Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 5 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Maths Set 5 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.
Consider the nonempty set consisting of children in a family and a relation R defined as aRb if a is the brother of b. Then, R is: [1]
(a) symmetric but not transitive
(b) transitive but not symmetric
(c) neither symmetric nor transitive
(d) both symmetric and transitive
Solution:
(b) transitive but not symmetric
Explanation:
aRb ⇒ a is the brother of b.
This does not mean b is also a brother of a as b can be sister of a.
Hence, R is not symmetric aRb
⇒ a is brother of b bRc
⇒ b is brother of c
So, a is brother of c
Hence, R is transitive.
Question 2.
Let f: R → R be defined as f(x) = 3x. Choose the correct answer. [1]
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
Solution:
(a) f is one-one onto
Explanation:
f: R → R is defined as f(x) = 3x
Let x, y ∈ R such that f(x) = f(y)
⇒ 3x = 3y
⇒ x = y
∴ f is one-one.
Also, for any real number y in co-domain R, there exists \(\frac{y}{3}\) in R such that \(f\left(\frac{y}{3}\right)=3\left(\frac{y}{3}\right)=y\)
∴ f is onto.
Hence, function f is one-one and onto.
Question 3.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\), then A + A’ = I, then the value of α is: [1]
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{3}\)
(c) π
(d) \(\frac{3\pi}{2}\)
Solution:
(b) \(\frac{\pi}{3}\)
Explanation:
Question 4.
If the matrix A is both symmetric and skew-symmetric then: [1]
(a) A is a diagonal matrix
(b) A is a zero matrix
(c) A is a square matrix
(d) None of these
Solution:
(b) A is a zero matrix
Explanation:
If A is both symmetric and skew-symmetric, then are have
A’ = A and A’ = A
⇒ A = -A
⇒ 2A = 0
⇒ A = 0
Question 5.
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: [1]
(a) 3
(b) 0
(c) -1
(d) 1
Solution:
(c) -1
Explanation:
\(\left|\begin{array}{lll}
2 & 3 & 2 \\
x & x & x \\
4 & 9 & 1
\end{array}\right|+3=0\)
On expanding along R1
2(x – 9x) – 3(x – 4x) + 2(ax – 4x) + 3 = 0
⇒ 2(-8x) – 3(-3x) + 2(5x) + 3 = 0
⇒ -16x + 9x + 10x + 3 = 0
⇒ 3x + 3 = 0
⇒ x = -1
Question 6.
If A = \(\left|\begin{array}{ccc}
2 & \lambda & -3 \\
0 & 2 & 5 \\
1 & 1 & 3
\end{array}\right|\), then A-1 exist if: [1]
(a) λ = 2
(b) λ ≠ 2
(c) λ ≠ -2
(d) None of these
Solution:
(d) None of these
Explanation:
Given that,
A = \(\left|\begin{array}{ccc}
2 & \lambda & -3 \\
0 & 2 & 5 \\
1 & 1 & 3
\end{array}\right|\)
Expanding along R1
|A| = 2(6 – 5) – λ(0 – 5) – 3(-2)
= 2 + 5λ + 6
= 8 + 5λ
We know that A-1 exists if A is a non-singular matrix
i.e. |A| ≠ 0
⇒ 5λ ≠ -8
∴ λ = \(\frac{-8}{5}\)
So, A-1 exists if and only if λ ≠ \(\frac{-8}{5}\)
Question 7.
If y = \(\log _e\left(\frac{x^2}{e^2}\right)\), then \(\frac{d^2 y}{d x^2}\) equals: [1]
(a) \(-\frac{1}{x}\)
(b) \(-\frac{1}{x^2}\)
(c) \(\frac{2}{x^2}\)
(d) \(\frac{-2}{x^2}\)
Solution:
(d) \(\frac{-2}{x^2}\)
Explanation:
Question 8.
The contentment obtained after eating x units of a new dish at a trial function is given by the function f(x) = x3 + 6x2 + 5x + 3. The marginal contentment when 3 units of dish are consumed is: [1]
(a) 60
(b) 68
(c) 24
(d) 48
Solution:
(b) 68
Explanation:
f(x) = x3 + 6x2 + 5x + 3
\(\frac{d(f(x))}{d x}\) = 3x2 + 12x + 5
At x = 3
Marginal contentment = 3(3)2 + 12 × 3 + 5
= 27 + 36 + 5
= 68 units
Question 9.
y = x(x – 3)2 decreases for the values of x given by: [1]
(a) 1 < x < 3
(b) x < 0
(c) x > 0
(d) 0 < x < \(\frac{3}{2}\)
Solution:
(a) 1 < x < 3
Explanation:
Given that y = x(x – 3)2
\(\frac{d y}{d x}\) = x . 2(x – 3) . 1 + (x – 3)2 . 1
= 2x2 – 6x + x2 + 9 – 6x
= 3x2 – 12x + 9
= 3(x2 – 3x – x + 3)
= 3(x – 3) (x – 1)
So y = x(x – 3)2 decreases for (1, 3)
Question 10.
