Kerala Plus Two Maths Question Paper March 2020 with Answers

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf March 2020 helps in understanding answer patterns.

Kerala Plus Two Maths Previous Year Question Paper March 2020

Time: 2 Hours
Total Score: 60 Marks

Question 1 to 8 carry 3 scores each. Answer any 7 questions.

Question 1.
(i) Let R be a relation in the set N of natural numbers given by R = {(a, b): a = b – 2 }.
Choose the correct answer.
(a) (2, 3) ∈ R
(b) (2, 3) ∈ R
(c) (6, 8) ∈ R
(d) (2, 3) ∈ R
(ii) Let * be a binary operation on the set Z of integers as a*b = a + b + 1. Then find the identity element:
Answer:
(i) (c) (6, 8) ∈ R

(ii) a * e = a
a + e + 1 = a
e + 1 = 0
e = -1

Question 2.
(i) Write two non-zero matrices A and B for which AB = 0
(ii) Express A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\) as the sum of a symmetric matrix and a skew symmetric matrix.
Answer:
(i) A = \(\left[\begin{array}{cc}
4 & 0 \\
8 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
0 & 0 \\
4 & 0
\end{array}\right]\)
Any two square matrices whose product is 0.

(ii) A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\) = A’ = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right]\)
We know that \(\frac{1}{2}\)(A + A’) + \(\frac{1}{2}\)(A – A’) = A
\(\frac{1}{2}\)(A + A’) = \(\frac{1}{2}\)\(\left[\begin{array}{cc}
2 & 1 \\
1 & 6
\end{array}\right]\)
\(\frac{1}{2}\)(A – A’) = \(\frac{1}{2}\)\(\left[\begin{array}{cc}
0 & -3 \\
3 & 0
\end{array}\right]\)
Then we have;
\(\frac{1}{2}\)\(\left[\begin{array}{cc}
2 & 1 \\
1 & 6
\end{array}\right]\) + \(\frac{1}{2}\)\(\left[\begin{array}{cc}
0 & -3 \\
3 & 0
\end{array}\right]\) = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\) = A

Question 3.
Using properties of determinants, prove that
\(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\) = (a – b) (b – c) (c – a)
Answer:
Apply R₂ – R₂ – R₁, R₃ – R₃ – R₁
LHS = \(\left|\begin{array}{ccc}
1 & a & a^2 \\
0 & b-a & b^2-a^2 \\
0 & c-a & c^2-a^2
\end{array}\right|\)
= \(\left|\begin{array}{ccc}
1 & a & a^2 \\
0 & b-a & (b-a)(b+a) \\
0 & c-a & (c-a)(c+a)
\end{array}\right|\)
(b – a)(c – a) = \(\left|\begin{array}{ccc}
1 & a & a^2 \\
0 & 1 & b+a \\
0 & 1 & c+a
\end{array}\right|\)
= (b – a)(c – a)((c + a) – (b + a))
= (b – a)(c – a)((c – b)
= (a – b)(b – c)(c – a)

Question 4.
(i) Which among the following is not true
(a) A polynomial function is always continuous.
(b) A continuous function is always differentiable.
(c) A differentiable function is always continuous.
(d) log x is continuous for all x greater than zero.
(ii) Find \(\frac{d y}{d x}\) if x² + y² + xy = 100
Answer:
(i) (b) A continuous function is always differentiable.

(ii) We have; x² + y² + xy = 100
Differentiating with respect to × we have;
2x + 2y\(\frac{d y}{d x}\) + x\(\frac{d y}{d x}\) + y(1) = 0
⇒ 2x + (2y + x)\(\frac{d y}{d x}\) + y = 0
⇒ (2y + x)\(\frac{d y}{d x}\) = – 2x – y
⇒ \(\frac{d y}{d x}\) = \(\frac{-2 x-y}{2 y+x}\)

Question 5.
(I) Identify the following function:

(a) sin x
(b) |sin x|
(c) sin |x|
(d) cos x
(ii) Is the above function differentiable? Why?
(iii) Find derivative of y = \(\sqrt{\tan x}\)
Answer:
(i) (b) |sin x|

(ii) Not differentiable since the graph is not a smooth curve. There is sharp corners at integer multiples of π

