Kerala Plus Two Maths Question Paper March 2024 with Answers

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Kerala Plus Two Maths Previous Year Question Paper March 2024

Time: 2 Hours
Total Score: 60 Marks

Answer any 6 questions from 1 to 8. Each carries 3 scores. (6 × 3 = 18)

Question 1.
Let R be a relation on a set A = {1, 2, 3, 4, 5, 6} defined as R = {(x, y : y = 2x – 1}.
(i) Write R in roster form and find its domain and range. (2)
(ii) Is R is an equivalence relation? Justify.(1)
Answer:
(i) R = {{x, y) : y = 2x – 1]
R = {(1, 1),(2, 3), (3, 5)}
Domain = {1,2, 3}
Range ={1,3,5}

(ii) R is not an equivalence relation
∵ (2, 2)∉ R and hence not reflexive.

Question 2.
A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\) show that
A² – 5A + 7I = O
(Where I is the identity matrix)
Answer:
A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\)
A² = A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2\end{array}\right]\)
\(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\)

∴ A² – 5A + 7I = O

Question 3.
(i) Check the continuity of the function
f (x) = 2x + 3 at x =1. (1)
(ii) Determine the value of k so that the function

is continuous at x = 5.
Answer:
(i) \(\lim _{x \rightarrow 1} f(x)[latex] = [latex]\lim _{x \rightarrow 1} 2 x+3[latex]
2 × 1 + 3 = 5
f(1) = 2 × 1 + 3 = 5
∴ [latex]\lim _{x \rightarrow 1} f(x)[latex] = f(1)
Hence, f is continuous at x = 1

(ii) Given f(x) is continuous at x = 5
∴ [latex]\lim _{x \rightarrow 5^{-}} f(x)\) = \(\lim _{x \rightarrow 5^{+}} f(x)\)
⇒ \(\lim _{x \rightarrow 5^{-}} k x+1\) = \(\lim _{x \rightarrow 5^{+}} 3 x-5\)
⇒ k × 5 + 1 = 3 × 5 – 5
⇒ 5k + 1 = 15 – 5 = 10
⇒ 5k = 9
k = \(\frac{9}{5}\)

Question 4.
(i) Find the principal value of sin^-1(\(\frac{1}{2}\)) (1)
(ii) Find the value of
tan^-1 [2 cos(2 sin^-1\(\frac{1}{2}\)) (2)
Answer:
(i) Let y = sin^-1(\(\frac{1}{2}\))
sin y = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
∴ y = \(\frac{\pi}{6}\)
Hence, the principal value of
sin^-1(\(\frac{1}{2}\)) = \(\frac{\pi}{6}\)

(ii) tan^-1[2 cos(2 sin^-1\(\frac{1}{2}\))]
= tan^-1[2 cos(2 × \(\frac{\pi}{6}\))]
= tan^-1[2 cos\(\frac{\pi}{3}\)]
= tan^-1 [2 × \(\frac{1}{2}\)]
= tan^-1(1)
= \(\frac{\pi}{4}\)

Question 5.
(i) Which of the following function is increasing in its domain:
(A) sin x
(B) cos x
(C) -2x
(D) log x (1)
(ii) Find the intervals in which the function f given by f(x) = x² – 4x + 6 is
(a) increasing
(b) decreasing. (2)
Answer:
(i) Option (D) log x

(ii) f(x) = x² – 4x + 6
f'(x) = 2x – 4
f'(x) = 0 ⇒ 2x – 4 = 0
⇒ 2x = 4
⇒ x = 2
∴ We have two intervals, (-∞, 2) and (2, ∞).
Consider (-∞, 2)
Put x = 0 ∈(-∞, 2) in f'(x)
f'(0) = -4 < 0
∴ f is strictly decreasing in (-∞, 2)
Consider (2, ∞),
Put x = 3 ∈ (2, ∞) in f'(x)
f'(3) = 2 × 3 – 4
= 2 > 0
∴ f is strictly increasing in (2, ∞)

Question 6.
(i) If θ is the angle between two non zero vectors \(\vec{a}\) and \(\vec{b}\) and |\(\vec{a}\) . \(\vec{b}\)| = |\(\vec{a}\) × \(\vec{a}\)| then θ = ___________ . (1)
(ii) Find the projection of the vector \(\vec{a}\) = î – ĵ on the vector \(\vec{b}\) = î + ĵ (2)
Answer:
(i) θ = \(\frac{\pi}{4}\)
∵ |\(\vec{a}\) . \(\vec{b}\)| = |\(\vec{a}\) × \(\vec{b}\)|
⇒ ab cos θ = ab sin θ
⇒ cos θ = sin θ
⇒ θ = \(\frac{\pi}{4}\)