\(\int_1^{\sqrt{3}} \frac{d x}{1+x^2}\) is equal to: [1]
(a) \(\frac{\pi}{3}\)
(b) \(\frac{2\pi}{3}\)
(c) \(\frac{\pi}{12}\)
(d) \(\frac{\pi}{6}\)
Solution:
(c) \(\frac{\pi}{12}\)
Explanation:
\(\int \frac{d x}{1+x^2}\) = tan-1 x = F(x)
By the second fundamental theorem of calculus, we obtain
\(\int_1^{\sqrt{3}} \frac{d x}{1+x}\) = F(√3) – F(1)
= tan-1(√3) – tan-1(1)
= \(\frac{\pi}{3}-\frac{\pi}{4}\)
= \(\frac{\pi}{12}\)
Question 11.
The area of the region bounded by the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) is: [1]
(a) 20π sq. units
(b) 20π2 sq. units
(c) 16π2 sq. units
(d) 25π sq. units
Solution:
(a) 20π sq. units
Explanation:
We have \(\frac{x^2}{5^2}+\frac{y^2}{4^2}=1\), which is an ellipse with its axes as coordinate axes.
Question 12.
The area of the region bounded by the curve x = 2y + 3 and the y-lines y = 1 and y = -1 is: [1]
(a) 4 sq. units
(b) \(\frac{1}{2}\) sq. units
(c) 6 sq. units
(d) 8 sq. units
Solution:
(c) 6 sq. units
Explanation:
From the figure, the area of the shaded region
A = \(\int_{-1}^1(2 y+3) d y\)
= \(\left(y^2+3 y\right)_{-1}^1\)
= 1 + 3 – 1 + 3
= 6 sq. units
Question 13.
The I.F. of \(\frac{d y}{d x}\) + y tan x = sec x is: [1]
(a) tan x
(b) sec x
(c) tan2x
(d) sec x . tan x
Solution:
(b) sec x
Explanation:
Given, a differential equation is:
\(\frac{d y}{d x}\) + y tan x = sec x dx
which is a linear differential equation.
Here, P = tan x, Q = sec x
∴ IF = \(e^{\int P \cdot d x}\)
= \(e^{\int \tan x \cdot d x}\)
= \(e^{\log s e c x d x}\)
= sec x
Question 14.
The order and degree of the differential equation \(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 4}+x^{1 / 5}=0\) respectively are: [1]
(a) 2 and 4
(b) 2 and 2
(c) 2 and 3
(d) 3 and 3
Solution:
(a) 2 and 4
Explanation:
Question 15.
If θ is the angle between two vectors \(\vec{a}\) and \(\vec{b}\), then \(\vec{a} \cdot \vec{b} \geq 0\) only when [1]
(a) 0 < θ < \(\frac{\pi}{2}\)
(c) 0 ≤ θ ≤ \(\frac{\pi}{2}\)
(b) 0 < θ < π
(d) 0 ≤ θ ≤ π
Solution:
(b) 0 ≤ θ ≤ \(\frac{\pi}{2}\)
Explanation:
Let θ be the angle between two vectors \(\vec{a}\) and \(\vec{b}\).
Then without loss of generality, \(\vec{a}\) and \(\vec{b}\) are non-zero vectors so that |\(\vec{a}\)| and |\(\vec{b}\)| are positive.
It is known that, |\(\vec{a}\)||\(\vec{b}\)| cos θ
∴ \(\vec{a} \cdot \vec{b} \geq 0\)
⇒ \(|\vec{a}||\vec{b}| \cos \theta \geq 0\) [∵ |\(\vec{a}\)| and |\(\vec{b}\)| are positive]
⇒ o ≤ θ ≤ \(\frac{\pi}{2}\)
Question 16.
The reflection of the point (α, β, γ) in the xy-plane is [1]
(a) (α, β, 0)
(b) (0, 0, γ)
(c) (-α, -β, γ)
(d) (α, β, -γ)
Solution:
(d) (α, β, -γ)
Explanation:
In the XY-plane, the reflection of the point (α, β, γ) is (α, β, -γ).
Question 17.
