# CBSE Class 12 Maths Question Paper 2023 (Series: EF1GH/5) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Maths with Solutions and CBSE Class 12 Maths Question Paper 2023 (Series: EF1GH/5) to familiarize themselves with the exam format and marking scheme.

## CBSE Class 12 Maths Question Paper 2023 (Series: EF1GH/5) with Solutions

Time Allowed: 3 hours

Maximum Marks: 80

General Instructions:

- This question paper contains 38 questions. All questions are compulsory.
- Question paper is divided into FIVE Sections – Section A, B, C, D and E.
- In Section A — Question no. 1 to 18 are Multiple Choice Questions (MCQ) type and Question no. 19 to 20 are Assertion-Reason based questions of 1 mark each.
- In Section B—Question no. 21 to 25 are Very Short Answer (VSA) type questions of 2 marks each.
- In Section C—Question no. 26 to 31 are Short Answer (SA) type questions, carrying 3 marks each.
- In Section D—Question no. 32 to 35 are Long Answer (LA) type questions carrying 5 marks each.
- In Section E—Question no. 36 to 38 are case study based questions carrying 4 marks each where 2 VSA type questions are of 1 mark each and 1 SA type question is of 2 marks. Internal choice is provided in 2 marks question in each case-study.
- 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.
- Use of calculator is NOT allowed.

Set – I Code No. 65/5/1

Section-A (Multiple Choice Questions)

Each Question Carries 1 Mark. Select the correct option out of the four given options:

Question 1.

Let A = {3, 5}. Then number of reflexive relations on A is

(a) 2

(b) 4

(c) 0

(d) 8

Solution:

(b) 4

Number of elements, n = 2

No. of reflexive relation on A is

= 2^{n(n – 1)}

= 2^{2(2 – 1)} = 2^{2} = 4

Question 2.

sin [\(\frac{\pi}{3}\) + sin^{-1}\(\left(\frac{1}{2}\right)\)] is equal to

(a) 1

(b) \(\frac{1}{2}\)

(c) \(\frac{1}{3}\)

(d) \(\frac{1}{4}\)

Solution:

(a) 1

Given, sin \(\left(\frac{\pi}{3}+\sin ^{-1}\left(\frac{1}{2}\right)\right)\)

= sin \(\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\)

= sin \(\frac{\pi}{2}\) = 1

Question 3.

If for a square matrix A, A^{2} – A + I = O, then A^{-1} equals

(a) A

(b) A + I

(c) I – A

(d) A – I

Solution:

(c) I – A

We have, A^{2} – A + I = 0

I = A – A^{2}

⇒ IA^{-1} = (A – A^{2}) A^{-1}

…[Post multiplied by A^{-1}

⇒ A^{-1} = AA^{-1} – A^{2}A^{-1}

∴ A^{-1} = I – A

Question 4.

If A = \(\left[\begin{array}{ll}

1 & 0 \\

2 & 1

\end{array}\right]\), B = \(\left[\begin{array}{ll}

x & 0 \\

1 & 1

\end{array}\right]\) and A = B^{2}, then x equals

(a) ±1

(b) -1

(c) 1

(d) 2

Solution:

(c) 1

Given, A = B^{2}

\(\left(\begin{array}{ll}

1 & 0 \\

2 & 1

\end{array}\right)\) = \(\left(\begin{array}{ll}

x & 0 \\

1 & 1

\end{array}\right)\left(\begin{array}{ll}

x & 0 \\

1 & 1

\end{array}\right)\)

⇒ \(\left(\begin{array}{ll}

1 & 0 \\

2 & 1

\end{array}\right)\) = \(\left(\begin{array}{ll}

x^2 & 0 \\

x+1 & 1

\end{array}\right)\)

⇒ x^{2} = 1; x + 1 = 2

∴ x = ±1; or x = 1

∴ Common solution of x = 1

Question 5.

If \(\left|\begin{array}{lll}

\alpha & 3 & 4 \\

1 & 2 & 1 \\

1 & 4 & 1

\end{array}\right|\) = 0, then the value of α is

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

(d) 4

Expanding through C_{1}; we have

α(2 – 4) – 1(3 – 16) + 1(3 – 8) = 0

⇒ -2α + 13 – 5 = 0

⇒ 8 = 2α

∴ α = 4

Question 6.