(ii) \(\vec{a}\) = î – ĵ
\(\vec{b}\) = î + ĵ
Projection of \(\vec{a}\) on \(\vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
= \(\frac{(\hat{i}-\hat{\mathrm{j}}) \cdot(\hat{i}+\hat{\mathrm{j}})}{\sqrt{1^2+1^2}}\)
= \(\frac{0}{\sqrt{2}}\) = 0

Question 7.
Let A and B are independent events with P(A) = 0.3, P(B) = 0.4, find
(i) P(A∩B)
(ii) P(A∪B)
(iii) P(A/B)
Answer:
(i) A and B are independent events.
∴ P (A∩B) = P(A) × P(B)
= 0.3 × 0.4
= 0.12

(ii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.3 + 4 – 0.12
= 0.58

(iii) P(A / B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{P(\mathrm{~B})}\)
= \(\frac{0.12}{0.4}\) = 0.3

Question 8.
Evaluate \(\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x\)
Answer:
\(\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x\)
Put u = x⁵ + 1
du = 5x⁴dx
x = 1 ⇒ w = 2
x = -1 ⇒ u = 0

Answer any 6 questions from 9 to 16. Each carries 4 scores. (6 × 4 = 24)

Question 9.
(i) What is the minimum number of ordered
pairs to form a reflexive relation on a set of 4 elements? (1)
(ii) Let A = R – {3}, B= R – {1}
Consider the function f : A B
defined by f(x) = \(\frac{x – 2}{x – 3}\)
Check whether f is one-one and onto. (3)
Answer:
(i) 4 ordered pairs

(ii) f(x) = \(\frac{x – 2}{x – 3}\)
f : A → B,
A = R – {3},B = R – {1}
For x₁, x₂ ∈ A
f(x₁) = f(x₂) ⇒ \(\frac{x_1-2}{x_1-3}\) = \(\frac{x_2-2}{x_2-3}\)
⇒ (x₁ – 2) (x₂ – 3) = (x₁ – 3)(x₂ – 2)
⇒ x₁ x₂ – 3x₁ – 2x₂ + 6 = x₁ x₂ – 2x₁ – 3x₂ + 6
⇒ -3x₁ – 2x₁ = -2x₁ – 3x₁
⇒ 3x₂ – 2x₂ = 3x₁ – 2x₁
⇒ x₁ = x₂

Question 10.
(i) If A is a skew symmetric matrix then A’= _____________ (1)
(ii) Express the matrix
A = \(\left[\begin{array}{ccc}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\)
as the sum of a symmetric and skew symmetric matrix. (3)
Answer:
(i) A¹ = -A

which is the symmetric matrix.

which is the skew symmetric matrix.

Question 11.
A wire of length 28 m is cut into two pieces, one of the pieces is to be made into a square and other into a circle. What should be the length of the two pieces so that the combined area of square and circle is minimum?
Answer:
Let length of the wire made into square be
x and made into circle is 28 – x
ie, perimeter of square = x
4a = x
a = \(\frac{x}{4}\)
⇒ Area, A₁ = \(\frac{x^2}{16}\)
Circumference of circle = 28 – x
2πr = 28 – x
r = \(\frac{28-x}{2 \pi}\)
⇒ Area, A₂ = π × (\(\frac{28-x}{2 \pi}\))² = (\(\frac{28-x}{2 \pi}\))²
Combained area A = A₁ + A₂
= \(\frac{x^2}{16}\) + \(\frac{(28-x)^2}{4 \pi}\)
\(\frac{d A}{d x}\) = \(\frac{2 x}{16}\) + \(\frac{2(28-x) \times(-1)}{4 \pi}\)
= \(\frac{x}{8}\) – \(\frac{28-x}{2 \pi}\)
\(\frac{d A}{d x}\) = 0 ⇒ \(\frac{x}{8}\) – \(\frac{28-x}{2 \pi}\) = 0
⇒ \(\frac{x}{8}\) = \(\frac{28-x}{2 \pi}\)
⇒ 2π x = 224 – 8x
⇒ 2π x + 8x = 224,
⇒ 2π x + 8x = 224
⇒ x = \(\frac{224}{2 \pi+8}\) = \(\frac{112}{\pi+4}\)
\(\frac{d^2 A}{d x^2}\) = \(\frac{1}{8}\) + \(\frac{1}{2 \pi}\) > 0
∴ \(\frac{d^2 A}{d x^2}\) > 0 at x = \(\frac{112}{\pi+4}\)
ie., Area is minimum at x = \(\frac{112}{\pi+4}\)
∴ Length of two pieces are,
\(\frac{112}{\pi+4}\) for square
28 – \(\frac{112}{\pi+4}\) = \(\frac{28 \pi+112-112}{\pi+4}\)
= \(\frac{28 \pi}{\pi+4}\) for circle

