CBSE Class 12 Maths Question Paper 2023 (Series: EF1GH/2) with Solutions

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CBSE Class 12 Maths Question Paper 2023 (Series: EF1GH/2) with Solutions

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. This question paper contains 38 questions. All questions are compulsory.
2. Question paper is divided into FIVE Sections – Section A, B, C, D and E.
3. In Section A – Question nos. 1 to 18 are Multiple Choice Questions (MCQ) and Question nos. 19 & 20 are Assertion-Reason based questions of 1 mark each.
4. In Section B – Question nos. 21 to 25 are Very Short Answer (VSA) type questions of 2 marks each.
5. In Section C – Question nos. 26 to 31 are Short Answer (SA) type questions, carryibg 3 marks each.
6. In Section D – Question nos. 32 to 35 are Long Answer (LA) type questions carrying 5 marks each.
7. In Section E – Question nos. 36 to 38 are source based / case based / passage based / integrated units of assessment questions carrying 4 marks each.
8. There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 3 questions in Section C, 2 questions in Section D and 2 questions in Section E.
9. Use of calculator is NOT allowed.

Set – I Code No. 65/2/1
Section-A (Multiple Choice Questions)

Each question carries 1 mark.

Question 1.
If A = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), then A ²⁰²³ is equal to
(a) \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
0 & 2023 \\
0 & 0
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
2023 & 0 \\
0 & 2023
\end{array}\right]\)
Solution:
(c) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)

Question 2.
If \(\left[\begin{array}{ll}
2 & 0 \\
5 & 4
\end{array}\right]\) = P + Q, where P is a symmetric and Q is a skew symmetric matrix, then Q is equal to
(a) \(\left[\begin{array}{cc}
2 & 5 / 2^{-} \\
5 / 2 & 4
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
0 & -\overline{5} / 2 \\
5 / 2 & 0
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
0 & 5 / 2 \\
-5 / 2 & 0
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
2 & -5 / 2 \\
5 / 2 & 4
\end{array}\right]\)
Solution:
(b) \(\left[\begin{array}{cc}
0 & -\overline{5} / 2 \\
5 / 2 & 0
\end{array}\right]\)

Question 3.
If \(\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
3 & a & 1
\end{array}\right]\) is non-singular matrix and a ∈ A, then the set A is
(a) R
(b) {0}
(c) {4}
(d) R – {4}
Solution:
(d) R – {4}

Here, | A | ≠ 0
(∵ A is a non-singular matrix)
∴ 1(3 – a) – 2(2 – 3) + 1(2a – 9) ≠ 0
⇒ 3 – a + 2 + 2a – 9 ≠ 0
⇒ a ≠ 4
All real numbers except 4.

Question 4.
If | A | = | kA |, where A is a square matrix of order 2, then sum of all possible values of k is
(a) 1
(b) -1
(c) 2
(d) 0
Solution:
(d) 0

Given.|A| = |kA|
| A | = k ² | A | ( Here order = 2)
k ² = 1
k = ±1
∴ Sum of all possible value of k
= 1 + (-1) = 0

Question 5.
If \(\frac{d}{d x}\)[f(x)] = ax + b and f(0) = 0, then f(x) is equal to
(a) a + b
(b) \(\) + bx
(c) \(\frac{a x^2}{2}\) + bx + c
(d) b
Solution:
(b) \(\) + bx

Question 6.
Degree of the differential equation sin x + cos \(\left(\frac{d y}{d x}\right)\) = y ² is
(a) 2
(b) 1
(c) not defined
(d) 0
Solution:
(b) 1

Question 7.
The integrating factor of the differential equation
(1 – y ² )dx + yx = ay,(-1 < y < 1) is
(a) \(\frac{1}{y^2-1}\)
(b) \(\frac{1}{\sqrt{y^2-1}}\)
(c) \(\frac{1}{1-y^2}\)
(d) \(\frac{1}{\sqrt{1-y^2}}\)
Solution:
(d) \(\frac{1}{\sqrt{1-y^2}}\)