The derivative of x^{2x} w.r.t. x is

(a) x^{2x-1}

(b) 2x^{2x}log x

(c) 2x^{2x}(1 + log x)

(d) 2x^{2x}(1 – log x)

Solution:

(c) 2x^{2x}(1 + log x)

Let y = \(x^{2 x}\) …(i)

Taking log on both sides

log y = log \(x^{2 x}\)

⇒ log y = 2xlog x …… [log x^{n} = n log x

Differentiating both sides w.r.t. x

Question 7.

The function f(x) = [x], where [x] denotes the greatest integer less than or equal to x, is continuous at

(a) x = 1

(b) x = 1.5

(c) x = -2

(d) x = 4

Solution:

(b) x = 1.5

Greatest integer function is discontinuous at all integral values of x.

So, discontinuous at x = 1, -2 and 4.

Question 8.

If x = A cos 4t + B sin 4t, then \(\frac{d^2 x}{d t^2}\) is equal to

(a) x

(b) -x

(c) 16x

(d) -16x

Solution:

(d) -16x

We have, x = A cos 4t + B sin 4t …….(i)

Differentiating the above w.r.t. t,

\(\frac{d x}{d t}\) = -4A sin 4t + 4B cos 4t

Again, differentiating the above w.r.t. t, we get.

\(\frac{d^2 x}{d t^2}\) = -16A cos 4t – 16B sin 4t

\(\frac{d^2 x}{d t^2}\) = -16(A cos 4t + B sin 4t)

= -16x …[From (i)

Question 9.

The interval in which the function f(x) = 2x^{3} + 9x^{2} + 12x – 1 is decreasing, is

(a) (-1, ∞)

(b) (-2, -1)

(c) (-∞, -2)

(d) [-1, 1]

Solution:

(b) (-2, -1)

Question 10.

\(\int \frac{\sec x}{\sec x-\tan x} d x\) equals

(a) sec x – tan x + c

(b) sec x + tan x + c

(c) tan x – sec x + c

(d) -(sec x – tan x) + c

Solution:

(b) sec x + tan x + c

Question 11.

\(\int_{-1}^1 \frac{|x-2|}{x-2} d x\), x ≠ 2 is equal to

(a) 1

(b) -1

(c) 2

(d) -2

Solution:

(d) -2

Question 12.

The sum of the order and the degree of the differential equation \(\frac{d}{d x}\left(\left(\frac{d y}{d x}\right)^3\right)\) is

(a) 2

(b) 3

(c) 5

(d) 0

Solution:

(b) 3

Question 13.

Two vectors \(\vec{a}\) = a_{1}\(\hat{i}\) + a_{2}\(\hat{j}\) + a_{3}\(\hat{k}\) and \(\vec{b}\) = b_{1}\(\hat{i}\) + b_{2}\(\hat{j}\) + b_{3}\(\hat{k}\) are collinear if

(a) a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3} = 0

(b) \(\frac{a_1}{b_1}\) = \(\frac{a_2}{b_2}\) = \(\frac{a_3}{b_3}\).

(c) a_{1} = b_{1}, a_{2} = b_{2}, a_{3} = b_{3}

(d) a_{1} + a_{2} + a_{3} = b_{1} + b_{2} + b_{3}

Solution:

(b) \(\frac{a_1}{b_1}\) = \(\frac{a_2}{b_2}\) = \(\frac{a_3}{b_3}\).

Since vectors are collinear, implies that scalar components are proportioned, i.e.,

\(\frac{a_1}{b_1}\) = \(\frac{a_2}{b_2}\) = \(\frac{a_3}{b_3}\)

Alternatively,

Question 14.

The magnitude of the vector 6\(\hat{i}\) – 2\(\hat{j}\) + 3\(\hat{k}\) is

(a) 1

(b) 5

(c) 7

(d) 12

Solution:

(c) 7

Let a\(\hat{i}\) = 6\(\hat{i}\) – 2\(\hat{j}\) + 3\(\hat{k}\)

\(|\vec{a}|\) = \(\sqrt{6^2+(-2)^2+(3)^2}\) = \(\sqrt{36+4+9}\) = 7

Question 15.

If a line makes angles of 90°, 135° and 45° with the x, y and z axes respectively, then its direction cosines are

(a) 0, –\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\)

(b) –\(\frac{1}{\sqrt{2}}\), 0, \(\frac{1}{\sqrt{2}}\)

(c) \(\frac{1}{\sqrt{2}}\), 0, –\(\frac{1}{\sqrt{2}}\)

(d) 0, \(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\)

Solution:

(a) 0, –\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\)

Given,

α = 90°, β = 135°, γ = 45°

Direction cosines are:

Question 16.

