Kerala Plus Two Maths Board Model Paper 2023 with Answers

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

Kerala Plus Two Maths Board Model Paper 2023 with Answers

Time: 2 Hours
Total Score: 60 Marks

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

Question 1.
Construct a 2 × 3 matrix whose elements are given by a_ij = 2i – 2j. (3)
Answer:
Given _ij = 2i – 2j
Let A = \(=\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]\)
a₁₁ = 2 × 1 – 2 × 1 = 0
a₁₂ = 2 × 1 – 2 × 2 = 2 – 4 = -2
a₁₃ = 2 × 1 – 2 × 3 = 2 – 6 = -4
a₂₁ = 2 × 2 – 2 × 1 = 4 – 2 = 2
a₂₂ = 2 × 2 – 2 × 2 = 4 – 4 = 0
a₂₃ = 2 × 2 – 2 × 3 = 4 – 6 = -2
∴ A = \(\left[\begin{array}{rrr}
0 & -2 & -4 \\
2 & 0 & -2
\end{array}\right]\)

Question 2.
Examine the relation R in the set A = {1, 2, 3….13, 14} defined as R = {(x, y) : 3x – y = 0} is reflexive, symmetric and transitive. (3)
Answer:
Given A = {1, 2, 3….13,14}
R = {(x, y) : 3x – y = 0}
= {(1, 3), (2, 6), (3, 9), (4, 12)}
R is not reflexive, since (a, a) ∉ R for all a ∈ A
R is not symmetric, since (1, 3) ∈ R but (3, 1) ∉ R
R is not transitive, since (1, 3) ∈ R, (3, 9) ∈ R but (1, 9) ∉ R

Question 3.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\) then show that |2A| = 4|A|.
Answer:
Given A = \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
2A = \(\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]\)
|2A| = \(\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]\) = 8 – 32 = -24
4|A| = 4\(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\) = 4(2 – 8) = 4 × -6 = -24
∴ |2A| = |4A|

Question 4.
Find the values of a and b such that the function defined by

is a continuous function. (3)
Answer:

At x = 2
LHL = RHL f(2)
2a + b = 5 ……………. (1)
At x =10
LHL = RHL = f(10)
10 a + b = 21
Solving (1) and (2)
a = 2 b = 1

Question 5.
Find the intervals in which the function f given by
f(x) = x² – 4x is increasing. (3)
Answer:
f(x) = x² – 4x
f'(x) = 0 ⇒ 2x – 4 = 0
x = 2

Intervals are (-∞, 2) (2, ∞)
At (-∞, 2), f'(0) = -4 < 0 At (2, ∞), f'(3) = 2 > 0
∴ f(x) is decreasing in (-∞, 2)
and increasing in (2, ∞)

Question 6.
Find the unit vector in the direction of the vector
\(\vec{a}\) = î + ĵ + 2k̂ (3)
Answer:
Given \(\vec{a}\) = î + ĵ + 2k̂
|\(\vec{a}\)|= \(\sqrt{(1)^2+(1)^2+(2)^2}\) = √6
Required unit vector = â = \(\frac{\vec{a}}{|\vec{a}|}\)
= \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{\sqrt{6}}\)
= \(\frac{1}{\sqrt{6}} \hat{\mathrm{i}}+\frac{1}{\sqrt{6}} \hat{\mathrm{j}}+\frac{2}{\sqrt{6}} \hat{\mathrm{k}}\)

Question 7.
Find the angle between the pair of lines given by
\(\vec{a}\) = 3î + 2ĵ – 4k̂ + λ(î + 2ĵ + 2k̂)
\(\vec{a}\) = 5î – 2ĵ + μ(3î + 2ĵ + 6k̂) (3)
Answer:
Let \(\vec{b}\)₁ = î + 2ĵ + 2k̂, \(\vec{b}\)₂ = 3î + 2ĵ + 6k̂
cos θ = \(\frac{\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{~b}}_2}{\left|\mathrm{~b}_1\right|\left|\mathrm{b}_2\right|}\) = \(\frac{3+4+12}{\sqrt{9} \cdot \sqrt{49}}\) = \(\frac{19}{3 \times 7}\) = \(\frac{19}{21}\)
θ = cos^-1(\(\frac{19}{21}\))

Question 8.
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6
(i) P (A/B) = _________ . (1)
(ii) Find P (A and not B) (2)
Answer:
(i) P(A/B) = P(A) [∵ A and B are independant]
= 0.3

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

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

Question 9.
(i) A function f : x → y is onto if and only if Range of f = ________ . (1)
(ii) If f : R → R defined by f(x) – 3 – 4x.
Check whether the function is bijective. (3)
Answer:
(i) Codomain of f

(ii) f(x) = 3 – 4x
f(x₁) = f(x₂) ⇒ 3 – 4x₁ = 3 – 4x₂
⇒ -4x₁ = -4x₂
x₁ = x₂
∴ f(x) is one – one (injective)
Let y = 3 – 4x
4x = 3 – y
x = \(\frac{3 – y}{4}\)
f(x) = f(\(\frac{3 – y}{4}\)) = 3 – 4(\(\frac{3 – y}{4}\))
= 3 – (3 – y) = y
∴ For every ‘y’ there exist x
such that f(x) = y
∴ f(x) is onto (surjective)
Hence f(x) is bijective

Question 10.
(i) If \(\frac{-\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\) then sin^-1 (sin x) = _________ . (1)
(ii) Find the principal value of sin^-1(\(\frac{1}{\sqrt{2}}\)) (1)
(iii) Find the value of sin^-1(sin(\(\frac{13 \pi}{6}\)))
Answer:
(i) x

(ii) \(\frac{\pi}{4}\)

(iii) sin^-1(sin \(\frac{13 \pi}{6}\))
= sin^-1(sin 2π + \(\frac{\pi}{6}\)
= sin^-1(sin \(\frac{\pi}{6}\)) = \(\frac{\pi}{6}\).

Question 11.
Express the matrix \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\) as the sum of a symmetric and skew-symmetric matrix. (4)
Answer:

\(\frac{\mathrm{A}+\mathrm{A}^{\mathrm{T}}}{2}\) + \(\frac{\mathrm{A}-\mathrm{A}^{\mathrm{T}}}{2}\) = A
where \(\frac{\mathrm{A}+\mathrm{A}^{\mathrm{T}}}{2}\) is a symmetric matrix and \(\frac{\mathrm{A}-\mathrm{A}^{\mathrm{T}}}{2}\) is a skew symmetric matrix.

Question 12.
Find the area enclosed by the circle x² + y² = a² (4)
Answer:
x² + y² = a²
y² = a² – x²
y = \(\sqrt{a^2-x^2}\)

Required area = 4 × area of shaded region.

Question 13.
F ind the general solution of the differential equation (4)
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
Answer:
Given \(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
which is of the form \(\frac{d y}{d x}\) + py = Q
where p = \(\frac{1}{x}\), Q = x²
Integrating factor (IF) = \(e^{\int P d x}=e^{\int \frac{1}{x} d x}\) = e^{log x} = x
∴ Solution is
y(IF) = ∫Q(IF)dx
yx² = ∫x². x dx
x²y = ∫x³dx + c
x²y = \(\frac{x^4}{4}\) + C

Question 14.
Find the area of a parallelogram whose adjacent sides are given by the vectors.
\(\vec{a}\) = 3î + ĵ + 4k̂
\(\vec{b}\) = î – ĵ + k̂ (4)
Answer:
\(\vec{a}\) = 3î + ĵ + 4k̂
\(\vec{b}\) = î – ĵ + k̂
\(\vec{a}\) × \(\vec{b}\) = \(\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & 1 & 4 \\
1 & -1 & 1
\end{array}\right|\)
= î(5) – ĵ(-1) + k̂(-4)
= 5î + ĵ – 4k̂
Area of parallelogram = \(\vec{a}\) × \(\vec{b}\)
= \(\sqrt{25 + 1 + 16}\)
= \(\sqrt{42}\) sq.units

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

Question 16.
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be black. Find the probability that the ball is drawn from the second bag.
Answer:

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

Question 17.
Solve by matrix method
x – y + z = 4
2x + y – 3z = 0
x + y+ z = 2 (6)
Answer:
Let A = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), B = \(\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]\)
∴ System of equations can be written as AX = B
|A| = 1(1 + 3) + 1(2 + 3) + 1(2 – 1)
= 4 + 5 + 1 = 10 ≠ 0
∴ System is consistent and the unique solution is
X = A^-1B

Question 18.
(i) Find \(\frac{d y}{d x}\) if 4x – 5y = sin x (3)
(ii) Find the rate of change of the area of a circle , with respect to its radius r when r = 3 cm. (3)
Answer:
(i) 4x – 5y = sin x
Differentiating with respect to x
4 – 5 . \(\frac{d y}{d x}\) = cos x
-5 \(\frac{d y}{d x}\) = cos x – 4
\(\frac{d y}{d x}\) = \(\frac{\cos x-4}{-5}\) = \(\frac{4-\cos x}{5}\)

(ii) Let area A = πr²
\(\frac{d A}{d r}\) = π × 2r = 2πr
\(\left.\frac{\mathrm{dA}}{\mathrm{dr}}\right]_{\mathrm{r}=3}\) = 2π × 3 = 6π

Question 19.
(i) Find \(\int \frac{x d x}{(x+1)(x+2)}\) (3)
(ii) Evaluate \(\int \frac{x d x}{(x+1)(x+2)}\) (3)
Answer:
(i) \(\frac{x}{(x+1)(x+2)}\) = \(\) + \(\)
x = A(x + 2) + B(x + 1)
Put x = -1, -1 = A(1) ⇒ A = -1
x = -2, -2 = B(-1) ⇒ B = 2

Question 20.
Solve Linear Programming Problem (LPP) graphically.
Maximize Z = 17.5x + 7y (6)
Subject to constraints
3x + y ≤ 12
x + 3y ≤ 12
x ≥ 0
y ≥ 0
Answer:
3x + y = 12

x	0	4
y	12	0

x + 3y =12

x	0	12
y	4	0

∴ Z has maximum when x = 3, y = 3
Maximum value = 73.5