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), than [1]
(a) P(B/A) = 1
(b) P(A/B) = 1
(c) P(B/A) = 0
(d) P(B/A) = 0
Solution:
(b) P(A/B) = 1
Explanation:
P(A) + P(B) – (P and B) = P(A)
⇒ P(A) + P(B) – P(A ∩ B) = P(A)
⇒ P(B) = P(A ∩ B)
⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{P(B)}{P(B)}\) = 1
Question 18.
The optimal value of the objective function is attainted at the points: [1]
(a) on X-axis
(b) on Y-axis
(c) any points of the feasible region
(d) None of these
Solution:
(c) any points of the feasible region
Explanation:
Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution.
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 ft. 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.
Consider the function f: R → R defined as f(x) = \(\frac{x}{x^2+1}\)
Assertion (A): f(x) is not one-one.
Reason (R): f(x) is not onto. [1]
Solution:
(b) A and R are true but R is not the correct explanation of A.
Explanation:
Given, f: R → R
f(x) = \(\frac{x}{1+x^2}\)
Taking x1 = 4, x2 = \(\frac{1}{4}\) ∈ R
f(x1) = f(4) = \(\frac{4}{17}\)
f(x2) = f(\(\frac{1}{4}\)) = \(\frac{4}{17}\) (x1 ≠ x2)
∴ f is not one-one.
A is true.
Let y ∈ R (co-domain)
f(x) = y
⇒ \(\frac{x}{1+x^2}\) = y
⇒ y(1 + x2) = x
⇒ yx2 + y – x = 0
x = \(\frac{1 \pm \sqrt{1-4 y^2}}{2 y}\)
Since x ∈ R
∴ 1 – 4y2 ≥ 10
⇒ \(-\frac{1}{2} \leq y \leq \frac{1}{2}\)
So Range (f) ∈ \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
Range (f) ≠ R (co-domain)
∴ f is not onto
R is true.
R is not the correct explanation of A.
Question 20.
Assertion (A): (A + B)2 = A2 + 2AB + B2
Reason (R): Generally, AB ≠ BA. [1]
Solution:
(a) Both A and R are true, and R is the correct explanation of A.
Explanation:
For two matrices A and B, generally AB ≠ BA
i.e., matrix multiplication is not commutative.
∴ R is true.
(A + B)2 = (A + B)(A + B)
= A2 + AB + BA + B2
= A2 + 2AB + B2
∴ A is true and R is the correct explanation of A.
Section – B (10 Marks)
This section comprises very short answer type-questions (VSA) of 2 marks each
Question 21.
Find the value of \(\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\) [2]
Solution:
Question 22.
Find \(\int \frac{x^4+1}{x^2+1} d x\)
OR
Find \(\int \frac{d x}{5-8 x-x^2}\) [2]
Solution:
Question 23.
Find the area bounded by the curve y = \(\frac{1}{x}\) the x-axis and between x = 1 and x = 4.
OR
Find the general solution of the differential equation log(\(\frac{d y}{d x}\)) = 2x + y. [2]
Solution:
Question 24.
If \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \quad \vec{b}=4 \hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{c}=\hat{i}-2 \hat{j}+\hat{k}\) find a vector of magnitude 6 units which is parallel to the vector \(2 \vec{a}-\vec{b}+3 \vec{c}\). [2]
Solution:
Question 25.
If \(\hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+5 \hat{i}, 3 \hat{i}+2 \hat{j}-3 \hat{k}\) and \(\hat{i}-6 \hat{j}-\hat{k}\) are position vectors of points A, B, C and D, then find the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\). Deduce that \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) is parallel or collinear. [2]
Solution:
Section – C (18 Marks)
This section comprises short answer type questions (SA) of 3 marks each
Question 26.
Let f: N → R be a function defined as f(x) = 4x2 + 12x + 5.
Show that f: N → S is invertible, where S is the range of f. [3]
Solution:
Question 27.
Determine the values of a, b, and c for which the function may be continuous at x = 0
OR
Discuss the continuity of the following at the indicated point(s): [3]
Solution:
Question 28.
Find \(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x\)
OR
Solve the differential equation:
\(\sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3\) [3]
Solution:
Question 29.
If \(y=\frac{\sqrt{1-x^2}(2 x-3)^{1 / 2}}{\left(x^2+2\right)^{2 / 3}}\), find \(\frac{d y}{d x}\) [3]
Solution:
Question 30.
Show that the lines \(\frac{x-a+d}{\alpha+\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta}\) and \(\frac{x-b+c}{\beta+\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}\) are coplanar. [3]
Solution:
Question 31.
There are two bags, one of which contains 3 black and 4 white balls, while the other contains 4 black and 3 white balls. A fair die is tossed, if face 1 or 3 turns up, a ball is taken from the first bag, and if any other face turns a ball is chosen from the second bag. Find the probability of choosing a black ball.