The angle between the lines 2x = 3y = -z and 6x = -y = -4z is

(a) 0°

(b) 30°

(c) 45°

(d) 90°

Solution:

(d) 90°

Question 17.

If for any two events A and B, P(A) = \(\frac{4}{5}\) and P(A ∩ B) = \(\frac{7}{10}\), then P(B|A) is equal to 1

(a) \(\frac{1}{10}\)

(b) \(\frac{1}{8}\)

(c) \(\frac{7}{8}\)

(d) \(\frac{17}{20}\)

Solution:

(c) \(\frac{7}{8}\)

As we know,

P(B | A) = \(\frac{P(A \cap B)}{P(A)}\) = \(\frac{\frac{7}{10}}{\frac{4}{5}}\)

= \(\frac{7}{10} \times \frac{5}{4}\) = \(\frac{7}{8}\)

Question 18.

Five fair coins are tossed simultaneously. The probability of the events that atleast one head comes up is

(a) \(\frac{27}{32}\)

(b) \(\frac{5}{32}\)

(c) \(\frac{31}{32}\)

(d) \(\frac{1}{32}\)

Solution:

(c) \(\frac{31}{32}\)

Since five coins are tossed.

∴ Total no. of ways = 2^{5} = 32

No head can be obtained as TTTT

i.e., 1 way.

∴ P(at least 1 head) = 1 – P(No Head)

= 1 – \(\left(\frac{1}{2}\right)^5\) = 1 – \(\frac{1}{32}\) = \(\frac{31}{32}\)

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, but (R) is not the correct explanation of (A).

(c) (A) is true and (R) is false.

(d) (A) is false, but (R) is true.

Question 19.

Assertion (A): Two coins are tossed simultaneously. The probability of getting two heads, if it is known that at least one head comes up, is \(\frac{1}{3}\).

Reason (R): Let E and F be two events with a random experiment then P(F | E) = \(\frac{P(E \cap F)}{P(E)}\)

Solution:

(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Let F : Getting two heads

E : Atleast one head

S = {HH, HT, TH, TT}

P(E) = 3/4 i.e., RH, HT, TH

P(F) = 1/4 i.e., HH

∴ P(F | E) = \(\frac{P(E \cap F)}{P(E)}\) = \(\frac{\frac{1}{4}}{\frac{3}{4}}\) = \(\frac{1}{3}\)

Question 20.

Assertion (A): \(\int_2^8 \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x\) = 3

Reason (R): \(\int_a^b f(x) d x\) = \(\int_a^b f(a+b-x) d x\)

Solution:

(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Section – B

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

Question 21.

Write the domain and range (principle value branch) of the following functions:

f(x) = tan^{-1} x

Solution:

Domain of tan^{-1} x is R

Range (principle value branch) of tan^{-1}x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

Question 22.

(a) If

then show that f is not differentiable at x = 1.

Solution:

= 1 + 1 = 2

Since LHD ≠ RHD

∴ f is not differentiable at x = 1.

Or

(b) Find the value(s) of ‘λ’, if the function

Solution:

Question 23.

Sketch the region bounded by the lines 2x + y =8, y = 2, y = 4 and the y-axis. Hence, obtain its area using integration.

Solution:

Question 24.

(a) If the vectors \(\vec{a}\) and \(\vec{b}\) are such that \(|\vec{a}|\) = 3, \(|\vec{b}|\) = \(\frac{2}{3}\) and \(\vec{a} \times \vec{b}\) is a unit vector, then find the angle between \(\vec{a}\) and \(\vec{b}\).

Solution:

Or

(b) Find the area of a parallelogram whose adjacent sides are determined by the vectors \(\vec{a}\) = \(\hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{b}\) = \(2 \hat{i}-7 \hat{j}+\hat{k}\).

Solution:

Question 25.

Find the vector and the cartesian equations of a line that passes through the point A(1, 2, -1) and parallel to the line 5x – 25 = 14 – 7y = 35z.

Solution:

Section – C

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

Question 26.

If A = \(\left[\begin{array}{ccc}

1 & 2 & 3 \\

3 & -2 & 1 \\

4 & 2 & 1

\end{array}\right]\), then show that A^{3} – 23A – 40I = O.

Solution:

Question 27.

(a) Differentiate sec^{-1}\(\left(\frac{1}{\sqrt{1-x^2}}\right)\) w.r.t. sin^{-1}(2x\(\sqrt{1-x^2}\)).

