Kerala Plus Two Maths Board Model Paper 2021 with Answers

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf Board Model Paper 2021 helps in understanding answer patterns.

Kerala Plus Two Maths Board Model Paper 2021 with Answers

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 value of x.
\(\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\) = \(\left|\begin{array}{cc}
2 & 4 \\
5 & 1
\end{array}\right|\) (3)
Answer:
Expanding on both sides
2x² – 24 = 2 – 20
2x² – 24 = -18
2x² = 24 – 18 = 6
x² = 3
x = ±√3

Question 2.
A = \(\left|\begin{array}{cc}
3 & 1 \\
-2 & 1
\end{array}\right|\)
(i) Find (adj.A) (1)
(ii) Also prove that A(adj A) = 5I (2)
Answer:

Question 3.
Determine the value of the constant k so that the function
is continuous at x = 2 (3)
Answer:
Given is continuous at x = 2
∴ At x = 2, LHL = RHL = f(2)
We have f(2) = k(2)² = 4 K _________ (1)
RHL = \(\lim _{x \rightarrow 2^{+}} f(x)\) = 3 _________ (2)
∴ From (1) and (2)
4K = 3
K = \(\frac{3}{4}\)

Question 4.
Verify Mean Value Theorem if f(x) = x² – 4x – 3, is the interval [1, 4]. (3)
Answer:
Since f(x) is a polynomial function, it is continuous in [1, 4]
f(x) = x² – 4x – 3
f'(x) = 2x – 4
∴ It is differentiable in (1, 4)
f(a) = f(1) = 1² – 4(1) – 3 = 1 – 4 – 3 = -6
f(b) = f(4) = 4² – 4(4) – 3 =16 – 16 – 3 = -3
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
2c – 4 = \(\frac{- 3 – 6}{ 4 – 1}\) = \(\frac{- 3 + 6}{3}\)
2c – 4 = \(\frac{3}{3}\) = 1
2c = 5
c = \(\frac{5}{2}\) ∈ (1, 4)
Hence Mean Value Theorem Verified.

Question 5.
The radius of a circle is increasing uniformly at the rate of 5 cm/s. Find the rate at which the area of the circle is increasing when the radius is 8 cm. (3)
Answer:
Let r be the radius and A be the area of the circle.
Then \(\frac{d r}{d t}\) = 5 cm/s
Let A = πr²
\(\frac{d A}{d t}\) = 2πr. \(\frac{d r}{d t}\) = 2πr × 5
= 10πr
\(\left.\frac{\mathrm{dA}}{\mathrm{dt}}\right]_{\mathrm{r}=8}\) = 10π × 8
= 80π cm²/s

Question 6.
Find the unit vector in the direction of vector \(\vec{\mathrm{PQ}}\), where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively. (3)
Answer:

Question 7.
Find the vector and cartesian equations of the planes that passes through the point (1, 4, 6) and the normal î – 2ĵ + k̂ (3)
Answer:
Given(x₁, y₁, z₁) = (1, 4, 6)
Direction ratios of normal <a, b, c> = <1, -2, 1>
∴ Equation is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
1(x – 1) – 2(y – 4) + 1(z – 6) = 0
x – 1 – 2y + 8 + z – 6 = 0
x – 2y + z + 1 = 0
Vector equation is \(\vec{r}\) . (î – 2ĵ + k̂) = -1

Question 8.
(i) Find the principal value of sin^-1(\(\frac{1}{\sqrt{2}}\)) (1)
(ii) Evaluate tan^-1(1) + cos^-1(\(\frac{-1}{2}\)) + sin^-1(\(\frac{-1}{2}\)) (2)
Answer:
(i) sin^-1(\(\frac{1}{\sqrt{2}}\)) = \(\frac{\pi}{4}\)

(ii) We have sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\)
∴ cos^-1(\(\frac{-1}{2}\)) + sin^-1(\(\frac{-1}{2}\)) = \(\frac{\pi}{2}\)
tan^-1(1) + cos^-1(\(\frac{-1}{2}\)) + sin^-1(\(\frac{-1}{2}\))
= \(\frac{\pi}{4}\) + \(\frac{\pi}{2}\) = \(\frac{3 \pi}{4}\)

Question 9.
Find the equation of line joining (1, 2) and (3, 6) using determinants. (3)
Answer:
Equation of line joining (x₁, y₁) and (x₂, y₂) is \(\left|\begin{array}{ccc}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x & y & 1
\end{array}\right|\) = 0
i.e., \(\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 6 & 1 \\
x & y & 1
\end{array}\right|\) = 0
Expanding along R₁,
1(6 – y) – 2(3 – x) + 1(3y – 6x) = 0
6 – y – 6 + 2x + 3y – 6x = 0
– 4x + 2y =0
ie., 2x – y = 0

