Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Maths Set 1 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 sub-parts.

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

Question 1.
If A = [aij] is a skew-symmetric matrix of order n, then [1]
(a) aij = \(\frac{1}{a_{j i}}\) ∀ i, j
(b) aij ≠ 0 ∀ i, j
(c) aij = 0, where i = j
(d) aij ≠ 0 where i = j
Solution:
(c) aij = 0, where i = j
Explanation:
In a skew-symmetric matrix, the (i, j)th element is negative of the (j, i)th element. Hence, the (i, j)th element = 0
By definition of the skew-symmetric matrix, diagonal elements of a skew-symmetric matrix are all 0.

Question 2.
If A is a square matrix of order 3, |A’| = -3, then |AA’| = [1]
(a) 9
(b) -9
(c) 3
(d) -3
Solution:
(a) 9
Explanation:
|AA’| = |A||A’| = (-3) (-3) = 9

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 3.
The area of a triangle with vertices A, B, C is given by [1]
(a) \(|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\)
(b) \(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\)
(c) \(\frac{1}{4}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AB}}|\)
(d) \(\frac{1}{8}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AB}}|\)
Solution:
(b) \(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\)
Explanation:
The area of the parallelogram with adjacent sides AB and AC = \(|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\).
Hence, the area of the triangle with vertices A, B, C = \(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\)

Question 4.
The value of ‘k’ for which the function f(x) = \(\left\{\begin{array}{cl}
\frac{1-\cos 4 x}{8 x^2} & \text { if } x \neq 0 \\
k, & \text { if } x=0
\end{array}\right.\) is continuous at x = 0 is [1]
(a) 0
(b) -1
(c) 1
(d) 2
Solution:
(c) 1
Explanation:
The function f is continuous at x = 0 if \(\lim _{x \rightarrow 0} f(x)=f(0)\)
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q4

Question 5.
If f'(x) = x + \(\frac{1}{x}\), then f(x) is [1]
(a) x2 + log|x| + C
(b) \(\frac{x^2}{2}\) + log |x| + C
(c) \(\frac{x}{2}\) + log |x| + C
(d) \(\frac{x}{2}\) – log |x| + C
Solution:
(b) \(\frac{x^2}{2}\) + log |x| + C
Explanation:
\(\frac{x^2}{2}\) + log |x| + C (∵ f(x) = \(\int\left(x+\frac{1}{x}\right) d x\))
Given, f'(x) = x + \(\frac{1}{x}\)
By integrating both sides, we get
f'(x) = \(\frac{x^2}{2}\) + log x + C

Question 6.
If m and n, respectively, are the order and the degree of the differential equation \(\frac{d}{d x}\left[\left(\frac{d y}{d x}\right)\right]^4\) = 0, then m + n = [1]
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3
Explanation:
The given differential equation is \(4\left(\frac{d y}{d x}\right)^3 \frac{d^2 y}{d x^2}=0\)
Here, m = 2 and n = 1
Hence, m + n = 3

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 7.
The solution set of the inequality 3x + 5y < 4 is [1]
(a) an open half-plane not containing the origin
(b) an open half-plane containing the origin
(c) the whole XY-plane not containing the Line 3x + 5y = 4
(d) a closed half-plane containing the origin
Solution:
(b) an open half-plane containing the origin.
Explanation:
The strict inequality represents an open half plane and it contains the origin as (0, 0) satisfies it

Question 8.
The scalar projection of the vector \(3 \hat{i}-\hat{j}-2 \hat{k}\) on the vector \(\hat{i}+2 \hat{j}-3 \hat{k}\) is [1]
(a) \(\frac{7}{\sqrt{14}}\)
(b) \(\frac{7}{14}\)
(c) \(\frac{6}{13}\)
(d) \(\frac{7}{2}\)
Solution:
(a) \(\frac{7}{\sqrt{14}}\)
Explanation:
Scalar Projection of \(3 \hat{i}-\hat{j}-2 \hat{k}\) on vector \(\hat{i}+2 \hat{j}-3 \hat{k}\) = \(\frac{(3 \hat{i}-\hat{j}-2 \hat{k}) \cdot(\hat{i}+2 \hat{j}-3 \hat{k})}{|\hat{i}+2 \hat{j}-3 \hat{k}|}=\frac{7}{\sqrt{14}}\)