Solution:

= sin^{-1}(2sinθ\(\sqrt{1-\sin ^2 \theta}\))

= sin^{-1} (2sin θ cos θ)

= sin^{-1} (sin 2θ)

= 2θ

Differentiating w.r.t θ

Or

(b) If y = tan x + sec x, then prove that \(\frac{d^2 y}{d x^2}\) = \(\frac{\cos x}{(1-\sin x)^2}\).

Solution:

Question 28.

(a) Evaluate: \(\int_0^{2 \pi} \frac{1}{1+e^{\sin x}} d x\)

Solution:

Or

(b) Find: \(\int \frac{x^4}{(x-1)\left(x^2+1\right)} d x\)

Solution:

Question 29.

Find the area of the following region using integration:

{(x, y) : y^{2} ≤ 2x and y ≥ x – 4}

Solution:

Question 30.

(a) Find the coordinates of the foot of the perpendicular drawn from the point P(0, 2, 3) to the line \(\frac{x+3}{5}\) = \(\frac{y-1}{2}\) = \(\frac{z+4}{3}\).

Solution:

Let \(\frac{x+3}{5}\) = \(\frac{y-1}{2}\) = \(\frac{z+4}{3}\) = q

General point lies on the given line

M(5q – 3, 2q + 1, 3q – 4)

Direction ratios of PM are:

5q – 3 – 0, 2q + 1 – 2, 3q – 4 – 3

5q – 3, 2q – 1, 3q – 7

Direction ratios of given line are 5, 2, 3

As PM ⊥ to given line …… [Given

Using a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

5(5q – 3) + 2(2q – 1) + 3(3q – 7) = 0

⇒ 25q – 15 + 4q – 2 + 9q – 21 = 0

⇒ 38q – 38 = 0

⇒ 38q = 38

⇒ q = 1

Putting the value of q in (i), we get

(5 – 3, 2 + 1, 3 – 4)

= (2, 3, -1)

Hence coordinates of M are (2, 3, -1)

Or

(b) Three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) satisfy the condition \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = \(\vec{0}\). Evaluate the quantity µp = \(\vec{a}\).\(\vec{b}\) + \(\vec{b}\).\(\vec{c}\) + \(\vec{c}\).\(\vec{a}\), if | \(\vec{a}\) | = 3, | \(\vec{b}\) | = 4 and | \(\vec{c}\) | = 2.

Solution:

Question 31.

Find the distance between the lines:

\(\vec{r}\) = (\(\hat{i}\) + 2\(\hat{j}\) – 4\(\hat{k}\)) + λ(2\(\hat{i}\) + 3\(\hat{j}\) + 6\(\hat{k}\)) ;

\(\vec{r}\) = (3\(\hat{i}\) + 3\(\hat{j}\) – 5\(\hat{k}\)) + µ(4\(\hat{i}\) + 6\(\hat{j}\) + 12\(\hat{k}\))

Solution:

∴ Shortest distance between parallel lines

Section – D

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

Question 32.

(a) The median of an equilateral triangle is increasing at the rate of 2\(\sqrt{3}\) cm/ s. Find the rate at which its side is increasing.

Solution:

Let x be the side of an equilateral ∆. Thus, median, m = \(\frac{\sqrt{3}}{2}\) side

Here, m = \(\frac{\sqrt{3}}{2}\)(x)

Differentiating both sides w.r.t time, (t) we have

\(\frac{d m}{d t}\) = \(\frac{\sqrt{3}}{2} \frac{d x}{d t}\)

⇒ \(2 \sqrt{3}\) = \(\frac{\sqrt{3}}{2} \frac{d x}{d t}\) …… [∵ \(\frac{d m}{d t}\) = \(2 \sqrt{3}\) Given]

⇒ \(2 \sqrt{3}\) × \(\frac{2}{\sqrt{3}}\) = \(\frac{d x}{d t}\)

\(\frac{d x}{d t}\) = 4 cm/s

Or

(b) Sum of two numbers is 5. If the sum of the cubes of these numbers is least, then find the sum of the squares of these numbers.