(iii) Given; y = \(\sqrt{\tan x}\)
\(\frac{d y}{d x}\) = \(\frac{1}{2 \sqrt{\tan x}}\) sec² x

Question 6.
(i) The slope of the tangent to the curve y = e^2xat (0, 1) is
(a) 1
(b) 2
(c) 0
(d) -1
(ii) Find the equation of a line perpendicular to the above tangent (tangent obtained in part(i)) and passing through (2, 3)
Answer:
(i) (b) 2
Given; y = e^2x = \(\frac{d y}{d x}\) ⇒ 2e^2x
Slope of tangent = 2e²⁽⁰⁾ = 2(1) = 2

(ii) Slope of the perpendicular line = –\(\frac{1}{2}\)
Equation of the perpendicular line is
y – 3 = –\(\frac{1}{2}\)(x – 2)
2y – 6 = – x + 2
x – 2y – 8 = 0

Question 7.
(i) The general solution of a differential equation contains 3 arbitrary constants. Then what is the order of the differential equation?
(a) 2
(b) 3
(c) 0
(d) 1
(ii) Check whether y = e^-3x is a solution of the differential equation \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) – 6y = 0
Answer:
(i) (b) 3

(ii) Given; y = e^-3x
⇒ \(\frac{d y}{d x}\) = -3e^-3x
⇒ \(\frac{d^2 y}{d x^2}\) = 9e^-3x
\(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) – 6y = 9e^-3x – 3e^-3x – 6e^-3x = 0

Question 8.
Consider the following figure:
(i) The equation of the plane PRQS is
(a) y = 0
(b) y = 2
(c) 2 = 4
(d) x = 3

(ii) Find the equation of the plane through the intersection of the planes PRQS and PSUT and the point (2, 1, 2)

Answer:
(i) (b) y = 2

(ii) The equation of the plane passing through PRQS is y – 2 = 0. The equation of the plane passing through PSQT is z – 4 = 0. Then equation of the passing through the intersection is
y – 2 + k(z – 4) = 0
Since it passes through (2, 1, 2)
1 – 2 + k(2 – 4) = 0
– 1 – 2k = 0 = 4 ⇒ 2k = -1 = 4 ⇒ k = –\(\frac{1}{2}\)
Therefore the equation of the plane is
y – 2 – \(\frac{1}{2}\)(z – 4) = 0
2y – 4 – z + 4 = 0 ⇒ 2y – z = 0

Questions 9 to 18 cany 4 scores each. Answer any 8.

Question 9.
Let A = R – {3} and B = R – {1} Consider the function f : A → B defined by f(x) = \(\frac{x – 2}{x – 3}\)
(i) Is f one-one and onto? Justify your answer.
(ii) Is it invertible? Why?
(iii) If invertible, find inverse of f(x)
Answer:
(i) f(x₁) = f(x₂)
\(\frac{x_1-2}{x_1-3}\) = \(\frac{x_2-2}{x_2-3}\)
⇒ (x₁ – 2)(x₂ – 3) = (x₂ – 2)(x₁ – 3)
⇒ x₁x₂ – 2x₂ – 3x₁ + 6 = x₁x₁ – 2x₁ – 3x₁ + 6
⇒ – 2x₂ – 3x₁ = – 2x₁ – 3x₂ ⇒ x₁ = x₂
So f is one-one.
Let y ∈ B
f(x) = y = \(\frac{x – 2}{x – 3}\)
y(x – 3) = x – 2)
xy – 3y = x – 2 ⇒ xy – x = 3y – 2
⇒ x(y – 1) = 3y – 2
⇒ x = \(\frac{3 y-2}{y-1}\) ∈ A
So f in onto.
(ii) Since f is one-one and onto, f Is invertible.
(iii) f^-1(x) = \(\frac{3 x-2}{x-1}\) is the inverse function.

