Kerala Plus Two Maths Board Model Paper 2020 with Answers

Reviewing Kerala Syllabus Plus Two Maths Previous Year Question Papers and Answers Pdf Board Model Paper 2020 helps in understanding answer patterns.

Kerala Plus Two Maths Board Model Paper 2020 with Answers

Time: 2 Hours
Total Score: 60 Marks

Question 1 to 8 carry 3 scores each. Answer any 7 questions.

Question 1.
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-3 & 1 & 0 \\
1 & 3 & -2
\end{array}\right]\)
(i) Which of the following is the order of the matrix A+B?
(a) 2 × 2
(b) 3 × 2
(c) 2 × 3
(d) 3 × 3
(ii) Find 3A
(iii) Evaluate 3A – B
Answer:
(i) (c) 2 × 3

(ii) 3A = \(\left[\begin{array}{lll}
3 & 6 & 9 \\
9 & 3 & 6
\end{array}\right]\)

(iii) 3A – B
= \(\left[\begin{array}{lll}
3 & 6 & 9 \\
9 & 3 & 6
\end{array}\right]\) – \(\left[\begin{array}{ccc}
-3 & 1 & 0 \\
1 & 3 & -2
\end{array}\right]\)
= \(\left[\begin{array}{lll}
6 & 5 & 9 \\
8 & 0 & 8
\end{array}\right]\)

Question 2.
(i) If y = sin^-1 x, Find \(\frac{d y}{d x}\)
(ii) Hence show that (1 – x²)\(\frac{d^2 y}{d x}\) – x\(\frac{d y}{d x}\) = 0
Answer:
(i) Given; y = sin^-1 x
\(\frac{d y}{d x}\) = \(\frac{1}{\sqrt{1-x^2}}\) …………….. (1)

Question 3.
(i) Which of the following is the solution of the differential equation \(\frac{d y}{d x}\) + sin x = 0
(a) y = C cos x
(b) y = cos x + C
(c) y = sin x + C
(d) y = C sin x
(ii) Form the differential equation representing the family of curves y = a sin(x + 6), where a and b are arbitrary constants.
Answer:
(i) (b) y = cos x + C

(ii) Given; y = a sin(x + b) …………… (1)
Differentiating (1) w.r.to x; we get
\(\frac{d y}{d x}\) = a cos (x + b)
Differentiating (2) w.r.to x; we get
\(\) = -a sin(x + b)
⇒ \(\frac{d^2 y}{d x^2}\) = -y

Question 4.
Using properties of determinants prove that
\(\left|\begin{array}{ccc}
b+c & a & a \\
b & c+a & b \\
c & c & a+b
\end{array}\right|\) = 4abc
Answer:
LHS = \(\left|\begin{array}{ccc}
b+c & a & a \\
b & c+a & b \\
c & c & a+b
\end{array}\right|\)
Apply R₁ → R₁ – R₂ – R₃
= \(\left|\begin{array}{ccc}
0 & -2 c & -2 b \\
b & c+a & b \\
c & c & a+b
\end{array}\right|\)
= 2c(ab + b² – bc) – 2 b(bc – c² – ac)
= 2abc + 2cb² – 2bc² – 2b²c + 2bc² + 2abc
= 4abc

Question 5.
(i) Find the maximum and minimum values of f, if any of the function f(x) = |x| + 3, x ∈ R
(ii) What is the absolute maximum value of the function f(x) – |x| + 3, x ∈ [-3, 2]?
Answer:

(i) The function is not differentiable at x = 0. But from the graph it is clear that it has only minimum at x = 0 and it is 3.

(ii) When the domain is restricted as [-3, 2], then f(x) has absolute maximum at x = -3 which is 6 and absolute minimum at x = 2 which is 5.

Question 6.
(i) Write a function which is not continuous at x = 0 and justify your answer
(ii) Check the continuity of the function

Answer:

Left limit and right limit are not equal. So not continuous.

(ii) For x < 0, f(x) – x + 2 and for x > 0 f(x) = – x + 2, which are polynomials so continuous. But the function is not defined at x = 0. Hence f(x) is continuous on its domain.

Question 7.
Consider the arrow diagram of the functions f and g.

(i) Check whether the functions f and g are bijective function? Justify.
(ii) Write the function go f.
(iii) Is go f a bijective function function? Justify.
Answer:
(i) f is not onto since d has no preimage, g is not one-one since c and d has same image z. Hence both f and g are not bijective.

(ii) go f = (1, x), (2, z), (3, y)

(iii) The domain of gof is {1, 2, 3} and range is {x, y, z}. So go f is a bijective function.

Question 8.
(i) If α, β, γ are the direction angles of a vector, then which the following can be α + β?
(a) 80°
(b) 60°
(c) 120°
(d) can’t be determined
(ii) Find a direction cosines of the line passing through the points (2, 8, 3) and (4, 5, 9).
Answer:
(i) (c) 120°. Since the sum any two direction angle should be greater than or equal to 90°.

(ii) The direction ratios of the line is 4 – 2, 5 – 8, 9 – 3.
2, -3, 6
Then direction cosines is
\(\frac{2}{\sqrt{4+9+36}}\), \(\frac{-3}{\sqrt{4+9+36}}\), \(\frac{6}{\sqrt{4+9+36}}\)
⇒ \(\frac{2}{7}\), \(\frac{-3}{7}\), \(\frac{6}{7}\)

Questions 9 to 18 carry 4 scores each. Answer any 8.

Question 9.
(i) If tan^-1 x = \(\frac{\pi}{10}\), then the value of cot^-1 x is
(a) \(\frac{\pi}{5}\)
(b) \(\frac{2\pi}{5}\)
(c) \(\frac{3\pi}{5}\)
(d) \(\frac{4\pi}{5}\)
(ii) Find the value of
sin (tan^-1\(\frac{2}{3}\)) + cos (tan^-1√3)
Answer:
(i) (b) \(\frac{2\pi}{5}\), since cot^-1 x = \(\frac{\pi}{2}\) – tan^-1 x

(ii) sin (2 tan^-1\(\frac{2}{3}\)) + cos (tan^-1√3)
= sin (tan^-1 \(\frac{2 \times \frac{2}{3}}{1-\frac{2}{3} \times \frac{2}{3}}\)) + cos (tan^-1√3)
= sin (tan^-1\(\frac{12}{5}\)) + cos(tan^-1√3)
Draw a triangle to find tan^-1\(\frac{12}{5}\) = sin^-1\(\frac{12}{13}\)
= sin (sin^-1\(\frac{12}{13}\)) + cos \(\frac{\pi}{3}\)
= \(\frac{12}{13}\) + \(\frac{1}{2}\) = \(\frac{37}{26}\)

Question 10.
Let A = {-1, 0, 1}
(i) Give reason why the operation defined by a ⊗ b = \(\frac{a}{b}\) is not a binary operation on A.
(ii) (a) Write a binary operation * on A
(b) Find (-1 * -1) * -1.
(iii) How many binary operations are possible on A?
Answer:
(i) Since the division by zero is not defined.

(ii) (a) a * b = a × b
(b) (-1 * -1) *-1 = (1) * -1 = -1

(iii) Binary operation is function A × A to A .
Therefore number of function is 39.

Question 11.
(i) Find the equation of a line L passing through the points (-1, 0, 2) and (2, 1, 3).
(ii) If \(\vec{c}\) = î + ĵ + λk̂ be a vector perpendicular to the above line, then find λ.
(iii) Find the equation of a plane on which the line L lies.
Answer:
(i) The equation of the line is
\(\frac{x + 1}{2 + 1}\) = \(\frac{y – 0}{1 – 0}\) = \(\frac{z – 2}{3 – 2}\)
⇒ \(\frac{x + 1}{3}\) = \(\frac{y}{1}\) = \(\frac{z – 2}{1}\)

(ii) Since vector is perpendicular, we have
3 × 1 + 1 × 1 + 1 × λ = 0
3 + 1 + A = 0 ⇒ λ = -4

(iii) Equation of the plane passing through (-1, 0, 2) and perpendicular vector = \(\vec{c}\) = î + ĵ – 4k̂ is
1(x + 1) + 1(y – 0) – 4(z – 2) = 0
x + y – 4z + 1 + 8 = 0
x + y – 4z + 9 = 0

Question 12.
Find \(\frac{d y}{d x}\) of the following:
(i) y = sec (tan x)
(ii) x^y = y^x
Answer:
(i) \(\frac{d y}{d x}\) = sec (tan x) tan (tan x) sec² x

(ii) Given x^y = y^x
Take log on both sides:
y log x = x log y
Differentiating w r to x

Question 13.
(i) Find \(\int \tan ^{-1} x d x\)
(ii) Hence find the area bounded by the curve
y = tan^-1 x with X – axis from x = 0 and x = 1
Answer:

Question 14.
Solve the differential equation \(\frac{d y}{d x}\) = \(\frac{x^2+y^2}{2 x y}\)
Answer:
Given ; \(\frac{d y}{d x}\) = \(\frac{x^2+y^2}{2 x y}\)
Put y = vx and \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\)
v + x\(\frac{d v}{d x}\) = \(\frac{x^2+(v x)^2}{2 x(v x)}\)
v + x\(\frac{d v}{d x}\) = \(\frac{1+v^2}{2 v}\)

Question 15.
Consider a plane which is equidistant from the points (1, 2, 1) and (3, 4, 7).
(i) Which of the following is a point on the plane?
(a) (1, 3, 1)
(b) (4, 2, 3)
(c) (2, 3, 4)
(d) (1, 3, 6)
(ii) Find the equation of the above plane.
Answer:
(i) (c) (2, 3, 4)

(ii) The plane will pass through the point (2, 3, 4) and
will a direction ratio 3 – 1, 4 – 2, 7 – 1 ⇒ 2, 2, 6
then dr’s can be taken as 1, 1, 3
Hence the equation is
1(x – 2) + 1(y – 3) + 3(z – 4) = 0
x + y + 3z – 17 = 0

Question 16.
Let \(\vec{a}\) = 2î + ĵ – 3k̂ and \(\vec{b}\) = 4î + ĵ + k̂ be two vectors
(i) Find \(\vec{a}\) vector \(\vec{c}\) perpendicular to \(\vec{a}\) and \(\vec{b}\).
(ii) Find the volume of the parallelepiped with coinitial vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
(iii) If \(\vec{c}\) is rotated in such a way that it makes 60° with its initial direction, then what is the volume of the new parallelepiped formed?
Answer:
(i) \(\vec{c}\) = \(\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 1 & -3 \\
4 & 1 & 1
\end{array}\right|\) = 4î – 14ĵ – 2k̂

(ii) volume = [\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]
= (\(\vec{a}\) × \(\vec{b}\) ×) . \(\vec{c}\) = \(\vec{c}\) . \(\vec{c}\) = |\(\vec{c}\)|²
= 16 + 196 + 4 = 216
When \(\vec{c}\) is rotated 60° the height of the parallelepiped will be reduced by half of the original volume.
Since Volume = \(\vec{c}\) × \(\vec{c}\)
= |\(\vec{c}\)| × |\(\vec{c}\)| cos(60°)
= |\(\vec{c}\)| × |\(\vec{c}\)| × \(\frac{1}{2}\)
Hence the new volume is 108.

Question 17.
(i) Evaluate \(\int_0^\pi x \sin x d x\)
(ii) Hence evaluate the area bounded by the curve y = x sin x between x = -π and x = π.
Answer:

Question 18.
In a ladies hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.
(i) Find the probability that she reads neither Hindi nor English newspapers.
(ii) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(iii) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Answer:
i) Let the events H- Hindi, E = English
P(H) = \(\frac{60}{100}\) = 0.6; P(E) = \(\frac{40}{100}\) = 0.4
P(H∩E) = \(\frac{20}{100}\) = 0.2
P(H∪E) = P(H) + P(E) – P(H∩E)
= 0.6 + 0.4 – 0.2 = 0.8
P(she reads neither Hindi nor English newspapers.)
= P(H’∩E’) = 1 – P(H∪E) = 1 – 0.8 = 0.2

(ii) P(E/H) = \(\frac{P(E \cap H)}{P(H)}\) = \(\frac{0.2}{0.6}\) = \(\frac{1}{3}\)

(iii) P(E/H) = \(\frac{P(E \cap H)}{P(E)}\) = \(\frac{0.2}{0.4}\) = \(\frac{1}{2}\)

Questions from 19 to 25 carry 6 scores each. Answer any 5.

Question 19.
(i) Construct a 3 × 3 matrix A, where elements are given by a_ij = 2i – j
(ii) Verify that C = A – A’ is a skew symmetric matrix.
(iii) Verify that C² is a symmetric matrix.
Answer:

Clearly (C²) = C² therefore symmetric.

Question 20.
Consider the matrix A = \(\left[\begin{array}{ccc}
1 & 3 & 6 \\
-1 & -1 & 2 \\
1 & 1 & 5
\end{array}\right]\)
(i) Find |A|
(ii) Verify that A × adj A = |A| I
(iii) Evaluate | A^-1|
Answer:
(i) |A| = \(\left[\begin{array}{ccc}
1 & 3 & 6 \\
-1 & -1 & 2 \\
1 & 1 & 5
\end{array}\right]\) = 14

(ii) C₁₁ = – 5 – 2 = -7; C₁₂ = -(5 – 2) = 7
C₁₃ = – 1 + 1 = 0; C₂₁ = -(15 – 6) = -9
C₂₂ = 5 – 6 = -1; C₂₃ = -(1 – 3) = 2
C₃₁ = – 6 + 6 = 12; C₃₂ = -(2 + 6) = -8
C₃₃ = – 1 + 3 = 2

(iii) A × A^-1 = I
|A × A^-1| = |I|
|A| × |A^-1| = 1
= \(\frac{1}{14}\)

Question 21.
Integrate the following with respect to x:
(i) \(\frac{1}{x^2-6 x+13}\)
(ii) \(\frac{\cos x}{(\sin x-1)(\sin x-2)}\)
Answer:

Question 22.
Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice.
(i) Write the probability distribution of X.
(ii) Find variance of X.
Answer:
(i) Sample space; n(S) = 36
X: Sum of the numbers obtained.
X = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Question 23.
(i) Consider the function f(x) = 2x³ – 6x² + 1
(a) Find the equation of tangents parallel to X- axis.
(b) Find the intervals in which the function f is decreasing.
(ii) The length x of a rectangle is decreasing at the rate of 5 cm/s and the width y is increasing at the rate of 4 cm/s. When x = 8 cm and y = 6 cm, find the rale of change of the area of the rectangle.
Answer:
(i) Given; f(x) = 2x³ – 6x² + 1
(a) f'(x) = 6x² – 12x
For point at which tangents parallel to x axis is when f'(x) – 0 ⇒ 6x² – 12x = 0 ⇒ x = 0, 2
Then the tangents lines are y = f(0) = 1 ⇒ y = 1 and y = f(2) = -7 ⇒ y = -7

(b) The turning point x = 0, 2 divides the domain into the following intervals (-∞, 0), (0, 2), (2, ∞) ‘ Clearly, f'(x) < 0, x ∈ (0, 2). Hence f(x) is decreasing in the interval (0, 2)

(ii) Given, x and y be the length and breadth of the rectangle at an instant t.
\(\frac{d y}{d t}\) = -5, \(\frac{d x}{d t}\) = 4
Given; x = 8 and y = 6
A = xy ⇒ \(\frac{d A}{d t}\) = x\(\frac{d y}{d t}\) + y\(\frac{d x}{t}\)
⇒ \(\frac{d A}{d t}\) = (8) (4) + (6) (-5) – 2 cm²/s

Question 24.
Let \(\vec{a}\) = 2î – 4ĵ + 5k̂ and \(\vec{b}\) = î – 2ĵ – 8k̂ be two vectors
(i) Find \(\vec{a}\) vector \(\vec{c}\) representing a diagonal of the parallelogram with \(\vec{a}\) and \(\vec{b}\) as the adjacent sides.
(ii) Find the projection of \(\vec{b}\) on \(\vec{c}\).
(iii) Find the angle between the vectors \(\vec{c}\) and \(\vec{a}\)
Answer:
(i) Here \(\vec{c}\) can be \(\vec{a}\) + \(\vec{b}\)
\(\vec{c}\) = 3î – 6ĵ – 3k̂
(ii) the projection of \(\vec{b}\) on \(\vec{c}\) = \(\frac{\vec{b} \cdot \vec{c}}{|\vec{c}|}\)

Question 25.
Maximise: Z = 600x + 400y
Subject to the constraints:
x + 2y ≤ 12 ;2x + y ≤ 12
4x + 5y ≥ 20 ; x ≥ 0, y ≥ 0
(i) Draw the feasible region.
(ii) Solve the LPP?
Answer:

(ii) The corner points are A(5, 0), B(6, 0), C(4, 4) D(0, 6), E(0, 4)

Corner Point	Value of Z z = 600x + 400y
A(5, 0)	z = 600x + 400y = 3000 + 0 = 3000
B(6, 0)	z = 600x + 400y = 3600 + 0 = 36 00
C(4, 4)	z = 600x + 400y = 2400 + 1600 = 4000
0(0, 6)	z = 600x + 400y = 0 + 2400 = 2400
E(0, 4)	z = 600x + 400y = 0 + 1600 = 1600

The maximum value of z is 4000 at C(4, 4).