Solution:

Let number be x and y

According to question, x + y = 5

y = 5 – x ….. (i)

Let s = x^{3} + y^{3}

s = x^{3} + (5 – x)^{3} ……..[From (i)

Differentiating both sides w.r.t x,

\(\frac{d s}{d x}\) = 3x^{2} + 3(5 – x)^{2}(-1)

= 3(x^{2} – 25 – x^{2} + 10x)

= 3(10x – 25)

For critical points, \(\frac{d s}{d x}\) = 0

10x – 25 = 0

∴ x = \(\frac{25}{10}\) = \(\frac{5}{2}\)

\(\frac{d^2 s}{d x^2}\) = 3(10) = 30 > 0 (+ve)

So, s is least when x = \(\frac{5}{2}\)

From (i), y = 5 – \(\frac{5}{2}\) = \(\frac{5}{2}\)

Required, x^{2} + y^{2} = \(\left(\frac{5}{2}\right)^2+\left(\frac{5}{2}\right)^2\)

= \(\frac{25}{4}\) + \(\frac{25}{4}\) = \(\frac{50}{4}\) = \(\frac{25}{2}\) or 12.5

Question 33.

Evaluate: \(\int_0^{\frac{\pi}{2}}\)sin 2x tan^{-1}(sin x)dx

Solution:

Question 34.

Solve the following Linear Programming Problem graphically:

Maximize: P = 70x + 40y

subject to: 3x + 2y ≤ 9,

3x + y ≤ 9,

x ≥ 0, y ≥ 0

Solution:

Question 35.

(a) In answering a question on a multiple choice test, a student either knows the answer or guesses. Let \(\frac{3}{5}\) be the probability that he knows the answer and \(\frac{2}{5}\) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \(\frac{1}{3}\). What is the probability that the student knows the answer, given that he answered it correctly?

Solution:

Let A be the event that he has answered correctly.

E_{1} be the event that he knows the answer.

E_{2} be the event that he guesses the answer.

P(E_{1}) = \(\frac{3}{5}\), P(A | E_{1}) = 1

P(E_{2}) = \(\frac{2}{5}\), P(A | E_{2}) = \(\frac{1}{3}\)

By Baye’s theorem

∴ P(E_{1}|A)

= \(\frac{\mathrm{P}\left(\mathrm{E}_1\right) \times \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \times \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \times \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_2\right)}\)

= \(\frac{\frac{3}{5}}{\frac{9+2}{15}}\) = \(\frac{3}{5} \times \frac{15}{11}\) = \(\frac{9}{11}\)

Or

(b) A box contains 10 tickets, 2 of which carry a prize of ₹8 each, 5 of which carry a prize of ₹4 each, and remaining 3 carry a prize of ₹2 each. If one ticket is drawn at random, find the mean value of the prize.

Solution:

Total tickets = 10

2 of which carry a prize of ₹8

∴ P(wins ₹8) = \(\frac{2}{10}\)

5 of which carry a ₹4

∴ P(wins ₹4) = \(\frac{5}{10}\)

5 carry a price of ₹2

∴ P(wins ₹2) = \(\frac{3}{10}\)

Section-E

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 (I), (II), (III) of marks 1, 1, 2 respectively. The third case study question has two sub-parts (I) and (II) of marks 2 each.

Case Study-I

Question 36.

An organisation conducted bike race under two different categories — Boys and Girls. There were 28 participants in all. Among all of them, finally three from category I and two from category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his college project.

Let B = {b_{1}, b_{2}, b_{3}} and G = {g_{1}, g_{2}}, where B represents the set of Boys selected and G the set of Girls selected for the final race.

Based on the above information, answer the following questions:

(i) How many relations are possible from B to G?

(ii) Among all the possible relations from B to G, how many functions can be formed from B to G?

(iii) Let R : B → B be defined by R = {(x, y): x and y are students of the same sex}. Check if R is an equivalence relation.

Solution:

Given. B = {b_{1}, b_{2}, b_{3}}, G = {g_{1}, g_{2}}

(i) n(B) = 3, n(G) = 2

∴ Number of relations from B to G

= \(2^{n(B) \times} n^{(G)}\) = 2^{3 × 2}

= 2^{6} = 64

(ii) Number to functions from B to G

= \(\left[(n(\mathrm{G})]^{n(\mathrm{~B})}\right.\)

= 2^{3} = 8

(iii) We have, R: B → B

R = ((x, y): x and y are students of the same sex}

Reflexive: (b_{1}, b_{1}) ∈ R;

∴ R is reflexive.

Symmetric: (b_{1}, b_{2}) ∈ R;

(b_{2}, b_{1}) ∈ R;

∴ R is Symmetric.

Transitive: (b_{1}, b_{2}) ∈ R; (b_{2}, b_{3}) ∈ R

Then (b_{1}, b_{3}) ∈ R;

∴ R is transitive.

Yes, R is an equivalence relation.