Question 12.
(i) Write the order and degree of the differential equation
xy\(\frac{d^2 y}{d x^2}\))³ + (\(\frac{d y}{d x}\))² – sin (\(\frac{d y}{d x}\)) = 0 (1)
(ii) Find the integrating factor of the differential equation
x\(\frac{d y}{d x}\) + 2y = x² (2)
(iii) Solve the differential equation
x\(\frac{d y}{d x}\) + 2y = x² (1)
Answer:
(i) Order = 2
Degree = not defined

(ii) x\(\frac{d y}{d x}\) + 2y = x²
\(\frac{d y}{d x}\) + \(\frac{2 y}{x}\) = x
which is of the form \(\frac{d y}{d x}\) + py = Q
P = \(\frac{2}{x}\) Q = x
Integrating Factor = \(e^{\int p d x}\)
= \(e^{\int \frac{2}{x} d x}\)
= \(e^{2 \log x}\)
= \(e^{\log x^2}\)
= x²

(iii) y.IF = \(\int Q \cdot I F d x\)
y.x² = \(\int x \cdot x^2 d x\)
x²y = \(\int x^3 d x\)
x²y = \(\frac{x^4}{4}\) + c

Question 13.
If \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = î + 2ĵ + 3k̂
(i) Find \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\). (1)
(ii) Find \(\vec{a}\) unit vector perpendicular to both \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\) (3)
Answer:
(i) \(\vec{a}\) = î + ĵ + k̂ \(\vec{b}\) = î + 2ĵ + 3k̂
\(\vec{a}\) + \(\vec{b}\) = (î + ĵ + k̂) + (î + 2ĵ + 3k̂) = 2î + 3j + 4k̂
\(\vec{a}\) – \(\vec{b}\) = {î + ĵ + k̂) + (î + 2ĵ + 3k̂)
= – ĵ – 2k̂

(ii) Vector perpendicular to both
\(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\)
(\(\vec{a}\) + \(\vec{b}\)) × (\(\vec{a}\) + \(\vec{b}\))
= \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
0 & -1 & -2
\end{array}\right|\)
= î(- 6 + 4) – ĵ(- 4 – 0) + k̂(- 2 – 0)
= – 2î + 4j – 2k̂
|(\(\vec{a}\) + \(\vec{b}\)) × (\(\vec{a}\) + \(\vec{b}\))| = \(\sqrt{(-2)^2+4^2+(-2)^2}\)
= \(\sqrt{4+16+4}\)
= \(\sqrt{24}\)
∴ Unit vector perpendicular to both \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\)
= \(\frac{-2 \hat{i}+4 \hat{j}-2 \hat{k}}{\sqrt{24}}\)

Question 14.
Find the shortest distance between the lines
\(\vec{r}\) = (î + ĵ + λ(2î – ĵ + k̂) and
\(\vec{r}\) = (2î + ĵ – k̂) + µ(3î + 5ĵ + 2k̂)
Answer:
\(\vec{r}\) = (î + ĵ) + λ(2î – ĵ + k̂)
\(\vec{r}\) = (2î + ĵ – k̂) + μ(3î – 5ĵ + 2k̂)
\(\vec{a}\)₁ = î + ĵ \(\vec{a}\)₂ = 2î + ĵ – k̂
\(\vec{b}\)₁ = 2î – ĵ + k̂ \(\vec{b}\)₂ = 3î – 5ĵ + 2k̂
\(\vec{b}\)₁ × \(\vec{b}\)₂ = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|\)
= î(- 2 + 5) – ĵ(4 – 3) + k̂(- 10 + 3)
= 3î – ĵ – 7k̂
|\(\vec{b}\)₁ × \(\vec{b}\)₂| = \(\sqrt{3^2+(-1)^2+(-7)^2}\)
= \(\sqrt{9+1+49}\)
= \(\sqrt{59}\)
∴ Shortest distance = \(\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)
= \(\left|\frac{(\hat{i}-\hat{k}) \cdot(3 \hat{i}-\hat{j}-7 \hat{k})}{\sqrt{59}}\right|\)
= \(\frac{3+0+7}{\sqrt{59}}\) = \(\frac{10}{\sqrt{59}}\)

Question 15.
In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 45% of the bolt. Of their output 5%, 4% and 2% are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that is manufactured by machine B?
Answer:
Let E₁, E₂, E₃ be the events that bolts produced by machines A, B, C respectively.
P(E₁) = \(\frac{25}{100}\), P(E₂) = \(\frac{35}{100}\), P(E₃) = \(\frac{45}{100}\)
Let F be the event that bolt is defective.
P(F|E₁) = \(\frac{5}{100}\)
P(F|E₂) = \(\frac{4}{100}\)
P(F|E₃) = \(\frac{2}{100}\)

Question 16.
Find the area of the region bounded by the ellipse \(\frac{x^2}{16}\) + \(\frac{y^2}{9}\) = 1 using integration.
Answer:

Answer any 3 questions from 17 to 20. Each carries 6 scores. (3 × 6 = 18)

Question 17.
Solve the following system of equations by matrix method:
2x – 3y + 5z = 11
3x + 2y – 4z = -5
x + y – 2z = -3
Answer:
A = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\) B = \(\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]\) X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
AX = B ⇒ x = A^-1B
A^-1 = \(\frac{{adj} \mathrm{A}}{|\mathrm{~A}|}\)
A₁₁ = 0 A₁₂ = 0 A₁₃ = 1
A₂₁ = -1 A₂₂ = -9 A₂₃ = -5
A₃₁ = 2 A₃₂ = 23 A₃₃ = 13

= 2[- 4 + 4] + 3(- 6 + 4) + 5 (3 – 2)
= 0 – 6 + 5
= -1

Question 18.
(i) sin x + cos y = xy find \(\frac{d y}{d x}\) (2)
(ii) x = a cos³ t ; y = a sin³ t find \(\frac{d y}{d x}\) (2)
(iii) If y = (sin^-1 x)² then show that
(1 – x²)\(\frac{d^2 y}{d x^2}\)) – x (\(\frac{d y}{d x}\)) = 2 (2)

Answer:
(i) sin x + cos y = xy
Differentiating with respect to x
cos x – sin y \(\frac{d y}{d x}\) = x\(\frac{d y}{d x}\) + y. 1
(x + sin y)\(\frac{d y}{d x}\) = cos x – y
\(\frac{d y}{d x}\) = \(\frac{\cos x-y}{x+\sin y}\)

(ii) x = a cos 3t y = a sin 3t
\(\frac{d x}{d t}\) = a × 3cos t × – sin t = -3a cos t sin t
\(\frac{d y}{d t}\) = a × 3sin t × cos t = 3a sin t cos t
∴ \(\frac{d y}{d x}\) = \(\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t}\)
= \(\frac{-\sin t}{\cos t}\) = – tan t

(iii) y = (sin^-1 x)²
\(\frac{d y}{d x}\) = 2 sin ^-1 × \(\frac{1}{\sqrt{1-x^2}}\)
\(\sqrt{1-x^2}\) \(\frac{d y}{d x}\) = 2 sin^-1 x
Squaring,
(1 – x²) (\(\frac{d y}{d x}\))² = 4(sin^-1 x)²
(1 – x²) (\(\frac{d y}{d x}\))² = 4y
Differentiating
(1 – x²)2\(\frac{d y}{d x}\) \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² × -2x = 4\(\frac{d y}{d x}\)
÷ by 2\(\frac{d y}{d x}\)
(1 – x²)\(\frac{d^2 y}{d x^2}\) – x\(\frac{d y}{d x}\) = 2

Question 19.
(i) Find \(\int \frac{x-1}{(x-2)(x-3)} d x\) (3)
(ii) Prove that \(\int_0^{-\frac{\pi}{4}} \log (1+\tan x) d x\) = \(\frac{\pi}{8}\)log 2 (3)
Answer:
(i) \(\int \frac{x-1}{(x-2)(x-3)} d x\)
\(\frac{x-1}{(x-2)(x-3)}\) = \(\frac{\mathrm{A}}{x-2}\) + \(\frac{\mathrm{B}}{x-3}\)
x – 1 = A(x – 3) + B(x – 2)
Put x = 3
3 – 1 = A(3 – 3) + B(3 – 2)
2 = B
Put x = 2
2 – 1 = A(2 – 3) + B(2 – 2)
1 = -A ⇒ A = -1
∴ \(\frac{x-1}{(x-2)(x-3)}\) = \(\frac{-1}{x-2}\) + \(\frac{2}{x-3}\)
Integrating on both sides,
\(\int \frac{x-1}{(x-2)(x-3)} d x\) = \(\int \frac{-1}{x-2} d x\) + \(\int \frac{2}{x-3} d x\)
= -log|x – 2| + 2log|x – 3| + C

Question 20.
Solve the following linear programming problem graphically:
Maximise Z = 60x + 15
Subject to the constraints
x + ≤ 50
3x + y ≤ 90
x ≥ 0,y ≥ 0
Answer:
z = 60x + 15y
x + y = 50

x	0	50
y	50	0

3x + y = 90

x	0	30
y	90	0

Corner points	z = 60x + 15y
A (0, 0) B (30, 0) C (20, 30) D (0, 50)	z = 0 z = 1800 z = 1650 z = 750

∴ Maximum value of z is 1800 at (30,0)