Kerala Plus Two Maths Question Paper March 2021 with Answers

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Kerala Plus Two Maths Previous Year Question Paper March 2021

Time: 2 Hours
Total Score: 60 Marks

Answer the following questions from 1 to 29 up to a maximum score of 60.

PART – A

Questions from 1 to 10 carry 3 scores each. (10 × 3 = 30)

Question 1.
Find the values of x for which \(\left|\begin{array}{ll}
3 & x \\
x & 1
\end{array}\right|\) = \(\left|\begin{array}{ll}
3 & 2 \\
4 & 1
\end{array}\right|\) (3)
Answer:
Given \(\left|\begin{array}{ll}
3 & x \\
x & 1
\end{array}\right|\) = \(\left|\begin{array}{ll}
3 & 2 \\
4 & 1
\end{array}\right|\)
Expanding on both sides
3 – x² = 3 – 8 = – 5
x² = 8
x = ±√8 = ±2√2

Question 2.
Let A = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right|\)
(i) Find adj A (2)
(ii) FindA.adjA. (1)
Answer:

Question 3.
Find the value of k so that the function is continuous at x = 5(3)
Answer:
Given f(x) =
Since f(x) is continuous at x = 5,
LHL = RHL = f(5)
f(5) = k(5) + 1 = 5k + 1 _______ (1)
RHL = \(\lim _{x \rightarrow 5^{+}} f(x)\) = \(\lim _{x \rightarrow 5^{+}} 3 x-5\) = 10 ________ (2)
From (1) and (2),
5k + 1 = 10.
5k = 9
k = \(\frac{9}{5}\)

Question 4.
Verify Rolle’s theorem for the function
f(x) = x² + 2x – 8, x ∈ [-4, 2] (3)
Answer:
Given f(x) = x² + 2x – 8, x ∈ [-4, 2]
Since f(x) is a polynomial function, it is continuous in [-4, 2]
f (x) = 2x + 2
∴It is differentiable in (-4, 2)
f(a) = f(-4)= 16 – 8 – 8 = 0
f(b) = f(2) = 4 + 4 – 8 = 0
∴ f(a) = f(b)
f (c) = 0 ⇒ 2c + 2 = 0 ⇒ c = -1 ∈ (-4, 2)
Hence Rolle’s theorem verified.

Question 5.
Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm. (3)
Answer:
Let A = π r²
\(\frac{d A}{d r}\) = 2 πr
\(\left.\frac{d A}{d r}\right]_{r=5}\) = 2π × 5 = 10 π

Question 6.
Find the projection of vector î + 3ĵ + 7k̂ on the vector 7î – ĵ + 8k̂ (3)
Answer:
Let \(\vec{a}\) = î + 3ĵ + 7k̂
\(\vec{b}\) = 7î – ĵ + 8k̂
\(\vec{a}\) . \(\vec{b}\) = 7 – 3 + 56 = 60
|\(\vec{b}\)| = \(\sqrt{49+1+64}\) = \(\sqrt{144}\)
Projection of \(\vec{a}\) on \(\vec{b}\) = \(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}}{|\overline{\mathrm{~b}}|}\) = \(\frac{60}{\sqrt{114}}\)

Question 7.
Find the equation of a plane passing through the point (1, 4, 6) and the normal to the plane is î – 2ĵ + k̂. (3)
Answer:
Given (x₁, y₁, z₁ ) = (1, 4, 6)
Direction ratios of normal,
<a, b, c> = <1, -2, 1>
Equation of plane is
a(x – x₁) + b(y – y₁) + (z – z₁) = 0
1(x – 1) – 2(y – 4) + 1(z – 6)= 0
x – 2y + z + 1 = 0

Question 8.
(i) Which of the following can be the domain of the function cos^-1 x?
(a) (0, π)
(b) (0, π)
(c) (-π, π)
(d) (\(\frac{-\pi}{2}\) \(\frac{\pi}{2}\)) (1)
(ii) Find the value of cos^-1 (-1/2) + 2 sin^-1(1/2). (2)
Answer:
(i) There is no correct option.
Answer is [-1, 1]

(ii) cos^-1(\(\frac{-1}{2}\)) + 2 sin^-1(\(\frac{1}{2}\))
= π – \(\frac{\pi}{3}\) + 2 × \(\frac{\pi}{6}\) = \(\frac{2 \pi}{3}\) + \(\frac{\pi}{3}\) = \(\frac{2 \pi}{3}\) = π

Question 9.
Find the are of a triangle with vertices (-2, -3), (3, 2) and (-1,-8). (3)
Answer:
Given vertices are (-2, -3, (3, 2), (-1, -8)
Area = \(\frac{1}{2}\)\(\left|\begin{array}{ccc}
-2 & -3 & 1 \\
3 & 2 & 1 \\
-1 & -8 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[|-2(2 + 8) + 3(3 + 1) + 1(-24 + 2)|]
= \(\frac{1}{2}\)[|-20 + 12 – 22|] = \(\frac{1}{2}\)|-30|
= 15 sq. units

Question 10.
Find the general solution of the differential equation \(\frac{d y}{d x}\) – y = cos x (3)
Answer:
Given \(\frac{d y}{d x}\) – y = cos x
Which is a linear differential equation of the form \(\frac{d y}{d x}\) +Py = Q
Where P = -1, Q = cos x
Integrating factor (I. F) = \(e^{\int P d x}\) = \(e^{\int-1 d x}\) = e^-x
∴ Solution is Y (I. F) = \(\int Q(I F) \mathrm{d} x\)
ye^-x = \(\int \cos x \cdot e^{-x} d x\)
ye^-x = \(\cos x \cdot \frac{e^{-x}}{-1}-\int-\sin x \cdot \frac{e^{-x}}{-1} \mathrm{~d} x\)
ye^-x = \(\frac{e^{-x}}{2}[\sin x-\cos x]+C\)

PART – B

Questions from 11 to 22 carry 4 scores each. (12 × 4 = 48)

Question 11.
Consider the matrices
A = \(\left[\begin{array}{cc}
3 & 4 \\
-5 & -1
\end{array}\right]\) and 3A + B = \(\left[\begin{array}{cc}
2 & 8 \\
3 & -4
\end{array}\right]\)
(i) Find the matrix B. (2)
(ii) Find AB. (2)
Answer:

Question 12.
If A = \(\left[\begin{array}{c}
-2 \\
4 \\
5
\end{array}\right]\) and B = [1, 3, -6]
(i) What is the order of AB? (1)
(ii) Verify (AB)’ = B’A’ (3)
Answer:
(i) Order of A= 3 × 1
Order of B = 1 × 3
∴ Order of AB = 3 × 3

Question 13.
(i) If xy < 1, tan^-1 x + tan^-1 y = __________
(a) tan^-1 \(\frac{x-y}{1+x y}\)
(b) tan^-1 \(\frac{1-x y}{x+y}\)
(c) tan^-1 \(\frac{x+y}{1-x y}\)
(d) tan^-1 \(\frac{x-y}{1+x y}\) (1)
(ii) Prove that tan^-1\(\frac{2}{11}\) + tan^-1\(\frac{7}{24}\) = tan^-1\(\frac{1}{2}\) (3)
Answer:
(i) (c) tan^-1(\(\frac{x+y}{1-x y}\))

Question 14.
Find \(\frac{d y}{d x}\)
(i) x² + xy + y² = 100 (2)
(ii) y = sin^-1(\(\frac{2 x}{1+x^2}\)), -1 ≤ x ≤
Answer:
(i) Given x² + xy + y² = 100
Differentiating with respect to x,
2x + x\(\frac{d y}{d x}\) + y + 2y \(\frac{d y}{d x}\) = 0
(x + 2y)\(\frac{d y}{d x}\) = – 2x – y
\(\frac{d y}{d x}\) = \(\frac{-(2 x+y)}{x+2 y}\)

(ii) We have sin^-1(\(\frac{2 x}{1+x^2}\)) = 2 tan^-1 x
∴ y = 2 tan^-1 x
\(\frac{d y}{d x}\) = 2 × \(\frac{1}{1+x^2}\) = \(\frac{2}{1+x^2}\)

Question 15.
Find the intervals in which the function f given by f(x) = 2x² – 3x is
(i) increasing
(ii) decreasing (4)
Answer:
Given f(x) = 2x² – 3x
f'(x) = 4x – 3
f'(x) = 0 ⇒ 4x – 3 = 0
x = \(\frac{3}{4}\)

Intervals are (-∞, 3/4), and (3/4, ∞)
In (-∞, 3/4), f'(0) = 0 – 3 = -3 < 0 In (\(\frac{3}{4}\), ∞) f'(1) = 4 – 3 = 1 > 0
∴ f(x) is decreasing in (-∞, 3/4)
and increasing in (3/4, ∞)

Question 16.
(i) Find the order and degree of the differential equation (\(\frac{d s}{d t}\))⁴ + \(\frac{3 \mathrm{~d}^2 \mathrm{~s}}{\mathrm{dt}^2}\) = 0. (1)
(ii) Find the general solution of the differential equation \(\frac{d y}{d x}\) = (1 + x²)(1 + y²). (3)
Answer:
(i) Order = 2
Degree = 1

(ii) Given \(\frac{d y}{d x}\) = (1 + x² )(1 + y²)
⇒ \(\frac{d y}{1+y^2}\) =(1 + x²)dx
Which is variable seperable.
∴ Solution is \(\int \frac{d y}{1+y^2}=\int\left(1+x^2\right) d x\)
⇒ tan^-1 y = x + \(\frac{x^3}{3}\) + C
or y = tan¹ x + \(\frac{x^3}{3}\) + C

Question 17.
Find a unit vector both perpendicular to the vectors if
\(\vec{a}\) = 2î + ĵ + 3k̂ and \(\vec{b}\) = 3î + 5ĵ – 2k̂ (4)
Answer:
Given \(\vec{a}\) = 2î + ĵ + 3k̂, \(\vec{b}\) = 3î + 5ĵ – 2k̂ vector perpendicular to both \(\vec{a}\) & \(\vec{b}\) is

Question 18.
Find the shortest distance between the skew lines
\(\vec{r}\) = î + 2ĵ + k + λ(î – ĵ + k) and
\(\vec{r}\) = 2î – ĵ – k̂+ μ(2î + ĵ + 2k̂) (4)
Answer:
\(\vec{a}\)₁ = î + 2ĵ + k̂, \(\vec{b}\)₁ = i – ĵ + k̂
\(\vec{a}\)₂ = 2î – ĵ – k̂, \(\vec{b}\)₂ = 2i + ĵ + 2k̂
\(\vec{a}\)₂ – \(\vec{a}\)₁ = î – 3ĵ – 2k̂

Question 19.
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4. Find
(i) P(A∩B) (2)
(ii) P(A | B) (1)
(iii) P(A∪B) (1)
Answer:
(i) P(B/A) = \(\frac{P(A \cap B)}{P(A)}\)
∴ P(A∩B) = P(A). P(B/A)
= 0.8 × 0.4 = 0.32

(ii) P(A/B) = \(\frac{P(A \cap B)}{P(B)}\) = \(\frac{0.32}{0.5}\) = 0.64

(iii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.8 + 0.5 – 0.32 = 0.98

Question 20.
(i) Let R be the relation on a set A = {1, 2, 3}, defined by R = {(1, 1), (2, 2), (3, 3), (1, 3)}. Then the ordered pair to be added to R to make it a
smallest equivalence relation is ___________
(a) (2, 1)
(b) (3, 1)
(c) (1, 2)
(d) (1, 3) (1)
(ii) Determine whether the relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x} is reflexive, symmetric and transitive. (3)
(a) (2, 1)
(b) (3, 1)
(c) (1, 2)
(d) (1, 3)
Answer:
(i) (b)(3, 1)

(ii) Given A = {1, 2, 3, 4, 5, 6}
R = {(x, y): y is divisible by x}
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4),
(5, 5), (6, 6)}
Since (a, a) ∈ R for all a ∈ A, R is reflexive.
(1, 2) ∈ R, but (2, 1) ∉ R,
∴ R is not symmetric.
(a, b) ∈ R,(b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c, ∈ A
∴ R is transitive.

Question 21.
Find \(\frac{d y}{d x}\)
(i) x^x (2)
(ii) x = 2at² ; y = at⁴ (2)
Answer:
(i) Let y = x³
Taking ‘log’ on both sides
Log y = x. log x
Differentiating with respect to x,
\(\frac{1}{y}\)\(\frac{d y}{d x}\) = x.\(\frac{1}{x}\) + logx.1
\(\frac{d y}{d x}\) = y[1 + log x] = x^x [1 + log x]

(ii) Given x = 2at², y = at⁴
\(\frac{d x}{d t}\) = 4 at, \(\frac{d y}{d x}\) 4a t³
∴ \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\) = \(\frac{4 a t^3}{4 a t}\) = t²

Question 22.
Integrate:
\(\int \frac{x}{(x+1)(x+2)} d x\) (4)
Answer:
Let \(\frac{x}{(x+1)(x+2)}\) = \(\frac{\mathrm{A}}{x+1}\) + \(\frac{\mathrm{B}}{x+2}\)
x = A(x + 2) + B(x + 1)
Put x = -1, -1 = A+ 0 ⇒ A = -1
x = -2, -2 = 0 + B(-1) ⇒ B = 2
\(\frac{x}{(x+1)(x+2)}\) = \(\frac{-1}{x+1}\) + \(\frac{2}{x+2}\)
\(\int \frac{x}{(x+1)(x+2)} d x\) = \(-\int \frac{1}{x+1} d x\) + \(2 \int \frac{1}{x+2} d x\)
= -log| x + 1| + 2log|x + 2| + C

Part – C

Questions from 23 to 29 carry 6 scores each. (7 × 6 = 42)

Question 23.
(i) construct a 3 × 2 matrix A = [a_ij] whose elements are given by a_ij = 3î – ĵ (2)
(ii) Express \(\left[\begin{array}{cc}
2 & 3 \\
1 & -4
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix. (4)
Answer:
(i) Given a_ij = 3î – ĵ
a₁₁= 2 a₁₂ = 1
a₂₁ = 5 a₂₂ = 4
a₃₁ = 8 a₃₂ = 7
∴ A = \(\left[\begin{array}{ll}
2 & 1 \\
5 & 4 \\
8 & 7
\end{array}\right]\)

Hence A = P + Q
where P is symmetric and Q is skew symmetric.

Question 24.
Solve the following system of equations by matrix method
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4 (6)
Answer:
Let A = \(\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right]\)
Here the system of equations can be written as AX = B.
|A| = \(\left|\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right|\)
= 3(2 – 3) + 2(4 + 4) + 3(- 6 – 4)
= – 3 + 16 – 30 = -17 ≠ 0
∴ System of eqns is consistent.

Question 25.
(i) Let f : {1, 3, 4} → {1, 2, 5} and g:{1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1))} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof. (3)
(ii) Consider f: R → R given by f(x) = 2x +1. Show that f is invertible. Find the inverse of f. (3)
Answer:
(i) Given f = {(1, 2), (3, 5), (4, 1)}, g = {(1, 3), (2, 3), (5, 1)}
∴ f(1) = 2, f(3) = 5, f(4) = 1, g(1) = 3, g(2) = 3, g(5) = 1
(gof)(1) = g(f(1)) = g(2) = 3
(gof) (3) = g(f(3)) = g(5) = 1
(gof) (4) = g(f(4))= g(1) = 3
∴ gof ={(1,3), (3,1), (4, 3)}

(ii) Given f(x) = 2x + 1
f(x₁) = f(x₂) ⇒ 2x₁ + 1 = 2x₂ + 1
2x₁ = 2x₂
x₁ = x₂
∴ is one one
Let y = 2x + 1 ⇒ 2x = y – 1
x = \(\frac{y-1}{2}\)
f(x) = f(\(\frac{y-1}{2}\)) = 2(\(\frac{y-1}{2}\)) + 1 = y – + 1 = y
∴ for every y, there exists x
such that f(x) = y
Hence f is onto
∴ f is bijective
Inverse of f = f^-1(x) = \(\frac{x-1}{2}\)

Question 26.
(i) Find the slope of the tangent to the curve
y = x³ – x at x = 2. (2)
(ii) Find the equation of tangent to the above curve. (2)
(iii) What is the maximum value of the function sin x + cos x ? (2)
Answer:
(i) y = x³
\(\frac{d y}{d x}\) = 3x²2 – 1
Slope of tangent = \(\left.\frac{d y}{d x}\right]_{x=2}\) = 3(4) – 1 = 11

(ii) Point is (2, 6)
Equation of tangent is
y – 6 = 11 (x – 2)
y – 6 = 11x – 22
11x – y – 16 = 0

(iii) f(x) = sin x + cos x, f'(x) = cos x – sin x
f”(x) = – sin – cos x = -(sin x + cos x)
f'(x) = 0 ⇒ cos x = sin x ⇒ x = \(\frac{\pi}{4}\)
f”(π/4) = -(\(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)) = -√2 < 0
∴ f is maximum, when x = π/4
Max. value = f(π/4) = \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = \(\frac{2}{\sqrt{2}}\) = √2

Question 27.
Integrate:
(i) \(\int \sin x \sin (\cos x) d x\) (3)
(ii) \(\int_0^1 \frac{\tan ^{-1} x}{1+x^2} \mathrm{~d} x\) (3)
Answer:
(i) ∫sin x sin(cos x)dx
|Put t = cos x
dt = -sin x dx
-dt = sinx dx|
= ∫sin t(-dt)
= -∫sin t dt = -(-cost) + C
= cos t + C = cos(cos x) + C

Question 28.
Solve the following problem graphically
Maximise: z = 3x + 2y
Subject to: x + 2y ≤ 10
3x + y ≤ 15,
x, y ≥ 0 (6)
Answer:
x + 2y = 10

x	0	10
y	5	0

3x + y = 15

x	0	5
y	15	0

∴ Z has maximum when x = 4, y = 3
Maximum value = 18

Question 29.
(1) Find the area of the region bounded by the curve y² = x and the line x = 1 & x = 4 and the x-axis. (3)
(ii) Find the area of the region bounded by two parabolas y = x² and y² = x. (3)
Answer:

(ii) Given y = x² (1)
f = x ________ (2)
Solving (1) and (2),