Question 8.
Unit vector along pQ, where coordinates of P and Q respectively are (2, 1, -1) and (4, 4, -7), is 1
(a) 2\(\hat{i}\) + 3\(\hat{j}\) – 6\(\hat{k}\)
(b) -2\(\hat{i}\) – 3\(\hat{j}\) + 6\(\hat{k}\)
(c) \(\frac{-2 \hat{i}}{7}-\frac{3 \hat{j}}{7}+\frac{6 \hat{k}}{7}\)
(d) \(\frac{2 \hat{i}}{7}+\frac{3 \hat{j}}{7}-\frac{6 \hat{k}}{7}\)
Solution:
(d) \(\frac{2 \hat{i}}{7}+\frac{3 \hat{j}}{7}-\frac{6 \hat{k}}{7}\)

Question 9.
Position vector of the mid-point of line segment AB is 3\(3 \hat{i}+2 \hat{j}-3 \hat{k}\). If position vector of the point A is \(2 \hat{i}+3 \hat{j}-4 \hat{k}\), then position vector of the point B is = 1
(a) \(\frac{5 \hat{i}}{2}+\frac{5 \hat{j}}{2}-\frac{7 \hat{k}}{2}\)
(b) \(4 \hat{i}+\hat{j}-2 \hat{k}\)
(c) \(5 \hat{i}+5 \hat{j}-7 \hat{k}\)
(d) \(\frac{\hat{i}}{2}-\frac{\hat{j}}{2}+\frac{\hat{k}}{2}\)
Solution:
(b) \(4 \hat{i}+\hat{j}-2 \hat{k}\)

Question 10.
Projection of vector \(2 \hat{i}+3 \hat{j}\) on the vector \(3 \hat{i}-2 \hat{j}\) is
(a) 0
(b) 12
(c) \(\frac{12}{\sqrt{13}}\)
(d) \(\frac{-12}{\sqrt{13}}\)
Solution:
(a) 0

Question 11.
Equation of a line passing through point (1, 1, 1) and parallel to z-axis is
(a) \(\frac{x}{1}=\frac{y}{1}=\frac{z}{1}\)
(b) \(\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{1}\)
(c) \(\frac{x}{0}=\frac{u}{0}=\frac{z-1}{1}\)
(d) \(\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{1}\)
Solution:
(d) \(\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{1}\)

Question 12.
If the sum of numbers obtained on throwing a pair of dice is 9, then the probability that number obtained on one of the dice is 4, is: 1
(a) \(\frac{1}{9}\)
(b) \(\frac{4}{9}\)
(c) \(\frac{1}{18}\)
(d) \(\frac{1}{2}\)
Solution:
(d) \(\frac{1}{2}\)

Let A: Number obtained on one of the dice is 4
Let B : Sum is 9
Two dice can be thrown as 6 × 6
= 36 ways
“Sum 9” can be obtained as (3, 6),
(6, 3), (4, 5), (5, 4) i.e., 4 ways

Question 13.
Anti-derivative of \(\frac{\tan x-1}{\tan x+1}\) with respect to x ¡s 1
(a) sec ² \(\left(\frac{\pi}{4}-x\right)\) + c
(b) \(-\sec ^2\left(\frac{\pi}{4}-x\right)+c\)
(c) \(\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)
(d) \(-\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)
Solution:
(c) \(\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+c\)

Question 14.
If (a, b), (c, d) and (e, f) are the vertices of ∆ABC and ∆ denotes the area of ∆ABC, then \(\left|\begin{array}{lll}
a & c & e \\
b & d & f \\
1 & 1 & 1
\end{array}\right|^2\) is equal to
(a) 2∆ ²
(b) 4∆ ²
(c) 2∆
(d) 4∆
Solution:
(b) 4∆ ²

Question 15.
The function f(x) = x|x| is
(a) continuous and differentiable at x = 0.
(b) continuous but not differentiable at x = 0.
(c) differentiable but not continuous at x = 0.
(d) neither differentiable nor continuous at x = 0.
Solution:
(a) continuous and differentiable at x = 0.

