angular node for p orbitals, two angular nodes for cf orbitals and so on. The total number of nodes are given by (n-1), i.e., sum of I angular nodes and (n-l-1) radial nodes.

Energies Of Orbitals
The order of energy of orbitals in single electron sys-tem are given below:
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f The orbitals having same energy are called degenerate.

Filling Of Orbitals In Atom
Aufbau principle: According to this principle in the ground state of an atom, an electron will occupy the orbital of lowest energy and orbitals are occupied by electrons in the order of increasing energy.

Pauli’s exclusiohn principle : Pauli’s exclusion principle states that ‘no two electrons in an atom can have the same values for all the four quantum numbers’

Since the electrons in an orbital must have the same n, I and m quantum numbers, if follows that an orbital can contain a maximum of two electrons provided their spin quantum numbers are different. This is an important consequence of Pauli’s exclusion principle which says that an orbital can have maximum two electrons and these must have opposite spins.

Hund’s rule of maximum multiplicity :
This rule states that electron pairing in orbitals of same energy will not take place until each available orbital of a given subshell is singly occupied (with parallel spin).
The rule can be illustrated by taking the example of carbon atom. The atomic number of carbon is 6 and its electronic configuration is 1s²2s²2p². The two electrons of the 2p subshell can be distributed in the following three ways.

According to Hund’s rule, the configuration in which the two unpaired electron occupying 2px, and 2py orbitals with parallel spin is the correct configuration of carbon.

Exceptional configurations of chromium and copper
The electronic configuration of Cr (atomic number 24) is expected to be [Ar] 4s² 3d4, but the actual configuration is [Ar] 4s¹ 3d5. Similarly, the actual configuration of Cu (At. No. 29) is [Ar] 4s¹ 3d10 instead of the expected configuration [Ar] 4s² 3d9.

This is because of the fact that exactly half filled or completely filled orbitals (i.e., d5, d10, f7, f14) have lower energy and hence have extra stability.

## Plus One Botany Chapter Wise Questions and Answers Chapter 1 Biological Classification

Students can Download Chapter 1 Biological Classification Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

## Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 1 Biological Classification

### Plus One Botany Biological Classification One Mark Questions and Answers

Plus One Botany Chapter Wise Questions And Answers Pdf Question 1.
In Whit takers, five-kingdom classification eukaryotes are distributed among
(a) two kingdoms
(b) three kingdoms
(c) four kingdoms
(d) all the five kingdoms
(c) four kingdoms

Plus One Botany Chapter Wise Questions And Answers Question 2.
Cyanobacteria are classified under which of the following kingdoms?
(a) Monera
(b) Protista
(c) Plantae
(d) Algae
(a) Monera

Plus One Botany Questions And Answers Question 3.
Main component of cell wall of fungi is
(a) cellulose
(b) chitin
(c) pectin
(d) silica
(b) chitin

Plus One Botany Chapter Wise Previous Questions And Answers Question 4.
Dinoflagellates are mostly
(a) marine and saprophytic
(b) freshwater and saprophytic
(c) marine and photosynthetic
(d) terrestrial and
(c) marine and photosynthetic

Plus One Biology Chapter Wise Questions And Answers Pdf Question 5.
Which of the following kingdoms do viruses belong to
(a) monera
(b) Protista
(c) fungi
(d) none of these
(d) none of these

Hsslive Plus One Botany Chapter Wise Questions And Answers Question 6.
Observe the relationship between the first pair and fill up the blanks.

1. Thermoacidophiles: Archaebacteria in hot spring
2. Ripening of fruits: …………….

Ethylene.

Plus One Botany Chapter Wise Previous Year Questions And Answers Question 7.
Fill in the blanks.

1. Rhizopus: Phycomycetes
Yeast: ………..
2. Holdfast: Anchorage
Heterocyst: ……….

1. Ascomycetes
2. N2 fixation

Plus One Botany Chapter Wise Questions And Answers Hsslive Question 8.
Who proposed Five kingdom classification?
R .H. Whittaker

Plus One Botany Previous Questions Chapter Wise Pdf Question 9.
Find out the correct sequence of taxonomical category.

1. Order → Kingdom → species → phylum
2. species → genus → order → phylum

2. species → genus → order → phylum

Biological Classification Important Questions Question 10.
In the five-kingdom system of Whittaker, how many kingdoms are eukaryotes?
Four kingdoms

Plus One Botany Previous Question Papers Chapter Wise Question 11.
Observe the relationship between the first pair and fill up the blanks.

1. Nostoc : Eubacteria:: methanogens: ………….
2. Yeast: ………………..:: Rhizopus: Phycomycetes :

1. Archaebacteria
2. Ascomycetes

Biology Classification Questions And Answers Question 12.
Find out the odd one.
a. Diatom, Gonyaulax, Yeast, Euglena, Plasmodium
Yeast

Hsslive Botany Previous Questions And Answers Question 13.
Vinod observed blooms in a polluted water body, his friend Kumar said that it might be nitrogen-fixing Nostoc or Anabaena. Can you suggest which type of cell can fix atmospheric nitrogen in these organisms?
Heterocyst

Botany Chapter Wise Questions And Answers Question 14.
Observe the relationship of the terms in the first pair and fill in the blanks:

1. Vibrio: Comma shaped
……….: Rod-shaped
2. Agaricus: Basidiomycetes
Penicillium: ………….

1. Bacillus
2. Ascomycetes

Biological Classification Previous Year Questions Question 15.
Difference between Virus and Viroid.
(a) Absence of protein coat in viroid but present in virus
(b) Presence of low molecular weight RNA in virus but absent in viroid
(c) Both a and b
(d) None of the above
(a) Absence of protein coat in viroid but present in virus

Question 16.
Viruses are non-cellular organisms but replicate themselves once they infect the host cell. To which of the following kingdom do viruses belong to?
(a) Monera
(b) Protista
(c) Fungi
(d) None of the above
(d) None of the above

Question 17.
A virus is considered as a living organism and an obligate parasite when inside a host cell. But virus is not classified along with bacteria or fungi. What are the characters of virus that are similar to nonliving objects?
Viruses are acellular and can be crystallized.

### Plus One Botany Biological Classification Two Mark Questions and Answers

Question 1.
The seven taxonomic categories are given below. Arrange them in the correct sequence starting from the smallest taxon.
Class → species → kingdom → order → family → division → genus.
Species → genus → family → order → class → division → kingdom.

Question 2.
“Two kingdom classification is inadequate one”. Comment on it.

1. It does not include organisms showing both plant and animal character.
2. It does not take into the consideration of nature of nucleus.

Question 3.
Five-kingdom classification of organism was given by R.H.Whittaker. State the criteria followed by Whittaker for his classification.

1. Nature of cell
2. Nature of nucleus
3. Mode of nutrition

Question 4.
Name the following;

1. A protist which can live both as an autotroph and as a heterotroph.
2. Name a protist group which consists of saprophytes.

1. Euglena
2. Slime mould

Question 5.
State two economic importance of

1. Heterotrophic bacteria
2. Archaebacteria

1. Major decomposers that help in the curdling of milk, production of antibiotic, fixing nitrogen and cause diseases like tetanus, typhoid, cholera etc.
2. Archaebacteria: production of biogas.

Question 6.
What is the nature of cell walls of diatoms?
Cell walls are made up of silica with two overlapping shells fit together like a soapbox.

Question 7.
Find out what do the terms algal blooms and red tides signify?

• Algal bloom: Excessive growth of blue-green algae causes pollution of water bodies with characteristic odour.
• Red tide: Dinoflagellates like gonyaulax are red in colour which imparts red colour to seawater.

Question 8.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify.
1. Algal bloom’: When colour of water changes due to profuse growth of coloured phytoplanktons, it is called algal bloom.

2. ‘Red tides: Redness of the red sea is due to the luxuriant growth of Trichodesmium erythrium, a member of cyanobacteria (blue-green algae)’

Question 9.
How are viroids different from virus?
Viroids are free RNA without protein coat. Viruses have protein coat which encloses either RNA or DNA.

Question 10.
Justify the physiological relationship between the algal and fungal component of lichen.
The fungus holds water, provides protection and ideal housing to the alga. The alga supplies carbohydrate food for the fungus. If the alga is capable of fixing nitrogen, it supplies fixed nitrogen to fungus. This association is called symbiosis.

Question 11.
Bacteria reproduce by various methods. Mention the type of reproduction given in the diagram. What are the other methods of reproduction occur in bacteria?

Binary fission
The other methods are sporulation and sexual reproduction.

Question 12.
Biological classification is essential. Comment.
The animals and plants vary greatly in their form, structure and mode of life. To find out an organism of known characters from the vast number of organism is simply impossible. So classification is important to divide into groups and subgroups.

Question 13.
Match the following:

 a. Produces a plant disease p. Saccharomyces cere visae b. is edible- light blight of potato. q. Phytophthora infestans c. is a source of antibiotic r. Agaricus campestris d. is used in the manufacture of ethanol s. Penicillium notatum

• a – Phytophthora infestans – light blight of potato.
• b- Agaricus campestris
• c – Penicillium notatum
• d – Saccharomyces cere visae

Question 14.
Plants are autotrophs. Can you think of some plants that are heterotrophs?
Generally all plants are autotrophs but plants like loranthus and cuscuta absorbs water & nutrients from other plants so they are called as heterotrophs.

Question 15.
What are the characteristic features of Euglenoides?
They have protein sheath is called pellicle instead of cell wall. They have two flagella – One long and other short. They are photosynthetic in the presence of light and behave as heterotrophs in the absence of sunlight.

Question 16.
Give 4 difference between Ascomycetes and Basidiomycetes:

 Ascomycetes Basidiomycetes 1. Mycelium consists of branched multicellular septate hyphae. 1. Mycelium may be primary, secondary (or) tertiary 2. The fruiting bodies are ascocarps 2. Fruiting bodies are basidiocarps. 3. Sexual reproduction leads to the formation of ascus 3. Formation of basidia formation of ascus.

Question 17.
Observe the cyanobacteria given below and answer the following.

1. Name the cyanobacteria, and the kingdom it belongs.
2. Label’s ‘P’ and mention its functions.

1. Nostoc-kingdom-Monera
2. Heterocyst – To fix nitrogen from the atmosphere.

Question 18.
What do the terms phycobiont and mycobiont signify?
Algal component of lichen is called phycobiont. It prepares food for fungus. Fungal partner is called mycobiont. It provides shelter and absorbs mineral nutrients for algae.

Question 19.
Prepare a comparative account of different classes of kingdom fungi by considering following statements.

1. Mode of nutrition
2. Mode of reproduction

Question 20.
The two-kingdom classification is introduced by Linnaeus. Why is the two kingdom classification inadequate?
There was no place of viruses and bacteriophages which can neither be considered as prokaryotes not eukaryotes.

In this classification, eukaryotes were put together with prokaryotes and non-photosynthetic fungi along with photosynthetic plants.

Question 21.
How is the five-kingdom classification advantageous over the two kingdom classification?
In this classification main criteria used by R H Whittaker include cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships. These characters were not considered in two kingdom classification.

Question 22.
Are chemosynthetic bacteria-autotrophic or heterotrophic?
Autotrophic, because they get energy from the oxidation of inorganic compounds. So the released energy is stored in the ATP molecules.

Question 23.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Cyanobacteria and other photosynthetic bacteria have thylakoids suspended freely in the cytoplasm (i.e., they are not enclosed in membrane), and they have bacteriochlorophyll

Question 24.
With respect to fungal sexual cycle, choose the correct sequence of events.

1. Karyogamy, Plasmogamy and Meiosis
2. Meiosis, Plasmogamy and Karyogamy
3. Plasmogamy, Karyogamy and Meiosis
4. Meiosis, Karyogamy and Plasmogamy

Question 25.
What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement?
It is due to the presence of special nitrogen-fixing cell called heterocyst present between the filaments. So it helps to increase N2 content in the soil.

Question 26.
Methane is the main component of biogas and it is produced by bacteria.

1. Name the bacteria.
2. Identify the group in which it belongs.

1. Methanogens
2. Archaebacteria

Question 27.
Based on the relationship, fill in the blanks.

1. Sac fungi: Ascomycetes
Imperfect fungi: …………
2. Thermoacidophiles: Archaebacteria in hot springs
…………………: Archaebacteria in Salty areas

1. Deuteromycetes
2. Halophiles

Question 28.
Name the kingdom in which euglena belongs. Give the special type nutrition.
Kingdom Protista, Mixotrophic nutrition (ie, both autotrophic and heterotrophic).

Question. 29
Some bacteria are different from others and they have the ability to survive in extreme conditions. Name it.
Archaebacteria (halophiles, thermoacidophiles and methanogens).

Question 30.
Mycoplasma are included in five kingdom classification but not viruses. Why?
Because mycoplasmas are living cellular organisms but viruses are acellular particles.

Question 31.
In which division of protista chief producers in ocean belongs. Give the cell wall composition of such organisms.
Chrysophytes, silicified cell wall.

Question 32.
Nitrobactor and nitrosomonas are free living nitrogen fixers and chemoautotrophs but their functions are different. Do you agree. Give reasons.
Yes. Nitrobactor converts nitrite into nitrate while nitrosomonas converts ammonia into nitrites.

Question 33.
Name the classes fungi shows exogenous and endogenous spore production. In which fruiting bodies they are found.

• Exogenous-Basidiomycetes. Its fruiting body is basidiocarp.
• Endogenous-Ascomycetes. Its fruiting body is ascocarp.

Question 34.
Rust and smut diseases are caused by the members of basidiomycetes. Name it.
Smut disease- Ustilago, Rust disease-Puccinia.

Question 35.
What are the events takes place in slime mould during favourable and unfavourable season?
During favourable condition the cells aggregate and form plasmodium while in unfavourable season plasmodium differentiates and produce fruiting bodies that bear spores at tip.

Question. 36
Suppose you accidentally find an old preserved permanent slide without a label. In your effort to identify it, you place the slide under microscope and observe the following features

1. Unicellular
2. Well defined nucleus
3. Biflagellate-one flagellum lying longitudinally and the other transversely.

What would you identify it as? Can you name the kingdom it belongs to?
Dinoflagellates, Kingdom protista

Question. 37
What would you identify it as? Can you name the kingdom it belongs to?
Dinoflagellates, Kingdom protista

Question. 38
Why lichens are called as dual organisms?
Lichens are said to be dual organisms because they show a symbiotic association between a fungus and alga.

Question 39.
Name the asexual, reproductive structure of penicillium and yeast.
Can penicilium reproduce through sexual method? If the yes or no Give reason.
Conidia – penicilium, buds – yeast
Yes, It is done by the production of ascospores in asci of Ascocarp.

Question 40.
Organise a discussion in your class on the topic virus. Are viruses living or non-living?
They are filterable and may becrystalised. They are inert outside their specific host and able to reproduce inside the living host cell, so they are considered as living. They use the protein synthesising machinery of the host.
Eg. AIDS virus, mumps virus etc.

Question 41.
How are viroids different from viruses?

 Virus Viroid 1. Their size is smaller than bacteria 1. Their size is smaller than viruses 2. Protein coat is Present 2. Protein coat is absent 3. Genetic material may be DNA or RNA 3. Genetic material is only RNA 4. They cause AIDS, smallpox etc. 4. They cause potato spindle tuber diseases

Question 42.
Some bacteria are specialised and live in extreme habitat.

1. Name the types of bacteria are specified in the above statement.
2. Which is the part of bacteria modified to live in that condition?

1. Types of bacteria

• Methanogens
• Halophiles
• Thermo acidophiles

2. Ceil wall structure

Question 43.
The two nuclei per cell can be seen in fungal cell but it later fuse in some members.

1. Name such type of fungal hyphae or mycelium.
2. Identify the classes of fungi.

1. Dikaryotic mycelium
2. Ascomycetes, Basidiomycetes

Question 44.
Classify the pathogenic microorganisms and disease in different groups based on the following symptoms mosaic disease, citrus canker .potato spindle tuber disease, sleeping sickness, malaria.
mosaic disease-virus, citrus canker-Bacteria, potato spindle tuber disease -viroids, sleeping sickness- Trypanosoma, malaria-Plasmodium vivax.

### Plus One Botany Biological Classification Three Mark Questions and Answers

Question 1.
Describe briefly the four major groups of protozoa.
Protozoans are heterotrophs act either as predators
or parasites. They are of four groups

1. Amoeboid protozoans: They capture their prey by using pseudopodia. They live in freshwater. Some are parasites eg: entamoeba.
2. Flagellated protozoans: They are free-living or parasites. They cause diseases, eg: Trypanosoma-sleeping sickness.
3. ciliated protozoans: They possess cilia in their body surface for locomotion. They have gullet for food intake. Eg: Paramecium
4. Sporozoans: They are spore-producing organism that causes diseases eg: plasmodium causing malaria.

Question 2.
Different types of fungi are given
1. Classify them into their specific classes.

 Groups Fungi Phycomycetes Trichoderma Ascomycetes Neurospora Basidiomycetes Albugo Deuteromycetes Mucor Agaricus Ustilago Alternaria Claviceps

2. Write the distinguishing characters of ascomycetes and basidiomycetes
3. The characteristic features of members of monera are given below.

Organisms lack cell wall, live without oxygen, smallest living cell and causes diseases. Identify the organism by analysing the above characters.
1. specific classes.

• Phycomycetes – Mucor, Albugo
• Ascomycetes – Neurospora, Claviceps
• Basidiomycetes-Agaricus, Ustilago
• Deuteromycetes – Altemaria, Trichoderma.

2. In ascomycetes, Asexual mode of reproduction is prominent by conidiospores. In Basidiomycetes asexual spores are not found. Sexual spores are arranged in ascus with Ascospores in ascomycetes, whereas sexual spores are arranged in basidium in basidiomycetes.

3. Mycoplasma

Question 3.
Give a brief account of virus with respect to their structure and nature of genetic material. Also, name four common viral diseases?
Viruses are organism having inert crystalline structure outside the living cell. They have genetic material RNA or DNA.which is either single-stranded/double-stranded. It is enclosed by protein capsid with subunits called capsomeres.

The viral genetic material takes control over the host cell mechanism during infection. Some common viral diseases are mumps, herpes, smallpox and influenza in animals and mosaic disease in plants.

Question. 4
In which groups are the following found- Sporangiophore, Conidia, zygospore and ascospore.

• Conidia are spores found in ascomycetes.
• These are haploid asexual spores produced in chains exogenously.
• Zygospores are the diploid resting spores found in mucor.
• Ascospores are haploid sexual spores found in sac-like structure (ascus).
• Sporangiophore is an aerial branch produced by hyphae in mucor that bear sporangia.

### Plus One Botany Biological Classification NCERT Questions and Answers

Question 1.
What is the nature of cell walls in diatoms?
The cell walls in diatoms are embedded with silica, which makes them indestructible. They form two thin overlapping shells which fit together as in a soapbox. Thus diatoms have left behind large amounts of cell wall deposits in their habitat.

Question 2.
How are viroids different from viruses?
Viroids are free RNAs without the protein coat, while virus have a protein coat encapsulating the RNA.

Question 3.
Describe briefly the four major groups of Protozoa.
Four major groups of Protozoa are as given below:
1. Amoeboid Protozoa:
They are found in freshwater, seawater or moist soil. They have pseudopodia, like amoeba, hence the name ameoboid protozoa.

2. Flagellated Protozoans:
They have flagella helps in locomotion. Some are parasite. Eg. Trypanosoma causes sleeping sickness.

3. Ciliated Protozoa:
They have thousands of cilia present all over the body. The cilia helps in locomotion and steering of food into the gullet.

4. Sporozoans:
Many protozoans have an infectious spore-like stage in the life cycle. The spore-like stage helps them get transferred from one host to another host.

Question. 4
Plants are autotrophic. Can you think of some plants that are partially heterotrophic?
Certain insectivorous plants, like bladderwort and venus fly trap, are partially heterotrophic.

Question. 5
What do the terms phycobiont and mycobiont signify?
Lichens are good examples of symbiotic life of algae and fungi. Phycobiont is the name of the part composed of algae and Mycobiont is the name of the part composed of fungi. Fungi provide minerals and support to the alage, while algae provide nutrition to the fungi.

Question 6.
What are the characteristic features of Euglenoids?
Features of Euglenoids.

• No cell wall.
• Protein-rich layer, called pellicle, which makes flexible body.
• Two flagella of different lengths.
• Autotrophs in sunlight, heterotrophs in the absence of sunlight. Example: Euglena.

Question 7.
Give a brief account of viruses with respect to their structure and nature of genetic material. Also name four common viral diseases.
Virus Structure:
Outside a host cell, virus is a crystalline structure, composed of protein. Inside the crystal, there is genetic material, which can be either RNA or DNA. No virus has both RNA and DNA. Viruses, infecting plants, have single-stranded RNA. Viruses, infecting animals, have either single or double-stranded RNA or double-stranded DNA.

The protein coat is called capsid. Capsid is made of smaller subunits, called capsomeres, it protects nucleic acid. Diseases caused by Virus; AIDS, Mumps, Influenza, Herpes.

Question 8.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify.
Dinoflagellates can be of different colours depending on the type of pigment present. The red dinoflagellate sometimes multiplies at a very rapid rate. This is called as algal bloom. This gives a red appearance to the part of affected sea. This is also known as ‘red tide’. Toxins released by them can kill other marine species.

### Plus One Botany Biological Classification Multiple Choice Questions and Answers

Question 1.
The life form used as indicators of pollution
(A) Lichens
(B) Protozoa
(C) Algae
(D) Agaricus
(A) Lichens

Question 2.
Kingdom monera comprises
(A) Amoeba, Bacteria,Trypanosoma
(B) Bacteria, Viruses,Virolds
(C) Archaebacteria, Eubacteria, Mycoplasma
(D) Mycoplasma, Viruses, Bacteria
(C) Archaebacteria, Eubacteria, Mycoplasma

Question 3.
Who discovered two-kingdom classification
(A) Ivanowsky
(B) Stanley
(C) leuwernhoek
(D) Linnaeus

Question 4.
Asexual reproduction takes place by Zoospores in
(A) Pythium
(B) Agaricus
(C) Rhizopus
(D) Ustilago
(D) Ustilago

Question 5.
Identify the organism used as bioweapon
(A) Bacillus thuringiensis
(B) Bacillus anthracis
(C) Pseudomonas citri
(D) Rhizobium tumefacient
(B) Bacillus anthracis

Question 6.
Reserve food in the form of glycogen and cell wall made up of chitin are characteristic of
(A) Protists
(B) bacteria
(C) Fungi
(D) protozoa
(C) Fungi

Question 7.
The fruiting body of club fungi is
(A) Basidium
(B) Ascus
(C) Ascocarp
(D) Basidiocarp
(D) Basidiocarp

Question 8.
RNA without protein coat are found in
(A) bacteria
(B) protozoa
(C) viruses
(D) viroides
(D) viroides

Question 9.
The phycobiont and mycobiont are found in
(A) bacteria
(B) lichen
(C) viroides
(D) fungi
(B) lichen

Question 10.
The organism which causing sleeping sickness belongs to
(A) Protists
(B) bacteria
(C) Fungi
(D) viruses
(A) Protists

Question 11.
In which of the following groups are neurospora and Penicillium included?
(A) Phycomycetes
(B) Basidiomycetes
(C) Zygomycetes
(D) Ascomycetes
(D) Ascomycetes

Question 12.
Occurrence of Dikaryon phase is characteristic feature of
(A) Bacteria
(B) Fungus
(C) Slime moulds
(D) Cyanobacteria
(B) Fungus

Question13.
Methane producers are belongs to
(A) Archaebacteria
(B) Cyanobacteria
(C) Eubactenia
(D) Actinomycetes
(A) Archaebacteria

Question 14.
Heterocyst are found in
(A) Nitrosomonas
(B) cyanobacteria
(C) fungi
(D) protozoa
(B) cyanobacteria

Question 15.
Colletotrichum falcatum is a fungus causing the following disease
(A) Smut of wheat
(B) Wilt disease of cotton
(C) Red rot of sugar cane
(D) Late blight of potato
(C) Red rot of sugar cane

## Plus One Maths Chapter Wise Questions and Answers Chapter 2 Relations and Functions

Students can Download Chapter 2 Relations and Functions Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 2 Relations and Functions

### Plus One Maths Relations and Functions Three Mark Questions and Answers

Plus One Maths Relations And Functions Previous Questions And Answers Question 1.
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by R = {(a, b): a, b ∈ A, b is exactly divisible by a}

1. Write R in roster form. (1)
2. Find the domain of R. (1)
3. Find the range of R. (1)

1. R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 4), (2, 2), (4,4), (6,6), (3,3), (3,6)}
2. Domain of R = {1, 2, 3, 4, 6}
3. Range of R = {1, 2, 3, 4, 6}

Plus One Maths Chapter Wise Questions And Answers Pdf Question 2.
Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9),(5, 10)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Relations And Functions Class 11 Important Questions Pdf Question 3.
A function f is defined as f(x) = 2x – 5, Write down the values of f(0), f(7), f(-3).
Given; f(x) = 2x – 5
f(0) = -5;
f(7) = 2(7) – 5 = 14 – 5 = 9
f(-3) = 2(-3) – 5 = -6 – 5 = -11

Hsslive Maths Textbook Answers Plus One Question 4.
Find the range of the following functions.

1. f(x) = 2 – 3x, x ∈ R, x>0 (1)
2. f(x) = x2 + 2, x is a real number. (1)
3. f(x) = x, x is a real number. (1)

1. Given; f(x) = 2 – 3x is a first degree polynomial function, therefore the range is R.
2. Given; f(x) = x2 + 2, The range of x2 is [0, ∞) , then the range of f(x) = x2 + 2 is [2, ∞)
3. Given; f(x) = x is the identity function, therefore the range is R.

### Plus One Maths Relations and Functions Four Mark Questions and Answers

Plus One Maths Chapter Wise Questions And Answers Question 1.
Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

1. A × (B ∩ C) = (A × B) ∩ (A × C) (2)
2. A × C is a subset of B × D (2)

1. A × (B ∩ C) ={1, 2} × Φ = Φ
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
(A × B) ∩ (A × C) = Φ
Hence; A × (B ∩ C) = (A × B) ∩ (A × C)

2. A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {1, 2, 3, 4} × {5, 6, 7, 8}
= {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3,6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Hence A × C is a subset of B × D.

Relations And Functions Class 11 Important Questions Question 2.
The arrow diagram given below shows a relation R from P to Q. Write the relation in roster form, set-builder form. Find its domain and range.

R – {(9, -3), (9, 3), (4, -2), (4, 2), (25, -5), (25, 5)}
R = {{x, y) : y2 = x}
Domain of R = {9, 4, 25}
Range of R = {5, 3, 2, -2, -3, -5}

Question 3.
Find the domain of the following.

1. f(x) = $$\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$$ (2)
2. f(x) = $$\frac{x^{2}+3 x+5}{x^{2}-5 x+4}$$ (2)

1. Given; f(x) = $$\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$$
The function is not defined at points where the denominator becomes zero.
x2 – 8x +12 = 0 ⇒ (x – 6)(x – 2) = 0 ⇒ x = 2, 6
Therefore domain of fis R – {2, 6}.

2. Given; f(x) = $$\frac{x^{2}+3 x+5}{x^{2}-5 x+4}$$
The function is not defined at points where the denominator becomes zero.
x2 – 5x + 4 = 0 ⇒ (x – 4)(x -1) = 0 ⇒ x = 1, 4
Therefore domain of f is R – {1, 4}.

Plus One Maths Questions And Answers Question 4.
Let f(x) = $$=\sqrt{x}$$ and g(x) = x be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and $$\left(\frac{f}{g}\right)(x)$$.
(f + g)(x) = f(x) + g(x) = $$=\sqrt{x}$$ + x
(f – g)(x) = f(x) – g(x) = $$=\sqrt{x}$$ – x
(fg)(x) = f(x) × g(x) = $$=\sqrt{x}$$ × x = $$x^{\frac{3}{2}}$$

Plus One Maths Relations And Functions Question 5.
Let f(x) = x2 and g(x) = 2x + 1 be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and $$\left(\frac{f}{g}\right)(x)$$.
(f + g)(x) = f(x) + g(x) = x2 + 2x + 1
(f – g)(x) = f(x) – g(x) = x2 – 2x – 1
f(fg)(x) = f(x) × g(x)
= x2(2x +1) = 2x3 + x2

Relations And Functions Questions And Answers Pdf Question 6.
A = {1, 2}, B = {3, 4}

1. Write A × B
2. Write relation from A to B in roster form. (1)
3. Represent all possible functions from A to B (Arrow diagram may be used) (2)

1. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

2. Any Subset of A × B (say R={(1, 3),(2, 4)})

3.

### Plus One Maths Relations and Functions Six Mark Questions and Answers

Relations And Functions Class 11 Important Questions With Solutions Question 1.
Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find

1. A × (B ∩ C) (1)
2. (A × B) ∩ (A × C) (2)
3. A × (B ∪ C) (1)
4. (A × B) ∪ (A × C) (2)

1. A × (B ∩ C) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}

2. (A × B) ∩ (A × C)
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ∩ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5) , (2, 6), (3, 4), (3, 5), (3, 6)}
= {(1, 4), (2, 4), (3, 4)}

3. A × (B ∪ C) = {1, 2, 3} × {3, 4, 5, 6}
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

4. (A × B) ∪ (A × C)
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ∪ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

Question 2.
Find the domain and range of the following

i) Given; f(x) = -|x|
D(f) = R, R(f) = (-∞, 0]

ii) Given; f(x) = $$\sqrt{9-x^{2}}$$
x can take values where 9 – x2 > 0
⇒ x2 ≤ 9 ⇒ -3 ≤ x ≤ 3 ⇒ x ∈ [-3, 3]
Therefore domain of f is [-3, 3]
Put $$\sqrt{9-x^{2}}$$ = y, where y ≥ 0
⇒ 9 – x2 = y2⇒ x2 = 9 – y2
⇒ x = $$\sqrt{9-x^{2}}$$
⇒ 9 – y2 ≥ 0 ⇒ y2 ≤ 9 ⇒ -3 ≤ y ≤ 3
Therefore range of fis [0, 3].

iii) Given; f(x) = |x – 1|
Domain of f is R
The range of |x| is [0, ∞) , then the range of
f(x) = |x -1| is [0, ∞)

iv) Given; f(x) = $$\sqrt{x-1}$$
x can take values where x – 1 ≥ 0
⇒ x ≥ 1 ⇒ x ∈ [1, ∞]
Therefore domain of fis [1, ∞]
The range of $$\sqrt{x}$$ is [0, ∞), then the range of
f(x) = $$\sqrt{x-1}$$ is [0, ∞).

### Plus One Maths Relations and Functions Practice Problems Questions and Answers

Question 1.
If (x + 1, y – 2) = (3, 1), find the values of x and y.
(x + 1, y – 2) = (3, 1) ⇒ x + 1 = 3, y – 2 = 1 ⇒ x = 2, y = 3.

Question 2.
If $$\left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)$$, find the values of x and y.

Question 3.
If G = {7, 8}; H = {2, 4, 5}, find G × H and H × G.

• G × H ={(7, 2), (7, 4), (7, 5), (8, 2), (8, 4), (8, 5)}
• H × G ={(2, 7), (2, 8), (4, 7), (4, 8), (5, 7), (5, 8)}

Question 4.
if A = {-1, 1} find A × A × A
A × A ={-1, 1} × {-1, 1}
= {(-1, -1), (-1, 1), (1, -1), (1, -1)}
A × A × A
= {(-1, -1), (-1, -1), (1,-1), (1, -1)} × {-1, 1}
= {(-1, -1, -1), (-1, 1, -1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1), (-1, 1, 1), (1, -1, 1), (-1, 1, 1)}.

Question 5.
Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.
2, 3, 5, 7 are the prime number less than 10.
R = {(2, 8),(3, 27),(5, 125),(7, 343)}

Question 6.
If f(x) = x2, find $$\frac{f(1.1)-f(1)}{(1.1-1)}$$?

Question 7.
Let $$\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right), x \in R\right\}$$ be a real function from R to R. Determine the domain and range of f.
Domain of f is R.
Let $$\frac{x^{2}}{1+x^{2}}$$ = y ⇒ x2 = y(1 + x2)
⇒ x2 = y + yx2 ⇒ x2 – yx2 = y
⇒ x2(1 – y) = y

⇒ y ≥ 0, 1 – y > 0
⇒ y ≥ 0, y < 1 ⇒ 0 ≤ y ≤ 1
Therefore range of f is [0, 1).

Question 8.
Graph the following real functions. (each carries 2 scores)

1. f(x) = |x – 2|
2. f(x) = x2
3. f(x) = x3
4. f(x) = $$\frac{1}{x}$$
5. f(x) = (x – 1)2
6. f(x) = 3x2 – 1
7. f(x) = |x| – 2

1. f(x) = |x – 2| = \left\{\begin{aligned}x-2, & x \geq 2 \\-x+2, & x<2 \end{aligned}\right.

2. f(x) = x2

3. f(x) = x3

4. f(x) = $$\frac{1}{x}$$

5. f(x) = (x – 1)2

6. f(x) = 3x2 – 1

7. f(x) = |x| – 2

Question 9.
Consider the relation, R = {(x, 2x – 1)/x ∈ A) where A = (2, -1, 3}

1. Write R in roster form. (1)
2. Write the range of R. (1)

1. x = 2 ⇒ 2x – 1 = 2(2) – 1 = 3
x = -1 ⇒ 2x – 1 = 2(-1) – 1 = -3
x = 3 ⇒ 2x – 1 = 2(3) – 1 = 5
R = {(2, 3), (-1, -3), (3, 5)}

2. Range of R = {3, -3, 5}

Question 10.
Let A = {1, 2, 3, 4, 6} and R be a relation on A defined by R = {(a, b): a, b ∈ A, b is exactly divisible by a}

1. Write R in the roster form. (1)
2. Find the domain and range of R. (1)

1. R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (4, 4), (5, 5), (6, 6)}
2. Domain = {1, 2, 3, 4, 6}; Range = {1, 2, 3, 4, 6}

Question 11.
Consider the real function

1. $$f(x)=\frac{x^{2}+2 x+3}{x^{2}-8 x+12}$$
2. Find the value of x if /(x) = 1
3. Find the domain of f.

1. Given; f(x) = 1 ⇒ 1 = $$\frac{x^{2}+2 x+3}{x^{2}-8 x+12}$$
⇒ x2 – 8x + 12 = x2 + 2x + 3
⇒ 10x = 9 ⇒ x = $$\frac{9}{10}$$

2. Find the value for which denominator is zero.
⇒ x2 – 8x + 12 = 0
⇒ (x – 6)(x – 2) = 0 ⇒ x = 6, 2
Therefore domain of f is R – {2, 6).

Question 12.
If f(x) = x3 + 5x and g(x) = 2x +1, find (f + g)(2) and {fg)(1).
(f + g)(2) = f(2) + g(2) = (2)3 + 5(2) + 2(2) + 1
= 8 + 10 + 4 + 1 = 23
(fg)(1) = f(1)g(1) = (1 + 5)(2 + 1) = 6 × 3 = 18.

Question 13.
Let A = {1, 2, 3, 4, 5} and R be a relation on A defined by R = {(a, b):b = a2}

1. Write R in the roster form.
2. Find the range of R.