Kerala Plus Two Maths Question Paper March 2023 with Answers

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf March 2023 helps in understanding answer patterns.

Kerala Plus Two Maths Previous Year Question Paper March 2023

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 A = {1, 2, 3}
B = {2, 3, 4, 5}
and f : A → B defined by f(x) = {(x, y) : y = x + 1}
(i) Write f in roster from. (1)
(ii) Check whether y is one-one and onto. (2)
A = {1, 2, 3}
B = {2, 3, 4, 5}
Answer:
f(x) = {(1, 2), (2, 3),(3, 4)}
(ii) f is one-one but not onto

Question 2.
Find the matrices X and Y so that
X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\) (3)
Answer:
Adding these two equations,

Question 3.
Find the equation of line through the points A (1, 3) and B(0, 0) using determinants. (3)
Answer:
Let (x, y) be a point on line AB. Then A, B, P are collinear.
∴ \(\left|\begin{array}{lll}
1 & 3 & 1 \\
0 & 0 & 1 \\
x & y & 1
\end{array}\right|\) = 0
1(0 – y) – 3(0 – x) + 1(0 – 0) = 0
– y + 3x + 0 = 0
3x – y = 0

Question 4.
Find the value of ‘a’ and ‘b’ if

is a continuous function. (3)
Answer:
At x = 3
LHL = RHL = f(3)
3a + b = 10 ________ (1)
At x = 4
LHL = RHL = f(4)
4a + b = 20 ________ (2)
Solving (1) and (2)
a = 10 , b = -20

Question 5.
Find the local maxima or local minima of the function f(x) = sin x + cos x, 0< x < \(\frac{\pi}{2}\) if it exists.
f(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\) (3)
Answer:
f(x) = sin x + cos x
f'(x) = cos x – sin x
f”(x) = – sin x – cos x
f'(x) = 0 ⇒ x = \(\frac{\pi}{4}\)
f”π/4 = –\(\frac{1}{\sqrt{2}}\)–\(\frac{1}{\sqrt{2}}\) = –\(\frac{-2}{\sqrt{2}}\) = -√2 < 0
∴ f(x) has a maximum at x = \(\frac{\pi}{4}\)
Max value = f(\(\frac{\pi}{4}\)) = sin\(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= √2

Question 6.
Consider the vectors:
\(\vec{a}\) = î + 2ĵ + 3k̂
\(\vec{b}\) = 3î + 2ĵ + k̂
(i) Find \(\vec{a}\) . \(\vec{b}\) (1)
(ii) Find the angle between \(\vec{a}\) and \(\vec{b}\) (2)
Answer:
(i) \(\vec{a}\) . \(\vec{b}\) = 3 + 4 + 3 = 10

(ii) cos = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\) = \(\frac{10}{\sqrt{1+4+9} \sqrt{9+4+1}}\)
= \(\frac{10}{\sqrt{14} \cdot \sqrt{14}}\) = \(\frac{10}{14}\) = \(\frac{5}{7}\)
θ = cos-1(\(\frac{5}{7}\))

Question 7.
Find the vector and Cartesian equation of the line passing through (1, 2, 3) and parallel to the vector 3î + 2ĵ – 2k̂ (3)
Answer:
Vector form:-
\(\vec{r}\) =(î + 2ĵ + 3k̂) + λ(3î +2ĵ – 2k̂)
Cartesian form:-
\(\frac{x – 1}{3}\) = \(\frac{y – 2}{2}\) = \(\frac{z – 3}{-2}\)= λ

Question 8.
If A and B are two events such that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\) and P(A’∪B’) = \(\frac{1}{4}\)
(i) FindP(A∩B) (1)
(ii) Check whether A and B are independent events. (2)
Answer:
Given P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\)
(i) P(A’∪B’) = \(\frac{1}{4}\)
(ie) P[(A∩B)’] = \(\frac{1}{4}\)
P(A∩B) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)

(ii) P(A).P(B) = \(\frac{1}{2}\) × \(\frac{7}{12}\) = \(\frac{7}{24}\) ≠ P(A∩B)
∴ A and B are not independant.

Answer any 6 questions from 9 to 16. Each carries 4 scores.

Question 9.
(i) Let R be a relation on the set Z, et of integers defined by
R = {(a, b): 2 divides (a – b)}
Choose the right answer:
(A) (2, 4) ∈ R
(B) (3, 8) ∈ R
(C)(7, 6) ∈ R
(D) (8, 7) ∈ R (1)
R = {(a, b): 2 divides (a – b)}
(ii) Check the above relation R is an equivalence relation. (3)
Answer:
(i) (A) (2, 4) ∈ R
(ii) R = {{a, b): 2 divides (a – b)}
2 divides 0,
(ie)2divides (a – a) ∀a ∈ Z
∴ (a, a) ∈ R ∀a ∈ Z
∴ R is reflexive.
Let (a, b) ∈ R
⇒ 2 divides (a – b)
⇒ 2 divides – (a – b)
⇒ 2 divides (b – a)
∴ (b, a) ∈ R
∴ R is symmetric
Let (a, b) ∈ R, (b, c) ∈ R
⇒ 2 divides (a – b), 2 divides (b – c)
⇒ 2 divides (a – b) + (b – c)
⇒ 2 divides (a – c)
⇒ (a, c) ∈ R
∴ R is transitive
Hence R is an equivalence relation.

Question 10.
(i) The principal value of sin-1(\(\frac{1}{2}\)) is ___________ (1)
(ii) Find tan-1[2 cos(2 sin-1\(\frac{1}{2}\))] (3)
Answer:
(i) \(\frac{\pi}{6}\)

(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 11.
(i) Which, among the following is not true?
(A) (A’)’ = A
(B) (A + B)’ = A’ + B’
(C) (AB)’ = A’.B’
(D) (kA)’ = k.A’ (1)
(ii) If A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), then verify that (A + A’) is symmetric and (A – A’) is skew -symmetric.
[A’ denotes the transposes of the matrix A] (3)
Answer:
(i) (C) (∵ (AB)’ = B’.A’)

(ii) A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), A’ = \(\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\)
A + A’ = \(\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right]\) it is symmetric.
A – A’ = \(\left[\begin{array}{ll}
0 & -1 \\
1 & 0\end{array}\right]\) it is skewsymmetric

Question 12.
Using integration find the area enclosed by the ellipse \(\frac{x^2}{9}\) + \(\frac{y^2}{4}\) = 1 (4)
Answer:
\(\frac{x^2}{9}\) + \(\frac{y^2}{4}\) = 1
\(\frac{y^2}{4}\) = 1 – \(\frac{x^2}{9}\)
= \(\frac{9-x^2}{9}\),
y² = \(\frac{4}{9}\) (9 – x²)
y = \(\frac{2}{3}\)\(\sqrt{9-x^2}\)

Question 13.
(i) The degree of the differential equation (1)
(\(\frac{d^2 y}{d x^2}\))³ + (\(\frac{d y}{d x}\))² + (\(\frac{d y}{d x}\)) + 1 = 0
(A) 1
(B) 2
(C) 3
(D) not defined
(ii) Solve the differential equation (3)
\(\frac{d y}{d x}\) = (1 + x²)(1 + y²).
Answer:
(i) (C) 3

(ii) Given \(\frac{d y}{d x}\) = (1 + x²) (1 + y² )
\(\frac{d y}{1+y^2}\) = (1 + x²)dx (which is variable separable)
∴ 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
y = tan^-1(x + \(\frac{x^3}{3}\) + C )

Question 14.
Consider the vectors:
\(\vec{a}\) = î – 7ĵ + 7k̂
\(\vec{b}\) = 3î – 2ĵ + 2k̂
(i) Find \(\vec{a}\) × \(\vec{b}\) (2)
(ii) Find the unit vector perpendicualr to both \(\vec{a}\) and \(\vec{b}\) (1)
(iii) Find the area of parallelogram whose adjacent sides are \(\vec{a}\) and \(\vec{a}\) (1)
Answer:

(iii) Area of parallelogram =| \(\vec{a}\) x \(\vec{b}\) |
= 19√2 sq. units

Question 15.
Find the shortest distance between the lines:
\(\vec{r}\) = (î + 2j + k̂) + λ(î + ĵ + k̂) and
\(\vec{r}\) = (2î – ĵ + 4k̂) + µ(2î + ĵ + k̂)
Answer:
Let \(\vec{a}\) = î + 2ĵ + k̂,
\(\vec{b}\)₁ = î + ĵ + k̂
\(\vec{a}\)₁ = 2î – ĵ + 4k̂,
\(\vec{a}\)₂ = 2î + ĵ + 2k̂
\(\vec{a}\)₂ – \(\vec{a}\)₁ = î – 3ĵ + 3k̂

Question 16.
Bag – I contains 3 red and 4 black balls, while Bag – II contains 5 red and 6 black balls. One of the bags is selected at random and a ball is drawn out of it. If the ball drawn is found to be red, find the probability that it was from Bag – II. (4)
Answer:

E₁ : bag II is selected
E₂ : bag II is selected
A : Getting a red ball

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

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

Question 18.
(i) If y = x^x find \(\frac{d y}{d x}\) (2)
(ii) If y = x = at² and y = 2at, find \(\frac{d y}{d x}\) (2)
(iii) The radius of a circle is icnreasing uniformly at the rate of 5 cm.sec. Find the rate at which the area of the circle is increasing when the aradius is 8 cm. (2)
Answer:
(i) y = x^x
log y = x.log x
Differentiating with respect to x,
\(\frac{1}{y} \frac{d y}{d x}\) = x × \(\frac{1}{x}\) + log x = 1 + log x
\(\frac{d y}{d x}\) = y(1 + log x) = x^x (1 + log x)

(ii) x = at² y = 2at
\(\frac{d x}{d t}\) = 2at \(\frac{d y}{d t}\) = 2a
\(\frac{d y}{d x}\) = \(\frac{d y / d t}{d x / d t}\) = \(\frac{2 a}{2 a t}\) = \(\frac{1}{t}\)

(iii) \(\frac{d r}{d t}\) = 5 cm / sec
A = πr²
\(\frac{d A}{d t}\) = 2πr.\(\frac{d r}{d t}\)
= 2πr × 5
= 10πr/
\(\left.\frac{d A}{d t}\right]_{\mathrm{r}=8}\) = 10 × π × 8 = 80 π cm²/sec

Question 19.
(i) \(\int \frac{1}{x^2-a^2} d x\) = ____________ (1)
(ii) Find : \(\int \frac{1}{x^2+4 x-5} d x\) (2)
(iii) Evaluate : \(\int_2^3 \frac{x}{1+x^2} d x\) (3)
Answer:
(i) \(\int \frac{1}{x^2-a^2} d x\) = \(\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\)

(ii) x² + 4x – 5 = x² + 4x + 4 – 4 – 5
=(x + 2)² – 9

= \(\frac{1}{2}\) log (\(\frac{10}{5}\))
= \(\frac{1}{2}\) log 2

Question 20.
Solve the LPP graphically: (6)
Maximize
Z = 250x + 75y
Subject to
5x + y ≤ 100
x + y ≤ 60
x ≥ 0
y ≥ 0
Answer:
5x + y = 100

x	0	20
y	100	0

x + y = 60

x	0	60
y	60	0

Z has max
When x = 10, y = 50
Max, Value = 6250