CBSE Sample Papers for Class 12 Maths Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Maths Set 3 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

• This Question paper contains five sections A, B, C, D, and E. Each section is compulsory. However, there are internal choices in some questions.
• Section A has 18 MCQs and 2 Assertion-Reason-based questions of 1 mark each.
• Section B has 5 Very Short Answer (VSA) type questions of 2 marks each.
• Section C has 6 Short Answer (SA) type questions of 3 marks each.
• Section D has 4 Long Answer (LA) type questions of 5 marks each.
• Section E has 3 sources based/case based/passage based/integrated units of assessment (4 marks each) with subparts.

Section – A (20 Marks)
(Multiple Choice Questions Each question carries 1 mark)

Question 1.
The number of reflexive relations possible in a set A whose n(A) = 3 are: [1]
(a) 3⁵
(b) 2⁶
(c) 2⁵
(d) 3²
Solution:
(b) 2⁶
Explanation:
Number of Reflexive relations on a set = \(2^{n^2-n}\)
Here, n = 3
∴ Number of reflexive relations = \(2^{3^2-3}\)
= 2^9-3
= 2⁶

Question 2.
A relation R in S = {1, 2, 3} is defined as R = {(1, 1), (1, 2), (2, 2), (3, 3)}. The element of relation R to be removed to make R an equivalence relation is: [1]
(a) (1, 1)
(b) (1, 2)
(c) (2, 2)
(d) (3, 3)
Solution:
(b) (1, 2)
Explanation:
An equivalence relation is one that is reflexive, symmetric, and transitive.
For reflexive, (a, a) ∈ R if (a, b) ∈ R, then (b, a)
For symmetric, if (a, b) ∈ R, then (b, a)
For transitive, if (a, b) ∈ R, (b, c) ∈ R, then (c, a)
Then, (1, 2) should be removed to make the given relation an equivalence relation.

Question 3.
An equivalence relation R in A divides it into equivalence classes A₁, A₂, A₃. The value of A₁ ∪ A₂ ∪ A₃ and A₁ ∩ A₂ ∩ A₃ respectively are: [1]
(a) A, φ
(b) A₁, A₂
(c) φ, φ
(d) A, A
Solution:
(a) A, φ
Explanation:
Since, R is an equivalence relation in A, which is divided into equivalence classes A₁, A₂, A₃.
Then, A₁ ∪ A₂ ∪ A₃ = A (since it is divided into 3 classes)
and A₁ ∩ A₂ ∩ A₃ = φ (as they have nothing in common)

Question 4.
If A and B are matrices of order 3 × n and m × 5 respectively, then the order of matrix 5A – 3B, given that it is defined is: [1]
(a) 3 × 5
(b) 5 × 3
(c) 3 × 3
(d) 5 × 5
Solution:
(a) 3 × 5
Explanation:
3 × 5
If 5A – 3B is defined, then the order of the A and B matrices will be the same.
Then 5A – 3B will be of the same order as A and B, which is 3 × 5.

Question 5.
Given that A is a square matrix of order 3 × 3 and |A| = -4. Then, |adj A| is: [1]
(a) 10
(b) -16
(c) 16
(d) -10
Solution:
(c) 16
Explanation:
Order of square matrix,
A = 3 × 3
Determinant of |A| = -4
Then, |adj A| = |A|^n-1
= (-4)^3-1
= (-4)²
= 16

Question 6.
Let A = [a_ij] be a square matrix of order 3 × 3 and |A| = -7. Then the value of a₁₁ A₂₁ + a₁₂ A₂₂ + a₁₃ A₂₃ where A_ij is the cofactor of element a_ij is: [1]
(a) 0
(b) 1
(c) -1
(d) -7
Solution:
(a) 0
Explanation:
“If elements of one row (or column) are multiplied by the cofactors of elements of any other row (or column) then their sum is zero.”

Question 7.
The value of ∫e^x (1 – cot x + cosec²x) dx is: [1]
(a) e^x cot x + c
(b) e^x (1 – cot x) + c
(c) e^x (1 – cosec x) + c
(d) e^x (cosec x – cot x) + c
Solution:
(b) e^x (1 – cot x) + c
Explanation:

Question 8.
The area (in sq. m) bounded by y = x², the x-axis and the lines x = -1 and x = 1 is: [1]
(a) \(\frac{2}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{5}{2}\)
(d) \(\frac{8}{7}\)
Solution:
(a) \(\frac{2}{3}\)
Explanation:
Here, y = x²

Question 9.
For what value of n is the following a homogeneous differential equation: [1]
\(\frac{d y}{d x}=\frac{x^3-y^n}{x^2 y+x y^2}\)
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3
Explanation:
In the homogeneous differential equation, the degree of terms is the same. So the value of n is 3.