Question 10.
(i) If xy < 1, tan^-1x + tan^-1y = …………………….
(a) tan^-1(\(\frac{x-y}{1+x y}\))
(b) tan^-1(\(\frac{x+y}{1-x y}\))
(c) \(\frac{\tan x+\tan y}{1-\tan x \tan y}\)
(d) \(\frac{\tan x-\tan y}{1+\tan x \tan y}\)
(ii) Solve tan^-1 2x + tan^-1 3x = \(\frac{\pi}{4}\)
Answer:
(i) (b) tan^-1(\(\frac{x+y}{1-x y}\))

(ii) tan^-1(\(\frac{2 x+3 x}{1-(2 x)(3 x)}\)) = \(\frac{\pi}{4}\)
\(\frac{2 x+3 x}{1-(2 x)(3 x)}\) = tan\(\frac{\pi}{4}\)
\(\frac{5 x}{1-6 x^2}\) = 1
5x = 1 – 6x²
⇒ 6x² + 5x – 1 = 0
x = -1, \(\frac{1}{6}\)
x = -1 does not satisfy the given equation so the solution x = \(\frac{1}{6}\)

Question 11.
(i) Find \(\frac{d y}{d x}\), if y = x^x + x^{sin x}
(ii) If y = x cos x, find \(\frac{d^2 y}{d x^2}\)
Answer:
(i) Given, y = x^x + x^{sin x}
Let, y = u + v
\(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) …………….. (1)
u = x^x take log on both sides,
logu = xlogx ⇒ \(\frac{1}{u}\)\(\frac{d u}{d x}\) = x\(\frac{1}{x}\) + logx
\(\frac{d u}{d x}\) = u (1 + logx) = x^x (1 + logx)
v = x^{sin x}, take log on both sides;

logv – sin xlogx ⇒ \(\frac{1}{v}\)\(\frac{d v}{d x}\) = sin x\(\frac{1}{x}\) + logx. cosx
\(\frac{d v}{d x}\) = v (\(\frac{\sin x}{x}\) + cos x log x)
= x^{sin x}(\(\frac{\sin x}{x}\) + cos x log x)
Then (1) implies;
\(\frac{d y}{d x}\) = x^x(1 + log x) + x^{sin x}(\(\frac{\sin x}{x}\) + cos x log x)

(ii) \(\frac{d y}{d x}\) = x(-sin x) + cos x
\(\frac{d^2 y}{d x^2}\) = -x(cos x) – sin x – sin x
\(\frac{d^2 y}{d x^2}\) = -x(cos x) – 2 sin x

Question 12.
(i) \(\int \frac{f(x)}{\tan x} d x\) = log |tan x| + c, then f(x) is
(a) cot x
(b) sec² x
(c) cosec² x
(d) cot² x
ii) if \(\frac{d f(x)}{d x}\) = 4x³ – \(\frac{3}{x^4}\), given f(2) = 0. Find f(x)
Answer:
(i) (b) sec² x

Question 13.
(i) Area bounded by the curve y = f(x), x axis and the lines x = a and x = b Is
(a) \(\int_a^b x d y\)
(b) \(\int_a^b y d x\)
(c) \(\int_a^b x^2 d y\)
(d) \(\int_a^b y^2 d y\)
(ii) From the following figure, find the area of the region bounded by the curves y = sin x, x = cos x and x axis as x varies from 0 to \(\frac{\pi}{2}\)

Answer:

Question 14.
(i) From the differential equation corresponding to the curve y = mx
(ii) Solve \(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
Answer:
(i) Given y = mx
Differentiating on both sides with respect to x
we get; \(\frac{d y}{d x}\) = m ⇒ y = \(\frac{d y}{d x}\)x

(ii) \(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
Then P = \(\frac{1}{x}\),Q = x²
Integrating factor = \(e^{\int P d x}=e^{\int \frac{1}{x} d x}\) = e^{log x} = x
The solution is
y.IF = ∫Q.IFdx + c
xy = ∫x. x²dx + c
xy = ∫x³dx + c
xy = \(\frac{x^4}{4}\) + c

Question 15.
Find a unit vector perpendicular to the plane ABC where A, B, C are points (1, 1, 2), (2, 3, 5) and (1, 5, 5)
Answer:

Question 16.
The Cartesian equation of two lines are \(\frac{x+1}{7}\) = \(\frac{y + 1}{-6}\) = \(\frac{z + 1}{1}\) and \(\frac{x – 3}{1}\) = \(\frac{y – 5}{-2}\) = \(\frac{z – 7}{1}\)
(i) Write the vector equations.
(ii) Find the shortest distance between these two lines.
Answer:
(i) \(\vec{r}\) = – î – ĵ – k̂ + λ(7î – 6ĵ + k̂)
\(\vec{r}\) = 3î + 5ĵ + 7k̂ + μ(î – 2ĵ + k̂)

(ii) \(\vec{a}\)₁ = – î – ĵ – k̂, \(\vec{a}\)₂ = 3î + 5ĵ + 7k̂
\(\vec{b}\)₁ = 7î – 6ĵ – k̂, \(\vec{b}\)₂ = î – 2ĵ + k̂
\(\vec{a}\)₂ – \(\vec{a}\)₁ = 4î + 6ĵ + k̂

Question 17.
(i) If a plane intersect the coordinate axes at a, b, c respectively, write the equation of the plane.
(ii) Find the distance of the plane obtained in part (i) from the origin.
(iii) Find the vector and Cartesian equations of the plane passing through (1, 0, -2) and normal to the plane is î + ĵ – k̂.
Answer:
(i) \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1

(ii) \(\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)

(iii) The perpendicular direction ratios required plane is 1, 1, -1.
The equation is
1(x – 1) + (y – 0) – 1(2 + 2) = 0
x – 1 + y – z – 2 = 0
x + y – 2 – 3 = 0

Question 18.
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6, find
(i) P(A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)
Answer:
P(A) = 0.3, P(B) = 0.6
P(A’) = 0.7, P(B’) = 0.4
(i) P(A and B) = P(A∩B) = P(A) × P(B)
= 0.3 × 0.6 = 0.18

(ii) P(A and not B) = P(A∩B’) == P(A) × P(B’)
= 0.3 × 0.4 = 0.12

(iii) P(A or B) = P(A∪B)
= P(A) + P(B) – P(A∩B)
= 0.3 + 0.6 – 0.18 = 0.72

(iv) P(neither A nor B) = P(A∩B’)
= P(A’) × P(B’)
= 0.7 × 0.4 = 0.28

Questions from 19 to 25 carry 6 scores each. Answer any 5.

Question 19.
(i) Let A = [a_ij]_{2 × 3} where a_ij = i + j. Construct A.
(ii) Find AA’ and hence prove that AA’ is symmetric.
(iii) For any square matrix A, prove that A + A’ is symmetric.
Answer:
(i) A = \(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5
\end{array}\right]\)

(ii) A’ = \(\left[\begin{array}{ll}
2 & 3 \\
3 & 4 \\
4 & 5
\end{array}\right]\)
AA’ = \(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5
\end{array}\right]\) × \(\left[\begin{array}{ll}
2 & 3 \\
3 & 4 \\
4 & 5
\end{array}\right]\)
= \(\left[\begin{array}{lll}
29 & 38 \\
38 & 50
\end{array}\right]\)
(AA’)’ = \(\left[\begin{array}{lll}
29 & 38 \\
38 & 50
\end{array}\right]\) = AA’

(iii) (A + A’)’ = A’ + (A’)’ = A’ + A
Therefore, A + A’ is symmetric.

Question 20.
(i) If A is a skew symmetric matrix of order 3. Then prove that it’s determinant is zero, (without using examples)
(ii) given \(\left|\begin{array}{ccc}
2+x & 3 & 4 \\
1 & -1 & 2 \\
x & 1 & 5
\end{array}\right|\) is a singular matrix.
Find the value of x.
(iii) Given A and B are square matrices of order
2 such that |A| = -1, |B| = 3. Find |3AB|.
Answer:
(i) Given A is skew symmetric matrix, then
A’ = -A
|A’| =|-A| = (-1)³|A| = -|A|
|A| = -|A|
2 |A| = 0 ⇒ |A| = 0
Given \(\left|\begin{array}{ccc}
2+x & 3 & 4 \\
1 & -1 & 2 \\
x & 1 & 5
\end{array}\right|\) = 0
(2 + x)(- 5 – 2) – 3(5 – 2x) + 4(1 + x) = 0
(2 + x)(- 7) + 15 + 6x + 4 + 4x = 0
– 14 – 1x + 15 + 6x + 4 + 4x = 0
3x – 25 = 0 ⇒ x = \(\frac{25}{3}\)