Or

A function f: B → G be defined by f = {(b_{1}, g_{1}), (b_{2}, g_{2}) (b_{3}, g_{1})}.

Check if f is bijective. Justify your answer.

Solution:

f = {(b_{1}, g_{1}), (b_{2}, g_{2}), (b_{3}, g_{1})}

As in the given function. Elements b_{1}, b_{3} have same image g_{1}.

So, f is not one-one.

Hence f is not bijective.

Case Study-II

Question 37.

Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of ₹160. From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of ₹190. Also Ankur buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of ₹250.

Based on the above information, answer the following questions:

(i) Convert the given above situation into a matrix equation of the form AX = B. 1

(ii) Find |A|.

(iii) Find A^{-1}.

Solution:

(i) Let the cost of a pen, a bag and a instrument box be ₹x, ₹y, ₹z respectively.

According to question,

5x + 3y + z = ₹160

2x + 1y + 3z = ₹190

1x + 2y + 4z = ₹250

On writing in matrix form, we get

(ii) | A | = 5(4 – 6) – 3(8 – 3) + 1(4 – 1)

= -10 – 15 + 3

= -22 ≠ 0

(iii)

Or

Determine P = A^{2} – 5A.

Solution:

Case Study – III

Question 38.

An equation involving derivatives of the dependent variable with respect to the inde-pendent variables is called a differential equation. A differential equation of the form \(\frac{d y}{d x}\) = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogeneous function of degree n if F(λx, λy) = λ^{n} F(x, y). To solve a homogeneous differential equation of the type \(\frac{d y}{d x}\) = F(x, y) = g\(\left(\frac{y}{x}\right)\), we make the substitution y = vx and then separate the variables.

Based on the above, answer the following questions:

(i) Show that (x^{2} – y^{2}) dx + 2xy dy = 0 is a differential equation of the type \(\frac{d y}{d x}\) = \(g\left(\frac{y}{x}\right)\).

Solution:

(ii) Solve the above equation to find its general solution.

Solution:

SET II Code No. 65/5/2

Section-A (Multiple Choice Questions)

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

Question 4.

If A = [a_{ij}] is a square matrix of order 2 such that

,

then A^{2} is

(a) \(\left[\begin{array}{ll}

1 & 0 \\

1 & 0

\end{array}\right]\)

(b) \(\left[\begin{array}{ll}

1 & 1 \\

0 & 0

\end{array}\right]\)

(c) \(\left[\begin{array}{ll}

1 & 1 \\

1 & 0

\end{array}\right]\)

(d) \(\left[\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right]\)

Solution:

(d) \(\left[\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right]\)

Question 5.

The value of the determinant \(\left|\begin{array}{ccc}

6 & 0 & -1 \\

2 & 1 & 4 \\

1 & 1 & 3

\end{array}\right|\) is

(a) 10

(b) 8

(c) 7

(d) -7

Solution:

(d) -7

Let D = \(\left|\begin{array}{ccc}

6 & 0 & -1 \\

2 & 1 & 4 \\

1 & 1 & 3

\end{array}\right|\)

Expanding along R_{1}, we get

6(3 – 4) – 0(6 – 4) -1(2 – 1)

= -6 – 0 – 1 = -7

Question 9.

The function f(x) = x | x |, x ∈ R is differentiable

(a) only at x = 0

(b) only at x = 1

(c) in R

(d) in R – {0}

Solution:

(d) in R – {0}

Question 11.

The value of \(\int_0^{\pi / 4}(\sin 2 x) d x\) is

(a) 0

(b) 1

(c) \(\frac{1}{2}\)

(d) –\(\frac{1}{2}\)

Solution:

(c) \(\frac{1}{2}\)

Question 14.

A unit vector \(\hat{a}\) makes equal but acute angles on the co-ordinate axes. The projection of the vector \(\hat{a}\) on the vector \(\vec{b}\) = \(5 \hat{i}+7 \hat{j}-\hat{k}\) is

(a) \(\frac{11}{15}\)

(b) \(\frac{11}{5 \sqrt{3}}\)

(c) \(\frac{4}{5}\)

(d) \(\frac{3}{5 \sqrt{3}}\)

Solution:

(a) \(\frac{11}{15}\)

Question 18.

If A and B are two independent events such that P(A) = \(\frac{1}{3}\) and P(B) = \(\frac{1}{4}\), then P(B’ | A) is

(a) \(\frac{1}{4}\)

(b) \(\frac{1}{8}\)

(c) \(\frac{3}{4}\)

(d) 1

Solution:

(c) \(\frac{3}{4}\)

Section – B

Question 21.