Question 16.
If tan \(\left(\frac{x+y}{x-y}\right)\) = k, then is equal to
(a) \(\frac{-y}{x}\)
(b) \(\frac{y}{x}\)
(c) \(\sec ^2\left(\frac{y}{x}\right)\)
(d) \(-\sec ^2\left(\frac{y}{x}\right)\)
Solution:
(b) \(\frac{y}{x}\)

Question 17.
The objective function Z = ax + by of an LPP has maximum value 42 at (4, 6) and minimum value 19 at (3, 2). Which of the following is true?
(a) a = 9, b = 1
(b) a = 5, b = 2
(c) a = 3, b = 5
(d) a = 5, b = 3
Solution:
(c) a = 3, b = 5

Question 18.
The corner points of the feasible region of a linear programming problem are (0, 4), (8, 0) and \(\left(\frac{20}{3}, \frac{4}{3}\right)\). If Z = 30x + 24 is the objective function, then (maximum value of Z – minimum value Z) is equal to
(a) 40
(b) 144
(c) 120
(d) 136
Solution:
(b) 144

Assertion—Reason Based Questions

In the following questions 19 and 20, 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 and (R) is not the correct explanation of (A).
(b) (A) is true, but (R) is false.
(b) (A) is false, hut (R) is true.

Question 19.
Assertion (A): Maximum value of (cos ^-1 x) ² is π ² .
Reason (R): Range of the principal value branch of cos ^-1 x is \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\).
Solution:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Range (Principal value branches) of cos ^-1 x is (0, π).
Maximum value of cos ^-1 x is π
∴ Maximum value of (cos ^-1 x) ² is π ² .

Question 20.
Assertion (A): If a line makes angles α, β, γ with positive direction of the coordinate axes, then sin ² α + sin ² β + sin ² γ = 2.
Reason (R): The sum of squares of the direction cosines of a line is 1.
Solution:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Given, l = cos α, m = cos β, n = cos γ‘
As we know, l ² + m ² + b ² = 1
cos ² α + cos ² β + cos ² γ = 1
Now, LHS. sin ² α + sin ² β + sin ² γ
= 1 – cos ² α + 1 – cos ² β + 1 – cos ² γ
= 3 – (cos ² α + cos ² β + cos ² γ)
= 3 – (1) ……..(From (i)
= 2 (Hence Proved)

Section – B

This section comprises of Very Short Answer (VSA) type questions of 2 marks each.

Question 21.
(a) Evaluate sin ^-1 \(\left(\sin \frac{3 \pi}{4}\right)\) + cos ^-1 (cos π) + tan ^-1 (1).
Solution:

Or

(b) Draw the graph of cos ^-1 x, where x ∈ [-1, 0]. Also, write its range.
Solution:

Graph of cos ^-1 x, x ∈ [-1, 0)
Range of cos ^-1 x is (0, π]
when x ∈ [-1, 1]
But when x ∈ [-1, 0]
∴ Range of cos ^-1 x is \(\left(\frac{\pi}{2}, \pi\right)\).

Question 22.
A particle moves along the curve 3y = ax ³ + 1 such that at a point with x-coordinate 1, y-coordinate is changing twice as fast at x-coordinate. Find the value of a.
Solution:
Since y-coordinate is changing twice as fast x-coordinate.
∴ \(\frac{d y}{d x}\) = 2\(\frac{d x}{d t}\) …….. (i)
Given. 3y = ax ³ + 1
y = \(\frac{a}{3} x^3+\frac{1}{3}\)
Differentiating both sides w.r.t t, we have
\(\frac{d y}{d t}\) = \(\frac{a}{3} \cdot 3 x^2 \frac{d x}{d t}\)
\(2 \frac{d x}{d t}\) = \(a x^2 \frac{d x}{d t}\) ……. [From (i)
2 = a(1) ² [∵ when x = 1
∴ a = 2

Question 23.
If \(\vec{a}, \vec{b}, \vec{c}\) are three non-zero unequal vectors such that \(\vec{a} \cdot \vec{b}\) = \(\vec{a} \cdot \vec{c}\), then find the angle between \(\vec{a}\) and \(\vec{b}-\vec{c}\).
Solution:

Question 24.
Find the coordinates of points on line \(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z+1}{2}\) which are at a distance of \(\sqrt{11}\) units
from origin.
Solution:
\(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z+1}{2}\) = q (let)
General point P(q, 2q + 1, 2q – 1) on the given line and on origin O (0, 0, 0).
OP = \(\sqrt{11}\) ……. [Given
OP ² = 11 ……[Squaring both sides
⇒ (q – 0) ² + (2q + 1 – 0) ² + (2q – 1 – 0) ² = 11
⇒ q ² + 4q ² + 1 + 4q + 4q ² + 1 – 4q = 11
⇒ 9q ² = 9
⇒ q ² = 1
∴ q = ±1
When q = 1, Point P is (1, 3, 1)
When q = -1, Point P is (-1, -1, -3)

Question 25.
(a) If y = \(\sqrt{a x+b}\), prove that \(y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2\) = 0.
Solution:
Given. y = \(\sqrt{a x+b}\) …….. (i)
Differentiating both sides w.r.t. x, we have

Or

(b) If

is a differentiable function in (0, 2), then find the values of a and b.
Solution:
f(x) is differentiable function in (0, 2) ………[Given
then it must be continuous in (0, 2)

Section-C

This section comprises of Short Answer type questions (SA) of 3 marks each.

Question 26.
(a) Evaluate \(\int_0^{\pi / 4} \log (1+\tan x) d x\).
Solution:

Or

(b) Find \(\int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}}\)
Solution:

Question 27.
Find \(\int e^{\cot ^{-1} x}\left(\frac{1-x+x^2}{1+x^2}\right) d x\).
Solution:

Question 28.
Evaluate \(\int_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{1}{\left(e^x+e^{-x}\right)\left(e^x-e^{-x}\right)} d x\)
Solution:

Question 29.
(a) Find the general solution of the differential equation:
(xy – x ² ) dy = y ² dx.
Solution:
\(\frac{d y}{d x}\) = \(\frac{y^2}{x y-x^2}\)
Dividing Numerator and Denominator
\(\frac{d y}{d x}\) = \(\frac{\left(\frac{y}{x}\right)^2}{\frac{y}{x}-1}\) = \(g\left(\frac{y}{x}\right)\) ……..(ii)
R.H.S. of differential equation (ii) is of the form \(g\left(\frac{y}{x}\right)\) and so it is a homogeneous.
Therefore, equation (i) is a homogeneous differential equation

Or

(b) Find the general solution of the differential equation:
(x ² + 1)\(\frac{d y}{d x}\) + 2xy = \(\sqrt{x^2+4}\)
Solution:

Question 30.
(a) Two balls are drawn at random one by one with replacement from an urn containing equal number of red balls and green balls. Find the probability distribution of number of red balls. Also, find the mean of the random variable.
Solution:
Let number of red balls be a,
then no. of green balls be = a
Total number of balls = a + a = 2a
Let p = P(a red ball) = \(\frac{a}{2 a}\) = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
Here random variable x can take values 0, 1, 2.
P(x = 0) = qq = q ² = \(\left(\frac{1}{2}\right)^2\) = \(\frac{1}{4}\)
P(x = 1) = pq + qp = 2qp = 2 × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{2}\)
P(x = 2) = p.p = p ² = \(\left(\frac{1}{2}\right)^2\) = \(\frac{1}{4}\)
∴ Probability distribution is

= 0 × \(\frac{1}{4}\) + 1 × \(\frac{1}{2}\) + 2 × \(\frac{1}{4}\)
= 0 + \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1

Or

(b) A and B throw a die alternately till one of them gets a ‘6’ and wins the game. Find their respective probabilities of wining, if A starts the game first.
Solution:
p = {1, 2, 3, 4, 5, 6}
Let p = P(a numbers 6) = \(\frac{1}{6}\),
q = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
A can win in 1, 3, 5, … throws

Question 31.
Solve the following linear programming problem graphically:
Minimize : Z = 5x + 10y
subject to constraints : x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0
Solution:

Section-D

This section comprises of Long Answer questions (LA) type of 5 marks each.