Question 9.
The value of \(\int_2^3 \frac{x}{x^2+1} d x\) is [1]
(a) log 4
(b) log\(\frac{3}{2}\)
(c) \(\frac{1}{2}\)log 2
(d) log\(\frac{9}{4}\)
Solution:
(c) \(\frac{1}{2}\)log 2
Explanation:
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q9

Question 10.
If A, B are non-singular square matrices of the same order, then (AB-1)-1 = [1]
(a) A-1B
(b) A-1B-1
(c) BA-1
(d) AB
Solution:
(c) BA-1
Explanation:
(AB-1)-1 = (B-1)-1 A-1 = BA-1

Question 11.
The corner points of the shaded unbounded feasible region of an LPP are (0, 4), (0.6, 1.6) and (3, 0) as shown in the figure. The minimum value of the objective function Z = 4x + 6y occurs at [1]
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q11
(a) (0.6, 1.6) only
(b) (3, 0) only
(c) (0.6, 1.6) and (3, 0) only
(d) at every point of the line segment joining the points (0.6, 1.6) and (3, 0)
Solution:
(d) at every point of the line segment joining the points (0.6, 1.6) and (3, 0)
Explanation:
The minimum value of the objective function occurs at two adjacent corner points (0.6, 1.6) and (3, 0) and there is no point in the half plane 4x + 6y < 12 in common with the feasible region. So, the minimum value occurs at every point of the line segment joining the two points.

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 12.
If \(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\), then the possible value(s) of ‘x’ is/are [1]
(a) 3
(b) √3
(c) -√3
(d) √3, -√3
Solution:
(d) √3, -√3
Explanation:
2 – 20 = 2x2 – 24
⇒ 2x2 = 6
⇒ x2 = 3
⇒ x = ±√3
Detailed Answer:
\(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)
⇒ 2 × 1 – 4 × 5 = 2x × x – 4 × 6
⇒ 2 – 20 = 2x2 – 24
⇒ 2x2 = -18 + 24
⇒ 2x2 = 6
⇒ x2 = 3
⇒ x = ±√3

Question 13.
If A is a square matrix of order 3 and |A| = 5, then |adj A| = [1]
(a) 5
(b) 25
(c) 125
(d) \(\frac{1}{5}\)
Solution:
(b) 25
Explanation:
|adj A| = |A|n-1
⇒ |adj A| = 25
Detailed Answer:
Given that square matrix of order 3 and |A| = 5.
We know that |adj A| = |A|n-1, where A is a square matrix of order n.
Therefore, |adj A| = |A|3-1 = |A|2 = 52 = 25.
(Because A is a square matrix of order 3)
Hence, |adj A| = 25

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 14.
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6 and P(A’ ∩ B) is [1]
(a) 0.9
(b) 0.18
(c) 0.28
(d) 0.1
Solution:
(c) 0.28
Explanation:
P(A’ ∩ B’) = P(A’) × P(B’) (As A and B are independent A’ and B’ are also independent)
= 0.7 × 0.4
= 0.28
Detailed Answer:
Two events A and B are independent if P(A ∩ B) = P(A) . P(B)
Given, P(A) = 0.3
and P(A’) = 1 – 0.3 = 0.7
P(B) = 0.6
P(B’) = 1 – 0.6 = 0.4
P(A and B) = P(A ∩ B)
= P(A’) . P(B’)
= 0.7 × 0.4
= 0.28

Question 15.
The general solution of the differential equation y dx – x dy = 0 is [1]
(a) xy = C
(b) x = Cy2
(c) y = Cx
(d) y = Cx2
Solution:
(c) y = Cx
Explanation:
y dx – x dy = 0
⇒ \(\frac{d y}{y}=\frac{d x}{x}\)
⇒ \(\int \frac{d y}{y}=\int \frac{d x}{x}\) + log K, K > 0
⇒ log|y| = log|x| + log K
⇒ log|y| = log|x| K
⇒ |y| = |x| K
⇒ y = ±Kx
⇒ y = Cx

Question 16.
If y = sin-1x, then (1 – x2) y2 is equal to [1]
(a) xy1
(b) xy
(c) xy2
(d) x2
Solution:
(a) xy1
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q16

Question 17.
If two vectors \(\vec{a}\) and \(\vec{b}\) are such that |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3 and \(\vec{a} \cdot \vec{b}\) = 4, then \(|\vec{a}-2 \vec{b}|\) is equal to [1]
(a) √2
(b) 2√6
(c) 24
(d) 2√2
Solution:
(b) 2√6
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q17

Question 18.
P is a point on the line joining points A(0, 5, -2) and B(3, -1, 2). If the x-coordinate of P is 6, then its z-coordinate is [1]
(a) 10
(b) 6
(c) -6
(d) -10
Solution:
(b) 6
Explanation:
The line through the points (0, 5, -2) and (3, -1, 2) is
\(\frac{x}{3-0}=\frac{y-5}{-1-5}=\frac{z+2}{2+2}\)
⇒ \(\frac{x}{3}=\frac{y-5}{-6}=\frac{z+2}{4}\)
Any point on the line is (3k, -6k + 5, 4k – 2), where k is an arbitrary scalar.
3k = 6
⇒ k = 2
The z-coordinate of the point P will be 4 × 2 – 2 = 6

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Assertion-Reason Based Questions
In the following questions, a statement of assertion (A) is followed by a statement of 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): The domain of the function sec-1 2x is (-∞, \(-\frac{1}{2}\)] ∪ [\(\frac{1}{2}\), ∞)
Reason (R): sec-1(-2) = \(-\frac{\pi}{4}\) [1]
Solution:
(c) A is true but R is false.
Explanation:
sec-1 x is defined if x ≤ -1 or x ≥ 1.
Hence, sec-1 2x will be defined if x ≤ \(-\frac{1}{2}\) or x ≥ \(\frac{1}{2}\)
Hence, A is true.
The range of the function sec-1 x is [0, π] – {\(\frac{\pi}{2}\)}
R is false.

Question 20.
Assertion (A): The acute angle between the line \(\vec{r}=\hat{i}+\hat{j}+2 \hat{k}+\lambda(\hat{i}-\hat{j})\) and the x-axis is \(\frac{\pi}{4}\)
Reason (R): The acute angle θ between the lines [1]
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q20
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
The equation of the x-axis may be written as \(\vec{r}=\hat{ti} .\).
Hence, the acute angle θ between the given line and the x-axis is given by
cos θ = \(\frac{|1 \times 1+(-1) \times 0+0 \times 0|}{\sqrt{1^2+(-1)^2+0^2} \times \sqrt{1^2+0^2+0^2}}\) = \(\frac{1}{\sqrt{2}}\)
⇒ θ = \(\frac{\pi}{4}\)

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

Question 21.
Find the value of \(\sin ^{-1}\left[\sin \left(\frac{13 \pi}{7}\right)\right]\) [2]
OR
Prove that the function f is surjective, where f: N → N such that
f(n) = \(\left\{\begin{array}{l}
\frac{n+1}{2} \text {, if } n \text { is old } \\
\frac{n}{2}, \text { if } n \text { is even }
\end{array}\right.\)
Is the function injective? Justify your answer.
Solution:
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q21
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q21.1
OR
Let y ∈ N (codomain). Then ∃ 2y ∈ N (domain) such that f(2y) = \(\frac{2y}{y}\) = y.
Hence, f is surjective.
1, 2 ∈ N (domain) such that f(1) = 1 = f(2)
Hence, f is not injective.

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 22.
A man 1.6 m tall walks at the rate of 0.3 m/s away from a street light that is 4 m above the ground. At what rate is the tip of his shadow moving? At what rate is his shadow lengthening? [2]
Solution:
Let AB represent the height of the street light from the ground. At any time t seconds.
let the man represented as ED of height 1.6 m be at a distance of x m from AB and the length of his shadow EC be y m.
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q22
Using similarity of triangles, we have
\(\frac{4}{1.6}=\frac{x+y}{y}\)
⇒ 3y = 2x
Differentiating both sides w.r.t t, we get
\(3 \frac{d y}{d t}=2 \frac{d x}{d t}\)
\(\frac{d y}{d t}=\frac{2}{3} \times 0.3\)
⇒ \(\frac{d y}{d t}\) = 0.2
At any time t seconds, the tip of his shadow is at a distance of (x + y) m from AB.
The rate at which the tip of his shadow moving \(\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\) m/s = 0.5 m/s
The rate at which his shadow is lengthening = \(\frac{d y}{d t}\) m/s = 0.2 m/s

Question 23.
If \(\vec{a}=\hat{i}-\hat{j}+7 \hat{k}\) and \(\vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}\), then find the value of λ, so that vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) are orthogonal. [2]
OR
Find the direction ratio and direction cosines of a line parallel to the line whose equations are 6x – 12 = 3y + 9 = 2z – 2.
Solution:
The equations of the line are 6x – 12 = 3y + 9 = 2z – 2, which when written in standard symmetric form, will be
\(\frac{x-2}{\frac{1}{6}}=\frac{y-(-3)}{\frac{1}{3}}=\frac{z-1}{\frac{1}{2}}\)
Since lines are parallel, we have \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the required direction ratios are \(\left(\frac{1}{6}, \frac{1}{3}, \frac{1}{2}\right)\) or (1, 2, 3)
and the required cosines are \(\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)\)

Question 24.
If \(y \sqrt{1-x^2}+x \sqrt{1-y^2}\) = 1, then prove that \(\frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}}\) [2]
Solution:
\(y \sqrt{1-x^2}+x \sqrt{1-y^2}\) = 1
Let sin-1 x = A and sin-1 y = B.
Then x = sin A and y = sin B
\(y \sqrt{1-x^2}+x \sqrt{1-y^2}\) = 1
⇒ sin B cos A + sin A cos B = 1
⇒ sin(A + B) = 1
⇒ A + B = sin-1 1 = \(\frac{\pi}{2}\)
⇒ sin-1 x + sin-1 y = \(\frac{\pi}{2}\)
Differentiating w.r. to x, we obtain
\(\frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}}\)

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 25.
Find |\(\vec{x}\)| if \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12\), where \(\vec{a}\) is a unit vector. [2]
Solution:
Since \(\vec{a}\) is a unit vector.
|\(\vec{a}\)| = 1
⇒ \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12\)
⇒ \(\vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\vec{a} \cdot \vec{a}=12\)
⇒ \(|\vec{x}|^2-|\vec{a}|^2=12\)
⇒ \(|\vec{x}|^2-1=12\)
⇒ \(|\vec{x}|^2=13\)
⇒ |\(\vec{x}\)| = √13

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

Question 26.
Find \(\int \frac{d x}{\sqrt{3-2 x-x^2}}\) [3]
Solution:
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q26

Question 27.
Three friends go for coffee. They decide who will pay the bill, by each tossing a coin and then letting the “odd person” pay. There is no odd person if all three tosses produce the same result. If there is no odd person in the first round, they make the second round of tosses and continue to do so until there is an odd person. What is the probability that exactly three rounds of tosses are made? [3]
OR
Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. Assume that the items are identical in shape and size.
Solution:
P(not obtaining an odd person in a single round) = P(All three of them throw tails or All three of them throw heads)
= \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times 2\)
= \(\frac{1}{4}\)
P(obtaining an odd person in a single round) = 1 – P(not obtaining an odd person in a single round) = \(\frac{3}{4}\)
The required probability = P(In the first round there is no odd person and ‘In the second round there is no odd person’ and ‘In the third round, there is an odd person)
= \(\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4}\)
= \(\frac{3}{64}\)
OR
Let X denote the Random Variable defined by the number of defective items.
P(X=0) = \(\frac{4}{6} \times \frac{3}{5}=\frac{2}{5}\)
P(X=1) = \(2 \times\left(\frac{2}{6} \times \frac{4}{5}\right)=\frac{8}{15}\)
P(X=2) = \(\frac{2}{6} \times \frac{1}{5}=\frac{1}{15}\)
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q27
Mean = Σpixi = \(\frac{10}{15}=\frac{2}{3}\)