Draw the graph of the principal branch of the function f(x) = cos^{-1} x.

Solution:

We have, f(x) = cos^{-1} x

Question 25.

Find the angle between the following two lines:

\(\vec{r}\) = 2\(\hat{i}\) – 5\(\hat{j}\) + \(\hat{k}\) + λ(3\(\hat{i}\) + 2\(\hat{j}\) + 6\(\hat{k}\));

\(\vec{r}\) = 7\(\hat{i}\) – 6\(\hat{k}\) + µ(\(\hat{i}\) + 2\(\hat{j}\) + 2\(\hat{k}\))

Solution:

Question 26.

Using determinants, find the area of ∆PQR with vertices P(3, 1), Q(9, 3) and R(5, 7). Also, find the equation of line PQ using determinants.

Solution:

Given. P(3, 1), Q(9, 3), R(5, 7)

∴ Area of ∆PQR = \(\frac{1}{2}\left|\begin{array}{lll}

x_1 & y_1 & 1 \\

x_2 & y_2 & 1 \\

x_3 & y_3 & 1

\end{array}\right|\) sq. units

= \(\frac{1}{2}\left|\begin{array}{lll}

3 & 1 & 1 \\

9 & 3 & 1 \\

5 & 7 & 1

\end{array}\right|\)

= \(\frac{1}{2}\)[3(3 – 7) – 1(9 – 5) + 1(63 – 15)]

= \(\frac{1}{2}\)(-12 – 4 + 48)

= 16 sq. units

∴ Equation of PQ is \(\left|\begin{array}{lll}

x & y & 1 \\

3 & 1 & 1 \\

9 & 3 & 1

\end{array}\right|\) = 0

⇒ x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0

⇒ -2x + 6y = 0

⇒ x – 3y = 0 or x = 3y

Question 28.

(a) Evaluate: \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos 2 x}{1+\cos 2 x} d x\)

Solution:

Or

(b) Find: \(\int e^{x^2}\left(x^5+2 x^3\right) d x\)

Solution:

Question 29.

Find the area of the minor segment of the circle x^{2} + y^{2} = 4 cut off by the line x = 1, using integration.

Solution:

Given. x^{2} + y^{2} = 4

⇒ y^{2} = 4 – x^{2}

∴ y = ±\(\sqrt{4-x^2}\)

Section – D

Question 32.

Evaluate: \(\int_0^\pi \frac{x}{1+\sin x} d x\)

Solution:

SET III Code No. 65/5/3

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.

Let R be a relation in the set N given by

R = {(a, b) : a = b – 2, b > 6}.

(a) (8, 7) ∈ R

(b) (6, 8) ∈ R

(c) (3, 8) ∈ R

(d) (2, 4) ∈ R

Solution:

(b) (6, 8) ∈ R

Given, a = b – 2

When b = 7, a = 7 – 2 = 5, (5, 7) ∉ R

When b = 8, a = 8 – 2 = 6, (6, 8) ∉ R

Question 2.

If A = \(\left[\begin{array}{ll}

5 & x \\

y & 0

\end{array}\right]\) and A = A^{T}, where A^{T} is the transpose of the matrix A, then 1

(a) x = 0, y = 5

(b) x = y

(c) x + y = 5

(d) x = 5, y = 0

Solution:

(b) x = y

Given. A = A^{T}

\(\left[\begin{array}{ll}

5 & x \\

y & 0

\end{array}\right]\) = \(\left[\begin{array}{ll}

5 & y \\

x & 0

\end{array}\right]\)

∴ x = y

Question 6.

If f(x) = | cos x |, then \(f\left(\frac{3 \pi}{4}\right)\) is 1

(a) 1

(b) -1

(c) \(\frac{-1}{\sqrt{2}}\)

(d) \(\frac{1}{\sqrt{2}}\)

Solution:

(d) \(\frac{1}{\sqrt{2}}\)

We have, f(x) = | cos x |

\(f\left(\frac{3 \pi}{4}\right)\) = \(\left|\cos \frac{3 \pi}{4}\right|\)

= \(\left|\cos \left(\pi-\frac{\pi}{4}\right)\right|\)

= \(\left|-\cos \frac{\pi}{4}\right|\) = \(\left|\frac{-1}{\sqrt{2}}\right|\) = \(\frac{1}{\sqrt{2}}\)

Question 9.