Question 10.
The unit vector in the direction opposite to \(-\frac{3}{4} \hat{j}\) is: [1]
(a) \(\hat{i}\)
(b) \(\hat{j}\)
(c) \(\hat{k}\)
(d) \(\frac{3}{4} \hat{j}\)
Solution:
(b) \(\hat{j}\)
Explanation:
A unit vector in a direction opposite to \(\frac{-3}{4} \hat{j}\) will be in the positive y-axis.
∴ Required vector is \(\hat{j}\)

Question 11.
The area of the triangle whose two sides are represented by the vectors \(2 \hat{i}\) and \(-3 \hat{j}\) is: [1]
(a) 3 sq. units
(b) 4 sq. units
(c) 5 sq. units
(d) 6 sq. units
Solution:
(a) 3 sq. units
Explanation:
Area of triangle

Question 12.
The angle between the unit vectors \(\hat{a}\) and \(\hat{b}\), given that \(|\hat{a}+\hat{b}|=1\) is: [1]
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{2 \pi}{3}\)
Solution:
(d) \(\frac{2 \pi}{3}\)
Explanation:
Given, \(\hat{a}\) and \(\hat{b}\) are unit vectors and
\(|\hat{a}+\hat{b}|=1\) …….(1)
On squaring both sides of equation (1), we get

Question 13.
If \(\vec{b}\) is a unit vector such that \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8\), then |\(\vec{a}\)| is: [1]
(a) 2
(b) 3
(c) 4
(d) 2√2
Solution:
(b) 3
Explanation:

Question 14.
If \(\vec{a}=x \hat{i}+2 \hat{j}-z \hat{k}\) and \(\vec{b}=3 \hat{i}-y \hat{j}+\hat{k}\) are two equal vectors, then the value of x + y + z is: [1]
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Explanation:
Given \(\vec{a}=\vec{b}\)
\(x \hat{i}+2 \hat{j}-z \hat{k}=3 \hat{i}-y \hat{j}+\hat{k}\)
⇒ x = 3, y = -2, z = -1
x + y + z = 3 – 2 – 1 = 0

Question 15.
The probabilities of A and B solving a problem independently are \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively. If both of them try to solve the problem independently, then the probability that the problem is solved is: [1]
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{3}{8}\)
Solution:
(a) \(\frac{1}{2}\)
Explanation:
P(A) = \(\frac{1}{3}\) and P(B) = \(\frac{1}{4}\)
The probability that the problem is solved = Probability that A solves the problem or B solves the problem
= P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
Since A and B are independent
P(A ∩ B) = P(A) P(B)
= \(\frac{1}{3} \times \frac{1}{4}\)
= \(\frac{1}{12}\)
∴ P(A ∪ B) = \(\frac{1}{3}+\frac{1}{4}-\frac{1}{12}\) = \(\frac{1}{2}\)
Hence, required probability is \(\frac{1}{2}\)

Question 16.
The probability that it will rain on any particular day is 50%. The probability that it rains only on the first 4 days of the week is: [1]
(a) \(\left(\frac{1}{5}\right)^6\)
(b) \(\left(\frac{1}{3}\right)^2\)
(c) \(\left(\frac{1}{2}\right)^7\)
(d) \(\left(\frac{1}{2}\right)^8\)
Solution:
(c) \(\left(\frac{1}{2}\right)^7\)
Explanation:
Here, the Probability of rain,
P = \(\frac{50}{100}=\frac{1}{2}\)
Probability of no rain, q = 1 – \(\frac{50}{100}\) = \(\frac{1}{2}\)
Then, required probability = \(\left(\frac{1}{2}\right)^4 \times\left(\frac{1}{2}\right)^3\) = \(\left(\frac{1}{2}\right)^7\)