Question 10.
Find the general solution of the differential equation.
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x² (3)
Answer:
Given \(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x²
It is a linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q
Where P = \(\frac{1}{x}\), Q = x²
Integrating factor (I. F) = \(e^{\int P d x}\)
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x}
= x
Solution is y (I. F) = ∫Q(IF) dx
yx = ∫x².x dx
ie., xy = ∫x³ dx
xy = \(\frac{x^4}{4}\) + C

Part – B

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

Question 11.
(i) Construct a 2 × 2 matrix A = [a_ij], whose elements are given by a_ij = \(\frac{(i+j)^2}{2}\) (2)
(ii) If \(\left[\begin{array}{cc}
a-b & d \\
2 a-b & c
\end{array}\right]\) = \(\left[\begin{array}{cc}
-1 & 4 \\
0 & 5
\end{array}\right]\) then find a,b,c and d. (2)
Answer:

(ii) Given \(\left[\begin{array}{cc}
a-b & d \\
2 a-b & c
\end{array}\right]\) = \(\left[\begin{array}{cc}
-1 & 4 \\
0 & 5
\end{array}\right]\)
a – b = -1 ________ (1) d = 4
2a – b = 0 ________ (2) c = 5
Solving (1) and (2), a = 1 b = 2

Question 12.
A = \(\left[\begin{array}{cc}
2 & 4 \\
3 & 2
\end{array}\right]\) B = \(\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\) C = \(\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right]\)
Find (i) 3A – C (2)
(ii) AB (2)
Answer:

Question 13.
(i) tan^-1 x + tan^-1 y = …………………… (1)
(ii) Prove that tan^-1\(\frac{2}{11}\) + tan^-1\(\frac{7}{24}\) = tan^-1\(\frac{1}{2}\) (3)
Answer:

Question 14.
Find \(\frac{d y}{d x}\).
(i) y = sin(cos(x²)) (2)
(ii) x² + xy + y² = 100 (2)
Answer:
(i) Given y = sin [cos(x²)]
\(\frac{d y}{d x}\) = cos[cos(x²)] . \(\frac{d}{d x}\)cos[cos(x²)]
= cos[cos(x²)] – sin(x²) \(\frac{d}{d x}\) .(x²)
= -cos[cos(x²)]. sin(x²).2x
= -2x sin(x²).cos[cos(x²)}

(ii) Given x² + xy – y² = 100
Differentiating with respect to x,
2x + \(\frac{x d y}{d x}\) + y + 2y.\(\frac{d y}{d x}\) = 0
(x + 2y)\(\frac{d y}{d x}\) = – 2x – 2y
\(\frac{d y}{d x}\) = \(\frac{-(2 x+y)}{x+2 y}\)

Question 15.
(i) Find the maximum value of the function f(x) = sin x + cos x in [0, π/2]. (2)
(ii) Find the intervals in which the function f(x) = x² + 2x – 5 is increasing or decreasing.
Answer:
(i) Given f(x) = sin x + cos x
f'(x) = cos x – sin x
f”(x) = – sin x – cos x
= – (sin x + cos x)
f'(x) = 0 ⇒ cos x – sin x = 0
cos x = sin x
x = \(\frac{\pi}{4}\); (∵ x ∈[0, \(\frac{\pi}{2}\)]
f”(\(\frac{\pi}{4}\)) = -(sin \(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\))
= – (\(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)) = -(\(\frac{2}{\sqrt{2}}\)) = -√2 < 0
∴ f(x) has a maximum value at x = \(\frac{\pi}{4}\)
Also f(0) = sin 0 + cos 0 = 1
f(\(\frac{\pi}{2}\)) = sin\(\frac{\pi}{2}\) + cos\(\frac{\pi}{2}\) = 1
f(\(\frac{\pi}{4}\)) = sin\(\frac{\pi}{4}\) + cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\) = √2
∴ Maximum value of f(x) = √2

(ii) Given f(x) = x² + 2x – 5
f'(x) = 2x + 2
f'(x) = 0 ⇒ 2x + 2 = 0
x = -1

∴ Intervals are (-∞, -1) and (-1, ∞)
At (-∞,-1), f'(-2) = -4 + 2 = -2 < 0 At (-1, ∞), f'(0) = 0 + 2 = 2 > 0
∴ f(x) is decreasing in (-∞, -1) and increasing in (-1, ∞)

Question 16.
(i) Write the order of the differential equation.
(\(\frac{d^2 s}{d t^2}\)) + 3(\(\frac{d s}{d t}\))³ + 4 = 0
(ii) Find the general solution of the differential equation.
sec² x . tan y dx + sec² y . tan x . dy = 0
Answer:
(i) Order = 2

(ii) Given sec² x tan y dx + sec² y tan x dy = 0
sec² x tan y dx = -sec² y tan x dy
\(\frac{\sec ^2 x}{\tan x}\)dx = \(\frac{\sec ^2 y}{\tan y}\)dy
Which is variable seperable
∴ Solution is \(\frac{\sec ^2 x}{\tan x}\)dx = \(\frac{\sec ^2 y}{\tan y}\)dy
ie., log |tan x| = – log |tan y| + log C
tan x = \(\frac{C}{\tan y}\)
tan x. tan y = C

Question 17.
Find a unit vector perpendicular to each of the vector \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\), where \(\vec{a}\) = 3î + 2ĵ + 2k̂ and \(\vec{b}\) = î + 2ĵ – 2k̂. (4)
Answer:
Let \(\vec{c}\) = \(\vec{a}\) + \(\vec{b}\) = 4î + 4ĵ
\(\vec{d}\) = \(\vec{a}\) – \(\vec{b}\) = 2î + 4k̂
Vector perpendicular to \(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – \(\vec{b}\) is \(\vec{c}\) × \(\vec{d}\)
ie., \(\vec{c}\) × \(\vec{d}\) = \(\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
4 & 4 & 0 \\
2 & 0 & 4
\end{array}\right|\) = 16î – 16ĵ – 8k̂
\(\vec{c}\) × \(\vec{d}\) = \(\sqrt{256+256+64}\)
= \(\sqrt{32}\)
= 24
∴ Required unit vector = \(\frac{\vec{c} \times \vec{d}}{|\vec{\mathrm{c}} \times \vec{\mathrm{d}}|}\)
= \(\frac{16 \hat{i}-16 \hat{j}-8 \hat{k}}{24}\)
= \(\frac{2}{3}\)î – \(\frac{2}{3}\)ĵ – \(\frac{1}{3}\)k̂

Question 18.
Find the shortest distance between the pair of lines
\(\vec{r}\) = (i + 2j + 3k) + λ(i – 3j + 2k)
\(\vec{r}\) = (4i + 5j + 6k) + µ(2i + 3j + k) (4)
Answer:
Let \(\vec{a}\)₁ = î + 2ĵ + 3k̂,
\(\vec{b}\)₁ = î – 3ĵ + 2k̂
\(\vec{a}\)₂ = 4î + 5ĵ + 6k̂,
\(\vec{b}\)₂ = 2î + 3ĵ + k̂
\(\vec{a}\)₂ – \(\vec{a}\)₁ = 3î + 3ĵ + 3k̂

Question 19.
Let A and B be two independent events such that P(A) = \(\frac{1}{7}\) and P(B) = \(\frac{1}{2}\). Find
() P(A∩B) (1)
(b) P(A∪B) (2)
(c) P[A∩B’)∩(B∩A’)] (1)
Answer:
Given P(A) = \(\frac{1}{7}\) P(B) = \(\frac{1}{5}\)
∴ P(A’) = \(\frac{6}{7}\) P(B) = \(\frac{4}{5}\)
(i) P(A∩B) = P(A).P(B) = \(\frac{1}{7}\) × \(\frac{1}{5}\) = \(\frac{1}{35}\)

(ii) P(A∪B) = 1 – P(A’). P(B’)
= 1 – \(\frac{6}{7}\) × \(\frac{4}{5}\) = 1 – \(\frac{24}{35}\) = \(\frac{11}{35}\)

(iii) P((A∩B’)∩(B∩A’))
Since A∩B’ and B∩A’ are,
mutually exclusive events,
(A∩B’)∩(B∩A’) = Φ
∴ P[(A∩B’)∩(B∩A’)] = 0

Question 20.
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even} is an equivalence relatIon. (4)
Answer:
We have
R = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1),
(3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (5, 5)}
Since (a, a) ∈ R ∀ a ∈ A, R is reflexive.
(a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ A
∴ R is symmetric.
(a, b) ∈ R, (b,c) ∈ R ⇒ (a, c) ∈ R ∀ a,b,c ∈ A
∴ R is transitive.
Hence R is an equivalence relation.

Question 21.
Find \(\frac{d y}{d x}\) of the function y^x = x^y. (4)
Answer:
Given y^x = x^y
taking ‘log’ on both sides
log y^x = log x^y
ie., x log y = y log x
Differentiating with respect to x,

Question 22.
Find
(i) ∫x log x dx (2)
(ii) ∫x² sin x dx (2)
Answer:

Part – C

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

Question 23.
Express the matrix A = \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix. (6)
Answer:
Given A = \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\) A’ = \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\)
Let P = \(\frac{1}{2}\)(A + A’) = \(\frac{1}{2}\)\(\left[\begin{array}{ccc}
12 & -4 & 4 \\
-4 & 6 & -2 \\
4 & -2 & 6
\end{array}\right]\) symmetric
Q = \(\frac{1}{2}\)(A – A’) = \(\frac{1}{2}\)\(\left[\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)Skew symmetric
∴ A = P + Q

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:

Question 25.
(i) State whether the function f : R → R defined by f(x) = 3 – 4x is a bijective function or not. Justify your answer. (4)
(ii) Let f : R → R and. g : R → R be defined by
f(x) = 2x +1 and g(x) = x². Then find gof and fog.(2)
Answer:
(i) Given f(x) = 3 – 4x
f(x¹) = f(x²) ⇒ 3 – 4x¹ = 3 – 4x²
⇒ -4x¹ = -4x²
⇒ x¹ = x²
∴ f is one – one
Let y = 3 – 4x
4x = 3 – y
x = \(\frac{3 – y }{4}\)
f(x) = f(\(\frac{3 – y }{4}\)) = 3 – 4(\(\frac{3 – y }{4}\)) = y
∴ For every y ∈ R, there exists x ∈ R such that f(x) = y
∴ f is onto
Hence f is a bijective function

(ii) Given f(x) = 2x +1, g(x) = x²
(gof)(x) = g[f(x)] = g(2x + 1)
= (2x + 1 )² = 4x² + 4x + 1
(fog)(x)= f[g(x)] = f(x²)
= 2x² + 1

Question 26.
Find the equation of the tangent line to the curve
y = x² – 2x – 7 which is
(i) Parallel to the line 2x – y + 9 = 0 (3)
(ii) Perpendicular to the line 5y – 15x = 13 (3)
Answer:
Given y = x² – 2x + 7
\(\frac{d y}{d x}\) = 2x – 2
(i) Tangent is parallel to 2x – y + 9 = 0
∴ Slope of tangent = slope of given line
\(\frac{d y}{d x}\) = \(\frac{-2}{-1}\)
2x – 2 = 2
2x = 4
x = 2
When x = 2, y = (2)² – 2 (2) + 7 = 7
∴ Point is (2, 7)
Equation of tangent is dy
y – y₀ = \(\frac{d y}{d x}\)(x – x₀)
y – 7 = 2(x – 2)
y – 7 = 2x – 4
2x – y + 3 = 0

(ii) Tangent is perpendicular to 5y – 15x = 13
ie., – 15 x + 5y – 13 = 0
∴ Slope of tangent × slope of given line = -1
\(\frac{d y}{d x}\) × 3 = -1
\(\frac{d y}{d x}\) = \(\frac{-1}{3}\)
2x – 2 = \(\frac{-1}{3}\)
2x = 2 – \(\frac{1}{3}\) = \(\frac{5}{3}\), 5. x = \(\frac{5}{6}\)
When x = \(\frac{5}{6}\), y = \(\frac{25}{36}\) – \(\frac{10}{6}\) + 7 = \(\frac{217}{36}\)
Point is (\(\frac{5}{6}\), \(\frac{217}{36}\))
∴ Equation of tangent is
y – y₀ = \(\frac{d y}{d x}\)(x – x₀)
y – \(\frac{217}{36}\) = –\(\frac{1}{3}\) (x – \(\frac{5}{6}\))
36y – 217 = -12x + 10
12x + 36y – 227 = 0

Question 27.
Find the following :
(i) \(\int \frac{1}{1+x^2} d x\) (1)
(ii) \(\int \frac{d x}{x^2-6 x+13}\) (2)
(iii) \(\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x\) (3)
Answer:

Question 28.
(i) Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant. (3)
(ii) Find the area of the region bounded by the two parabolas y = x² and y2 = x. (3)
Answer:
(i) Required area

Given y² = x _________ (1)
y = x² __________ (2)
Point of intersection is given by solving (1) & (2)
ie., √x = x² ⇒ x = 0, 1
When x = 0, y = 0
x = 1, y = 1
∴ Points of intersections are (0, 0) and (1, 1)

Question 29.
Solve Linear Programming Problem (LPP) graphically.
Maximize : z = 3x + 2y
Subject to constraints: x + 2y ≤ 10 (6)
3x + y ≤ 15
x, y ≥ 0
Answer:
Corresponding equation are
x + 2y = 10 ________ (1)

x	0	10
y	5	0

3x + 2y = 15 _________ (2)

x	0	5
y	15	0

Solving (1) and (2)
The point of intersection is (4, 3)

∴ z has maximum when x = 4 and y = 3
Maximum value = 18