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CBSE Class 12 Maths Question Paper 2020 (Series: HMJ/4) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Maths with Solutions and CBSE Class 12 Maths Question Paper 2020 (Series: HMJ/4) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Maths Question Paper 2020 (Series: HMJ/4) with Solutions

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

  1. This question paper comprises four sections- A, B, C and D. This question paper carries 36 questions. All questions are compulsory.
  2. Section A – Question no. 1 to 20 comprises of 20 questions of one mark each.
  3. Section B – Question no. 21 to 26 comprises of 6 questions of two marks each.
  4. Section C – Question no. 27 to 32 comprises of 6 questions of four marks each.
  5. Section D – Question no. 33 to 36 comprises of 4 questions of six marks each.
  6. There is no overall choice in the question paper. However, an internal choice has been provided in 3 questions of one mark, 2 questions of two marks, 2 questions of four marks and 2 questions of six marks. Only one of the choices in such questions have to be attempted.
  7. In addition to this, separate instructions are given with each section and question, wherever necessary.
  8. Use of calculators is not permitted.

SET I Code No. 65/4/1 
Section-A

Q.No. 1 to Q.10 are multiple choice type questions of 1 mark each. Select the correct option:

Question 1.
The value of sin-1 cos \(\left(\frac{3 \pi}{5}\right)\) is
(a) \(\frac{\pi}{10}\)
(b) \(\frac{3 \pi}{5}\)
(c) \(\frac{-\pi}{10}\)
(d) \(\frac{-3 \pi}{5}\)
Solution:
(c) \(\frac{-\pi}{10}\)

Question 2.
If A = [2 -3 4], B = \(\left[\begin{array}{l}
3 \\
2 \\
2
\end{array}\right]\), then AB + XY equals
(a) [28]
(b) [24]
(c) 28
(d) 24
Solution:
(a) [28]

Question 3.
If \(\left|\begin{array}{lll}
2 & 3 & 2 \\
x & x & x \\
4 & 9 & 1
\end{array}\right|\) + 3 = 0, then the value of x is
(a) 3
(b) 0
(c) -1
(d) 1
Solution:
(c) -1

\(\left|\begin{array}{lll}
2 & 3 & 2 \\
x & x & x \\
4 & 9 & 1
\end{array}\right|\) + 3 = 0
⇒ 2(x – 9x) – 3(x – 4x) + 2(9x – 4x) = -3
⇒ -16x + 9x + 10x = -3
⇒ 3x = -3 ∴ x = -1

Question 4.
\(\int_0^{\pi / 8} \tan ^2(2 x)\) dx is equal to
(a) \(\frac{4-\pi}{8}\)
(b) \(\frac{4+\pi}{8}\)
(c) \(\frac{4-\pi}{4}\)
(d) \(\frac{4-\pi}{2}\)
Solution:
(a) \(\frac{4-\pi}{8}\)

Question 5.
If \(\vec{a} \cdot \vec{b}\) = \(\frac{1}{2}|\vec{a}||\vec{b}|\), then the angle between \(\vec{a}\) and \(\vec{a}\) is
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Solution:
(c) 60°

Question 6.
The two lines x = ay + b, z = cy + d; and x = a’y + b’, z = c’y + d’ are perpendicular to each other other, if
(a) \(\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}\) = 1
(b) \(\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}\) = -1
(c) aa’ + cc’ = 1
(d) aa’ + cc’ = -1
Solution:
(d) aa’ + cc’ = -1

Question 7.
The equation of the line in vector form passing through the point (-1, 3, 5) and parallel to line \(\frac{x-3}{2}\) = \(\frac{y-4}{3}\), z = 2 is

Solution:
(b);

Direction ratio’s of the normal of given plane are 2, 3, 0,
∴ Direction ratio’s of the normal of required plane are 2, 3, 0
Vector equation of required plane,
\(\begin{gathered}
\vec{r}
\end{gathered}\) = \(\vec{A}+\lambda \vec{B}\)
Here, Direction Ratio’s i.e.,
\(\vec{B}\) = \(2 \hat{i}+3 \hat{j}+0 \hat{k}\)
We have, Point(-1, 3, 5) i.e.,
\(\vec{A}\) = \(-1 \hat{i}+3 \hat{j}+5 \hat{k}\)
∴ \(\vec{r}\) = \((-\hat{i}+3 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+3 \hat{j})\)

Question 8.
In an LPP, if the objective function z – ax + by has the same maximum value on two corner points of the feasible region, then the number of points at which zmax occurs is
(a) 0
(b) 2
(c) finite
(d) infinite
Solution:
(d) infinite

Infinite, If two corner points of the feasible region are both optimal solutions of the same type, i.e., both produce the same maximum or minimum, then any point on the line segment joining these two points is also an optimal solution of the same type.

Question 9.
From the set {1, 2, 3, 4, 5}, two numbers a and b(a≠b) are chosen at random. The probability that \(\frac{a}{b}\) is an integer is:
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{3}{5}\)
Solution:
(b) \(\frac{1}{4}\)

Question 10.
A bag contains 3 white, 4 black and 2 red balls. If 2 balls are drawn at random (without replacement), then the probability that both the balls are white is
(a) \(\frac{1}{18}\)
(b) \(\frac{1}{36}\)
(c) \(\frac{1}{12}\)
(d) \(\frac{1}{24}\)
Solution:
(c) \(\frac{1}{12}\)

In Q. Nos. 11 to 15, fill in the blanks with correct word/sentence:

Question 11.
Vector of magnitude 5 units and in the direction opposite to \((2 \hat{i}+3 \hat{j}-6 \hat{k})\) is ………….
Solution:

Question 12.
If \(\left[\begin{array}{cc}
x+y & 7 \\
9 & x-y
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 7 \\
9 & 4
\end{array}\right]\) then x, y = ______
Solution:

Question 13.
The number of points of discontinuity of f defined by f(x) = |x| – |x + 1| is ______
Solution:
Zero, there is no point of discontinuity.

Question 14.
The rate of change of the area of a circle with respect to its radius r, when r = 3 cm is ____.
Solution:
Let Area of circle, A = πr2
Differentiating w.r.t. r, we have
\(\frac{d \mathrm{~A}}{d r}\) = 2πr
∴ \(\frac{d \mathrm{~A}}{d r}\) where r = 3 cm = 2π(3) = 6π cm2

Question 15.
If \(\vec{a}\) is a non-zero vector, then \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}\) equals ______
Solution:

Or

The projection of the vector \(\hat{i}-\hat{j}\) on the vector \(\hat{i}+\hat{j}\) is ____
Solution:

Questions number 16 to 20 are very short answer type questions:

Question 16.
Find adj A, if A = \(\left[\begin{array}{cc}
2 & -1 \\
4 & 3
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{cc}
2 & -1 \\
4 & 3
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{cc}
3 & 1 \\
-4 & 2
\end{array}\right]\)

Question 17.
Find \(\int \frac{2^{x+1}-5^{x-1}}{10^x} d x\)
Solution:

Question 18.
Evaluate \(\int_0^{2 \pi}|\sin x| d x\)
Solution:

Question 19.
If \(\int_0^a \frac{d x}{1+4 x^2}\) = \(\frac{\pi}{8}\), then find the value of a.
Solution:

Or

Find \(\int \frac{d x}{\sqrt{x}+x}\)
Solution:

Question 20.
Show that the function y = ax + 2a2 is a solution of the differential equation
\(2\left(\frac{d y}{d x}\right)^2\) + x\(\left(\frac{d y}{d x}\right)\) – y = 0
Solution:

Section – B

Q.No. 21 to 26 carry 2 marks each.

Question 21.
Check if the relation R on the set A = {1, 2, 3, 4, 5, 6} defined as R = {(x, y):y is divisible by x} is
(i) symmetric
(ii) transitive.
Solution:
R = {(x, y); y is divisible by x}

(i)

(ii)
Transitive:
(a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R
Let (x, y) ∈ R and (y, z) ∈ R
To prove (x, z) ∈ R
Here, y is divisible by x …….. (A)
z is divisible by y ……….. (B)
From (A) and (B), we have
z is divisible by x.
∴ (x, z) ∈ R
It is transitive.

Question 22.
Find the value of \(\frac{d y}{d x}\) at θ = \(\frac{\pi}{3}\), if x = cos θ – cos 2θ, y = sin θ – sin 2θ.
Solution:

Question 23.
Show that the function f defined by f(x) = (x – 1)ex + 1 is an increasing function for all x > 0.
Solution:
We havef(x) (x – 1) ex + 1
Differentiating both sides w,r.t x, we have
f’(x) = (x – 1)ex + ex.1
= ex(x – 1 + 1) = xex
When x > 0, x > 0
ex > 0 […ex > 0 for all x ∈ R
then xex > 0
f’(x) > 0 ……[for all x > 0
∴ f(x) is an increasing function for all x > 0.

Question 24.
Find | \(\vec{a}\) | and | \(\vec{b}\) |, if | \(\vec{a}\) | = 2| \(\vec{b}\) | and (\(\vec{a}\) + \(\vec{b}\)). (\(\vec{a}\) – \(\vec{b}\)) = 12.
Solution:

Or

Find the unit vector perpendicular to each of the vectors \(\vec{a}\) = 4\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) and \(\vec{b}\) = 2\(\hat{i}\) – \(\hat{j}\) + 2\(\hat{k}\).
Solution:

Question 25.
Find the distance between the lines l1 and l2 giveh by

Solution:

Question 26.
Find [P(B | A) + P(A | B)], if P(A) = \(\frac{3}{10}\), P(B) = \(\frac{2}{5}\) and P (A ∪ B) = \(\frac{3}{5}\).
Solution:

Section – C

Q. Nos. 27 to 32 carry 4 marks each.

Question 27.
Prove that the relation R on Z, defined by R {(x, y):(x – y) is divisible by 5} is an equivalence relation.
Solution:
R = {(x, y): x, y ∈ Z, and (x – y) is divisible by 5 .. [Given
Reflexive:
R is reflexive, for x ∈ Z
x – x = 0, which is divisible by 5
(x, x) ∈ R ∴ R is reflexive
Symmetric:
R is symmetric, (x, y) ∈ R, where x, y ∈ Z
x – y is divisible by 5
Let (x – y) = 5λ, where λ ∈ Z
y – x = -5λ
∴ y – x is divisible by 5
(y, x) ∈ R ;. R is symmetric
Transitive:
R is transitive, (x, y) ∈ R where x, y ∈ Z
x – y is divisible by 5
x – y = 5λ, where λ ∈ Z ………. (i)
For (y, z) ∈ R, where y, z ∈ Z
y – z is divisbie by 5
y – z = 5β, where β ∈ Z ……(ii)
Adding (i) and (ii),
(x – y) + (y – z) = 5λ + 5
x – z = 5(λ + 13)
(x – z) is divisible by 5
(x, y) ∈ R, (y, z) e R
∴ (x, z) ∈ R where x, y, z ∈ Z
∴ R is transitive

Question 28.
If y = sin-1\(\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)\), then show that \(\frac{d y}{d x}\) = \(\frac{-1}{2 \sqrt{1-x^2}}\)
Solution:

Question 29.
Evaluate: \(\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\)
Solution:

Question 30.
For the differential equation given below, find a particular solution satisfying the given condition
(x + 1)\(\frac{d y}{d x}\) = 2e-y + 1; y = 0 when x = 0.
Solution:

Question 31.
A manufacturer has three machines I, II and III installed in his factory. Machine I and II are capable of being operated for almost 12 hours whereas machine III must be operated for atleast 5 hours a day. He produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit of M and N on three machines are given in the following table:

He makes a profit of ₹600 and ₹400 on one unit of items M and N respectively. How many units of each item should he produce so as to maximize his profit assuming that he can sell all the items that he produced. What will be the maximum profit?
Solution:
Let x and y be the number of items M and N respectively.
Maximise profit, Z = ₹(600x + 400y)
Subjects to the constraints,
x + 2y ≤ 12 ……….(i)
2x + y ≤ 12 ………(ii)
x + 1.25y ≥ 5 or x + \(\frac{5}{4}\)y ≥ 5 …(iii)
x ≥ 0, y ≥ 0

Checking at Origin (0, 0):
(i) x + 2y ≤ 12 ⇒ 0 + 0 ≤ 12,
(True shading towards origin)
(ii) 2x + y ≤ 12 ⇒ 0 + 0 ≤12
(True shading towards origin)
(iii) x + \(\frac{5}{4}\)y ≥ 5,
⇒ 0 + 0 ≥ 5,
(False shading against opposite to origin.)

∴ The manufacturer has to produce 4 units of each item to get the maximum profit of ₹4,000.

Question 32.
A coin is biased so that the head is three times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails. Hence find the mean of the number of tails.
Solution:
Given. P(H) = 3P(T)
As we know, P(H) + P(T) = 1
3P(T) + P(T) = 1
4P(T) = 1
P(T) = 1/4

Or

Question 32.
Suppose that 5 men out of 100 and 25 women out of 1000 are good orators. Assuming that there are equal number of men and women, find the probability of choosing a good orator.
Solution:

Section – D

Q. Nos. 33 to 36 carry 6 marks each.

Question 33.
If A = \(\left[\begin{array}{ccc}
1 & 3 & 2 \\
2 & 0 & -1 \\
1 & 2 & 3
\end{array}\right]\), then show that A3 – 4A2 – 3A + 111 = O. Hence find A-1.
Solution:

Question 34.
Find the intervals on which the function f(x) = (x – 1)3 (x – 2)2 is
(a) strictly increasing
(b) strictly decreasing.
Solution:

Or

Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its side. Also, find the maximum volume.
Solution:
Let breadth and length of rectangle be a x cm and y cm.
Given. Perimeter of Rectangle = 36 cm
⇒ 2(x + y) = 36
⇒ x + y = 18
⇒ y = 18 – x ….. (i)

When revolved about one of its length y.

⇒ 36x = 3x2
⇒ 3x2 – 36x = 0
⇒ 3(x2 – 12x) = 0
⇒ 3x(x – 12) = 0
x = 0 or x = 12
Again, differentiating w.r.t. x we get

At x = 12, volume of the resultant cylinder is the maximum.
So, the dimensions of rectangles are 12 cm and 6 cm, respectively.
∴ Maximum volume of resultant cylinder,
Vx=12 = π[18.(12)2 – (12)3]
= π[122(18 – 12)]
= π × 144 × 6 = 864πcm3

Question 35.
Find the area of the region lying in the first quadrant and enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32.
Solution:

The area of the shaded region

Question 36.
Find whether the lines \(\vec{r}\) = (\(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)) + λ(2\(\hat{i}\) + \(\hat{j}\)) and \(\vec{r}\) = (2\(\hat{i}\) – \(\hat{j}\)) + µ(\(\hat{j}\) + \(\hat{j}\) – \(\hat{k}\)) intersect or not. If intersecting, find their point of intersection.
Solution:

Hence, the given two lines are intersecting
Point of intersection
(1 + 2λ, -1 + λ, -1) or (2 + µ, -1 + µ, -µ)
(1 + 2(1), -1 + 1, -1) or (2 + 1, -1 + 1, -1)
(3, 0, -1) or (3, 0, -1)

SET II Code No. 65/4/2 

Note: Except for the following questions, all the remaining questions have been asked in Set-I.

Question 8.
Let A = \(\left[\begin{array}{rr}
200 & 50 \\
10 & 2
\end{array}\right]\), then | AB | is equal to
(a) 460
(b) 2000
(c) 3000
(d) -7000
Solution:
(d) -7000

| A | = 400 – 500 = -100;
∴ | B | = 150 – 80 = 70
∴ | AB | = |A| |B|
= (-100) (70) = -7000

Question 9.
Let \(\vec{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\). If \(\vec{b}\) is a vector such that \(\vec{a} \cdot \vec{b}\) = \(|\vec{b}|^2\) and \(|\vec{a}-\vec{b}|\) = \(\sqrt{7}\), then \(|\vec{b}|\) equals
(a) 7
(b) 14
(c) \(\sqrt{7}\)
(d) 21
Solution:
(c) \(\sqrt{7}\)

Question 10.
Three dice are thrown simultaneously. The probability of obtaining a total score of 5 is
(a) \(\frac{5}{216}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{36}\)
(d) \(\frac{1}{49}\)
Solution:
(c) \(\frac{1}{36}\)

Three dice can be thrown as 6 × 6 × 6 = 216 ways
“A total of 5” can be obtained as (1, 1, 3) (1, 3, 1), (3, 1, 1), (2, 2, 1) (2, 1, 2), (1, 2, 2), i.e. 6 ways
∴ P(a total of 5) = \(\frac{6}{216}\) = \(\frac{1}{36}\)

Question 15.
If f(x) = 2|x| + 3 |sin x| + 6, then the right hand derivativ of f(x) at x = 0 is _____
Solution:

Question 19.
Find \(\int \sin ^5\left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right) d x\).
Solution:

Question 20.
If A = \(\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]\), then find A3.
Solution:

Question 25.
Find the derivative of xlog x w.r.t. log x.
Solution:

Question 32.
Evaluate: \(\int_{-1}^2\left|x^3-x\right|\) dx
Solution:

Question 36.
Using integration find the area of the region: {(x, y): 0 ≤ y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 2}
Solution:

SET I Code No. 65/4/3 

Note: Except for the following questions, all the remaining questions have been asked in Set-I & Set-II.

Question 8.
The value of tan-1\(\left[\frac{1}{2} \cos ^{-1}\left(\frac{\sqrt{5}}{3}\right)\right]\) is
(a) \(\frac{3+\sqrt{5}}{2}\)
(b) \(\frac{3-\sqrt{5}}{2}\)
(c) \(\frac{-3+\sqrt{5}}{2}\)
(d) \(\frac{-3-\sqrt{5}}{2}\)
Solution:
(b) \(\frac{3-\sqrt{5}}{2}\)

Question 9.
If A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\), then det (adj A) equals
(a) a27
(b) a9
(c) a6
(d) a2
Solution:
(c) a6

Question 15.
If f(x) = x|x|, then f’(x) = __________
Solution:

Question 19.
Find: \(\int \frac{1}{x\left(1+x^2\right)} d x\)
Solution:

Question 20.
If [x] denotes the greatest integer function, then find \(\int_0^{3 / 2}\left[x^2\right] d x\)
Solution:

Question 25.
Show that the function f(x) = \(\frac{x}{3}+\frac{3}{x}\) decreases in the intervals (-3, 0) ∪ (0, 3).
Solution:

Question 26.
Three distinct numbers are chosen randomly from the first 50 natural numbers. Find the probability that all the three numbers are divisible by both 2 and 3.
Solution:
Total numbers, S = {1, 2, 3, … 50} = 50
Numbers are divisible by both 2 and 3 are 6, 12, 18, 24, 30, 36, 42, 48 i.e. 8 numbers.

Question 31.
Evaluate \(\int_0^1 \sqrt{3-2 x-x^2}\)dx
Solution:

Question 32.
Find the general solution of the differential equation \(\frac{d y}{d x}+\frac{1}{x}\) = \(\frac{e^y}{x}\).
Solution:


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