Question 32.
If A = \(\left[\begin{array}{ccc}
-3 & -2 & -4 \\
2 & 1 & 2 \\
2 & 1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 2 & 0 \\
-2 & -1 & -2 \\
0 & -1 & 1
\end{array}\right]\), then find AB and use it to solve the following system of equations:
x – 2y = 3
2x – y – z = 2
-2y + z = 3
Solution:

Or

(b) If f(α) = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\), prove that f(α) . f(-β) = f(α – β)
Solution:

Question 33.
(a) Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, -6), Q(5, -3,1), R(12, 4, 5) and S(11, 9, -2). Use these equations to find the point of intersection of diagonals.
Solution:
Diagonals PR, P(4, 2, -6), R(12, 4, 5)

Or

(b) A line 1 passes through point (-1, 3, -2) and is perpendicular to both the lines \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) and \(\frac{x+2}{-3}\) = \(\frac{y-1}{2}\) = \(\frac{z+1}{5}\). Find the vector equation of the line l.
Hence, obtain its distance from origin.
Solution:
Let DR’s of line l are a, b, c.
Equation of line, l \(\frac{x-x_1}{a}\) = \(\frac{y-y_1}{b}\) = \(\frac{z-z_1}{c}\)
\(\frac{x+1}{a}\) = \(\frac{y-3}{b}\) = \(\frac{z+2}{c}\) ……. (i)
Line (i) is ⊥ to given I ^st line
a + 2b + 3c = 0 …[Using a ₁ a ₂ + b ₁ b ₂ + c ₁ c ₂ = 0
Line (i) is ⊥ to give 2 ^nd line
\(\frac{a}{10-6}\) = \(\frac{-b}{5+9}\) = \(\frac{c}{2+6}\) = λ(cot)
a = 4λ, b = -14λ, c = 8λ
or a = 2, b = -7, c = 4
Putting the value of a, b, c in (i),
\(\frac{x+1}{2}\) = \(\frac{y-3}{-7}\) = \(\frac{z+2}{4}\) = (let)

Question 34.
Using integration, find the area of region bounded by line y = \(\sqrt{3} x\), the curve y = \(\sqrt{4-x^2}\) and y-axis in first quadrant.
Solution:

Question 35.
A function f : [-4, 4] → [0, 4] is given by f (x) = \(\sqrt{16-x^2}\). Show that f is an onto function but not a one-one function. Further, find all possible values of ‘a’ for which f (a) = \(\sqrt{7}\) . 5
Solution:
f(x) = \(\sqrt{16-x^2}\)
one-one,
When x = -4 ∈ [-4, 4], f(-4) = \(\sqrt{16-16}\) = 0
When x = -4 ∈ [-4, 4], f(4) = \(\sqrt{16-16}\) = 0
since, f(-4) = f(4)
But -4 ≠ 4
∴ f is not one-one
Onto, Let y = f(x)

Section – E

This section comprises of 3 source based / case-based / passage based / integrated units of assessment questions of 4 marks each.

Case Study-1

Question 36.
Engine displacement is the measure of the cylinder volume swept by all the pistons of a piston engine. The piston moves inside the cylinder bore.

The cylinder bore in the form of circular cylinder open at the top is to be made from a metal sheet of area 75 π cm ² .
Based on the above information, answer the following questions:
(i) If the radius of cylinder is r cm and height is h cm, then write the volume V of cylinder in terms of radius r.
Solution:
Surface area of open cyclinder
= 75π cm ²
2πrh + πr ² = 75π
2rh + r ² = 75
2rh = 75 – r ²
Volume of cylinder, V = πr ² h
= πr ² \(\left(\frac{75-r^2}{2 r}\right)\) ……[From (i)
= \(\frac{\pi r}{2}\)(75 – r ² )
V = \(\frac{\pi}{2}\) (75r – r ³ )

(ii) Find \(\frac{d \mathrm{~V}}{d r}\)
Solution:
Differentiating w.r.t. r, we have
\(\frac{d V}{d r}\) = \(\frac{\pi}{2}\)(75 – 3r ² )

(iii) (a) Find the radius of cylinder when its volume is maximum.
Solution:

Or

(b) For maximum volume, h > r. State true or false and justify.
Solution:
From (i), h = \(\frac{75-25}{2(5)}\) ……….[∵ r = 5 cm
= \(\frac{50}{10}\) = 5 cm
Here, r = h = 5 cm
Therefore, h > r is false.

Case Study – II

Question 37.
Recent studies suggest that roughly 12% of the world population is left handed.

Depending upon the parents, the chances of having a left handed child are as follows:
A: When both father and mother are left handed:
Chances of left handed child is 24%.
B: When father is right handed and mother is left handed:
Chances of left handed child is 22%.
C: When father is left handed and mother is right handed:
Chances of left handed child is 17%.
D: When both father and mother are right handed:
Chances of left handed child is 9%.
Assuming that P(A) = P(B) = P(C) = P(D) = and L denotes the event that child is left handed.

Based on the above information, answer the following questions:
(i) Find P(L/C)
(ii) Find P(\(\overline{\mathbf{L}}\) / A)
(iii) (a) Find P(A/L)
Solution:

Or

(b) Find the probability that a randomly selected child is left handed given that exactly one of the parents is left handed.
Solution:
Required probability, = P(L | B) + P(L | C)
= \(\frac{22}{100}+\frac{17}{100}\) = \(\frac{39}{100}\) = 0.39

Case Study-III

Question 38.
The use of electric vehicles will curb air pollution in the long run.
The use of electric vehicles is increasing every year and estimated electric vehicles in use at any time t is given by the function V:

V(t) = \(\frac{1}{5} t^3\) – \(\frac{5}{2} t^2\) + 25t – 2
where t represents the time and t = 1, 2, 3… corresponds to year 2001, 2002, 2003, …….. respectively.
Based on the above information, answer the following questions:

(i) Can the above function be used to estimate number of vehicles in the year 2000? Justify.
Solution:
Given. V(t) = \(\frac{1}{5} t^3-\frac{5}{2} t^2+25 t-\)
Put t = 0, V(0) = 0 – 0 + 0 – 2
= -2 (Negative)
No, number of vehicles cannot be negative. Therefore, the given statement is not possible.

(ii) Prove that the function V(t) is an increasing function.
Solution:

Set – II Code No. 65/2/2
Section-A (Multiple Choice Questions)

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

Question 1.
If \(\frac{d}{d x}\)f(x) = 2x + \(\frac{3}{x}\) and f(1) = 1, then f(x) is
(a) x ² + 3 log |x| + 1
(b) x ² + 3 log |x|
(c) 2 – \(\frac{3}{x^2}\)
(d) x ² + 3 log |x| – 4
Solution:
(b) x ² + 3 log |x|

Question 5.
If in ∆ABC, \(\overrightarrow{\mathbf{B A}}\) = 2\(\vec{a}\) and \(\overrightarrow{B C}\) = 3\(\vec{b}\), then \(\overrightarrow{\mathrm{AC}}\) is
(a) \(2 \vec{a}+3 \vec{b}\)
(b) \(2 \vec{a}-3 \vec{b}\)
(c) \(3 \vec{b}-2 \vec{a}\)
(d) \(-2 \vec{a}-3 \vec{b}\)
Solution:
(c) \(3 \vec{b}-2 \vec{a}\)

Question 6.
If \(|\vec{a} \times \vec{b}|\) = \(\sqrt{3}\) and \(\vec{a} \cdot \vec{b}\) = -3, then angle between \(\) and \(\) is
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{5 \pi}{6}\)
Solution:
(d) \(\frac{5 \pi}{6}\)

Question 7.
Equation of line passing through origin and making 30°, 60° and 90° with x, y, z axes respectively is
(a) \(\frac{2 x}{\sqrt{3}}\) = \(=\frac{y}{2}\) = \(\frac{z}{0}\)
(b) \(\frac{2 x}{\sqrt{3}}\) = \(\frac{2 y}{1}\) = \(\frac{z}{0}\)
(c) 2x = \(\frac{2 y}{\sqrt{3}}\) = \(\frac{z}{1}\)
(d) \(\frac{2 x}{\sqrt{3}}\) = \(\frac{2 y}{1}\) = \(\frac{z}{1}\)
Solution:
(b) \(\frac{2 x}{\sqrt{3}}\) = \(\frac{2 y}{1}\) = \(\frac{z}{0}\)

Question 8.
If A and B are two events such that P(A | B) = 2 × P(B | A) and P(A) + P(B) = \(\frac{2}{3}\), then P(B) is equal to
(a) \(\frac{2}{9}\)
(b) \(\frac{7}{9}\)
(c) \(\frac{4}{9}\)
(d) \(\frac{5}{9}\)
Solution:
(a) \(\frac{2}{9}\)

Question 15.
If A is a 2 × 3 matrix such that AB and AB’ both are defined, then order of the matrix B is
(a) 2 × 2
(b) 2 × 1
(c) 3 × 2
(d) 3 × 3
Solution:
(d) 3 × 3

Section – B

Question 23.
If the equation of a line is x = ay + b, z = cy + d, then find the direction ratios of the line and a point on the line.
Solution:

Equation of line \(\frac{x-b}{a}\) = \(\frac{y-0}{1}\) = \(\frac{z-d}{c}\),
………..[∵ x ₁ = b, y ₁ = 0, z ₁ = d
∴ DR’s of the line are a, 1, c, and A point on the line (b, 0, d)

Question 25.
If the circumference of circle is increasing at the constant rate, prove that rate of change of area of circle is directly proportional to its radius.
Solution:
Let c, r and A be the circumference, radius and area of the circle.

Section – C

Question 29.
Solve the following linear programming problem graphically:
Maximize : Z = x + 2y
subject to constraints:
x + 2y ≥ 1oo,
2x – y ≤ 0,
2x + y ≤ 200,
x ≥ 0, y ≥ 0.
Solution:

Question 30.
(a) Evaluate: \(\int_{-1}^1\left|x^4-x\right| d x\)
Solution:

Or

(b) Find: \(\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x\)
Solution:

Question 31.
Find: \(\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x\)
Solution:

Section-D

Question 33.
Using Integration, find the area of triangle whose vertices are (-1, 1), (0, 5) and (3, 2).
Solution:

Now Area. of ∆ABC

Set – II Code No. 65/2/3
Section-A (Multiple Choice Questions)

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

Question 1.
If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{3}\) and \(|\vec{a} \times \vec{b}|\) = 3\(\sqrt{3}\), then the value of \(\vec{a} \cdot \vec{b}\) is
(a) 9
(b) 3
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{3}\)
Solution:
(b) 3

Question 2.
The position vectors of three consecutive vertices of a parallelogram ABCD are A(4\(\hat{i}\) + 2\(\hat{j}\) – 6\(\hat{k}\)), B(5\(\hat{i}\) – 3\(\hat{j}\) + \(\hat{k}\)) and C(12\(\hat{i}\) + 4\(\hat{j}\) + 5\(\hat{k}\)). The position vector of D is given by I
(a) -3\(\hat{i}\) – 5\(\hat{j}\) – 10\(\hat{k}\)
(b) 21\(\hat{i}\) + 3\(\hat{j}\)
(c) 11\(\hat{i}\) + 9\(\hat{j}\) – 2\(\hat{k}\)
(d) -11\(\hat{i}\) – 9\(\hat{j}\) + 2\(\hat{k}\)
Solution:
(c) 11\(\hat{i}\) + 9\(\hat{j}\) – 2\(\hat{k}\)

Let the position vector of ∆ is Mid point of AC = Mid point of BD
…….[Diagonal of a ||gm bisect each other
\(\frac{4 \hat{i}+2 \hat{j}-6 \hat{k}+12 \hat{i}+4 \hat{j}+5 \hat{k}}{2}\) = \(\frac{5 \hat{i}-3 \hat{j}+\hat{k}+\vec{d}}{2}\)
= 16\(\hat{i}\) + 6\(\hat{j}\) – \(\hat{k}\) – 5\(\hat{i}\) + 3\(\hat{j}\) – \(\hat{k}\) = \(\vec{d}\)
∴ \(\vec{d}\) = 11\(\hat{i}\) + 9\(\hat{j}\) – 2\(\hat{k}\)

Question 3.
If for two events A and B, P(A – B) = \(\frac{1}{5}\) and P(A) = \(\frac{3}{5}\), then \(\mathbf{P}\left(\frac{\mathbf{B}}{\mathbf{A}}\right)\) is equal to
(a) \(\frac{1}{2}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{2}{3}\)
Solution:
(d) \(\frac{2}{3}\)