Question 17.
The value of \(\tan ^{-1}\left\{\tan \frac{15 \pi}{4}\right\}\) is: [1]
(a) \(\frac{\pi}{4}\)
(b) \(\frac{-\pi}{4}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{-\pi}{2}\)
Solution:
(b) \(\frac{-\pi}{4}\)
Explanation:
We have,

Question 18.
A vector in the direction of a vector \(\vec{a}=\hat{i}-\hat{j}+\hat{k}\) which has magnitude 8 units is: [1]
(a) \(\frac{\hat{i}-\hat{j}+\hat{k}}{8}\)
(b) \(8(\hat{i}-\hat{j}+\hat{k})\)
(c) \(\frac{8(\hat{i}-\hat{j}+\hat{k})}{\sqrt{3}}\)
(d) \(\frac{(\hat{i}-\hat{j}+\hat{k})}{\sqrt{3}}\)
Solution:
(c) \(\frac{8(\hat{i}-\hat{j}+\hat{k})}{\sqrt{3}}\)
Explanation:
Given \(\vec{a}=\hat{i}-\hat{j}+\hat{k}\)
Now, the unit vector in the direction of \(\vec{a}\) is
\(\hat{a}=\frac{\hat{a}}{|\vec{a}|}=\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{(1)^2+(-1)^2+(1)^2}}\)
= \(\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}\)
Vector of magnitude 8 units in direction of \(\vec{a}\) = \(\frac{8(\hat{i}-\hat{j}+\hat{k})}{\sqrt{3}}\)

Assertion-Reason Based Questions
In the following questions, a statement of assertion (A) is followed by a statement of the reason (R).
Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 19.
Assertion (A): A relation R = {(1, 1), (1, 3), (3, 1), (3, 3), (3, 5)}, defined on the set A = {1, 3, 5} is reflexive.
Reason (B): A relation R on the set A is said to be transitive if for (a, b) ∈ R and (b, c) ∈ R we have (a, c) ∈ R. [1]
Solution:
(d) A is false but R is true.
Explanation:
Given R = {(1, 1) (1, 3) (3, 1) (3, 3) (3, 5)}
We know that relation ‘R’ is reflexive on set A if ∀ a ∈ A, (a, a) ∈ R
Here, Set A = {1, 3, 5}
(1, 1) ∈ R, (3, 3) ∈ R but (5, 5) ∉ R
∴ R is not reflexive

Question 20.
Assertion (A): The points with position vectors a, b, c are collinear if 2a – 7b + 5c = 0
Reason (R): The points with position vectors \(\vec{a}, \vec{b}, \vec{c}\) are collinear if \(\overrightarrow{l a}+m \vec{b}+n \vec{c}=0\) [1]
Solution:
(c) A is true but R is false.
Explanation:
\(\vec{a}, \vec{b}, \vec{c}\) are collinear.
⇒ l, m, n are all zeroes such that
\(\overrightarrow{l a}+m \vec{b}+n \vec{c}=0\), if \(2 \vec{a}-7 \vec{b}+5 \vec{c}=0\)
Then \(\vec{a}, \vec{b}, \vec{c}\) are collinear since 2 – 7 + 5 = 0

Section – B (10 Marks)
This section comprises very short answer type-questions (VSA) of 2 marks

Question 21.
If A is a square matrix of order 3 such that A² = 2A, then find the value of |A|. [2]
OR
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), show that A² – 5A + 7I = 0. Hence find A^-1. [2]
Solution:
A is a square matrix of order 3. So, n = 3.
Also, A² = 2A (Given)
⇒ |AA| = |2A|
⇒ |A||A| = 8 |A| (∵ |AB| = |A| |B| and |2A| = 2ⁿ |A|)
⇒ |A| (|A| – 8) = 0
⇒ |A| = 0 or 8
OR
Here, A = \(\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\)

Question 22.
Find the value(s) of k so that the following function is continuous at x = 0 [2]
\(f(x)=\left\{\begin{array}{cl}
\frac{1-\cos k x}{x \sin x} & \text { if } x \neq 0 \\
\frac{1}{2} & \text { if } x=0
\end{array}\right.\)
Solution:

Question 23.
Find \(\int \frac{1}{\cos ^2 x(1-\tan x)^2} d x\) [2]
OR
Evaluate \(\int_0^1 x(1-x)^n d x\)
Solution:

Question 24.
Find the area of the parallelogram whose one side and a diagonal are represented by coinitial vectors \(\hat{i}-\hat{j}+\hat{k}\) and \(4 \hat{i}+5 \hat{j}\) respectively. [2]
Solution:

Question 25.
A refrigerator box contains 2 milk chocolates and 4 dark chocolates. Two chocolates are drawn at random. Find the probability distribution of the number of milk chocolates. What is the most likely outcome? [2]
Solution:
Let x be the number of milk chocolates drawn. Then x = 0, 1 or 2.
So, the probability distribution of the number of milk chocolates is

The most likely outcome is getting one chocolate of each type.

Section – C (18 Marks)
This section comprises short answer type questions (SA) of 3 marks

Question 26.
Check whether the relation R in the set Z of integers defined as R = {(a, b) : a + b is “divisible by 2”} is reflexive, symmetric, or transitive. Write the equivalence class containing 0 i.e. [0]. [3]
Solution:
(A) Reflexive
Since, a + a = 2a which is even
∵ (a, a) ∈ R ∀ a ∈ Z
Hence, R is reflexive.
(B) Symmetric
If (a, b) ∈ R, then
a + b = 2λ
⇒ b + a = 2λ
⇒ (b, a) ∈ R
Hence, R is symmetric
(C) Transitive
If (a, b) ∈ R and (b, c) ∈ R
Then a + b = 2λ ……..(1)
and b + c = 2µ ……..(2)
On adding (1) and (2) we get
a + 2b + c = 2(λ + µ)
⇒ a + c = 2(λ + µ – b)
⇒ a + c = 2k
where λ + µ – b = k
⇒ (a, c) ∈ R
Hence, R is transitive.
Equivalence class containing
[0] = {…. -4, -2, 0, 2, 4 ….}

Question 27.
If y = \(e^{x \sin ^2 x}+(\sin x)^x\), find \(\frac{d y}{d x}\) [3]
Solution:

Question 28.
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 2 is not differentiable at x = 1. [3]
OR
If x = a sec θ, y = b tan θ find \(\frac{d^2 y}{d x^2} \text { at } x=\frac{\pi}{6}\)
Solution:

Question 29.
Find \(\int \frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)} d x\) [3]
OR
Find \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log \tan x d x\)
Solution:

Question 30.
Find the area of the region bounded by the curves x² + y² = 4, y = √3x, and the x-axis in the first quadrant. [3]
OR
Find the area of the ellipse x² + 9y² = 36 using integration.
Solution:
Given, curves are
x² + y² = 4, y = √3x

Question 31.
Find the general solution of the following differential equation: [3]
x dy – (y + 2x²) dx = 0
Solution:

Section – D (20 Marks)
This section comprises long answer type questions (LA) of 5 marks

Question 32.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 0 \\
-2 & -1 & -2 \\
0 & -1 & 1
\end{array}\right]\), find A^-1. Hence Solve the system of equations;
x – 2y = 10
2x – y – z = 8
-2y + z = 7
OR
Evaluate the product AB, where
A = \(\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\). Hence solve the system of linear equations
x – y = 3
2x + 3y + 4z = 17
y + 2z = 7 [5]
Solution:
Here, A = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
-2 & -1 & -2 \\
0 & -1 & 1
\end{array}\right]\)
Now, |A| = (-1 – 2) – 2(-2 – 0) + 0(2 + 0)
= 1(-3) + 4 + 0
= 1
Then, A is non-singular, therefore A^-1 exists.

Question 33.
Find the shortest distance between the lines \(\vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})\) and \(\vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})\). If the lines intersect find their point of intersection. [5]
Solution:

Question 34.
Solve the following linear programming problem (L.P.P) graphically.
Maximize Z = x + 2y subject to constraints;
x + 2y ≥ 100
2x – y ≤ 00
2x + y ≤ 200
x, y ≥ 0
OR
The corner points of the feasible region determined by the system of linear constraints are as shown below:

Answer each of the following:
(A) Let Z = 3x – 4y be the objective function. Find the maximum and minimum value of Z and also the corresponding points at which the maximum and minimum value occurs.
(B) Let Z = px + qy, where p, q > 0 be the objective function. Find the condition on p and q so that the maximum value of Z occurs at B(4, 10) and C(6, 8). Also mention the number of optimal solutions in this case. [5]
Solution:
Max Z = 3x + y
Subject to constraints
X + 2y ≥ 100 ………(1)
2x – y ≤ 0 ……..(2)
2x + y ≤ 200 ……….(3)
x, y ≥ 0

Change equations into equations.
For equation (1), x + 2y = 100

For equation (2), 2x – y = 0

For the equation of (3), 2x + y = 200

∴ Z is maximum at x = 0, y = 400
OR
(A)

Max Z = 12 at E(4, 0)
Min Z = -32 at A(0, 8)
(B) Since, the maximum value of Z occurs at B(4, 10) and C(6, 8)
and objective function, Z = px + qy
then, 4p + 10q = 6p + 8q
⇒ 2q = 2p
⇒ p = q
The number of the optimal solution is infinite.

Question 35.
(A) Find the intervals in which the function f is given by:
f(x) = tan x – 4x, x ∈ (0, \(\frac{\pi}{2}\)) is
(a) Strictly increasing
(b) Strictly decreasing
(B) Evaluate ∫tan(x – θ) tan(x + θ) tan 2x dx [5]
Solution:
(A) f(x) = tan x – 4x
f'(x) = sec²x – 4
(a) for f(x) to be strictly increasing f'(x) > 0


(B) We know that
2x = (x – θ) + (x + θ)
tan 2x = tan{(x – θ) + (x + θ)}
⇒ tan 2x = \(\frac{\tan (x-\theta)+\tan (x+\theta)}{1-\tan (x-\theta) \tan (x+\theta)}\)
⇒ tan 2x – tan(x – θ) tan(x + θ) tan 2x = tan(x – θ) + tan(x + θ)
⇒ tan(x – θ) tan(x + θ) tan 2x = tan 2x – tan(x – θ) – tan(x + θ)
I = ∫tan(x – θ) tan(x + θ) tan 2x dx
= ∫{tan 2x – tan(x – θ) – tan(x + θ)} dx
= \(-\frac{1}{2}\) log|cos 2x|+ log|cos(x – θ)| + log|cos(x + θ)| + c

Section – E (12 Marks)

This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-parts. First, two case-study questions have three sub-parts (A), (B), (C) of marks 1, 1, 2 respectively. The third case-study question has two subparts of 2 marks each.

Question 36.
An architect designs a building for a multi¬national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200m as shown below:

Based on the above information answer the following:
(A) If x and y represents the length and breadth of the rectangular region, then what is the relation between the variables? [1]
(B) What is the area of the rectangular region A expressed as a function of x? [1]
(C) What is the maximum value of area A?
OR
The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semicircular ends. For this to happen, what is the value of x? [2]
Solution:

Question 37.
In an office, three employees Vmay, Sonia, and Iqbal process incoming copies of a certain form. Vinay process 50% of the forms. Sonia processes 20% and Iqbal the remaining 30% of the forms. Vinay has an error rate of 0.06, Sonia has an error rate of 0.04 and Iqbal has an error rate of 0.03

Based on the above information answer the following:
(A) What is the conditional probability that an error is committed in processing given that Sonia processed the form? [1]
(B) Find the total probability of committing an error in processing the form? [1]
(C) The manager of the company wants to do a quality check. During the inspection he selects a form at random from the day’s output of processed forms, if the form selected at random has an error, What is the probability that the form is NOT processed by Vinay?
OR
Let A be the event of committing an error in processing the form and let E₁, E₂, and E₃ be the events that Vinay, Sonia, and Iqbal processed the form. What is the value of \(\sum_{i=1}^3 \mathrm{P}\left(\mathrm{E}_i \mid \mathrm{A}\right)\) is: [2]
Solution:
(A) E₁ = Vinay processed the form
E₂ = Sonia processed the form
E₃ = Iqbal processed the form 50
P(E₁) = \(\frac{50}{100}=\frac{1}{2}\)

Question 38.
A general election of Lok Sabha is a gigantic exercise. About 911 million people were eligible to vote and voter turnout was about 67%, the highest ever.

Let f(x) be the set of all citizens of India who were eligible to exercise their voting right in the general election held in 2019. A relation ‘R’ is defined on I as follows:
If f(x) is a continuous function defined on [a, b] then \(\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\) on the basis of the above information answer the following equations:
(A) \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x\) [2]
(B) The value of \(\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x\), a > 0. [2]
Solution: