MATHEMATICS – Code No.

041
SAMPLE QUESTION PAPER
CLASS - XII (2025-26)

Maximum Marks: 80 Time: 3 hours

General Instructions:
Read the following instructions very carefully and strictly follow them:
1. This Question paper contains 38 questions. All questions are compulsory.
2. This Question paper is divided into five Sections - A, B, C, D and E.
3. In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs) with only one
correct option and Questions no. 19 and 20 are Assertion-Reason based questions of 1
mark each.
4. In Section B, Questions no. 21 to 25 are Very Short Answer (VSA)-type questions, carrying
2 marks each.
5. In Section C, Questions no. 26 to 31 are Short Answer (SA)-type questions, carrying 3
marks each.
6. In Section D, Questions no. 32 to 35 are Long Answer (LA)-type questions, carrying 5
marks each.
7. In Section E, Questions no. 36 to 38 are Case study-based 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 one subpart each in 2
questions of Section E.
9. Use of calculator is not allowed.

SECTION-A
This section comprises of multiple choice questions (MCQs) of 1 mark each.
Select the correct option (Question 1 - Question 18)

Q.No. Questions Marks

1. Identify the function shown in the graph 1

𝑥
(A) sin−1 𝑥 (B) sin−1(2𝑥) (C) sin−1 (2) (D) 2 sin−1 𝑥

For Visually Impaired:

1 1
1. Inverse Trigonometric Function, whose domain is [− 3 , 3] , is …
𝑥
(A) cos−1 𝑥 (B) cos −1 (3)
(C) cos−1 (3𝑥) (D) 3 cos−1 𝑥
*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 1 of 10
 2. If for three matrices 𝐴 = [𝑎𝑖𝑗 ]𝑚×4 , B = [𝑏𝑖𝑗 ]𝑛×3 𝑎𝑛𝑑 C = [𝑐𝑖𝑗 ]𝑝×𝑞 products 𝐴𝐵 and 1
𝐴𝐶 both are defined and are square matrices of same order, then value of 𝑚, 𝑛, 𝑝
and 𝑞 are:

(A) 𝑚 = 𝑞 = 3 𝑎𝑛𝑑 𝑛 = 𝑝 = 4 (B) 𝑚 = 2, 𝑞 = 3 𝑎𝑛𝑑 𝑛 = 𝑝 = 4

(C) 𝑚 = 𝑞 = 4 𝑎𝑛𝑑 𝑛 = 𝑝 = 3 (D) 𝑚 = 4, 𝑝 = 2 𝑎𝑛𝑑 𝑛 = 𝑞 = 3
3. 0 𝑟 −2 1
𝑞+𝑡
If the matrix 𝐴 = [3 𝑝 𝑡 ] is skew-symmetric, then value of is….
𝑝+𝑟
𝑞 −4 0
(A)−2 (B) 0 (C) 1 (D) 2

4. If 𝐴 is a square matrix of order 4 and |𝑎𝑑𝑗 𝐴| = 27, then 𝐴 (𝑎𝑑𝑗 𝐴) is equal to 1

(A) 3 (B) 9 (C) 3 𝐼 (D) 9 𝐼

5. 3 0 0 1
The inverse of the matrix [0 2 0] is…
0 0 5
1
0 0
0 0 3 3
1
(A) [0 2 0] (B) 0 2
0
5 0 0 1
[0 0 5]
1
−3 0 0
−3 0 0
1
(C) 0 −2 0 (D) [ 0 −2 0 ]
1 0 0 −5
[0 0 − 5]

6. cos 67𝑜 sin 67𝑜 1

Value of the determinant | | is
sin 23𝑜 cos 23𝑜
1 √3
(A) 0 (B) (C) (D) 1
2 2

7. If a function defined by 𝑓(𝑥) = {

𝑘𝑥 + 1, 𝑥 ≤ 𝜋 1
cos 𝑥 , 𝑥 > 𝜋
is continuous at 𝑥 = 𝜋, then the value of 𝑘 is
−1 −2
(A) 𝜋 (B) (C) 0 (D)
𝜋 𝜋