Question 4.
If \(\int_0^{2 \pi} \cos ^2 x d x\) = \(k \int_0^{\pi / 2} \cos ^2 x d x\), then the value of k is
(a) 4
(b) 2
(c) 1
(d) 0
Solution:
(a) 4

Question 10.
Number of symmetric matrices of order 3 × 3 with each entry 1 or -1 is
(a) 512
(b) 64
(c) 8
(d) 4
Solution:
(b) 64

a ₁₁ can be filled by 2 ways
(i.e., 1 or -1)
a ₂₂ can be filled by 2 ways
a ₃₃ can be filled by 2 ways
a ₁₂ can be filled by 2 ways
a ₁₃ cart be filled by 2 ways
a ₂₃ can be filled by 2 ways
∴ Total number of ways = 2 ⁶ = 64

Question 18.
Equation of a line passing through point (1, 2, 3) and equally inclined to the coordinate axis, is
(a) \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)
(b) \(\frac{x}{1}=\frac{y}{1}=\frac{z}{1}\)
(c) \(\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{3}\)
(d) \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}\)
Solution:
(d) \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}\)

Section – B

Question 21.
If points A, B and C have position vectors 2\(\hat{i}\), \(\hat{j}\) and 2\(\hat{k}\) respectively, then show that ∆ABC is an isosceles triangle.
Solution:

Question 23.
If equal sides of an isosceles triangle with fixed base 10 cm are increasing at the rate of 4 cm/sec, how fast is the area of triangle increasing at an instant when all sides become equal?
Solution:

Section – C

Question 26.
Solve the following Linear Prgramming problem graphically:
Maximize : Z = 3x + 3.5 y
subject to constraints: x + 2y ≥ 240,
3x + 1.5y ≥ 270
1.5x + 2y ≤ 310,
x ≥ 0, y ≥ 0.
Solution:

Question 27.
(a) Find \(\int \frac{x+2}{\sqrt{x^2-4 x-5}} d x\)
Solution:

Or

(b) Evaluate \(\int_{-a}^a f(x) d x\), where f(x) = \(\frac{9^x}{1+9^x}\)
Solution:

Question 31.
(a) Two numbers are selected from first six even natural numbers at random without replacement. If X denotes the greater of two numbers selected, find the probability distribution of X.
Solution:
First 6 even natural numbers are 2, 4, 6, 8, 10, 12.

When greater no. is 4, then (2, 4), (4, 2) = 2 ways
When greater no. is 6, then (2, 6), (6, 2), (4, 6), (6, 4) = 4 ways
When greater no. is 8, then (2, 8), (8, 2), (4, 8), (8, 4), (6, 8), (8, 6) = 6 ways
When greater no. is 10, then (2, 10), (10, 2), (4, 10), (10, 4), (6, 10), (10, 6), (8, 10), (10, 8) = 8 ways
When greater non is 12, then (2, 12), (12, 2), (4, 12), (12, 4), (6, 12), (12, 6), (8, 12), (12, 8), (10, 12), (12, 10) = 10 ways
P(X = 4) = P(larger number is 4) = \(\frac{2}{30}\) = \(\frac{1}{15}\)
P(X = 6) = P(larger number is 6) = \(\frac{4}{30}\) = \(\frac{2}{15}\)
P(X = 8) = P(larger number is 8) = \(\frac{6}{30}\) = \(\frac{1}{5}\)
P(X = 10) = P(larger number is 10) = \(\frac{8}{30}\) = \(\frac{4}{15}\)
P(X = 12) = P(larger number is 12) = \(\frac{10}{30}\) = \(\frac{1}{3}\)
∴ Probability distribution is

Or

(b) A fair coin and art unbiased die are tossed. Let A be the event, “Head appears on the coin” and B be the event, “3 comes on the die”. Find whether A and B are independent events or not.
Solution:

Section – D

Question 35.
Find the area of the smaller region bounded by the curves \(\frac{x^2}{25}+\frac{y^2}{16}\) = 1 and \(\frac{x}{5}+\frac{y}{4}\) = 1, using integration.
Solution:
Given. \(\frac{x^2}{25}+\frac{y^2}{16}\) = 1

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