The function f(x) = x^{3} + 3x is increasing in interval

(a) (-∞, 0)

(b) (0, ∞)

(c) R

(d) (0, 1)

Solution:

(c) R

We have f(x) = x^{3} + 3x

f’(x) = 3x^{2} + 3

= 3(x^{2} + 1) > 0

for all x ∈ R, f’(x) > 0

Question 12.

The order and the degree of the differential equation \(\left(1+3 \frac{d y}{d x}\right)^2\) = 4\(4 \frac{d^3 y}{d x^3}\) respectively are:

(a) 1, \(\frac{2}{3}\)

(b) 3, 1

(c) 3, 3

(d) 1, 2

Solution:

(b) 3, 1

Order = 3; Degree = 1

Question 13.

If \(\vec{a} \cdot \hat{i}\) = \(\vec{a} \cdot(\hat{i}+\hat{j})\) = \(\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1, then \(\vec{a}\) is

(a) \(\hat{\boldsymbol{k}}\)

(b) \(\hat{\boldsymbol{i}}\)

(c) \(\hat{j}\)

(d) \(\hat{i}+\hat{j}+\hat{k}\)

Solution:

(b) \(\hat{\boldsymbol{i}}\)

Section – B

Question 21.

(a) Find the value of k for which the function f given as

is continuous at x = o.

Solution:

Or

(b) If x = a cos t and y = b sin t, then find \(\frac{d^2 y}{d x^2}\).

Solution:

Question 22.

Find the value of tan^{-1} [2cos(2sin^{-1}\(\frac{1}{2}\))] + tan^{-1} 1.

Solution:

Section – C

Question 26.

Show that the determinant \(\left|\begin{array}{ccc}

x & \sin \theta & \cos \theta \\

-\sin \theta & -x & 1 \\

\cos \theta & 1 & x

\end{array}\right|\) is independent of θ.

Solution:

\(\left|\begin{array}{ccc}

x & \sin \theta & \cos \theta \\

-\sin \theta & -x & 1 \\

\cos \theta & 1 & x

\end{array}\right|\)

Expanding along C_{1}

= x(-x^{2} – 1) + sin θ(x sin θ – cos θ) + cos θ(sin θ + x cos θ)

= -x^{3} – x sin^{2}θ – sin θ cos θ + sin θ cos θ + x cos^{2} θ

= – x^{3} – x + x (sin^{2}θ + cos^{2}θ)

= -x^{3} – x + x (1) ….. [∵ sin^{2}θ + cos^{2}θ = 1

= -x^{3} which is independent of θ.

Question 27.

Using integration, find the area of the region bounded by y = mx (m > 0), x = 1, x = 2 and

the x-axis.

Solution:

Question 28.

(a) Find the coordinates of the foot of the perpendicular drawn from point (5, 7, 3) to the line \(\frac{x-15}{3}\) = \(\frac{y-29}{8}\) = \(\frac{z-5}{-5}\).

Solution:

Given line \(\frac{x-15}{3}\) = \(\frac{y-29}{8}\) = \(\frac{z-5}{-5}\) = q(let)

General point on given line

M(3q + 15, 8q + 29, -5q + 5) ……..(i)

DR’s of PM are

3q + 15 – 5; 8q + 29 – 7; -5q + 5 – 3

3q + 10; 8q + 22; -5q + 2

DR’s of given line are 3, 8, -5

Using a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

[∴ PM ⊥ given line

3(3q + 10) + 8(8q + 22) – 5(-5q + 2) = 0

⇒ 9q + 30 + 64q + 176 + 25q – 10 = 0

⇒ 98q = -196

∴ q = \(\frac{-196}{98}\) = -2

Putting the value of q in (1),

Point M (-6 + 15, -16 + 29, 10 + 5)

∴ Required foot of perpendicular = (9, 13, 15)

Or

(b) If \(\vec{a}\) = \(\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}\) = \(\hat{i}+2 \hat{j}+3 \hat{k}\) then find a unit vector perpendicular to both \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\).

Solution:

Section – D

Question 32.

Solve the following Linear Programming Problem graphically:

Minimise : Z = 60x + 80y

subject to constraints:

3x + 4y ≥ 8

5x + 2y ≥ 11

x, y ≥ 0

Solution:

Hence the feasible region is unbounded.

Now, 60x + 80y < 160

Let 3x + 4y < 8

to check whether the resulting open half plane has any point common with the feasible region.

On checking, It has no points in common.

Minimum value Z = 160 at all the points on the line segment joining the points B(\(\frac{8}{3}\), 0) and C(2, \(\frac{1}{2}\)).