Question 28.
Evaluate \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}\) [3]
OR
Evaluate \(\int_0^4|x-1| d x\)
Solution:
Let I = \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}\)
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q28
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q28.1
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q28.2

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 29.
Solve the differential equation: y dx + (x – y2) dy = 0 [3]
OR
Solve the differential equation: x dy – y dx = \(\sqrt{x^2+y^2} d x\)
Solution:
y dx + (x – y2) dy = 0
Reducing the given differential equation to the form \(\frac{d x}{d y}\) + Px = Q
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q29
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q29.1

Question 30.
Solve the following Linear Programming Problem graphically: [3]
Maximize Z = 400x + 300y subject to x + y ≤ 200, x ≤ 40, x ≥ 20, y ≥ 0
Solution:
We have Z = 400x + 300y subject to x + y ≤ 200, x ≤ 40, x ≥ 20, y ≥ 0
The corner points of the feasible region are C(20, 0), D(40, 0), B(40, 160), A(20, 180)
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q30
Maximum profit occurs at x = 40, y = 160
and the maximum profit = ₹ 64,000

Question 31.
Find \(\int \frac{\left(x^3+x+1\right)}{\left(x^2-1\right)} d x\) [3]
Solution:
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q31
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q31.1

Section – D (20 marks)
(This Section comprises of Long answer type question (LA) of 5 marks)

Question 32.
Make a rough sketch of the region {(x, y): 0 ≤ y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 2} and find the area of the region using integration. [5]
Solution:
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q32
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q32.1

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 33.
Define the relation R. in the set N × N as follows: [5]
For (a, b), (c, d) ∈ N × N, (a, b) R (c, d) if ad = bc. Prove that R is an equivalence relation in N × N.
OR
Given a non-empty set X, define the relation R in P(X) as follows:
For A, B ∈ P(X), (A, B) ∈ R if A ⊂ B. Prove that R s reflexive, transitive, and not symmetric.
Solution:
Let (a, b) ∈ N × N. Then we have
ab = ba (by commutative property of multiplication of natural numbers)
⇒ (a, b) R (a, b)
Hence, R is reflexive.
Let (a, b), (c, d) ∈ N × N such that (a, b) R (c, d).
Then ad = bc
⇒ cb = da (by commutative property of multiplication of natural numbers)
⇒ (c, d) R (a, b)
Hence, R is symmetric.
Let (a, b), (c, d), (e, f) ∈ N × N such that
(a, b) R (c, d) and (c, d) R (e, f).
Then ad = bc, cf = de
⇒ adcf = bcde
⇒ af = be
⇒ (a, b) R (e, f)
Hence, R is transitive.
Since, R is reflexive, symmetric, and transitive, R is an equivalence relation on N × N.
OR
Let A ∈ P(X). Then A ⊂ A
⇒ (A, A) ∈ R
Hence, R is reflexive.
Let A, B, C ∈ P(X) such that
(A, B), (B, C) ∈ R
⇒ A ⊂ B, B ⊂ C
⇒ A ⊂ C
⇒ (A, C) ∈ R
Hence, R is transitive.
Φ, X ∈ P(X) such that Φ ⊂ X.
Hence (Φ, X) ∈ R. But X ⊄ Φ
which implies that (X, Φ) ∉ R
Thus, R is not symmetric.

Question 34.
An insect is crawling along the line \(\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\) and another insect is crawling along the line \(\vec{r}=-4 \hat{i}-k+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})\). At what points on the lines should they reach so that the distance between them is the shortest? Find the shortest possible distance between them. [5]
OR
The equations of motion of a rocket are x = 2t, y = -4t, z = 4t, where the time t is given in seconds, and the coordinates of a moving point in km. what is the path of the rocket? At what distances will the rocket be from the starting point O(0, 0, 0) and from the following line in 10 seconds?
\(\vec{r}=20 \hat{i}-10 \hat{j}+40 \hat{k}+\mu(10 \hat{i}-20 \hat{j}+10 \hat{k})\)
Solution:
The given lines are non-parallel lines. There is a unique line segment PQ (P lying on one and Q on the other, which is at right angles to both lines.
PQ is the shortest distance between the lines. Hence, the shortest possible distance between the insects = PQ
The position vector of P lying on the line
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q34
Since PQ is perpendicular to both lines
(-10 + 3µ – λ) + (-2µ – 2 + 2λ)(-2) + (-3 – 2µ – 2λ) . 2 = 0
i.e, µ – 3λ = 4 …..(i)
and (-10 + 3µ – λ)3 + (-2µ – 2 + 2λ) (- 2) + (-3 – 2µ – 2λ) (-2) = 0
i.e., 17µ – 3λ = 20 ……..(ii)
solving (i) and (ii) for λ and µ, we get µ = 1, λ = -1.
The position vector of the points, at which they should be so that the distance between them is the shortest are \(5 \hat{i}+4 \hat{j}\) and \(-\hat{i}-2 \hat{j}-3 \hat{k}\)
\(\overrightarrow{P Q}=-6 \hat{i}-6 \hat{j}-3 \hat{k}\)
The shortest distance = \(|\overrightarrow{P Q}|=\sqrt{6^2+6^2+3^2}=9\)
OR
Eliminating t between the equations, we obtain the equation of the path \(\frac{x}{2}=\frac{y}{-4}=\frac{z}{4}\), which are
the equations of the line passing through the origin having direction ratios < 2, -4, 4 >. This line is the path of the rocket.
When t = 10 seconds, the rocket will be at the point (20, -40, 40).
Hence, the required distance from the origin at 10 seconds = \(\sqrt{20^2+40^2+40^2}\) km
= 20 × 3 km
= 60 km
The distance of the point (20, -40, 40) from the given line
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q34.1

Question 35.
If A = \(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\), find A-1. Use A-1 to solve the following system of equations
2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3. [5]
Solution:
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q35

Section – E (12 marks)

This section comprises 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 Sub-Parts of 2 marks each.

Question 36.
Case Study 1: Read the following passage and answer the questions given below. [4]
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q36
The temperature of a person during an intestinal illness is given by f(x) = -0.1x2 + mx + 98.6, 0 ≤ x ≤ 12, m being a constant, where f(x) is the temperature in °F at x days.
(A) Is the function differentiable in the interval (0, 12)? Justify your answer.
(B) If 6 is the critical point of the function, then find the value of the constant m.
(C) Find the intervals in which the function is strictly increasing/strictly decreasing.
OR
Find the points of Local maximum/local minimum, if any, in the interval (0, 12) as well as the points of absolute maximum/absolute minimum in the interval [0, 12]. Also, find the corresponding local maximum/local minimum and the absolute maximum/absolute minimum values of the function.
Solution:
(A) f(x) = -0.1x2 + mx + 98.6, is a polynomial function that is differentiable everywhere, hence, differentiable in (0, 12).
(B) f'(x) = -0.2x + m
Since 6 is the critical point.
(C) f(x) = -0.1x2 + 1.2x + 98.6
f(x) = -0.2x + 1.2 = -0.2(x – 6)
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q36.1
OR
f(x) = -0.1x2 + 1.2x + 98.6,
f(x) = -0.2x + 1.2,
f(6) = 0,
f'(x) = -0.2
f'(6) = -0.2 < 0
Hence, by the second derivative test 6 is a point of the local maximum. The local maximum value = f(6)
= -0.1 × 62 + 1.2 × 6 + 98.6
= 102.2
We have f(0) = 98.6, f(6) = 102.2, f(12) = 98.6
6 is the point of absolute maximum and the absolute maximum value of the function = 102.2.
0 and 12 are both the points of absolute minimum and the absolute minimum value of the function = 98.6.

CBSE Sample Papers for Class 12 Maths Set 1 with Solutions

Question 37.
Case Study 2: Read the following passage and answer the questions given below. [4]
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q37
For an elliptical sports field, the authority wants to design a rectangular soccer field with the maximum possible area. The sports field is given by the graph of \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).
(A) If the length and the breadth of the rectangular field be 2x and 2y respectively, then find the area function in terms of x.
(B) Find the critical point of the function.
(C) Use the first derivative test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area.
OR
Use the second derivative test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area.
Solution:
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q37.1
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q37.2
(C) For the values of x less than \(\frac{a}{\sqrt{2}}\) and close to \(\frac{a}{\sqrt{2}}, \frac{d A}{d x}>0\) and for the values of x greater than \(\frac{a}{\sqrt{2}}\) and close to \(\frac{a}{\sqrt{2}}, \frac{d A}{d x}<0\).
Hence, by the first derivative test, there is a local maximum at the critical point x = \(\frac{a}{\sqrt{2}}\)
Since there is only one critical point, therefore, the area of the soccer field is maximum at this critical point x = \(\frac{a}{\sqrt{2}}\)
Thus, for the maximum area of the soccer field, its length should be a√2 and its width should be b√2.
OR
A = 2x × \(2 \frac{b}{a} \sqrt{a^2-x^2}\), x ∈ (0, a)
Squaring both sides, we get
Z = A2 = \(\frac{16 b^2}{a^2} x^2\left(a^2-x^2\right)=\frac{16 b^2}{a^2}\left(x^2 a^2-x^4\right)\), x ∈ (0, a)
A is maximum when Z is maximum.
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q37.3
Hence, by the second derivative test, there is a local maximum value of Z at the critical point x = \(\frac{a}{\sqrt{2}}\).
Since there is only one critical point, therefore, Z is maximum at x = \(\frac{a}{\sqrt{2}}\),
hence, A is maximum at x = \(\frac{a}{\sqrt{2}}\)
Thus, for the maximum area of the soccer field, its length should be a√2 and its width should be b√2.

Question 38.
Case Study 2: Read the following passage and answer the questions given below. [4]
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q38
There are two antiaircraft guns, named A and B. The probabilities that the shell fired from them hit an airplane are 0.3 and 0.2 respectively.
Both of them fired one shell at an airplane at the same time.
(A) What is the probability that the shell fired from exactly one of them hit the plane?
(B) If it is known that the shell fired from exactly one of them hit the plane, then what is the probability that it was fired from B?
Solution:
(A) Let P be the event that the shell fired from A hits the plane and Q be the event that the shell fired from B hits the plane.
The following four hypotheses are possible before the trial with the guns operating independently:
\(E_1=P Q, E_2=\vec{P} \vec{Q}, E_3=\vec{P} Q, E_4=P \vec{Q}\)
Let E = The shell fired from exactly one of them hits the plane.
P(E1) = 0.3 × 0.2 = 0.06
P(E2) = 0.7 × 0.8 = 0.56
P(E3) = 0.7 × 0.2 = 0.14
P(E4) = 0.3 × 0.8 = 0.24
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q38.1
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q38.2
Note: The four hypotheses form the partition of the sample space and it can be seen that the sum of their probabilities is 1. The hypotheses E1 and E2 are actually eliminated as \(P\left(\frac{E}{E_1}\right)=P\left(\frac{E}{E_2}\right)=0\)

An alternative way of writing the solution:
(A) P(Shell fired from exactly one of them hits the plane) = P[(Shell from A hits the plane and Shell from B does not hit the plane) or (Shell from A does not hit the plane and Shell from B hits the plane)]
= 0.3 × 0.8 + 0.7 × 0.2
= 0.38
(B) P(Shell fired from B hit the plane/Exactly one of them hit the plane)
CBSE Sample Papers for Class 12 Maths Set 1 with Solutions Q38.3