OR
Three cards are drawn successively with replacement from a well-shuffled deck of 52 cards. A random variable denotes the number of hearts in the three card’s drawn. Find the mean of X. [3]
Solution:
Let E1 be the event that a ball is drawn from the first bag.
E2 be the event that a ball is drawn from the second bag.
E be the event a black ball is choosen.
P(E1) = \(\frac{2}{6}=\frac{1}{3}\)
OR
Let E = Event of drawing a heart
Then, P(E) = \(\frac{13}{52}=\frac{1}{4}\) and
\(P(\bar{E})=\left(1-\frac{1}{4}\right)=\frac{3}{4}\)
Let X = the number of hearts in a draw.
Then X = 0, 1, 2 or 3
Section – D (20 Marks)
This section comprises of long answer type questions (LA) of 5 marks each
Question 32.
If A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & -2 & 4
\end{array}\right]\), 6A-1 = A2 + cA + dI then find the values of c and d.
OR
If A-1 = \(\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\), find (AB)-1. [5]
Solution:
Question 33.
(A) Differentiate \(\tan ^{-1}\left(\frac{a-x}{1+a x}\right)\) with respect to x.
(B) If x = asin3t and y = acos3t, then find \(\left.\frac{d y}{d x}\right|_{t=\frac{3 \pi}{4}}\). [5]
Solution:
Question 34.
Find the particular solution of the differential equation:
\(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\)
given that y = \(\frac{\pi}{2}\), when x = 1
OR
Solve the following differential equation
\(x y \log \left[\frac{y}{x}\right] d x+\left[y^2-x^2 \log \left|\frac{y}{x}\right|\right] d y=0\) [5]
Solution:
Question 35.
Write the vector equations of the following lines and hence find the distance between them
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}, \frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}\) [5]
Solution:
Section – E (12 Marks)
This section comprises 3 case-study/passage-based questions of 4 marks each with two Sub-part, 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.
In a parliament election, a political party hired a public relation firm to promote its candidates in three ways-telephone, house calls, and letters. The cost per contact (in paise) given in matrix A is
The number of contacts of each type made in two cities X and Y is given in matrix B as
Based on the above information, answer the following questions:
(A) Write the matrix for finding the total amount spent by the party for doing the campaign. [1]
(B) What is the order of the resultant matrix. [1]
(C) What is the total amount spent by the party in the two cities? [2]
OR
If A = \(\left[\begin{array}{ll}
1 & -1 \\
2 & -1
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
a & 1 \\
b & -1
\end{array}\right]\) and (A + B)2 = A2 + B2, then find the values of a and b.
Solution:
Question 37.
A teacher asks a question to three of their students Ravi, Mohit, and Sonia. The probability of solving the question by Ravi, Mohit, and Sonia are 30%, 25%, and 45% respectively.
The probability of making errors by Ravi, Mohit, and Sonia is 1%, 1.2%, and 2% respectively.
Based on the above information, answer the following questions:
(A) What is the conditional probability that an error is committed to solving the question, given that question is solved by Sonia? [1]
(B) What is the probability that the question is solved and Sonia committed an error? [1]
(C) What is the total probability of committing an error in solving a question? [1]
OR
If the solution to the question is checked by the teacher and has some error, then is the probability it is not solved by Ravi? [2]
Solution:
(A) E: Event that question has some error
E1: Event that question is solved by Ravi
E2: Event that question is solved by Mohit
E3: Event that question is solved by Sonia
Then P(E1) = \(\frac{30}{100}\), P(E2) = \(\frac{25}{100}\), P(E3) = \(\frac{45}{100}\)
P(A/E1) = \(\frac{1}{100}\), P(A/E2) = \(\frac{12}{100}\), P(A/E3) = \(\frac{2}{100}\)
Required probability = P(A/E3) = \(\frac{2}{100}\) = 0.02
(B) Required probability = P(A ∩ E3)
= P(E3) P(A/E3)
= \(\frac{45}{100} \times \frac{2}{100}\)
= \(\frac{90}{10000}\)
= 0.09
(C) Total probability = P(A)
Question 38.
A company is into creating a packaging of items. They were into making a cylindrical-shaped container. They tried different options i.e. different sizes of the right circular cylinder to get the best option. The cylinder is of the volume 432π ml and should have a minimum surface area.
Based on the above information, answer the following questions:
(A) What is the radius r of the cylinder for the minimum surface area? [2]
(B) What is the relation between the radius of the base and the height of the cylinder of minimum surface area? [2]
Solution:
