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Kerala State Board New Syllabus Plus Two Maths Chapter Wise Previous Questions and Answers Chapter 1 Relations and Functions.

Kerala Plus Two Maths Chapter Wise Previous Questions Chapter 1 Relations and Functions

Plus Two Maths Relations and Functions 3 Marks Important Questions

Question 1.
Consider the set A = {1, 2, 3, 4, 5}, and B = {1, 4, 9, 16, 25} and a function ƒ : A → B defined by f(1) = 1, f(2) = 4, f(3) = 9, f(4) = 16 and f(5) = 25
(i) Show that f is one-to-one
(ii) Show that f is onto.
(iii) Does ƒ^-1 exists? Explain (May – 2013)
Answer:
(i) ƒ = {(1,1), (2,4), (3,9), (4,16), (5,25)}
Every element in A is mapped to different elements in B. Therefore one-to-one.
(ii) R (ƒ) = {1, 4, 9, 16, 25} = B. Therefore onto.
(iii) Since f is one-to-one and onto function, ƒ^-1 exists.
ƒ^-1 = {(1,1), (4,2), (9,3), (16,4), (25,5)}

Question 2.
a) When a relation R on a set A is said to be reflexive
b) Show that ƒ : [-1, 1] → R given by \(f(x)=\frac{x}{x+2}\) is one-one (May – 2015)
Answer:

Question 3.
a) The function ƒ :N → N given by ƒ(x) = 2x
i) one-one and onto
ii) one-one and not onto
iii) not one-one and not onto
iv) onto but not one-one
b) Find goƒ(x), if ƒ(x) = 8x³ and g(x) = x^1/2
c) Let * be an operation such that a*b= LCM of a and b defined on the set A = {1,2,3,4,5}. Is * a binary operation? Justify your answer. (March-2016)
Answer:
a) ii) one-one and not onto
b) Answered in previous years questions No. 2(ii) (6 Mark question)
c) LCM of 2 and 3 is 6 ∉ A, therefore not a binary operation.

Plus Two Maths Relations and Functions 4 Marks Important Questions

Question 1.
(i) ƒ : {1,2,3,4} → {5} defined by ƒ = {(1,5), (2,5), (3,5), (4,5)} Does the function is invertible?
(ii)
(iii) Let A = Nx N, N-set of natural numbers and * 1be a binary operation on A defined by (a,b) * (c,d) = (ac—bd,ud +bc). Show that* is commutative on A. (March -2011)
Answer:
(i) Inverse does not exists because fis not one-one.

Hence cummutative.

Question 2.
Let N be the set of Natural numbers. Consider the function ƒ: N → N defined by ƒ(x) = x + l, x ∈ N
(i) Prove that f is not onto
(ii) \(If g(x)=\left\{\begin{array}{ll}x-1, & x>1 \\ 1, & x=1\end{array}\right. then find g o f\)
(iii) Check whether goƒ is an onto function. (May 2011)
Answer:

(iii) Since f is not onto goƒ is also not onto.

Question 3.
(i) Give a relation on a set A = {1,2,3,4} which is reflexive , symmetric and not transitive.
(ii) Show that ƒ : [-1,1] → R given by \(f(x)=\frac{x}{x+2}\) is one-one.
(iii) Let ‘*’ be a binary operation on Q⁺ defined by a*b = \(a * b=\frac{a b}{6}\) ’.Find the inverse of 9 with respect to ’ * ’. (March -2012)
Answer:
(i) Given A = {1,2,3,4}
R = {(1,1)(2,2),(3,3),(4,4),(1,2),(2,1),(1,3),(3,1)}

Question 4.
(i) *:R x R → Ris given by a * b = 3a2 – b
Find the value of 2 * 3. Is ‘*’ commutative? Justify your answer.
(ii) ƒ :R → R is defined by ƒ(x) = x² – 3x + 2 Find ƒoƒ (x) and ƒoƒ. (May 2012)
Answer:

Question 5.
(i) Consider ƒ : R → R given by ƒ(x) = 5x + 2
(a) Show that f is one-one.
(b) Is f invertible? Justify your answer.

(ii) Let * be a binary operation on N defined by a * b = HCF of a and b
(a) Is * commutative?
(b) Is * associative? (March-2013)
Answer:
(i) (a) Let x₁, x₂, ∈ R
ƒ(x₁) = ƒ(x₂) ⇒ 5x₁ + 2 = 5x₂ + 2
⇒ 5x₂ = 5x₂ ⇒ x₁ = x₂
Therefore fis one-one.

(b) Yes.
Let y e range of ƒ
⇒ ƒ(x) = y ⇒ 5x + 2 = y
\(\Rightarrow x=\frac{y-2}{5} \in R\)
Therefore corresponding to every y ∈ R there existsa real number \(\frac{y-2}{5}\) Therefore f is onto.
Hence bijective, so invertible.

(ii) (a) Yes.
a * b = HCF (a,b) = HCF (b,a) = b * a
Hence commutative.

(b) Yes.
a * (b * c) = a* HF(b,c) = HCF(a,b,c)
(a*b) * c =HCF(a,b) * c HCF(a,b,c)
a * (b * c) = (a * b) * c
Hence associative.

Question 6.
(a) Let f: R → R be given by ƒ (x) = \(\frac{2 x+1}{3}\) find ƒoƒ and show that f is invertible.
(b) Find the identity element of the binary operation * on N defined by a * b = ab². (May 2014)
Answer:

Therefore f is onto.
Hence f is bijective and invertible.

(b) let ‘e’ be the identity element, then

Since e is not unique, this operation has no identity element.

Question 7.
a) What is the minimum number of pairs to form a non-zero reflexive relation on a set of n elements?
b) On the set R of real numbers, S is a relation defined as S = {(x,y)/X∈R, y ∈ R, x + y = xy}. Find a ∈ R such that ‘a’ is never the first element of an ordered pair in S. Also find b ∈ R such that ‘b’ is never the second element of an ordered pair in S.
c) Consider the function \(f(x)=\frac{3 x+4}{x-2}, x \neq 2\) Find a function on a suitable domain such that goƒ(x) = x = ƒog(x). (March 2015)
Answer:

Question 8.
(i) If ƒ: R → R and g: R → R defined by ƒ(x) = x² and g(x) = x + 1, then goƒ (x) is
(a) (x + 1)²
(b) x³ + l
(c) x² + l
(d) x + l

(ii) Consider the function ƒ: N → N, given by ƒ(x) = x³. Show that the function ‘ƒ’ is injective but not surjective.
(iii) The given table shows an operation on A = {p,q}

(a) Is * a binary operation?
(b) * commutative? Give reason. (May 2016)
Answer:
(i) (C) x² + 1
(ii) ƒ : N → N , given by ƒ(x) = x³
for x,y ∈ N ⇒ ƒ(x) = ƒ(y)
x³ = y³ ⇒ x = y

There fore f is injective.
Now 2 ∈ N, but there does not exists any element x in domain N such that ƒ(x) = x³ = 2 their fore f is not surjective.

(iii) (a) Yes
(b) No, because p*q = q; q*p = p
⇒ p*q ≠ q*p

Question 9.
(i) Let R be a relation defined on A{1,2,3} by R = {(13),(3,1),(2,2)} is
(a) Reflexive
(b) Symmetric
(C) Transitive
(d) Reflexive but not transitive.
(ii) Find fog and gof if ƒ(x) = |x+1| and g(x) = 2x – 1
(iii) Let * be a binary operation defined on N x N by (a,b) * (c,d.) = (a + c, b + d)
Find the identity element for * if it exists. (March – 2017)
Answer:
(i) (b) Symmetric

(ii) ƒog(x) = |g(x) + 1| = |2x – 1 + 1| = |2x|
goƒ(x) = 2 ƒ(x) – 1 = 2 |x + 1| – 1

(iii) Let e =(e₁, e₂) be the identity element of the operation in ? N x N then, (a,b)*(e₁, e₂) = (a + e₁, b + e₂) ≠ (a,b) Since, e₁ ≠ 0, e₂ ≠ 0

Therefore identity element does not existš.

Question 10.
(i) If R = {(x,y) : x, y ∈ Z, x – y ∈ Z}, then the relation R is
(a) Reflexive but not transitive
(b) Reflexive but not symmetric
(C) Symmetric but not transitive
(d) An equivalence relation.

(ii) Let * be a binary operation on the set Q of rational numbers by a*b = 2a + b. Find 2 * (3 * 4) and (2 * 3) * 4.
(iii) Let ƒ : R → R, g : R → R be two one-one funçtions. Check whether gof is one-one or not. (May- 2017)
Answer:
(i) (d) An equivalance relation.
(ii) 2* (3 * 4) = 2 * 10 = 14
(2 * 3)* 4 = 7 * 4 = 18
(iii) ƒ : R → R, g : R → R
Let x₁, x₂, ∈ R
goƒ(x₁) = g(ƒ(x₁)) = g(ƒ(x₂)) = g(ƒ(x₂))
⇒ x₁ = x₂

Plus Two Maths Relations and Functions 6 Marks Important Questions

Question 1.
(i) (a) A function ƒ : X → Y is onto if range of ƒ = ………….
(b) Let ƒ : {1, 3, 4} {3, 4, 5} and
g: {3, 4, 5} → {6, 8, 10} be functions defined by
ƒ (1) = 3, ƒ (3) = 4, ƒ (4) = 5;
g (3) = 6, g(4) = 8, g(5) = 8 ,then (goƒ) (3) = …………..

(ii) Let Q be the set of Rational numbers and ‘*’ be the binary operation on Q defined by \(a * b=\frac{a b}{4}\) for all a,b in Q
(a) What is the identity element of ‘ * ’on Q?
(b) Find the inverse element of * ’ on Q.
(c) Show that a * (b * c) = (a * b) * c, ∀a,b,c ∈ Q.
Answer:

Question 2.
(i) Let R be the relation on the set N of natural numbers given by
R = {(a,b): a – b > 2, b>3}
Choose the correct answer
(a) (4, 1) ∈ R
(b) (5, 8) ∈ R
(c) (8, 7) ∈ R
(d) (10, 6) ∈ R

(ii) If ƒ(x) = 8x³ and g(x) = x^1/3, findg(ƒ(x)) and ƒ(g(x))
(iii) Let * be a binary operation on the set Q of rational numbers defined by a*b = \(\frac{a}{b}\). Check whether * is commutative and associative? (March – 2014, May – 2015, March – 2016)
Answre:

Question 3.
Let \(f(x)=\frac{x-1}{x-3}, x \neq 3\) and \(g(x)=\frac{x-3}{x-1}, x \neq 1\) be two functions defined on R.

(i) Find ƒog(x), x ≠ 0
(ii) Find ƒ^-1 (x) and g^-1 (x), x ≠ 1
(iii) Find (goƒ)^-1 (x) (May-2010)
Answer:

Kerala State Board New Syllabus Plus Two Maths Previous Year Question Papers and Answers.

Kerala Plus Two Maths Previous Year Question Paper Say 2018 with Answers

Time : 2 1/2 Hours
Cool off time : 15 Minutes
Maximum : 80 Score

General Instructions to Candidates :

• There is a ‘Cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘Cool off time’ to get familiar with questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Question 1 to 7 carry 3 scores each. Answer any 7 questions.

Question 1.
a) Construct a 2 × 2 matrix whose elements are given a_ij = 2i + j
b) Find A².
Answer:

Question 2.
a) If \(\int \frac{f(x)}{x^{2}+1} d x\) = log | x² + 1 | + C, then f(x) = …………
b) Find ∫ xe^x dx
Answer:
a) f(x) = 2x
b) ∫ xe^x dx = x∫e^x dx – ∫1 × e^x dx
= xe^x – e^x + c = e^x (x – 1) + c

Question 3.
Form the differential equation of the family of all circles touching the y-axis at origin.
Answer:
Equation of the family of circle which touches the y-axis at origin is of the form.
(x – a)² + y² = a² ……… (1)

Question 4.
Consider the relation in the set N of Natural numbers defined as R = { (a,b): ab is a factor of 6}. Determine whether the relation is reflexive, symmetric or transitive.
Answer:
(2, 2) ∉ R, Not reflexive
(x, y) ∈ R ⇒ (y, x) ∈ R ⇒ xy = yx, Symmetric
(3, 2) ∈ R, (2, 3) ∈ R ⇒ (3, 3) ∉ R,
Since 3 × 3 = 9, not transitive.

Question 5.
Find the area bounded by the curve y= cos x and x axis between x = 0 and x = π.
Answer:
Area = 2 \(\int_{0}^{\pi / 2}\) cos xdx
= 2 \([\sin x]_{0}^{\pi / 2}\) = 2[1 + 0] = 2

Question 6.

A rectangular plot is to be fenced using a rope of length 20 meters with one of its sides is a wall as shown in the figure. Find the maximum area of such rectangle.

Length = 20 – 2x, breadth = x
A = x(20 – 2x) = 20x – 2x²
A'(x) = 20 – 4x ⇒ A'(x) = 0 ⇒ x = 5
A”(x) = -4 < 0
Hence A is maximum at x = 5
Maximum area = 5 × 10 = 50

Question 7.
A manufacture produces nuts and bolts. The time required to produce one packet of nuts and one packet of bolts on machines A and B is given in the following table

He earns a profit of Rs. 25 per packet of nuts and Rs. 12 per packet of bolts. He operates his machines for almost 15 hours a day. Formulate a linear programming problem to maximise his profit.
Answer:
Maximise: Z = 25x + 12y
Subject to
2x + 3y ≤ l5; 3x + y ≤ 5; x, y ≥ 0

Questions 8 to 17 carry 4 scores each. Answer any 8.

Question 8.
Consider the curve y = x³ + 8x + 3
a) Find the point on the curve at which the slope of the tangent is 20.
b) Does there exist a tangent to the curve with negative slope? Justify your answer.
Answer:
a) \(\frac{d y}{d x}\) = 3x² + 8
Slope is given as 20
20 = 3x² + 8 ⇒ x = ±2
Therefore points (2, 27), (-2, -21)
b) No. 3x² + 8 ≥ 0 (Always positive for any value of x.)

Question 9.
a) Which of the following functions is not continuous at zero?
i) f(x) = sin x

b) Find the Values of a and b such that the function defined by
f(x) = \(\left\{\begin{array}{cc}
10, & x \leq 3 \\
a x+b, & 3<x<4 \\
20, & x \geq 4
\end{array}\right.\)
[Here f(x) is oscillating between -1 and 1 as x approaches to 0. In other cases the limit value and function value are same. So continuous.]
b) \(\lim _{x \rightarrow 3^{+}}\)f(x) = 3a + b ⇒ 3a + b = 10
\(\lim _{x \rightarrow 4^{-}}\) f(x) = 4a + b ⇒ 4a + b = 20
Solving both equations we get a = 10, b = -20

Question 10.
Consider the plane 2x – 3y + z = 5
a) Find the equation of the plane passing through the point (1, 1, 3) and parallel to the above plane.
b) Find the distance between above planes
Answer:
a) Equation of a plane parallel to the
2x – 3y + z = 5 is of the form 2x – 3y + z = k .
Since it passes through the point (1, 1, 3),
we have 2 – 3 + 3 = k ⇒ k = 2
⇒ 2x – 3y + z = 2

b) Distance between the planes
= \(\left|\frac{5-2}{\sqrt{4+9+1}}\right|=\frac{3}{\sqrt{14}}\)

Question 11.
Consider the vectors
\(\vec{a}\) = 2i + j + 3k; \(\vec{b}\) = i + 4j – k
a) Find the projection of \(\vec{a}\) on \(\vec{b}\)
b) If \(\vec{a}\) is perpendicular to a vector \(\vec{c}\) then projection of \(\vec{a}\) on \(\vec{c}\)
c) Write a vector \(\vec{d}\) such that the projection of \(\vec{a}\) on \(\vec{d}\) = |\(\vec{a}\)|
Answer:

b) Projection will be zero.
c) Projection of \(\vec{a}\) on \(\vec{d}\) = |\(\vec{a}\)|, means the angle between \(\vec{a}\) and \(\vec{d}\) is zero. Hence both are parallel. So any vector parallel to \(\vec{a}\) is \(\vec{d}\).

Question 12.
a)

In the figure ABCD is a Parallelogram. If
\(\overrightarrow{A B}\) = 3i – j + 2k; \(\overrightarrow{A D}\) = i + j + 2k, find \(\overrightarrow{A C}\) and \(\overrightarrow{D B}\)
b) If \(\vec{a}\) and \(\vec{b}\) are adjacent sides of any parallelogram \(\vec{c}\) and \(\vec{d}\) are diagonals,
then show that |\(\vec{c}\) × \(\vec{d}\)| = 2|\(\vec{a}\) × \(\vec{b}\)|
Answer:
a) \(\overrightarrow{A C}\) = \(\overrightarrow{A B}\) + \(\overrightarrow{A D}\) = 4i + 4j
\(\overrightarrow{B D}\) = \(\overrightarrow{A B}\) – \(\overrightarrow{A D}\) = 2i – 2j

b) Let \(\vec{c}\) = \(\vec{a}\) + \(\vec{b}\) and d = \(\vec{a}\) – \(\vec{b}\)
\(\vec{c}\) × \(\vec{d}\) = (\(\vec{a}\) + \(\vec{b}\)) × (\(\vec{a}\) – \(\vec{b}\))
= \(\vec{a}\) × \(\vec{a}\) – \(\vec{a}\) × \(\vec{b}\) + \(\vec{b}\) × \(\vec{a}\) – \(\vec{b}\) × \(\vec{b}\)
= 0 – \(\vec{a}\) × \(\vec{b}\) – \(\vec{a}\) × \(\vec{b}\) – 0 = -2(\(\vec{a}\) × \(\vec{b}\))

Question 13.
Find ∫(4x + 7)\(\sqrt{x^{2}+4 x+13}\)dx
Answer:
4x + 7 = A(2x + 4) + B ⇒ A = 2, B = -1
I = 2∫\(\sqrt{x^{2}+4 x+13}\)dx – ∫\(\sqrt{x^{2}+4 x+13}\)dx
I = 2I₁ – I₂

Question 14.
a) Write integrating factor of the linear differential equation \(\frac{d y}{d x}+\frac{y}{x}\) = sin x
b) Slope of the tangent to a curve at any point is twice the x coordinate of the point. If the curve passes through the point (1, 4), find its equation.
Answer:
IF = \(e^{\int P d x}\) = \(e^{\log x}\) = x
b) \(\frac{d y}{d x}\) = 2x ⇒ dy = 2x dx
Integrating we get;
∫dy = ∫ 2x dx + C ⇒ y = x² + C
Since passes through (1, 4) we have;
4 = 1 + C ⇒ C = 3
Hence equation is y = x² +3

Question 15.
Solve the linear programming problem graphically Maximise Z = 3x + 5y
Subject to the constraints
x + 3y ≤ 3
x + y ≤ 2
x, y ≥ 0

Maximum 7 at B = \(\left(\frac{3}{2}, \frac{1}{2}\right)\)

Question 16.
a) If cos^-1 \(\frac{12}{13}\) = tan^-1 x then find x.
b) Show that cos^-1 \(\frac{4}{5}\) + cos^-1 \(\frac{12}{13}\) = tan^-1 \(\frac{14}{33}\)
Answer:

Question 17.
Consider the binary operation * on the set R of real numbers, defined by a*b = \(\frac{a b}{4}\)
a) Show that * is commutative and associative.
b) Find the identity element for * o R.
c) Find the inverse of 5.
Answer:

Questions from 18 to 24 carry 6 scores each. Answer any 5.

Question 18.
Consider the matrix A = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 1 & 2 \\
0 & 4 & 9
\end{array}\right]\)
a) Find A^-1 using elementary row operations.
b) Find the solution of the system of equations given below:
(A^-1 obtained above may be used)
x + 2z + 2; y + 2z + 1; 4y + 9z = 3
Answer:
a) A = IA
\(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 1 & 2 \\
0 & 4 & 9
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] A\)

Question 19.
a) Show that
\(\left|\begin{array}{lll}
1 & a & b c \\
1 & b & a c \\
1 & c & a b
\end{array}\right|\) = (a – b)(b – c)(c – a)
b) If A = \(\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\) verift that A × adj A = |A|I
Answer:

Question 20.
a) If f is a function such that f(-x) = f(x), then \(\int_{-a}^{a}\) f(x) dx = ………
b) Evaluate \(\int_{-\pi / 2}^{\pi / 2}\) cos x dx
c) Evaluate \(\int_{0}^{1}\) (x² + 1)dx as the limit of a sum.
Answer:

Question 21.
a) Verify mean value theorem for the function f(x) = x² – 4x – 3 in the interval [1, 4].
b) Consider the function
f(x) = sin^-1 2x \(\sqrt{1-x^{2}}\), \(\frac{-1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\)
i) Show that f(x) = 2sin^-1 x
ii) Find f'(x)
Answer:
a) f(x) is a continuous function in [1, 4], since it is a polynomial.
f'(x) = 2x – 4, f is differentiable in (1, 4).
f(4) = 16 – 16 – 3 = -3, f(1) = 1 – 4 – 3 = -6

Hence mean value theorem is verified.

b) i) Put x = sinθ
f(x) = sin^-1 (2 sin θ\(\sqrt{1-\sin ^{2} \theta}\))
= sin^-1 (2 sin θ cos θ)
= sin^-1 (sin 2θ) = 2θ = 2 sin^-1 x
ii) f'(x) = \(\frac{2}{\sqrt{1-x^{2}}}\)

Question 22.
a) Show that the lines
\(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{1} ; \frac{x-3}{2}=\frac{y-1}{1}=\frac{z-4}{2}\) are coplanar.
b) Find the equation of the plane that contains the above lines.
c) Show that the above lines intersect at the point (3, 1, 4).
Answer:
a) Points (2, -1, 3) and (3, 1, 4)

Normal direction ratios are 3, 0, -3.
Therefore equation of the plane is
3(x – 2) -3 (z – 3) = 0
3x – 6 – 3z + 9 = 0
x – z + 1 = 0

c) (3, 1, 4) is a point on the second line. Substitute the point in the first line
\(\frac{3-2}{1}=\frac{1+1}{2}=\frac{4-3}{1} \Rightarrow \frac{1}{1}=\frac{2}{2}=\frac{1}{1}\)
Therefore the point (3, 1, 4) satisfies the first line. Hence both interest at (3, 1, 4).

Question 23.
a) A coin is tossed 3 times. Find the probability distribution of the number of heads.
b) A bag contains 5 black and 6 white balls, 4 balls of the same colour (Black or white) are added to the bag, shuffled well and one ball is drawn. If the ball obtained is white. What is the probability that the balls added are black?
Answer:
a) Let X be the random variable denoting the
number of heads appears. Then X = {0, 1, 2, 3}
P(X = x) = ⁿC_x P^xq^n-x; p = \(\frac{1}{2}\), q = \(\frac{1}{2}\), n = 3

E₁ = Balls added are black.
E₂ = Balls added are white.
A = Ball drawn is white.

Question 24.

In a circle of radius 2 a square is inscribed as shown in the figure. Using integration, find the area of the shaded region (Area of a square may be calculated using any convenient method)
Answer:

Equation of the circle is x² + y² = 4
Area of the sector in the first quadrant

Area of the triangle = \(\frac{1}{2}\) (2)(2) = 2
Area of the shaded region in the first quadrant = π – 2
Hence the area of the required region = 4(π – 2) = 4π – 8

Kerala State Board New Syllabus Plus Two Maths Previous Year Question Papers and Answers.

Kerala Plus Two Maths Previous Year Question Paper March 2019 with Answers

Time : 2 1/2 Hours
Cool off time : 15 Minutes
Maximum : 80 Score

General Instructions to Candidates :

• There is a ‘Cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘Cool off time’ to get familiar with questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Question 1 to 7 carry 3 scores each. Answer any 6 questions. (6 × 3 = 18)

Question 1.
a) If f(x) = sinx, g(x) = x², x∈R, them find (fog)(x)
b) Let u and v be two functions defined on R as u(x) = 2x – 3 and v(x) = \(\frac{3+x}{2}\). Prove that u and v are inverse to each other.
Answer:
a) f(x) = sinx, g(x) = x², x∈R
fog(x) = f(g(x)) = f(x²) = sin(x²)

b) uov(x) = u(v(x))
= \(u\left(\frac{3+x}{2}\right)=\frac{2(3+x)}{2}-3\) = X
vou(x) = v(u(x))
v(2x – 3) = \(\frac{3+2 x-3}{2}\) = x

Question 2.
a) For the symmetric matrix
A = \(\left[\begin{array}{lll}
2 & x & 4 \\
5 & 3 & 8 \\
4 & y & 9
\end{array}\right]\)
Find the values of x and y.
b) From Part (a), verify AA’ and A + A’ are symmetric matrices.
Answer:
x = 5, y = 8
b)

Question 3.
a) Find the slope of tangent line to the curve y = x² – 2x + 1
b) Find the equation to the above curve which is parallel to the line 2x – y + 9 = 0.
Answer:
a) y = x² – 2x + 1 ⇒ \(\frac{d y}{d x}\) = 2x – 2
⇒ slope = 2x – 2

b) Since the tangent is parallel to the line 2x – y + 9 = 0 , both have same slope.
Slope of the line 2x – y + 9 = 0 is 2.
⇒ 2x – 2 = 2 ⇒ X = 2 ⇒ y = 1
Therefore the point is (2, 1)
Hence the equation of the tangent line is
y – 1 = 2 (x – 2) ⇒ y – 2x + 3 = 0

Question 4.
a) If ∫ f(x) dx = log |tan x| + C . Find f(x).
b) Evaluate ∫ \(\frac{1}{\sqrt{1-4 x^{2}}} d x\)
Answer:

Question 5.
a) Area bounded by the curve y = f(x) and the lines x = a, x = b and the x axis = ………..

b) Find area of the shaded region using integration.

Answer:
a) i) \(\int_{a}^{b} x d y\)
b) Here the slope of the line is 3 and passes through the origin. So its equation is y = 3x.

Question 6.
a) The order of the differential equation formed by y = A sin x + B cos x + c, where A and B are arbitrary constants is
i) 1   ii) 2   iii) 0   iv) 3
b) Solve the differential equation
sec²x tan ydx + sec²y tan xdy = 0
Answer:
a) ii) 2
b) sec²x tan ydx + sec²y tan xdy = 0

⇒ log |tan x| + log |tan y| = log c
⇒ log |tan x tan y| = log c
⇒ tan x tan y = c

Question 7.
A factory produces three items P, Q and R at two plants A and B. The number of items produced and operating cost per hour is as follows:

It is desired to produce at least 500 items of type P, at least 400 items of type Q and 300 items of type R per day.
a) Is it a maximisation case or a minimisation case? Why?
b) Write the objective function and constraints.
Answer:
a) Cost of operation should be minimum for a factory.
Hence this is a minimisation problem.

b) Maximise : Z = 1000 x + 800 y
Subject to
20x + 30y ≥ 500; 15x +12y ≥ 400;
25x + 23y ≥ 300; x, y ≥ 0

Questions 8 to 17 carry 4 scores each. Answer any 8. (8 × 4 = 32)

Question 8.
a) The function P is defined as “to each person on the earth is assigned a date of birth.” Is this a function one-one? Give reason.
b) Consider the function f: \(\left[0, \frac{\pi}{2}\right]\) → R given by f(x) = sin x and g: \(\left[0, \frac{\pi}{2}\right]\) → R given by g(x) = cos x.
i) Show that f and g are one-one functions.
ii) Is f + g one-one? Why?
c) The number of one-one functions from a set containing 2 elements to a set containing 3 elements is ………..
i) 2   ii) 3   iii) 6   iv) 8
Answer:
a) Not one-one. Since different persons have same birthdays.

b) i) f(x) = sin x and g(x) = cos x are one-one in the domain value in the domainone \(\left[0, \frac{\pi}{2}\right]\). Since for each value in domain \(\left[0, \frac{\pi}{2}\right]\) both have only one image.

ii) (f + g)(x) = sin x + cos x
(f + g)(0) = sin0 + cos0 = 0 + 1 = 1

Hence not one-one.

c) ³P₂ = 3 × 2 = 6

Question 9.
If A = sin^-1 \(\frac{2 x}{1+x^{2}}\) ,B = cos^-1 \(\frac{1-x^{2}}{1+x^{2}}\), C = tan^-1 \(\frac{2 x}{1+x^{2}}\) satisfies the condition 3A – 4B + 2C = \(\frac{\pi}{3}\). Find the value of x.
Answer:
a) 3A – 4B + 2C = \(\frac{\pi}{3}\)
3sin-1 \(\frac{2 x}{1+x^{2}}\) – 4cos-1 \(\frac{1-x^{2}}{1+x^{2}}\) + 2tan-1 \(\frac{2 x}{1+x^{2}}\)
3 × 2 tan^-1 x – 4 × 2 tan^-1 x + 2 × 2 tan^-1 x = \(\frac{\pi}{3}\)
⇒ 6tan^-1 x – 8tan^-1 x + 4tan^-1 x = \(\frac{\pi}{3}\)
⇒ 2tan^-1 x = \(\frac{\pi}{3}\) ⇒ tan^-1 x = \(\frac{\pi}{6}\)
⇒ x = \(\frac{1}{\sqrt{3}}\)

Question 10.
a) Write the function whose graph is shown below.

b) Discuss the continuity of the function obtained in part (a).
c) Discuss the differentiability of the function obtained in part (a).
Answer:
a) f(x) = \(\left\{\begin{array}{ll}
x^{2}, & x \leq 0 \\
x, & x>0
\end{array}\right.\)

b)

For x > 0, f(x) = x which is a polynomial, hence continuous.
For x < 0, f(x) = x² which is a polynomial, hence continuous. Therefore the function is continuous.

c) Since the function has a sharp corner at x = 0.
The function is not differentiable at x = 0.
Hence the function is not differentiable.

Therefore left derivative is not equal to right derivative. Hence not differentiable at x = 0.

Question 11.
A cuboid with a square base and given volume ‘V’ is shown in the figure:

a) Express surface area ‘S’ as a function of x.
b) Show that the surface area is minimum when it is a cube.
Answer:

Question 12.
a) If 2x + 4 = A(2x + 3) + B, find A and B.
b) Using part (a) evaluate ∫\(\frac{2 x+4}{x^{2}+3 x+1} d x\)
Answer:
a) A = 1, B = 1


Question 13.
Consider the Differential equation cos² x \(\frac{d y}{d x}\) + y = tan x. Find
a) its degree
b) the integrating factor
c) the general solution.
Answer:
a) One.

Question 14.
The position vectors of three points A, B, C are given to be i + 3j + 3k, 4i + 4k, -2i + 4j + 2k respectively
a) Find \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\)
b) Find the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\)
c) Find a vector which is perpendicular to both \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) having magnitude 9 units.
Answer:
a) \(\overrightarrow{A B}\) = 3i – 3j + k, \(\overrightarrow{A C}\) = -3i + j – k


Question 15.
a) If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are coplanar vectors, write the vector perpendicular to \(\bar{a}\)
b) If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are coplanar, prove that [\(\bar{a}\) + \(\bar{b}\) \(\bar{b}\) + \(\bar{c}\) \(\bar{c}\) + \(\bar{a}\)] are coplanar.
Answer:
a) Cross product of \(\bar{a}\) with any of the vectors \(\bar{b}\) or \(\bar{c}\).

b) Given,
[\(\bar{a}\), \(\bar{b}\), \(\bar{c}\)] = 0

Question 16.
a) Write all the direction cosines of x-axis.
b) If a line makes α, β, γ with x, y, z axis respectively, then prove that sin² α + sin² β + sin² γ = 2
c) If a line makes equal angles with the coordinate axes, find the direction cosines of the lines.
Answer:
a) 1, 0, 0

b) LHS = sin² α + sin² β + sin² γ
= 1 – cos² α + 1 — cos² β + 1 — cos² γ
= 3 – (cos² α + cos² β + cos² γ) = 3 – 1 = 2

c) Given α, β, γ are equal. Then
⇒ cos² α + cos² α + cos² α = 1
⇒ 3 cos² α = 1
⇒ cos α = \(\frac{1}{\sqrt{3}}\)
⇒ α = cos^-1 \(\frac{1}{\sqrt{3}}\)

Question 17.
The activities of a factory are given in the following table:

Solve the linear programming problem graphically and find the maximum profit subject to the above constraints.
Answer:
Maximise: Z= 5x + 8y
x + 4y ≤ 24; 3x + y ≤ 21; x + y ≤ 9; x, y ≥ 0

Maximum is at (4, 5); Z = 60

Questions from 18 to 24 carry 6 scores each. Answer any 5. (5 × 6 = 30)

Question 18.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\). Show that A² – 5A + 7I = 0. Hence find A⁴ and A^-1
Answer:

A² – 5A + 7I = 0
Multiplying by A^-1 we have;
A^-1 (A² – 5A + 7I) = 0
⇒ A – 5I + 7A^-1 = 0

Question 19.
If A = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\), then
a) Find A^-1
b) Using A^-1 from part (a) solve the system of equations.
Answer:
|A| = \(\left|\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right|\) = -1
C₁₁ = 0, C₁₂ = 2, C₁₃ = 1
C₂₁ = -1, C₂₂ = -9, C₂₃ = -5
C₃₁ = 2, C₃₂ = 23, C₃₃ = 13

b) X = A^-1 B

Question 20.
Find for the following:
a) sin² x + cos² y = 1
b) y = x^x
c) x = a(t – sin t), y = a(1 + cos t)
Answer:
a) sin² x + cos² y = 1
Differentiating w.r.to x we have;
2 sinx cosx + 2 cos y(-sin y) \(\frac{d y}{d x}\) = 0
sinx cosx = cosy siny \(\frac{d y}{d x}\)
\(\frac{d y}{d x}=\frac{\sin x \cos x}{\cos y \sin y}=\frac{\sin 2 x}{\sin 2 y}\)

b) y = x^x Take log on both sides;
logy = x log x
Differentiating w.r to x

Question 21.
Evaluate the following:

Answer:
i)



Question 22.
a) Find the area bounded by the curve y = sin x and the lines x = 0, x = 2π and x axis.
b) Two fences are made in a grass field as shown in the figure. A cow is tied at the point O with a rope of length 3m.

i) Using integration, find the maximum area of grass that cow graze within the fences. Choose D as origin.
ii) If there is no fences find the maximum area of grass that cow can graze.
Answer:

b) i) The Area the cow grazes in the sector of the circle with radius 3 and centered at origin.
x² + y² = 9 ⇒ y = \(\sqrt{9-x^{2}}\)
Area = \(\int_{a}^{b}\) ydx = \(\int_{0}^{3} \sqrt{9-x^{2}} d x\)

ii) The required area is area inside the full circle = 4 × \(\frac{9 \pi}{4}\) = 9π

Question 23.
a) Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).
b) The Cartesian equation of two lines are given by \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}, \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\). Write the vector equation of these two lines.
c) Find the shortest distance between the lines mentioned in part (b).
Answer:
a) (3x – y + 2z – 4) + k (x + y + z – 2) = 0
It passes through the point (2, 2, 1)
(3(2) – 2 + 2(1) -4)+ k (2 + 2 + 1 – 2) = 0
⇒ 2 + k(3) = 0 ⇒ k = \(-\frac{2}{3}\)
(3x – y + 2z – 4) –\(\frac{2}{3}\) (x + y + z – 2) = 0
⇒ 9x -3y + 6z – 12 – 2x – 2y – 2z + 4 = 0
⇒ 7x – 5y + 4z – 8 = 0

b) \(\vec{r}\) = (-i – j – k) + λ (7i – 6j + k)
\(\vec{r}\) = (3i + 5j + 7k) + μ (i – 2j + k)

c) \(\overline{a_{1}}\) = -i -j -k; \(\overline{b_{1}}\) =7i – 6j + 2k
\(\overline{a_{2}}\) = 3i + 5j + 7k; \(\overline{b_{2}}\) =i – 2j + k
\(\overline{a_{2}}\) – \(\overline{a_{1}}\) = 4i + 4j + 8k

Question 24.
a) A bag contains 4 red and 4 black balls. Another bag contains 2 red and 5 black balls. One of the two bags is selected at random and a ball is drawn from the bag and which is found to be red. Find the probability that the ball is drawn from first bag.
b) A random variable X has the following distribution function:

i) Find k.
ii) Find the mean and the variance of the random variable.
Answer:
E₁ = Event of choosing bag I
E₂ = Event of choosing bag II
A = Event of drawing a red ball.
P(E₁) = P(E₂) = \(\frac{1}{2}\)

b) i) ΣP_i = 1
⇒ k + 3k + 5k + 7k + 4k = 1
⇒ 20k = 1 ⇒ k = \(\frac{1}{20}\)

ii)

Kerala State Board New Syllabus Plus Two Computer Application Previous Year Question Papers and Answers.

Kerala Plus Two Computer Application Previous Year Question Paper Say 2018 with Answers

Time: 2 Hours
Cool off time : 15 Minutes

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Part – A

Answer all questions from 1 to 5. Each question carries 1 score. (5 × 1 = 5)

Question 1.
Write the name of the built-in function of C++ to convert the given character into its lower case.
Answer:
tolower()

Question 2.
Which is the tag used to create a line break in an HTML page?
Answer:
<br>

Question 3.
A candidate key that is not a primary key is called the _______ key.
Answer:
alternate key

Question 4.
Which is the keyword used in the SQL SELECT command to eliminate duplicate values in the selection.
Answer:
distinct

Question 5.
Expand the term CDMA.
Answer:
Code Division Multiple Access

Part – B

Answer any 9 questions from 6 to 16. Each question carries 2 scores. (9 × 2 = 18)

Question 6.
Write the function prototype for the following function:

1. A function sum( ) takes two integer arguments and returns integer value.
2. A function print() has no argements and nonreturn value.

Answer:

1. int sum(int, int);
2. void print();

Question 7.
Differentiate actual arguments and formal arguments in C++.
Answer:
The argument that present in the called function is called the formal argument and it is present in the calling function is called an actual argument. The data type of both is the same.

Question 8.
Write the output of the following HTML code:
<OL Type = “I” Start = “10”>
<LI> keyboard </LI>
<LI> mouse </LI>
<LI> light pen </LI>
Answer:
10. keyboard
11. mouse
12. light pen

Question 9.
Describe any four values of Type attributes of the <INPUT> Tag in HTML.
Answer:
<Input> It is used to create input controls. Its type of attribute determines the control type.
Main values of the type attribute are given below.

1. Text – To create a text box.
2. Password – To create a password text box.
3. Checkbox – To create a check box.
4. Radio – To create a radio button.
5. Reset – To create a Reset button.
6. Submit-To creates a submit button.
7. Button – To create a button

Question 10.
Write a short note on a virtual private server.
Answer:
Virtual Private Server (VPS): A VPS is a virtual machine sold as a service by an Internet hosting Service. A VPS runs its own copy of an OS (Operating System) and customers have super level access to that OS instance, so they can install almost any s/w that runs on that OS. This type is suitable for websites that require more features than shared hosting but fewer features than dedicated hosting.

Question 11.
Define primary key and alternate key.
Answer:
Primary key – It is a set of one or more attributes used to uniquely identify a row.
Alternate key – A candidate key other than the primary key.

Question 12.
Write a short note on UNION operation in Relational algebra.
Answer:
UNION operation: This operation returns a relation consisting of all tuples appearing in either or both of the two specified relations. It is denoted by U. duplicate tuples are eliminated. Union operation can take place between compatible relations only, i.e., the number and type of attributes in both the relations should be the same and also their order.
e.g. SCIENCE U COMMERCE gives all the tuples in both COMMERCE and SCIENCE.

Question 13.
Differentiate the data type CHAR and VARCHAR in SQL.
Answer:
Char – It is used to store a fixed number of characters. It is declared as char (size).
Varchar – It is also used to store characters but it uses only enough memory.
the char data type is fixed length. It allocates maximum memory and maybe there is a chance of memory wastage. But Varchar allocates only enough memory to store the actual size.

Question 14.
Write a short note on Supply Chain Management.
Answer:
Supply Chain Management (SCM): This deals with moving raw materials from suppliers to the company as well as finished goods from the company to customers. The activities include are inventory (raw materials, work in progress, and finished goods) management, warehouse management, transportation management, etc.

Question 15.
Write a short note on the mobile operating system.
Answer:
Mobile Operating System: It is an OS used in handheld devices such as smartphones, tablets, etc. It manages the hardware, multimedia functions, Internet connectivity, etc. Popular OSs are Android from Google, iOS from Apple, BlackBerry OS from BlackBerry, and Windows Phone from Microsoft.

Question 16.
Define the following term:

1. SIM
2. MMS

Answer:

1. The network is identified using the SIM (Subscriber Identity Module).
2. Multimedia Messaging Service (MMS): It allows sending Multi-Media (text, picture, audio, and video file) content using mobile phones. It is an extension of SMS.

Part – C

Answer any 9 questions from 17 to 27. Each carries 3 scores. (9 × 3 = 27)

Question 17.
Rewrite the following C++ code using the if…else statement:

switch (choice)
{
Case 1:
cout<<"one";
break;
case 0:
cout<<"zero";
break;
default;
cout<<"End"
break;
}
Answer:
if(choice==1)
cout<<“One”;
else if (choice==0)
cout<<“Zero”;
else
cout<<“End”;

Question 18.
Write the output of the following C++ code. Justify your answer.

for(i=1; i<5; i++)
{
cout<<“\t”<<i;
if(i==3)
break;
}

Answer:
This prints 1 2 3. This is because the value of i becomes ‘3’ then the break statement executes, it terminates the loop and hence the output.

Question 19.
Consider the following C++ code :
a) char name [20];
cin>>name;
cout<<name;
b) char name [20];
gets (name);
cout<<name;
Write the output in both cases if the string entered value is “NEW DELHI”. Justify your answer.
Answer:
a) The output is New. This is because of cin operator reads up to the delimiter space. The characters after space will not be read.
b) The output is New Delhi. This is because of the gets() function reads characters upto the user press the enter key, including space.

Question 20.
Define array traversal with an example.
Answer:
Traversal: All the elements of an array is visited and processed is called traversal
Eg:

#include<iostream>
using namespace std;
int main()
{
int n[10], i, sum=0;
for(i=0; i<10; i++)
{
cout<<“Enter value for number”<<i+1<<":";
cin>>n[i];
if(n[i]%5==0)
sum+=n[i];
}
cout<<"The sum of numbers which are exact multiple of 5 is "<<sum;
}

Question 21.
Consider the following function definition in C++:

void sum (int a, int b=10, int c=20)
{
int sum = a + b + c;
cout<<sum:
}

Write the output of the above code for the following function call:
(a) sum (1, 2, 3);
(b) sum (2, 3);
(c) sum (3);
Answer:
a) 6
Here a = 1, b = 2 and c = 3
So the answer is 6

b) 25
Here a = 2, b = 3 and c = 20 (The default value)
So the answer is 25

c) 33
Here a = 3, b = 10 and c = 20 (Default values for b and c)
So the answer is 33

Question 22.
Compare client-side scripting and server-side scripting.
Answer:

Question 23.
Write the HTML code to generate the following table:

Answer:

<html>
<head>
<title>
Table creation
</title>
</head>
<body bgcolor="red">
<table border="1">
<tr align="center">
<th>Roll No</th>
<th> Name</th>
<th> Class</th>
</tr>
<tr align="center">
<td>100</td>
<td> Vishnu</td>
<td> C1</td>
</tr>
<tralign="center">
<td>101</td>
<td> Anupama</td>
<td> C2</td>
</tr>
<tralign="center">
<td> 102</td>
<td> Biju</td>
<td> A1 </td>
</tr>
</table>
</body>
</html>

Question 24.
Classify the following values in JavaScript into suitable data type:
“Hello”, False, 125.0, 148, “True”, True
Answer:
String – “Hello”, “True”
Numeric – 125.0,148
Boolean – False, True

Question 25.
What is Content Management System? Write any two popular CMS software.
Answer:
Content Management System(CMS): CMS is a collection of programs that is used to create, modify, update, and publish website contents. CMS can be downloaded freely and is useful to design and manage attractive and interactive websites with the help of templates that are available in CMS. WordPress, Joomla, etc. are examples of CMS.

Question 26.
Define the following terms:

1. Cardinality
2. Schema
3. Tuple

Answer:

1. Cardinality – The number of rows.
2. Schema – The structure of the table is called the schema.
3. Tuple means the rows.

Question 27.
Explain any three benefits of the ERP system.
Answer:
Benefits of ERP system
1. Improved resource utilization: Resources such as Men, Money, Material, and Machine are utilized maximum hence increase productivity and profit.

2. Better customer satisfaction: Without spending more money and time all the customer’s needs are considered well. Because the customer is the king of the market. Nowadays a customer can track the status of an order by using the docket number through the Internet.

3. Provides accurate information: Right information at the right time will help the company to plan and manage the future cunningly. A company can increase or reduce production based upon the right information hence increase productivity and profit.

4. Decision-making capability: Right information at the right time will help the company to take a good decisions.

5. Increased flexibility: A good ERP will help the company to adopt good things as well as avoid bad things rapidly. It denotes flexibility.

6. Information integrity: A good ERP integrates various departments into a single unit. Hence reduce the redundancy, inconsistency, etc.

Part – D

Answer any 2 questions from 28 to 30. Each question carries 5 scores. (2 × 5 = 10)

Question 28.
Consider the following HTML code and answer the following:
<EM> COMPUTER </EM> <BR>
<STRONG> APPLICATION </STRONG> <BR> <HR>
(a) Name the tag used to make the text as italics and bold in the above code. (1)
(b) What is the purpose of <HR> tag? Explain its any two attributes. (2)
(c) Write the HTML statement to scroll the text given in <EM> from top to bottom. (2)
Answer:
a) for Italics <I> or <i> is used
for bold <strong> or <b> is used
b) <HR> is used to draw a horizontal line. Its attributes are size, width, shade, and color.

c) <html>
<head>
<title>
Demo of Marquee
</title>
</head>
<body bgcolor="red">
<marquee direction="down">
<em>hi welcome to BVM</em>
</marquee>
</body>
</html>

Question 29.
Consider the following JavaScript code:

function print ()
{
var i,
for (i=1; i<=10; ++i)
{
document.write(i);
document.write("<<BR>");
}
}

(i) Write the output of the above code. (1)
(ii) Rewrite the above code using a while loop. (2)
(iii) Modify the above code to find the sum of first 10 counting numbers. (2)
Answer:
i) It prints 1 to 10 line by line

ii) function Print()
{
var i;
i=1;
while(i<= 10)
{
document.write(i);
document.write (" <BR> ");
i++;
}
}

iii) function Print()
{
var i, sum=0;
for(i=1; i<=10; i++)
sum=sum+i;
document.write("The sum of first 10 countimg numbers is "+sum);
}

Question 30.
Define constrain. Explain any four-column constraints.
Answer:
Constraints are used to ensure database integrity.

1. Not Null – It ensures that a column can never have NULL values.
2. Unique – It ensures that no two rows have the same value in a column.
3. Primary key – Similar to unique but it can be used only once in a table.
4. Default – We can set a default value.
5. Auto_increment – This constraint is used to perform auto_increment the values in a column. That automatically generates serial numbers. Only one auto_increment column per table is allowed.

Kerala State Board New Syllabus Plus Two Computer Application Previous Year Question Papers and Answers.

Kerala Plus Two Computer Application Previous Year Question Paper March 2019 with Answers

Time: 2 Hours
Cool off time : 15 Minutes

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Part – A

Answer all questions from 1 to 5. Each carries 1 score. (5 × 1 = 5)

Question 1.
The input operator in C++ is ___________
Answer:
>> or cin >>

Question 2.
_________ character is stored at the end of the string.
Answer:
NULL or ‘\0’

Question 3.
The process of breaking large program into smaller sub-programs is called __________
Answer:
Modularization

Question 4.
Name the keyword used to declare variables in JavaScript.
Answer:
var

Question 5.
Expand MIS.
Answer:
Management Information System

Part – B

Answer any 9 questions from 6 to 16. Each carries 2 scores. (9 × 2 = 18)

Question 6.
List the type modifiers in C++.
Answer:
Type modifiers used in C++ are signed, unsigned, short and long.

Question 7.
Rewrite the following code using for loop:

int x = 1;
start:
cout<<x;
x = x + 5;
if (x < = 50)
goto start;

Answer:

for(x=1; x<=50; x+=5)
cout<<x;

Question 8.

1. Define an Array.
2. Initialize an integer array with 5 elements.

Answer:

1. Array: An array is a collection of elements with the same data type store in contiguous memory location.
2. int mark[] = {40, 42, 44, 46, 48, 50};

Question 9.
Write the port number for the following web services:

1. Simple Mail Transfer Protocol.
2. HTTP secure (HTTPS)

Answer:

1. 25
2. 443

Question 10.
What is the use of frame tag in HTML? What is its limitation?
Answer:
frame tag helps to view multiple web pages in a single window. The main limitation is that all browsers not supporting the frame tag.

Question 11.
Write the HTML code to display the following using list tag:
i) Biology Science
ii) Commerce
iii) Humanities
Answer:

<html>
<head>
<title>list demo
</title>
</head>
<body bgcolor="red">
<ol type="i">
<li> Biology Science</li>
<li> Commerce</li>
<li> Humanities</li>
</ol>
</body>
</html>

Question 12.
What is the difference between isNaN() and Number() functions in JavaScript?
Answer:
isNaN() function checks the given value is a number or not. If it is not a number it returns a true value otherwise false.
Number() function converts the data into numerical type.

Question 13.
What is CMS? Give two examples.
Answer:
CMS means Content Management System. It is a collection of programs that are used to create, modify, update, and publish website content. CMS can be downloaded freely and is useful to design and manage attractive and interactive websites with the help of templates that are available in CMS. WordPress, Joomla, etc. are examples of CMS.

Question 14.
First table containing 4 rows and 3 columns, the second table contains 5 rows and 2 columns, then the Cartesian product table contains ______ rows and ______ columns.
Answer:
The number of rows is the product of rows, i.e. 4 × 5 = 20 rows
The number of columns is the sum of columns, i.e. 3+2 = 5 columns

Question 15.
How Business Process Re-Engineering (BPR) is related to Enterprise Resource Planning (ERP)?
Answer:
ERP and BPR will not make much change if they are in stand-alone. To improve the efficiency of an enterprise integrate both ERP and BPR because they are the two sides of a coin. For better results conducting BPR before implementing ERP, will help an enterprise to avoid unnecessary modules from the software.

Question 16.
Define the terms:
i) Cyber Forensics
ii) Infomania
Answer:
i) Cyber Forensics: Critical evidence of a particular crime is available in electronic format with the help of computer forensics. It helps to identify the criminal with help of blood, skin or hair samples collected from the crime site. DNA, polygraph, finger prints are another effective tool to identify the accused person is a criminal or not.

ii) Infomania: Right information at the right time is considered as the key to success. The information must be gathered, stored, managed and processed well. Infomania is the excessive desire (Infatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media, Instant Message Application (WhatsApp) and Smart Phones. Due to this, the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired. They give first preference to the Internet than others. They create their own Cyber World and no interaction to the surroundings and the family.

Part – C

Answer any 9 questions from 17 to 27. Each carries 3 scores. (9 × 3 = 27)

Question 17.
Compare the selection statements ‘if’ and ‘switch’.
Answer:
Following are the difference between the switch and if-else if ladder.

1. Switch can test only for equality but if can evaluate a relational or logical expression.
2. If else is more versatile.
3. If else can handle floating values but switch cannot
4. If the test expression contains more variable if else is used.
5. Testing a value against a set of constants switch is more efficient than if-else.

Question 18.
Write a program in C++ to accept a string with white space like “good morning” from the keyboard and display the same string.
Answer:

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char str[80];
cout<<"Enter a string:";
gets(str);
puts(str);
}

Question 19.
Compare static webpage and dynamic webpage.
Answer:

Question 20.
i) What is the use of reserved characters for HTML entities? (1)
ii) List any four reserved characters and their use. (2)
Answer:
(i) HTML entities are used to print reserved characters in HTML.
(ii)

Question 21.
Write the built-in JavaScript functions used for the following situation:

1. Display warning message in the screen.
2. Character at a particular position.
3. Convert uppercase to lowercase.

Answer:

1. alert()
2. charAt()
3. toLowerCase()

Question 22.
Write the merits and demerits of free Webhosting.
Answer:
The name implies it is free of cost service and the expense is met by the advertisements. Some service providers allow limited facility such as limited storage space, do not allow multimedia (audio and video) files.

Question 23.
What is the key? Explain any two keys in a relational database management system.
Answer:
Key is used to identify or distinguish a tuple in a relation.

• Candidate key – It is used to uniquely identify the row.
• Primary key – It is a set of one or more attributes used to uniquely identify a row.
• Alternate key – A candidate key other than the primary key.
• Foreign key – A single attribute or a set of attributes, which is a candidate key in another table is called a foreign key.

Question 24.
Define the term Data independence. Explain different levels of data independence.
Answer:
Data Independence – It is the ability to modify the schema definition in one level without affecting the scheme definition at the next higher level.

• Physical Data Independence – It is the ability to modify the physical scheme without causing application programs to be rewritten.
• Logical Data Independence – It is the ability to modify the logical scheme without causing application programs to be rewritten.

Question 25.
Explain any three situations to modify the structure of a table with the help of alter command in SQL.
Answer:
We can alter the table in two ways.
We can add a new column to the existing table using the following syntax,
ALTER TABLE <tablename>ADD(<cloumnname> <type> <constraint>);
We can also change or modify the existing column in terms of type or size using the following syntax,
ALTER TABLE<tablename>MODIFY(<column> <newtype>);

Question 26.
Explain the merits of ERP system.
Answer:
Benefits of ERP system
1. Improved resource utilization: Resources such as Men, Money, Material and Machine are utilized maximum hence increase productivity and profit.

2. Better customer satisfaction: Without spending more money and time all the customer’s needs are considered well. Because the customer is the king of the market. Nowadays a customer can track the status of an order by using the docket number through the Internet.

3. Provides accurate information: Right information at the right time will help the company to plan and manage the future cunningly. A company can increase or reduce production based upon the right information hence increase productivity and profit.

4. Decision-making capability: Right information at the right time will help the company to take a good decision.

5. Increased flexibility: A good ERP will help the company to adopt good things as well as avoid bad things rapidly. It denotes flexibility.

6. Information integrity: A good ERP integrates various departments into a single unit. Hence reduce the redundancy, inconsistency, etc.

Question 27.
Compare GPRS and EDGE.
Answer:
GPRS (General Packet Radio Services): It is a packet-oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards Institute (ETSI) GPRS usage is typically^fiarged based on the volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

EDGE (Enhanced Data rates for GSM Evolution): It is three times faster than GPRS. It is used for voice communication as well as an internet connection.

Part – D

Answer any 2 questions from 28 to 30. Each carries 5 scores. (2 × 5 = 10)

Question 28.
Identify the built-in C++ function for the following cases:

1. to convert -25 to 25.
2. compare ‘computer’ and ‘COMPUTER’ ignoring cases.
3. to check the given character is a digit or not.
4. to convert the character from ‘B’ to ‘b’.
5. to find the square root of 64 or a number.

Answer:

1. abs()
2. strcmpi()
3. isdigit()
4. tolower()
5. sqrt()

Question 29.
(i) Write the name of the tag used to group related data in an HTML form. (1)
(ii) Write the HTML code to display the following webpage: (4)

Answer:
(i) <fieldset> tag

(ii) <html>
<head>
<title>
login page
</title>
</head>
<BODY BGCOLOR="cyan">
<FORM NAME="frmlogin">
<center>
User Name
<input type="text" name="txtname">
<br><br>
Password
<input type="password" name="txtpass">
<br><br>
<input type="Submit" value="Submit">
<input type="Reset" value="Reset">
</center>
</FORM>
</body>
</html>

Question 30.
Consider the table student with attribute admno, Name, course, percentage. Write the SQL statements to do the following:

1. Display all the student details. (1)
2. Modify the course’Commerce1 to’Science1. (1)
3. Remove the student details with a percentage below 35. (1)
4. Create a view from the above table with a percentage greater than 90. (2)

Answer:

1. select * from student;
2. update student set course=”Science” where course=”Commerce”;
3. delete from student where percentage<35;
4. create view stud view as select * from student where percentage > 60;

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 11 Trends and Issues in ICT.

Kerala Plus Two Computer Application Notes Chapter 11 Trends and Issues in ICT

Mobile Computing
The drawbacks of Desk computers are, it is heavy and power consumption rate is high and it is not portable(not mobile).
The advancements in computing technology, lightweight and low power consumption have led to the developments of more computing power in handheld devices like laptops, tablets, smartphones, etc.

Nowadays instead of desktops, lightweight and low power consumption devices are used because they are cheap and common. Moreover people are able to connect to others through the internet even when they are in motion.

Mobile Communication
The term ‘mobile’ help people to change their lifestyles and become the backbone of society. Mobile communication networks do not require any physical connection.

Generations in mobile communication
The mobile phone was introduced in the year 1946. Early-stage it was expensive and limited services hence its growth was very slow. To solve this problem, cellular communication concept was developed in 1960’s at Bell Lab. 1990’s onwards cellular technology became a common standard in our country.
The various generations in mobile communication are
a) First Generation networks(1G):
It was developed around 1980, based on analog. system and only voice transmission were allowed.

b) Second Generation networks (2G):
This is the next-generation network that was allowed voice and data transmission. Picture message and MMS(Multimedia Messaging Service) was introduced. GSM and CDMA standards were introduced by 2G.

i) Global System for Mobile(GSM):
It is the most successful standard. It uses narrowband TDMA (Time Division Multiple Access), allows simultaneous calls on the same frequency range of 900 MHz to 1800 MHz. The network is identified using the SIM(Subscriber Identity Module).

GPRS (General Packet Radio Services): It is a packet-oriented mobile data service on the 2G on GSM. GPRS was originally standardized by European Telecommunications Standards

Institute (ETSI) GPRS usage is typically charged based on the volume of data transferred. Usage above the bundle cap is either charged per megabyte or disallowed.

EDGE (Enhanced Data rates for GSM Evolution): It is three times faster than GPRS. It is used for voice communication as well as an internet connection.

ii) Code Division Multiple Access (CDMA):
It is a channel access method used by various radio communication technologies. CDMA is an example of multiple access, which is where several transmitters can send information simultaneously over a single communication channel. This allows several users to share a band of frequencies To permit this to be achieved without undue interference between the users and provide better security.

c) Third Generation networks (3G):
It allows a high data transfer rate for mobile devices and offers high-speed wireless broadband services combining voice and data. To enjoy this service 3G enabled mobile towers and handsets required.

d) Fourth Generation networks (4G): It is also called Long Term Evolution(LTE) and also offers ultra-broadband Internet facility such as high quality streaming video. It also offers good quality image and videos than TV.

e) Fifth Generation networks (5G): This is the next-generation network and expected to come into practice in 2020. It is more faster and cost-effective than the other four generations. More connections can be provided and more energy efficient.

Mobile communication services

a) Short Message Service(SMS): It allows transferring short text messages containing up to 160 characters between mobile phones. The sent message reaches a Short Message Service Center(SMSC), that allows ‘store and forward’ systems. It uses the protocol SS7 (Signaling System No7). The first SMS message ‘Merry Christmas’ was sent on 03/12/1992 from a PC to a mobile phone on the Vodafone GSM network in the UK.

b) Multimedia Messaging Service (MMS): It allows sending Multi-Media(text, picture, audio, and video file) content using mobile phones. It is an extension of SMS.

c) Global Positioning System(GPS): It is a space-based satellite navigation system that provides location and time information in all weather conditions, anywhere on or near the Earth where there is an unobstructed line of sight to four or more GPS satellites. The system provides critical capabilities to military, civil, and commercial users around the world. It is maintained by the United States government and is freely accessible to anyone with a GPS receiver. GPS was created and realized by the U.S. Department of Defense (DoD) and was originally run with 24 satellites. It is used for vehicle navigation, aircraft navigation, ship navigation, oil exploration, Fishing, etc. GPS receivers are now integrated with mobile phones.

d) Smart Cards: A smart card is a plastic card with a computer chip or memory that stores and transacts data. A smart card (may be like your ATM card) reader used to store and transmit data. The advantages are it is secure, intelligent and convenient.
The smart card technology is used in SIM for GSM phones. A SIM card is used as identification proof.

Mobile operating system: It is an OS used in hand held devices such as smart phone, tablet, etc. It manages the hardware, multimedia functions, Internet connectivity,etc. Popular OSs are Android from Google, iOS from Apple, BlackBerry OS from BlackBerry and Windows Phone from Microsoft.

Android OS: It is a Linux-based OS for Touch screen devices such as smartphones and tablets.lt was developed by Android Inc. founded in Palo Alto, California in 2003 by Andy Rubin and his friends. In 2005, Google acquired this. A team led by Rubin devel¬oped a mobile device platform powered by the Linux Kernel. The interface of Android OS is based on touch inputs like swiping, tapping, pinching in, and out to manipulate on-screen objects. In 2007 onwards this OS is used in many mobile phones and tablets. An¬droid SDK(Software Development Kit) is available to create applications(apps) like Google Maps, FB, What’s App,etc. It is of open-source nature and many Apps are available for free download from the Android Play Store hence increase the popularity. Different Android Versions are shown below:

Version – Code name
4.4 – KitKat
4.1 – Jelly Bean
4.0.3 – Ice Cream Sandwich
3.1 – Honeycomb
2.3 – Gingerbread
2.2 – Froyo
2.0 – Eclair
1.6 – Donut
1.5 – Cupcake

ICT in business: Drastic developments in ICT have changed the shopping habits of people. Earlier people shops traditionally. But nowadays people buy products and services online. A study reveals that online shopping habits of people are increased. Aftersale service is also good, delivery of the products is prompt and safe. The status of the product can be tracked easily hence increase the confidence level of the online customers.

Social networks and big data analytics: Earlier before buying a product people may consult two or three shop keepers or local friends and take decisions. But nowadays before taking decisions, people search shopping sites, social network groups(Facebook, WhatsApp, Instagram, Twitter, etc), web portals, etc. for the best prices. Almost all online sites have product comparison menus. By this, we can compare the price, features, etc. Earlier a product is created and customers are forced to buy. But today customer is the King of the market, so products are created for the choices of the customers.

So companies gathering information about the customers from various sources such as social media like Internet forums, social blogs, Microblogs, etc. The volume of such data is very large and considered big data in business. With the help of an s/w analysis this big data and generate a report that contains all the information such as choices, taste, needs, status etc of a customer.

Business logistics: It is the management of the flow(transportation) of resources such as food, consumer goods, services, animals etc in a business between the point of origin(source) and the point of consumption (destination) in order to meet the needs of companies and customers. Business logistics consists of many more complexities. The effective use of hardware and software reduces the complexities faced in Business logistics.

For this the hardware used is RFID(Radio Frequency Identification) tag and the reader. It is like the bar code. The RFID tag contains all the details of a product and it consists of a combination of a transmitter and a receiver. The data stored in the RFID tag can be accessed by a special reader and to read the data no need for an RFID tag and reader in a line of site instead both are within a range.

This tag is used in Vehicles as a prepaid tag and makes the payments easier in Toll booths. Similarly, it is useful to take the Census of wild animals also.

Information Security: The most valuable to a company(An enterprise ora Bank, etc) is their data base hence it must be protected from accidental or unauthorised access by unauthorised persons

Intellectual Property Right: Some people spend lots of money, time body, and mental power to create some products such as a classical movie, album, artistic work, discoveries, invention, software, etc. These types of Intellectual properties must be protected from unauthorized access by law. This is called Intellectual Property right(IPR).
Paris convention held in 1883 protects Industrial Property
Berne Convention held in 1886 protects Literary and Artistic work.
World Intellectual Property Organisation(WIPO) in 1960, Guided by the United Nations(UN) ensures/protects the rights of creators or owners and rewarded for their creation.

A person or an organization can register their Intellectual property such as creations, trademarks, designs, etc.

Intellectual property is divided into two categories

1. Industrial Property
2. Copyright

1) Industrial property: It ensures the protection of industrial inventions, designs, Agricultural products etc from unauthorized copying or creation or use. In India, this is done by the Controller of Patents Designs and Trademarks.

Patents: A person or organization that invented a product or creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India, the validity of the right is up to 20 years. After this anybody can use it freely.

Trademark: This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or^opy by anybody else.

Industrial designs: A product or article is designed so beautifully to attract customers. This type of design is called industrial design. This is a prototype and used as a model for large scale production.

Geographical indications: Some products are well known by the place of its origin. Kozhikkodan Halwa, Marayoor Sharkkara (Jaggery), Thirupathi Ladoo, etc are examples.

B) Copyright: The trademark is ©, copyright is the property right that arises automatically when a person creates a new work on his own, and by Law, it prevents the others from the unauthorized or intentional copying of this without the permission of the creator for 60 years after the death of the author.

Infringement (Violation): Unauthorized copying or use of Intellectual property rights such as Patents, Copyrights, and Trademarks are called intellectual property Infringement(violation). It is a punishable offense.

Patent Infringement: It prevents others from unauthorized or intentional copying or use of Patent without the permission of the creator.

Piracy: It is the unauthorized copying, distribution, and use of a creation without the permission of the creator. It is against the copy right act and hence the person committed deserves the punishment.

Trademark Infringement: It prevents others from unauthorized or intentional copying or use of Trademark without the permission of the creator.

Copy right Infringement: It prevents others from unauthorized or intentional copying or use of Copy right without the permission of the creator.

Cyberspace: Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too. Later telephones were used for voice communication. Nowadays telephone systems and computer systems are integrated and create a virtual(un real) environment. This is called cyberspace. The result of this integration is that tremendous speed and it is very cheap. The various departments of Govt, are providing speed, reliable and convenient online service hence increase productivity. Online shopping, Online banking, Online debate, Online Auction etc. are the various services offered by the Internet.

Through this one can transfer funds from our account to another account, hence one can pay bills such as telephone, electricity, purchase tickets(Flight, Train, Cinema, etc). As much as CyberSpace helps us that much as it gives us troubles.

Cyber Crimes: Just like normal crimes (theft, trespassing private area, destroy, etc,) Cybercrimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography etc) also increased significantly. Due to cybercrime, the victims lose money, reputation,etc and some of them commit suicide.

A) Cybercrimes against individuals
i) Identity theft: The various information such as personal details(name, Date of Birth, Address, Phone number etc), Credit / Debit Card details(Cand number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing this information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

ii) Harassment: Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of the Internet is called Cyberstalking (Nuisance). This is a kind of torturing and it may lead to spoiling friendship, career, self-image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

iii) Impersonation and cheating: Fake accounts are created in Social media and act as the original one for the purpose of cheating or misleading others. Eg: Fake accounts in Social Medias (Facebook, Twitter, etc), fake SMS, fake emails etc.

iv) Violation of privacy: Trespassing into another person’s life and try to spoil life. It is a punishable offense. A hidden camera is used to capture the video or picture and blackmailing them.

v) Dissemination of obscene material: With the help of hidden camera capture unwanted video or picture. Distribute or publish these obscene clips on the Internet without the consent of the victims may mislead people specifically the younger ones.

B) Cybercrimes against property: Stealing credit card details, hacking passwords of social media accounts or mail account or Net banking, uploading the latest movies etc, are considered as cyber crimes against property.

i) Credit card fraud: Stealing the details such as credit card number, company name, expiry date, CVV number, password etc. and use these details to make payment for purchasing goods or transfer funds also.
ii) Intellectual property theft: The violation of Intellectual Property Right of Copyright, Trademark, Patent, etc. In the film industry crores of investment are needed to create a movie. Intellectual Property thieves upload the movies on the Releasing day itself. Hence the revenue from the theatres is less significant and undergoes huge loss. (Eg: Premam, Bahubali, etc)
Copying a person’s creation and present as a new creation is called plagiarism. This can be identified as some tools(programs) available in the Internet

iii) Internet time theft: This is deals with the misuse of WiFi Internet facilities. If it is not protected by a good password there is a chance of misuse of our devices (Modem/Router) to access the Internet without our consent by unauthorized persons. Hence our money and volume of data(Package) will lose and we may face the consequences if others make any crimes.

C) Cybercrimes against the government: The cyber crimes against Govt, websites is increased significantly. For example in 2015 the website of the Registration Department of Kerala is hacked and destroys data from 2012 onwards.

i) Cyber terrorism: It deals with attacks against very sensitive computer networks like computer-controlled atomic energy power plants, air traffic controls, Gas line controls, telecom, Metro rail controls, Satellites, etc.. This is a very serious matter and may lead to a huge loss (money and life of citizens). So Govt is very conscious and give tight security mechanism for their services.

ii) Website defacement: It means to spoil or hacking websites and posting bad comments about the Govt.

iii) Attacks against e-governance websites: Its main target is a Web server. Due to this attack, the Web server/ computer forced to restart and this results in refusal of service to the genuine users. If we want to access a website first you have to type the web site address in the URL and press the Enter key, the browser requests that page from the webserver. Dos attacks send a huge number of requests to the webserver until it collapses due to the load and stop functioning.

Cyberethics
Guidelines for using computers over the internet

• Emails may contain Viruses so do not open any unwanted emails
• Download files from reputed sources(sites)
• Avoid clicking on pop-up Advt.
• Most of the Viruses spread due to the use of USB drives so use cautiously.
• Use a firewall on your computer
• Use anti-virus and update regularly
• Use spam blocking software
• Take backups in regular time intervals
• Use strong passwords, i.e a mixture of characters (a-z & A-Z), numbers, and special characters.
• Do not use bad or rude language in social media and emails.
• Untick ‘Remember Me’ before login.

CyberLaws: It ensures the use of computers and the Internet by people safely and legally. It consists of rules and regulations like the Indian Penal Code(IPC) to stop crimes and for the smooth functions of Cyberworld. Two Acts are IT Act 2000 and IT Act Amended in 2008

Information Technology Act 2000(amended in 2008)
IT Act 2000 controls the use of Computer(client), Server, Computer Networks, data, and Information in Electronic format and provides the legal infrastructure for E-commerce, in India.

This is developed to promote the IT industry, control e-commerce also ensures the smooth functioning of E-Governance and it prevents cyber crimes.

The person who violates this will be prosecuted. In India, the IT bill introduced in the May 2000 Parliament Session and it is known as the Information Technology Act 2000. Some exclusions and inclusions are introduced in December 2008

Cyber Forensics: Critical evidence of a particular crime is available in electronic format with the help of computer forensics. It helps to identify the criminal with help of blood, skin, or hair samples collected from the crime site?: DNA, polygraph, fingerprints are other effective tools to identify the accused person is the criminal or not.

Info mania: Right information at the right time is considered the key to success. The information must be gathered, stored, managed, and processed well. Infomania is excessive desire(infatuation) for acquiring knowledge from various modem sources like the Internet, Email, Social media, Instant Message applications (WhatsApp), and Smart Phones. Due to this, the person may neglect daily routines such as family, friends, food, sleep, etc. hence they get tired. They give first preference to the Internet than others. They create their own Cyber World and no interaction with the surroundings and the family.

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 10 Enterprise Resource Planning.

Kerala Plus Two Computer Application Notes Chapter 10 Enterprise Resource Planning

The goal(aim) of the management of an enterprise(Proprietor of a Company or a Venture or an organization) is to handle the resources in a good manner and .make good profit. The resources include the employees, customers, raw materials, finished goods machinery etc… Hence an enterprise handles large amount of data(DataBase) such as employee data, customer data, raw material purchase, sales data, financial data etc. The size of data to be handle is large and hence the complexity is also high. To solve this problem .organizations use ERP packages

Overview of an enterprise
Let us consider a production unit in an enterprise. The activities involved are planning, purchasing raw material, production, storing finished goods(warehouse), sales, finance etc. These activities are performed by different departments and theirduties are interlinked. Altogether the resources are classified into four M’s, That is Man, Material, Money and Machine.

Concepts of Enterprise Resource Planning
An enterprise(organization) is considered as a system(A system is an orderly grouping of interdependent components linked together to achieve an objective, according to a plan. Human body is an example for System). All the departments of an enterprise are connected to a centralized data base. ERP consists of single database and a collection of programs to handle the database hence handle the enterprise efficiently and hence enhance the productivity.

Functional units of ERP
Different modules are given below:
Financial Module: It is the core. This is used to generate financial report such as balance sheet, general ledger, trial balances, financial statements etc.

Manufacturing Module: It provides information for the production and capable to change the methods in the manufacturing sector.

Production planning Module: This module ensures the effective use of resources and helps the enterprise to enhance productivity hence increase profit.

HR (Human Resource) Module: This model ensures the effective use of Human resources and Human capital.

Inventory control Module: This model is useful to maintain the appropriate level of stock(includes raw material, work in progress and finished goods)

Purchasing Module: This module is useful to make available the required raw materials in good condition and in the right time and price.

Marketing Module: It is used for handle the orders of customers.

Sales and distribution Module: The existence of a company is based on the income from sales. This module will help to handle the sales enquiries, order placement ans scheduling, dispatching and invoicing.

Quality (Ql & QC) management module: The quality of a product or service is very much important to a company.This module helps to maintain the quality of the product. Quality planning, inspection and control are the main activities involved in this module.

Business Process Re-engineering (BPR)
In this world, tight competition is based on price, quality, wide variety of selection and quick service. To increase the business and hence increase the profit of a Business firm various activities are involved. IT and Re-engineering play major roles to increase productivity.

In general, BPR is a series of activities such as rethinking and redesign the business process to enhance the enterprise’s performance such as reducing the cost(expenses), improve the quality, prompt, and speed(time-bound) service.

BPR enhances the productivity and profit of an enterprise.

A business process consists of three elements

1. Input – Supply data for processing
2. Processing – Series of activities to convert the input into output
3. Outcome – After processing, we will get the result as output.

The connection between ERP and BPR
ERP and BPR will not make much change if they are stand-alone. To improve the efficiency of an enterprise integrate both ERP and BPR because they are the two sides of a coin. For better results conducting BPR before implementing ERP, will help an enterprise to avoid unnecessary modules from the software.

Implementation of ERP
Wonderful changes are shown if you select and implement the correct ERP. Right ERP implemented at the right time will enhance the productivity and profit of an enterprise.

The different phases of ERP implementation are given below
Pre-evaluation screening: Many ERP packages are available in the markets. At most care should be taken before implementing an ERP. Select a few from the available ERP packages.

Package selection: The selection of the right ERP to our enterprise is a laborious task and it needs huge investment. Various factors should be kept in mind before you purchase an ERP that should meet our complete needs.

Project planning: Good planning is essential to implement an ERP. From the beginning to the end activities are depicted in this phase.

Gap analysis: A cent percent(100%) problem-solving ERP is not available in the market. Most of them solve a maximum of 70% to 80% problems. The rest (30% to 20%) of the problems and their solutions are mentioned here.

Business Process Reengineering: In general BPR is the series of activities such as rethinking and redesign of the business process to enhance the enterprise’s performance such as reducing the cost(expense), improve the quality, prompt and speed(time-bound) service.
BPR enhances the productivity and profit of an enterprise

Installation and configuration: In this phase the new system are installing, before implementing the whole system a miniature of the actual system is going to be implemented as a test dose. Then check the reactions if it is good it is the time to install the whole system completely.

Implementation team training: In this phase the company trains its employees to implement and run the system.

Testing: This phase is very important. It determines whether the system produces proper result. Errors in design and logic are identified.

Going live: Here a change over is taken place to new system from old system. It is not an easy process without the support and service from the ERP vendors.

End-user training: This phase will start familiarising the users with the procedures to be used in the new system. It is very important.

Post-implementation: Once the system is implemented maintenance and review begin. In this phase repairing or correct previously ill-defined problems and upgrade or adjust the performance according to the company needs.

ERP solution providers / ERP packages
The selection of right ERP is a difficult task. Many ERP packages are available in the market. Most of them are too expensive and cannot afford by small enterprises. The reason behind the expensiveness is that the ERP companies investing huge amount of time, money and effort in the research and development of ERP packages.

Popular ERP packages are given below
Oracle
American based company famous in database(Oracle 9i-SQL) packages situated in Redwood shores, California.
ERP package is a solution for finance and accounting problems. Their other products are

1. Customer Relationship Management (CRM)
2. Supply Chain Management (SCM) Software

SAP
SAP stands for Systems, Applications and Products for data processing.
It is a German MNC in Walldorf and founded in 1972. Earlier they developed ERP packages for large MNC. But nowadays they developed for small scale industries also.

The other software products they developed are

1. Customer Relationship Management(CRM)
2. Supply Chain Management(SCM)
3. Product Life cycle Management(PLM)

Odoo
Formerly known as OpenERP.
It is an open-source code ERP. Unlike other companies, their source code is available and can be modified as and when the need arises.

Microsoft Dynamics

• American MNC in Redmond, Washington
• ERP for midsized companies.
• This ERP is more user friendly
• Another s/w is Customer Relationship Management (CRM)

Tally ERP
Indian company situated in Bangalore.
This ERP provides a total solution for accounting, inventory and Payroll.

Benefits and risks of ERP
ERP packages have a lot of advantages as well as many drawbacks also.

Benefits of ERP system

1. Improved resource utilization: Resources such as Men, Money, Material, and Machine are utilized maximum hence increase productivity and profit.

2. Better customer satisfaction: Without spending more money and time all the customer’s needs are considered well. Because the customer is the king of the market. Nowadays a customer can track the status of an order by using the docket number through the Internet.

3. Provides accurate information: Right information at the right time will help the company to plan and manage future cunningly. A company can increase or reduce production based upon the right information hence increase productivity and profit.

4. Decision-making capability: Right information at the right time will help the company to make good decisions.

5. Increased flexibility: A good ERP will help the company to adopt good things as well as avoid bad things rapidly. It denotes flexibility.

6. Information integrity: A good ERP integrates various departments into a single unit. Hence reduce the redundancy, inconsistency, etc.

Risks of ERP implementation

1. High cost: Very huge investment is required to purchase and configure an ERP. Moreover, it requires up gradation or replacement of hardware(Man, computer, or machine) is an additional investment. So small-scale enterprises cannot afford this.

2. Time consuming: The full-fledged implementation of the ERP package needs one or two years. That is highly time-consuming.

3. Requirement of additional trained staff: The existing staff may not capable to work with ERP. To overcome this give proper training to them otherwise appoint trained and experienced employees to cop up.

4. Operational and maintenance issues: The first major problem is that the resistance from the existing employees. To overcome this give awareness to the existing employees. The second problem is that the ERP package is a cyclic process-oriented package. It is a continuous process and should be maintained well otherwise the correct output will not available.

ERP and related technologies
It is an all in one system. It integrates various functions such as raw material purchase, production planning, marketing, financial etc, into a single application.

Product Life Cycle Management (PLM): It manages the entire life cycle of a product. PLM consists of programs to increase the quality and reduce the price by the efficient use of resources.

Customer Relationship Management (CRM): As we know the customer is the king of the market. The existence of a company mainly the customers. CRM consists of programs to enhance the customer’s relationship with the company.

Management Information System (MIS): Management is the decision and policymakers. Good management can make a good decision and that will help to do the business well. A good relationship between Management and employees is a key to success. MIS will collect relevant data from inside and outside of a company. Based on this information produce reports and take appropriate decisions.

Supply Chain Management (SCM): This is deals with moving raw materials from suppliers to the company as well as finished goods from the company to customers. The activities include are inventory(raw materials, work in progress, and finished goods) management, warehouse management, transportation management, etc.

Decision Support System (DSS): It is a computer-based system that takes inputs as business data and after processing it produces good decisions as output that will make the business easier.

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 9 Structured Query Language.

Kerala Plus Two Computer Application Notes Chapter 9 Structured Query Language

SQL – Structured Query Language developed at IBM’s San Jose Research Lab.

The result of the compilation of DDL statements is a set of tables, which are stored in a special file called a data dictionary.

Creating a database in Mysql
CREATE DATABASE <database_name>;
Eg: mysql>CREATE DATABASE BVM;

Opening a database
USE command used to use a database
USE <database_name>;
Eg: mysql>USE BVM;

SHOW command is used to list entire database in our system.
mysql>SHOW DATABASES;

Data Types

1. Char – It is used to store fixed number of characters. It is declared as char(size).

2. Varchar – It is used to store characters but it uses only enough memory.

3. Dec or Decimal – It is used to store numbers with decimal point. It is declared as Dec (size, scale). We can store a total of size number of digits.

4. Int or Integer – It is used to store numbers with¬out decimal point. It is declared as int. It has no argument. Eg: age int.

5. Smallint – Used to store small integers.

6. Date – It is used to store date. The format is yyyy-mm-dd.
Eg: ‘1977-05-28’.

7. Time – It is used to store time. The format is

DDL commands (3 commands)

• Create table
• Avertable
• Drop table

DML commands (4 commands)

• Select
• Insert
• Delete
• Update

DCL (Data Control Language) commands

• Grant
• Revoke

Rules for naming tables and columns

• The name may contain alphabets(A-Z, a-z), digits(0-9), underscore(_) and dollar ($) symbol
• The name must contain at least one character.
• Special characters cannot be used except _ and $
• Cannot be a keyword
• The name must be unique.

Constraints are used to ensure database integrity.

• Not Null
• Unique
• Primary key
• Default
• Auto_increment

Order By – Used to sort rows either in ascending (asc) or descending (desc) order.

Aggregate functions

• Sum() – find the total of a column.
• Avg() – find the average of 3 column.
• Min() – find the smallest value of a column.
• Max() – find the largest value of the column.
• Count() – find the number of values in a column.

Group by clause is used to group the rows. Having clause is used with Group By to give conditions.

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 8 Database Management System.

Kerala Plus Two Computer Application Notes Chapter 8 Database Management System

DBMS means Data Base Management System. It is a tool used to store a large volume of data, retrieve and modify the data as and when required. DBMS consists of data and programs.

Advantages of DBMS

1. Data Redundancy
2. Inconsistency can be avoided
3. Data can be shared
4. Standards can be enforced
5. Security restrictions can be applied
6. Integrity can be maintained
7. Efficient data access
8. Crash recovery

Structure of DBMS

1. Fields – Smallest unit of data. Eg: Name, age, sex, …
2. Record – Collection of related fields.
3. File – Collection of records

Components of DBMS

1. Databases – It is the main component.
2. Data Definition Language (DDL) – It is used to define the structure of a table.
3. Data Manipulation Language (DML) – It is used to add, retrieve, modify and delete records in a database.
4. Users – With the help of programs users interact with the DBMS.

Database Abstraction – Abstraction means hiding, it hides certain details of how data is stored and main-tained.

Levels of Database Abstraction:

1. Physical Level (Lowest Level) – It describes how the data is actually stored in the storage medium.
2. Logical Level (Next Higher Level) – It describes what data are stored in the database.
3. View Level (Highest level) – It is closest to the users. It is concerned with the way in which the individual users view the data.

Data Independence – It is the ability to modify the scheme definition in one level without affecting the scheme definition at the next higher level.

1. Physical Data Independence – It is the ability to modify the physical scheme without causing application programs to be rewritten.
2. Logical Data Independence – It is the ability to modify the logical scheme without causing application programs to be rewritten.

Users of Database

1. Database Administrator
2. Application Programmer
3. New users

Data models – It is a collection of tools for describing data, data relationship, data semantics and consistency problem. 3 models.

1. Hierarchical model
2. Network model
3. Relational model

RDBMS – Relational DataBase Management System. It consists of a collection of relations as database.

Relation means table.

Domain – A pool of possible values from which col-umn values are drawn. ‘

Tuple means rows.

Attributes means columns.

Cardinality – The number of rows.

Degree – The number of columns

View – A view is a virtual table derived from one or more base tables.

Key is used to identify or distinguish a tuple in a relation.

Candidate key – It is used to uniquely identify the row.

Primary key – It is a set of one or more attributes used to uniquely identify a row.

Alternate key – Acandidate key other than the primary key.

Foreign key – A single attribute ora set of attributes, which is a candidate key in another table is called foreign key.

Relational Algebra – It consists of a set of opera¬tions that takes one or two relations as input and produces a new relation as a result.

1. Select operation (σ)
2. Project Operation (π)
3. Cartesian Product
4. Union Operation (∪)
5. Intersection operation (∩)
6. Set difference operation (-)

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 7 Web Hosting.

Kerala Plus Two Computer Application Notes Chapter 7 Web Hosting

Web hosting
Buying or renting storage space to store website in a web server and provide service(made available 24×7) to all the computers connected to the Internet. This is called web hosting. Such service providing companies are called web hosts. Programming languages used are PHP, ASP.NET, JSP.NET, etc.

Types of web hosting
Various types of web hosting services are available. We can choose the web hosting services according to our needs depends upon the storage space heeded for hosting, the number of visitors expected to visit, etc.
1) Shared Hosting
2) Dedicated Hosting
3) Virtual Private Server (VPS)

Buying hosting space
We designed a website of our school and we decide our school website to be made available to all over the world, we have to place the website files on a web server for that we have to purchase hosting space(memory space) in a web server.
Following factors to be considered
1) Buying sufficient amount of memory space for storing ourwebsite files
2) If the web pages contain programming contents supporting technology must be consider
3) Based upon the programs select Windows hosting or Linux hosting

Domain Name System(DNS) Registration
Millions of websites are available over Internet so that ourwebsite must be registered with a suitable name. Domain Name registration is used to identify a website over Internet. A domain name must be unique(i.e. no two website with same name is available). So you have to check the availability of domain name before you register it, for this www.whois.net website will help. If the domain name entered is available then we can register it by paying the Annual registration fees through online.

FTP (File Transfer Protocol) client software When a client requests a website by entering website address. Then FTP client software helps to establish a connection between client computer and remote server computer. Unauthorised access is denied by using username and password hence secure our website files forthat SSH(Secure Shell) FTP simply SFTP is used. Instead of http://, it uses ftp://.
By using FTP client s/w we can transfer(upload) the files from our computer to the web server by using the ‘drag and drop’ method. The popular FTP client software are FileZilla, CuteFTP, SmartFTP, etc.

Free hosting
The name implies it is free of cost service and the expense is meet by the advertisements. Some service providers allow limited facility such as limited storage space, do not allow multimedia(audio and video) files.
A paid service website’s address is as follows
eg: www.bvmhsskalparamba.com

Usually two types of free web hosting services as follows
1) as a directory service.
Service provider’s website address/ ourwebsite address
eg: www.facebook.com / bvm hss kalparambu
2) as a Subdomain
Our website address, service providers website address
eg: bvmhsskalparamba.facebook.com

Earlier web hosting services are expensive but nowadays it is cheaper hence reduced the need for free web hosting.
Example for free web hosting.

Content Management System(CMS)
Do you heard about Data Base Management System(DBMS). DBMS is a software(collection of programs) used to create, alter, modify, delete and retrieve records of a DataBase. Similarly, CMS is a collection of programs that is used to create, modify, update and publish website contents. CMS can be downloaded freely and is useful to design and manage attractive and interactive websites with the help of templates that are available in CMS. WordPress, Joomla, etc are examples of CMS.

Responsive web design
The home page is displayed differently according to the screen size of the browser window(different screen sized devices-mobile phone, palmtop, tablet, laptop, and desktop) we used. The website is designed dynamically(flexibly) that suit the screen size of a different device introduced by Ethan Marcotte. Before this, companies have to design different websites for different screen sized devices. By responsive web design, companies have to design only one website that suitably displayed according to the screen size of the devices. It is implemented by using a flexible grid layout, images, and media queries

Flexible grid layouts: It helps to set the size of the web page to fit the screen size of the device.

Flexible image and video: It helps to set the image or video dimension to fit the screen size of the device.

Media queries: There is an option(settings) to select the size of the web page to match our device, this can be done by using media queries inside the CSS file.

A well known Malayalam daily Malayala Manorama launched their responsive website.

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 6 Client-Side Scripting Using Java Script.

Kerala Plus Two Computer Application Notes Chapter 6 Client-Side Scripting Using Java Script

JavaScript(Original name was Mocha) was developed by Brendan Eich for the Netscape Navigator browser later all the browsers support this.

Getting Started With Javascript

Scripts are small programs embedded in the HTML pages, to write scripts <SCRIPT> tag is used.

Two types of scripts
1. Client scripts – These are scripts executed . by the browser(client) hence reduces network traffic and workload on the server.
2. Server scripts – These are scripts executed by the server and the results as a webpage returned to the client browser.

The languages that are used to write scripts are known as scripting languages. Eg: VB Script, Javascript etc.

Javascript and VB Script are the two client-side scripting languages.

Java script developed by Brendan Eich for the Netscape browser) is a platform-independent scripting language. Means It does not require a particular browser. That is it runs on any browser hence it is mostly accepted scripting language. But VB Script(developed by Microsoft) is a platform-dependent scripting language. Means it requires a particular browser(MS Internet Explorer) to work which is why it is not a widely accepted scripting language.

Attribute makes the tags meaningful

Language attribute specifies the name of the scripting language used.

Example:
<SCRIPT Language=”JavaScript”>
</SCRIPT>

The identifiers are case sensitive (means Name and NAME both are treated as different)

CamelCase: An identifier does not use special characters such as space hence a single word is formed using multiple words. Such a naming method is called CamelCase(without space between words and all the words first character is in upper case letter). These are two types
1) UpperCamelCase : when the first character of each word is capitalised.
Eg. Date Of Birth, JoinTime, etc….
2) LowerCamelCase: when the first character of each word except the first word is capitalised.
Eg. dateOfBirth, joinTime, etc,…

To write anything on the screen the following function is used document.write(string);
eg. document.writefWelcome to BVM HSS, Kalparamba”);

Note: Like C++ each and every statement in javascript must be end with semicolon(;).

To create a web page using javascript

<HTML>
<HEAD><TITLE>JAVASCRIPT- WELCOME</
TITLE></HEAD>
<BODY>
<SCRIPT Language=”JavaScript”>
document.write(“welcome to my first javascript page”);
</SCRIPT>
</BODY>
</HTML>

Its output is as follows

Creating Functions in Javascript

Function: Group of instructions(codes) with a name, declared once can be executed any number of times. There are two types
1) built in and
2) user defined

To declare a function, the keyword function is used.

A function contains a function header and function body

Even though a function is defined within the body section, it will not be executed, if it is not called.
Syntax:

function <function name>()
{
Body of the function;
}
Eg: function print()
{
document.write(“Welcome to JS”);
}

Here function is the keyword.
print is the name of the user defined function
To execute(call) the above function namely print do as follows:
print();
Eg:
<HTML>
<HEAD><TITLE>JAVASCRIPT- functions</
TITLE></HEAD>
<SCRIPT Language=”JavaScript”>
function print()
{
document.write(“welcome to my first javascript page using print function”);
}
</SCRIPT>
<BODY>
<SCRIPT Language=”JavaScript”>
print();
</SCRIPT>
</BODY>
</HTML>

Data Types in Javascript

Unlike C++ it uses only three basic data types
1) Number: Any number(whole or fractional) with or without sign.
Eg: +1977, -38.0003, -100, 3.14157,etc
2) String: It is a combination of characters enclosed within double quotes.
Eg: “BVM”, “[email protected]”, etc
3) Boolean: We can store either true or false.lt is case sensitive. That means can’t use TRUE OR FALSE

Variables in Javascript

For storing values you have to declare a variable, for that the keyword var is used. There is no need to specify the data type.
Syntax:
var<variable name1> [, <variable name2>, <variable name3>, etc…]
Here square bracket indicates optional.
Eg: var x, y, z;
x = 11;
y = “BVM”;
z = false;
Here x is of number type, y is of string and z is of Boolean type.
typeof(): this function is used to return the data type
undefined: It is a special data type to represent variables that are not defined using var.

Operators in Javascript
Operators are the symbols used to perform an operation

Arithmetic operators
It is a binary operator. It is used to perform addition(+), subtraction(-), division(/), multiplication(*), modulus (%-gives the remainder) , increment(++) and decrement(–) operations.
Eg. If x = 10 and y = 3 then

If x = 10 then
document.write(++x); -> It prints 10 + 1 = 11
If x = 10 then
document.write(x++); -> It prints 10 itself.
If x = 10 then
document.write(–x); It prints 10 – 1 = 9
If x = 10 then
document.write(x–); -> It prints 10 itself.

Assignment operators
If a = 10 and b = 3 then a = b.
This statement sets the value of a and b are the same, i.e. it sets a to 3.
It is also called shorthands
If X = 10 and Y = 3 then

Relational(Comparison) operators
It is used to perform a comparison or relational operation between two values and returns either true or false.
Eg:
If X = 10 and Y = 3 then

Logical operators
Here AND(&&), OR(||) are binary operators and NOT(!) is a unary operator. It is used to combine relational operations and it gives either true or false
If X = true and Y = false then

Both operands must be true to get a true value in the case of AND(&&) operation
If X = true and Y = false then

Either one of the operands must be true to get a true value in the case of OR(||) operation
If X = true and Y = false then

String addition operator(+)
This is also called a concatenation operator. It joins(concatenates) two strings and forms a string.
Eg:
var x, y, z;
x = “BVM HSS”;
y = “Kalparamba”;
z = x + y;

Here the variable z becomes “BVM HSS Kalparamba”.

Note: If both the operands are numbers then the addition operator(+) produces a number as a result otherwise it produces a string as a result.

Consider the following

Eg:
1) 8(number) + 3(number) = 11 (Result is a number)
2) 8 (number) + “3”( string) = “83” (Result is a string)
3) “8” (string) + 3 (number) = “83”(Result is a string)
4) “8” (string) + “3” (string) = “83” (Result is a string)

Control Structures in JavaScript
In general, the execution of the program is sequential, we can change the normal execution by using the control structures.

Simple if Syntax:

if(test expression)
{
statements;
}
First the test expression is evaluated,

if it is true then the statement block will be executed otherwise not.

if-else Syntax:

if(test expression)
{
statement block1;
}
else
{
statement block2;
}

First the test expression is evaluated, if it is true then the statement block1 will be executed otherwise statement block2 will be evaluated.

Switch
It is a multiple bratich statement. Its syntax is given below.

switch(expression)
{
case value1: statements;break;
case value2: statements;break;
case value3: statements;break;
case value4: statements;break;
case value5: statements;break;
..................................
default: statements;
}

First expression evaluated and selects the statements with matched case value.

for loop
The syntax of for loop isgiven below

For(initialisation; testing; updation)
{
Body of the for loop;
}

while loop
It is an entry controlled loop The syntax is given below

Loop variable initialised
while(expression)
{
Body of the loop;
Update loop variable;
}

Here the loop variable must be initialised outside the while loop. Then the expression is evaluated if it is true then only the body of the loop will be executed and the loop variable must be updated inside the body. The body of the loop will be executed until the expression becomes false.

Built-in Functions (methods)
1) alert(): This is used to display a message (dialogue box) on the screen.
eg: alert(“Welcome to JS”);
2) isNaN(): To check whether the given value is a number or not. It returns a Boolean value.
If the value is not a number(NaN) then this function returns a true value otherwise it returns a false value.
Eg.

1. isNaN(“BVM”); returns true
2. isNaN(8172); returns false
3. isNaN(“680121″); returns false
4. alert(isNaN(8172); displays a message box as false

3. toUpperCase(): This is used to convert the text to uppercase.
Eg: var x=”bvm”;
alert(x.toUpperCase());

4. toLowerCase(): This is used to convert the text to lowercase.
Eg: var x=”BVM”;
alert(x.toLowerCase());

5. charAt(): It returns the character at a particular position.
Syntax: variable.charAt(index);
The index of first character is 0 and the second is 1 and so on.
Eg.var x=”HIGHER SECONDARY”;
alert(x.charAt(4));
Eg 2.
var x=”HIGHER SECONDARY”;
alert(“The characters @ first position is “+x.charAt(O));

6. length property: It returns the number of characters in a string.
Syntax: variable.length;
Eg.
var x=”HIGHER SECONDARY”;
alert(“The number of characters is “+ x.length);
Output is as follows(note that space is a character)

Accessing Values in a Textbox Using JavaScript.

Name attribute of FORM, INPUT, etc is very important for accessing the values in a textbox.

Consider the following program to read a number and display it

<HTML>
<HEAD><TITLE>JAVASCRIPT- read a value from the console</TITLE>
<SCRIPT Language=”JavaScript”>
function print()
{
var num;
num=document.frmprint.txtprint. value;
document.write(“The number you entered is ” + num);
}
</SCRIPT>
</HEAD>
<BODY>
<FORM Name=”frmprint”>
<CENTER>
Enter a number
< IN PUT Type=”text” name=”txtprint”>
<INPUT Type=”button” value=”Show” onClick= “print()”>
</CENTER>
</FORM>
</BOD Y>
</HTML>

In the above code,
print() is the user-defined function.
onClick is an event(lt is a user action). The function print() is executed when the user clicks the show button. Here code is executed as a response to an event.
frmprintisthe name of the form.
txtprint is the name of the text.

Ways to Add Scripts to a Web Page.

Inside <BODY> section
Scripts can be placed inside the <BODY> section.

Inside <HEAD> section
Scripts can be placed inside the <HEAD> section.
This method is a widely accepted method

External (another) JavaScript file
We can write scripts in a file and save it as a separate file with the extension .js. The advantage is that this file can be used across multiple HTML files and can be enhance the speed of page loading.

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 5 Web Designing Using HTML.

Kerala Plus Two Computer Application Notes Chapter 5 Web Designing Using HTML

3 types of Lists in HTML.

1. Unordered List (<UL>) – Items are displayed with square, circle or disc in front.

2. Ordered List (<OL>) – Items are displayed with the following type values.

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

3. Definition List (<DL>) – It is formed by definitions.
<L/> – It is used to specify List items.
<DT> – It is used to specify Definition Term.
<DD> – Used to specify the description
<A> is used to provide hyperlinks. Two types of linking. Its attribute is HREF.

1. External link – Used to connect 2 different web pages.
2. Internal link – Used to connect different locations of same page.

Concept of URL
URL means Uniform Resource Locator.
Two types of URL
a) Relative URL – Here we explicitly give the web site address
Eg: <A href=http://www.hscap.kerala.gov.in>

b) Absolute URL – Here we implicitly give the website address. The path is not specified here.
Eg: Consider the web pages index.html and school.html saved in the folder C:\BVM.
The file indexs.html contains the following.
<A href=”school.html”>.

Here we did not specify the full path of the file school.html. But this implicitly points to the file stored in C:\BVM

Creating Graphical hyperlinks
It can be achieved by using the <img> tag inside the <a> tag.
Eg: <A href=”school.html”><img src=”schoo|.jpg”></A>

Creating E- mail linking
It can be achieved by using the key word mailto as a value to href attribute
Eg: <A href=mailto:”[email protected]”> SPARK</A>

Insert music and videos
<embed> tag is used to add music or video to the page

Attributes

• src – specifies the file to play
• width – Specifies the width of the player
• height – Specifies the height of the player
• hidden – Used to specifies the player is visible or not
• <noembed> – Used to specifies an alternate when the browser does not support the <embed> tag.

Attribute

• src – Used to specify the image file
• alt – Used to specify the alternate text

Eg:
<html>
<head>
</head>
<body>
Here is a tag embed to play music
<embed src=”c:\alvis.wma” width=”500″ height=”500″ hidden=”true”> </embed>
</body>
</html>

<bgsound> tag
This tag is used to play back ground song or music
Eg:
<html>
<head>
</head>
<body>
<bgsound src=”c:\alvis.wma” loop=”infinite”>
</body>
</html>

• <Table> is used to create a table.
• <TR> is used to create a row.
• <TH> is used to create heading cells.
• <TD> is used to create data cells.

<Table> Attributes

1. Border – It specifies the thickness of the borderlines.
2. Bordercolor – Color for borderlines.
3. Align – Specifies the table alignment in the window.
4. Bgcolor – Specifies background colour.
5. Cellspacing – Specifies space between table cells.
6. Cellpadding – Specifies space between cell border and content.
7. Cols – Specifies the number of columns in the table.
8. Width – Specifies the table width.
9. Frame – Specifies the border lines around the table.
10. Rules – Specifies the rules (lines) and it overrides the border attribute. Values are given below:
    • none – display no rules
    • cols – display rules between columns only(vertical lines)
    • rows – display rules between rows only(horizontal lines)
    • groups – display rules between row group and column groups only
    • all – rules between all rows and columns

<TR> attributes

1. align – specifies the horizontal alignment. Its val¬ues are left, right, centre or justify.
2. Valign – Specifies the vertical alignment. Its values are top, middle, bottom or baseline.
3. Bgcolor – Used to set background-color

<TH> and <TD> attributes

1. Align – specifies a horizontal alignment. Its values are left, right, centre or justify.
2. Valign – Specifies vertical alignment. Its values are top, middle, bottom or baseline.
3. Bgcolor – Specifies border color for the cell.
4. Colspan – Specifiesthenumberofcolumnsspan for the cell.
5. Rowspan – Specifies the number of rows span for the cell.

Frameset – It is used to divide the window into more than one pane. It has no body section.

<Frameset> attributes

1. cols – It is used to divide the window vertically.
2. rows – It is used to divide the window horizontally.
3. border – specifies the thickness of the frame border.
4. bordercolor – specifies the color of the frame border.

Frame – It specifies the pages within a frameset.

<Frame> attributes

1. SRC – specifies the web page.
2. Scrolling – Scroll bar is needed or not its values are yes, no or auto.
3. Noresize – It stops the resizing of the frame.
4. Margin width and Marginheight – Sets margins
5. Name – To give a name for the frame.
6. Target – specifies the target.

<Noframe> – It is used to give content when some browsers that do not support frameset.
Nesting of framesets
Step 6: Finally execute the frame.html file

<Form> – It is used to take data from the users and send to the server.

<Input> – It is used to create input controls. Its type attribute determines the control type.

Main values of the type attribute are given below.

1. Text – To create a text box.
2. Password – To create a password text box.
3. Checkbox – Tq^teate a check box.
4. Radio – To create a radio button.
5. Reset – To create a Reset button.
6. Submit-To creates a submit button.
7. Button – To create a button

To create a group of radio buttons, then the name attribute must be the same.

<Textarea> is used to create a multiline text box. <Label> It is used to give labels.

<Select> It is used to create a list box or combo box. The items must be given by using <option> tag.

Attribute

Name – Specifies the name of the object to identify

Size – If it is 1, the object is a combo box otherwise it is a list box.

Multiple – Allows selecting multiple items

<Form> attributes

1) Action – Here we give the name of the program (including the path) stored in the Webserver.
2) Method – There are 2 types of methods get and post.

3) Target – Specifies the target window for displaying the result. Values are given below.

• _blank – Opens in a new window
• _self – Opens in the same frame
• _parent – Opens in the parent frameset
• _top – Opens in the main browser window
• name – Opens in the window with the specified name.
• <Fieldset> tag

This tag is helpful to divide a form into different subsections and form groups. <legend> tag used to give a caption forthe <fieldset> section.

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 4 Web Technology.

Kerala Plus Two Computer Application Notes Chapter 4 Web Technology

Website – It is a collection of web pages contained text and multimedia(image, audio, video, graphics, animation etc) files.

A webpage is created by HTML tags

The first web page of a website is known as the home page.

www – means world wide web.

Portals – Rediff, Hotmail, Yahoo, etc are called portals from which the user can do multiple activities.

Communication on the Web
Following are the steps that happened in between the user’s click and the page being displayed

1. The browser determines the URL selected.
2. The browser asks the DNS for URLS corresponding IP address (Numeric address)
3. The DNS returns the address to the browser.
4. The browser makes a TCP connection using the IP address.
5. then it sends a GET request for the required file to the server.
6. The server collects the file and sends it back to the browser.
7. The TCP connection is released.
8. The text and the images in the web pages are displayed in the browser.

Client to Web Server Communication
This communication is carried out between the client to the webserver (shopping site). The technology used to protect data that are transferred from client to web server is HTTPS (HyperText Transfer Protocol Secure). This encrypts user name, password etc., and sent to the server. HTTPS works using Secure Sockets Layer (SSL) ensures privacy as well as prevents it from unauthorized access (changes) from other websites. Following are the steps

1. The browser requests a web page to the server.
2. The server returns its SSL certificate.
3. The browser checks the genuinity of the certificate by the authorised certification authority
   (Eg: Veri sign)
4. The certificate authority certifies whether it is valid or not.
5. If it is valid the browser encrypts the data and transmits it. The certificate can be viewed by click on the lock symbol.

Web Server to Web Server Communication
This communication is usually carried out between web server (seller) to another web server (normally bank). For the safe transactions Digital certificate issued by third party web sites are used.
Payment gateway is a server (Computer) that acts as a bridge (interface) between merchant’s server and bank’s server to transfer money.

Web Server Technologies

Web server: A computer with high storage capacity, high speed and processing capabilities is called a web server.

Software ports: The computer is not a single unit. It consists of many components. The components are connected to the computer through various ports. Two types of ports Hardware and Software.

Hardware ports: Monitors are connected through VGA ports and the keyboard or mouse are connected through PS/2 ports.

Software ports: It is used to connect client computers to servers to access different types of services. For example HTTP, FTP, SMTP etc. Unique numbers are assigned to software ports to identify them. It is a 16-bit number followed by an IP address.

DNS Servers
A DNS server is a powerful computer with networking software. It consists of domain names and their corresponding IP addresses. A string address is used to represent a website, it is familiar to humans. The string address is mapped back to the numeric address using a Domain Name System (DNS). It may consist of 3 or 4 parts. The first part is www., the second part is the website name, the third top-level domain, and the fourth geographical top-level domain.
eg.- http://www.nic.kerala.gov.in / results.html.

http – http means hypertext transfer protocol. It is a protocol used to transfer hypertext.
www – World Wide Web. With an email address, we can open our mailbox from anywhere in the world.
nic.kerala – It is a unique name. It is the official website name of the National Informatics Centre.
<script> in – It is the geographical top-level domain. It represents the country, in is used for India.
results.html – It represents the file name.

Web Designing
Any text editor can be used for web designing. Besides that many software tools are available in the market to make the web pages more attractive and interactive, some of the popular softwares are Adobe dream weaver, Microsoft Expression web, Blue fish, Bootstrap etc.

Static and Dynamic Web Pages
Some pages are displaying same content(same text, images,etc) every time. Its content are not changing. This type of web pages are called static page. Conventional wep pages display static pages and has some limitations.
Advanced tools are used to create web pages dynamic, which means pages are more attractive and interactive. For this JavaScript, VBScript, ASP, JSP, PHP, etc are used.
Following are the differences

Scripts
Scripts are small programs embedded in the HTML pages.
<script> tag is used to write scripts The attributes used are
Type – To specify the scripting language
Src – Specify the source file

Two types of scripts
1. Client scripts – These are scripts executed by the browser.
Eg: VB Script, Javascript etc.
2. Server scripts – These are scripts executed by the server.
Eg: ASP, JSP, PHP, Perl, etc.

The languages that are used to write scripts are known as scripting languages.

Scripting Languages
a. JavaScript: Java script(developed by Brendan Eich for the Netscape browser) is a platform independent scripting language. Means It does not require a particular browser. That is it runs on any browser hence it is mostly accepted scripting language.
Ajax: It is a technology to take data from the server and filled in the text boxes without refreshing (without reloading the entire page) the web page. Ajax is Asynchronous JavaScript and Extensible Mark up Language (XML). XML is an Extensible Mark up Language, it allows to create our own new tags. This technology uses JavaScript to perform this function. When we turned off JavaScript features in the browser, the Ajax application will not work.

b. VB Script: VB Script(developed by Microsoft) is a platform dependent scripting language. Means it requires a particular browser(MS Internet Explorer) to work that is’why it is not widely accepted scripting language.

c. PHP (PHP Hypertext Preprocessor)

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

d. Active Server Pages (ASP)

• ASP introduced by Microsoft
• ASP stands for Active Server Page.
• ASP’s are web pages that are embedded with dynamic contents, such as text, HTML tags and scripts.
• An ASP file uses .asp extension.
• In ASP, the script executes in the server and the effect will be sent back to the client computer.
• Here a real time communication exists between the client and server.
• ASP applications are very small.
• The only server used is Microsoft Internet Information Server(IIS), hence it is platform dependant

e. Java Server Pages (JSP)

• JSP introduced by Sun Micro System
• JSP stands for Java Server Page.
• An JSP file uses .jsp extension
• It is platform-independent
• It uses Apache Tomcat webserver
• JSP binds with Servlets (Servlets are Java codes run in Server to serve the client requests).

Cascading Style Sheet (CSS)
It is a style sheet language used for specifying common format like colour of the text, font, size, etc. other than the HML codes. That is CSS file used to separate HTML content from its style.
It can be written in 3 ways as follows:

1. Inline CSSIn the body section of the HTML file
2. Embedded CSS In the head section of the HTML file
3. Linked CSS A separate file(extemal file, eg. bvm.css) with extension .css and can be linked in the web page

Code reusability(just like a function in C++) is the main advantage of CSS and can be used in all the pages in a website

• HTML – Hyper Text Markup Language. Used to create webpage.
• A website is a collection of web pages.
• It was developed by Tim Berners – Lee in 1980 at CERN.
• Lynx, a text only browser for unix.
• Mosaic it is a graphical browser.
• Netscape’ Navigator, Microsoft Internet Explorer, Opera, Ice Weasel,Mozilla FireFox etc. are dif¬ferent browsers.
• Java, C#are programming languages used for web applications.
• HTML files are saved with .htm or .html.
• A web browser is a piece of software used to view web pages.

Structure of an HTML Document

<HTML>
<HEAD>
<TITLE>
give title to the web page here
</TITLE>
</HEAD>
<BODY>
This is the body section.
</BODY>
</HTML>

Tags are keywords used to define the HTML document. Two types of tags Empty and container. The container tag has both an opening and closing tags. But empty tag has an opening tag only, no closing tag.
Eg: empty tag: <hr>, <br> etc.
container tag: <html>, </html>, etc.

Attributes are parameters used for providing additional information within a tag.

An HTML document has 2 sections. Head section and body section.

Attributes of <HTML> tag
1. Dir – This attribute specifies the direction of text displayed on the webpage, values are ltr(left to right), rtl(right to left)
2. Lang – This attribute specifies the language values are En(English), Hi(Hindi), Ar(Arabic), etc
Eg: <HTML dir=”ltr” lang=”Hi”>

The title tag is given in the head section.

Web page contents are given in the body section.

Attributes of the Body tag.
Bgcolor, Background, Text, Link, ALink, VLink, LeftMargin andTopmargin

Heading Tags(6 tags)
<H1 >,<H2>,<H3>,<H4>,<H5> and <H6>.

<H1> provides big heading and <H6> provides smallest

<HR> is used to draw a horizontal line. Its attributes are size, width, no shade and color.

<BR> is used to break a line.

Six Heading tags are used in HTML <H1 > to <H6>.

<B> to make the content Bold.

<I> to make the content in Italics.

<U> to underline the content.

<S> and <STRIKE> – These two are used for striking out the text

<BIG> To make the text size bigger than the normal text

<SMALL> To make the size smaller than the normal text.

<STRONG> The effect is same as <B> tag. That is to emphasize a block of text

<EM> – The effect is same as <i> tag

<SUB> – create a subscript

<SUP> create a superscript

<BLOCKQUOTE> – It is used to give indentation(giving leading space to a line)

<Q> It is used to give text within double quotes

<PRE> (Preformatted text) – This tag is used to display the content as we entered in the text editor.

<ADDRESS> This tag is used to provide information of the author or owner.

<MARQUEE> – This tag is used to scroll a text or image vertically or horizontally.

Attributes of <MARQUEE>

Height – Sets the height of the Marquee text

Width – Sets the width of the Marquee text

Direction – Specifies the scrolling direction of the text such as up, down, left or right

Behavior- Specifies the type such as Scroll, Slide(Scroll and stop)and altemate(to and fro).
<marquee behavior=”scroirscrollamount=”100″> hello</marquee>
<manquee behavior=”slide” scrollamount=”100″> hello</manquee>
<marquee behavior=”alternate” scrollamount= “100”>hello</manquee>

Scrolldelay – Specifies the time delay in seconds between each jump.

scrollamount- Specifies the speed of the text

loop – This specifies the number of times the marquee scroll. Default infinite.

bgcolor – Specifies the back ground colour.

Hspace – Specifies horizontal space around the marquee

Vspace – Specifies vertical space around the marquee

<Div> – Used to define a section or a block of text with the same format.

Attributes
align – Sets the horizontal alignment. Values are left, right, center and justify
Id – Used to give a unique name
Style – Specify a common style to the content for example
<Font> used to specify the font characteristics. Its attributes are size, face, and color.

Special Characters

<IMG> tag is used to insert an image. Its important attributes are align, height, width and alt.

Comments are given by using <!– and → symbols.

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 3 Functions.

Kerala Plus Two Computer Application Notes Chapter 3 Functions

String handling using arrays: A string is a combination of characters hence char data type is used to store the string. A string should be enclosed in double-quotes. In C++ a variable is to be declared before it is used.Eg. “BVM HSS KALPARAMBU”.

Memory allocation for strings: To store “BVM” an array of char type is used. We have to specify the size. Remember each and every string is end with a null (\0) character. So we can store only size-1 characters in a variable. Please note that \0 is treated as a single character. \0 is also called as the delimiter, char school_name[4]; By this, we can store a maximum of three characters.

Consider the following declarations
char my_name[10]=”Andrea”;
char my_name2[]=”Andrea”;
char str[ ]=”Hello World”

In the first declaration 10 Bytes will be allocated but it will use only 6+1 (one for ‘\0’) = 7 Bytes the remaining 3 Bytes will be unused. But in the second declaration the size of the array is not mentioned so only 7 Bytes will be allocated and used hence no wastage of memory. Similarly in the third declaration the size of the array is also not mentioned so only 12( one Byte for space and one Byte for‘\0’) Bytes will be allocated and used hence no wastage of memory.

Input / output operations on strings

Consider the following code

# include<iostream>
using namespace std;
int main()
{
charname[20];
cout<<“Enter your name:”; cin>>name;
cout<<“Hello “<<name;
}

If you run the program you will get the prompt as follows:
Enter your name: Alvis Emerin
The output will be displayed as follows and the “Emerin” will be truncated.
Hello Alvis

This is because of cin statement that will take upto space. Here space is the delimiter. To resolve this gets() function can be used. To use gets() and puts() function the header file stdio.h must be included. gets() function is used to get a string from the keyboard including spaces.
puts() function is used to print a string on the screen. Consider the following code snippet that will take the input including the space.

# include<iostream>
#include<cstdio>
using namespace std;
int main()
{
charname[20];
cout<<“Enter your name:”;
gets(name);
cout<<“Hello”<<name;
}

More console functions
Input functions

Output functions

Stream functions for I/O operations: Some functions that are available in the header file iostream.h to perform I/O operations on character and strings(stream of characters). It transfers streams of bytes between memory and objects. Keyboard and monitor are considered as the objects in C++.

Input functions: The input functions like get( ) (to read a character from the keyboard) and getline() (to read a line of characters from the keyboard) is used with cin and dot(.) operator.

Eg.

# include<iostream>
using namespace std;
int main()
{
char str[80], ch= 'z';
cout<<“enter a string that end with z:”;
cin.getline(str, 80, ch);
cout<<str;
}

If you run the program you will get the prompt as follows:
Enter a string that end with z: Hi I am Jobi. I am a teacherz. My school is BVM HSS
The output will be displayed as follows and the string after ‘z’ will be truncated.
Hi I am Jobi. I am a teacher

Output function: The output functions like put() (to print a character on the screen) and write() (to print a line of characters on the screen) is used with cout and dot(.) operator.

Complex programs are divided into smaller subprograms. These subprograms are called functions.
Eg. main(), clrscr(), sqrt(), strlen(),…

Concept of modular programming: The process of converting big and complex programs into smaller programs is known as modularisation. These small programs are called modules or subprograms or functions. C++ supports modularity in programming called functions.

Merits of modular programming

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

Demerits of modular programming
While dividing the program into smaller ones extra care should be taken otherwise the ultimate result will not be right.

Functions in C++
Some functions that are already available in C++ are called pre defined or built in functions.
In C++, we can create our own functions for a specific job or task, such functions are called user-defined functions.
A C++ program must contain a mainO function. A C++ program may contain many lines of statements(including so many functions) but the execution of the program starts and ends with main() function.

Predefined functions
To invoke a function that requires some data for performing the task, such data is called parameter or argument. Some functions return some value back to the called function.

String functions
To manipulate string in C++ a header file called string.h must be included.
a) strlen() – to find the number of characters in a string(i.e. string length).
Syntax: strlen(string);
Eg.
cout<<strlen(“Computer”); It prints 8.

b) strcpy() – It is used to copy the second string into the first string.
Syntax: strcpy(string1, string2);
Eg.
strcpy(str, “BVM HSS”);
cout<<str; It prints BVM HSS.

c) strcat() – It is used to concatenate the second string into first one.
Syntax: strcat(string1, string2)
Eg.
strcpy(str1, “Hello”);
strcpy(str2, “World”);
strcat(str1, str2);
cout<<str1; It displays the concatenated string “Hello World”

d) strcmp() – it is used to compare two strings and returns an integer.
Syntax: strcmp(string1, string2)

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

Eg.

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
char str1[10], str2[10];
strcpy(str1, “Kiran”);
strcpy(str2, “Jobi”);
cout<<strcmp(str1, str2);
}

It returns a +ve integer.

e) strcmpi() – It is same as strcmp() but it is not case sensitive. That means uppercase and lowercase are treated as same.
Eg. “ANDREA” and “Andrea” and “andrea” these are same.

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
char str1[10], str2[10];
strcpy(str1, "Kiran”);
strcpy(str2, "KIRAN”);
cout<<strcmpi(str1, str2);
}

It returns 0. That is both are the same.

Mathematical functions.
To use mathematical functions a header file called math.h must be included
a) abs() – To find the absolute value of an integer.
Eg. cout<<abs(-25); prints 25.
cout<<abs(+25); prints 25.

b) sqrt() – To find the square root of a number.
Eg. cout<<sqrt(49); prints 7.

c) pow() – To find the power of a number.
Syntax. pow(number1, number2)
Eg. cout<<pow(2, 10); It is equivalent to 210. It prints 1024.

Character functions
To manipulate the character in C++ a header file called ctype.h must be included.
a) isupper() – To check whether a character is in uppercase or not. If the character is in uppercase it returns a value 1 otherwise it returns 0.
Syntax: isupper(char ch);

b) islower() – To check whether a character is in lowercase or not. If the character is in lowercase it returns a value 1 otherwise it returns 0.
Syntax: islower(char ch);

c) isalpha() – To check whether a character is an alphabet or not. If the character is an alphabet it returns a value 1 otherwise it returns 0.
Syntax: isalpha(char ch);

d) isdigit() – To check whether a character is a digit or not. If the character is a digit it returns a value 1 otherwise it returns 0.
Syntax: isdigit(char ch);

e) isalnum() – To check whether a character is an alphanumeric or not. If the character is an alphanumeric it returns a value 1 otherwise it returns 0.
Syntax: isalnum(char ch);

f) toupper() – It is used to convert the given character into uppercase.
Syntax: toupper(char ch);

g) tolower() – It is used to convert the given character into lowercase.
Syntax: tolower(char ch);

User defined functions

Syntax:

Return type Function_name(parameter list)
{
Body of the function
}

1. Return type: It is the data type of the value returned by the function to the called function;
2. Function name: A name given by the user.

Different types of User-defined functions.

1. A function with arguments and return type.
2. A function with arguments and no return type.
3. A function with no arguments and with the return type.
4. A function with no arguments and no return type.

Prototype of functions

Consider the following codes

Method 1

# include<iostream>
using namespace std;
int sum(int n1, int n2)
{
return(n1+n2);
}
int main()
{
int n1, n2;
cout<<“Enter 2 numbers:”; cin>>n1>>n2;
cout<<“The sum is “<<sum(n1, n2);
}

Method 2

#include<iostream>
using namespace std;
int main()
{
int n1, n2;
cout<<“Enter 2 numbers:”; cin>>n1>>n2;
cout<<“The sum is “<<sum(n1, n2);
}
int sum(int n1, int n2)
{
retum(n1+n2);
}

In method 1 the function is defined before the main function. So there is no error.
In method 2 the function is defined after the main function and there is an error called “function sum should have a prototype”. This is because the function is defined after the main function. To resolve this a prototype should be declared inside the main function as follows.

Method 2

# include<iostream>
using namespace std;
int main()
{
int n1, n2;
int sum(int, int);
cout<<“Errter 2 numbers:"; cin>>n1>>n2;
cout<<“The sum is “<<sum(n1, n2);
}
int sum(int n1, int n2)
{
return(n1+n2);
}

Functions with default arguments
We can give default values as arguments while declaring a function. While calling a function the user doesn’t give a value as arguments the default value will be taken. That is we can call a function with or without giving values to the default arguments.

Methods of calling functions: Two types call by value and call by reference.
1. Call by value: In the call by value method, the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
2. Call by reference: In the call by reference method, the address of the original value is passed to the function, if the function makes any change will affect the original value.

Scope and life of variables and functions
a) Local scope – A variable declared inside a block can be used only in the block. It cannot be used in any other block.
Eg.

#include<iostream>
using namespace std;
int sum(int n1, int n2)
{
int s;
s=n1+n2;
return(s);
}
int main()
{
int n1, n2;
cout<<“Enter 2 numbers:"; cin>>n1>>n2;
cout<<“The sum is “<<sum(n1, n2);
}

Here the variable s is declared inside the function sum and has a local scope;

b) Global scope – A variable declared outside of all blocks can be used anywhere in the program.

# include<iostream>
using namespace std;
int s;
int sum(int n1, int n2)
{
s=n1+n2;
return(s);
}
int main()
{
int n1, n2;
cout<<“Enter 2 numbers:"; cin>>n1>>n2;
cout<<“The sum is “<<sum(n1, n2);
}

Here the variable s is declared outside of all functions and we can use variable s anywhere in the program.

Kerala State Board New Syllabus Plus Two Computer Application Notes Chapter 2 Arrays.

Kerala Plus Two Computer Application Notes Chapter 2 Arrays

An array is a collection of elements with same data type Or with the same name we can store many elements, the first or second or third etc can be distinguished by using the index(subscript). The first element’s index is 0, the second element’s index is 1, and so on.

Declaring arrays
Suppose we want to find the sum of 100 numbers then we have to declare 100 variables to store the values. It is laborious work. Hence the need for an array arises.

Syntax: data_type array_name[size];

To store 100 numbers the array declaration is as follows
int n[100]; By this, we store 100 numbers. The index of the first element is 0 and the index of last element is 99.

Memory allocation for arrays
The amount of memory requirement is directly related to its type and size.
int n[100]; It requires 2 Bytes (for each integer)*100 = 200 Bytes.
float d[100]; It requires 4 Bytes(for each float)*100 = 400 Bytes.

Array initialization
Array can be initialized in the time of declaration.
Eg. int age[4] = {16, 17, 15, 18};

Accessing elements of arrays
Normally loops are used to store and access elements in an array.
Eg.

int mark[50], i;
for(i=0; i<50; i++)
{
cout<<“Enter value formark”<<i+1; cin>>mark[i];
}
cout<<“The marks are given below:’’;
for(i=0; i<50; i++)
cout<<mark[i];