Question 21.
(i) Find the intervals in which the function f(x) = x² + 2x – 5 strictly increasing or decreasing.
(ii) Find the equation of tangent and normal for the curve y = x³ at (1, 1)..
(iii) Find local maximum and local minimum if. any for the function h(x) = sin x + cos x ,
0 < x < \(\frac{\pi}{2}\).
Answer:
(i) f(x) = x² + 2x – 5
f'(x) = 2x + 2 = 0 ⇒ x = -1
The turning point divide the domain into the following intervals (-∞, -1); (-1, ∞)
f'(-2) = 2(-2) + 2 < 0 So decreasing in (-∞, -1) f'(0) = 2(0) + 2 > 0
So increasing in (-1, ∞)

(ii) y = x³ ⇒ \(\frac{d y}{d x}\) = 3x²
Slope at (1, 1) = 3(1)² = 3
Equation of tangent is
y – 1 = 3(x – 1) ⇒ y – 3x + 2 = 0
Equation of normal is
y – 1 = \(\frac{1}{3}\)(x – 1) ⇒ 3y + x – 4 = 0

(iii) h(x) = sin x + cos x
h²(x) = (sin x + cos x)²
= sin² x + cos² x + 2 sin x cos x
= 1 + sin2x
The maximum value of sin 2x = 1
Maximum value of h²(x) is 2
Hence maximum value of h(x) = √2.

Question 22.
Integrate the following
(i) \(\int \frac{d x}{1+\frac{x^2}{4}}\)
(ii) \(\int \frac{x}{(x-1)(x-2)} d x\)
(iii) \(\int_0^{\frac{\pi}{2}} x \cos x d x\)
Answer:

Question 23.
(i) If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three coplanar vectors, then [\(\vec{a}\), \(\vec{b}\), \(\vec{c}\)] is
(a) 1
(b) 0
(c) -1
(d) not defined.

(ii) If |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3 and θ is the angle between \(\vec{a}\) and \(\vec{b}\). Then the maximum value of \(\vec{a}\) . \(\vec{b}\) occurs when θ = …………………
(a) \(\frac{\pi}{2}\)
(b) π
(c) 0
(d) \(\frac{\pi}{4}\)
(iii) If \(\vec{b}\) = 2î + ĵ – k̂, \(\vec{c}\) = î + 3k̂ and \(\vec{a}\) is a unit vector. Find the maximum value of scalar triple product [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
Answer:
(i) (b) 0

(ii) (c) 0, maximum value is obtained when angle between them is 90°.

(iii) \(\vec{b}\) × \(\vec{c}\) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & -1 \\
1 & 0 & 3
\end{array}\right|\) = 3î – 7ĵ – k̂
The maximum value of [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] is obtained when \(\vec{a}\) is perpendicular to \(\vec{b}\) × \(\vec{c}\). Since \(\vec{a}\) is a unit vector the maximum value of [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)] is |\(\vec{b}\) × \(\vec{c}\)| = \(\sqrt{9+49+1}\) = \(\sqrt{59}\)

Question 24.
Solve the linear programming problem graphically:
Max: Z = 3z + 2y
Subject to:
x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, y ≥ 0
Answer:

Solving
x + 2y = 10; 3x + y = 15
we get B(4, 3)

Maximum Z = 18 at (4,3) x = 4, y = 3

Question 25.
The probability distribution of a random variable X in given in the following table:

X	0	1	2	3	4
P(X)	0.1	k	2k	2k	k

(i) Find k
(ii) Find the probability that X lies between 1 and 4.
(iii) Find the mean of X.
(iv) Find Variance of X.
Answer:
(i) 6k + 0.1 = 1
6k = 0.9
k = \(\frac{0.9}{6}\) = 0.15

(ii) P(1 < x < 4) = P(2) + P(3)
2k + 2k = 4k = 4 × 0.15 = 0.6

(iii)

x	P(x)	x P(x)	x∧2P(x)
0	0.1	0	0
1	0.15	0.15	0.15
2	0.3	0.6	1.2
3	0.3	0.9	2.7
4	0.15	0.6	2.4
		2.25	6.45

Mean = 2.25
Variance = Σx²P(x) – (ΣxP(x))²
= 6.45 – (2.25)² = 1.3875