8. If 𝑓(𝑥) = 𝑥 tan−1 𝑥 , then 𝑓 ′ (1)is equal to 1

𝜋 1 𝜋 1 𝜋 1 𝜋 1
(A) 4 − (B) 4 + (C) − − (D) − 4 +
2 2 4 2 2

9. A function 𝑓(𝑥) = 10 − 𝑥 − 2𝑥 2 is increasing on the interval 1

1 1 1 1 1
(A) (−∞, − 4] (B) (−∞, 4) (C) [− 4 , ∞) (D)[− 4 , 4]

10. The solution of the differential equation 𝑥𝑑𝑥 + 𝑦𝑑𝑦 = 0 represents a family of 1
(A) straight lines (B) parabolas (C) Circles (D) Ellipses

*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 2 of 10
 11. 𝑏
If 𝑓(𝑎 + 𝑏 − 𝑥) = 𝑓(𝑥), then ∫𝑎 𝑥 𝑓(𝑥)𝑑𝑥 is equal to 1
𝑎+𝑏 𝑏 𝑎+𝑏 𝑏
(A) ∫𝑎 𝑓(𝑏 − 𝑥)𝑑𝑥 (B) ∫𝑎 𝑓(𝑎 − 𝑥)𝑑𝑥
2 2
𝑏−𝑎 𝑏 𝑎+𝑏 𝑏
(C) ∫𝑎 𝑓(𝑥) 𝑑𝑥 (D) ∫𝑎 𝑓(𝑥)𝑑𝑥
2 2

12. If ∫ 𝑥 3 𝑠𝑖𝑛4 (𝑥 4 ) cos(𝑥 4 ) 𝑑𝑥 = 𝑎 𝑠𝑖𝑛5 (𝑥 4 ) + C, then 𝑎 is equal to 1

1 1 1 1
(A) − (B) (C) (D)
10 20 4 5

13. A bird flies through a distance in a straight line given by the vector 𝑖̂ + 2𝑗̂ + 𝑘̂ . A 1
man standing beside a straight metro rail track given by 𝑟⃗ = (3 + λ)𝑖̂ + (2λ −
1)𝑗̂ + 3λ𝑘̂ is observing the bird. The projected length of its flight on the metro
track is
6 14 8 5
(A) units (B) units (C) units (D) units
√14 √6 √14 √6

14. The distance of the point with position vector 3𝑖̂ + 4𝑗̂ + 5𝑘̂ from the y-axis is 1

(A) 4 units (B) √34 units (C) 5 units (D) 5√2 units

15. If 𝑎⃗ = 3𝑖̂ + 2𝑗̂ + 4𝑘̂ , 𝑏⃗⃗ = 𝑖̂ + 𝑗̂ − 3𝑘̂ and 𝑐⃗ = 6𝑖̂ − 𝑗̂ + 2𝑘̂ are three given vectors, 1
then (2𝑎⃗. 𝑖̂)𝑖̂ − (𝑏⃗⃗. 𝑗̂)𝑗̂ + (𝑐⃗. 𝑘̂)𝑘̂ is same as the vector

(A) 𝑎⃗ (B) 𝑏⃗⃗ + 𝑐⃗ (C) 𝑎⃗ − 𝑏⃗⃗ (D) 𝑐⃗

16. A student of class XII studying Mathematics comes across an incomplete 1

question in a book.

Maximise 𝑍 = 3𝑥 + 2𝑦 + 1
Subject to the constraints 𝑥 ≥ 0, 𝑦 ≥ 0, 3𝑥 + 4𝑦 ≤ 12,

He/ She notices the below shown graph for the said LPP problem, and finds that
a constraint is missing in it:
Help him/her choose the required constraint from the graph.

The missing constraint is

(A) 𝑥 + 2𝑦 ≤ 2 (B) 2𝑥 + 𝑦 ≥ 2
(C) 2𝑥 + 𝑦 ≤ 2 (D) 𝑥 + 2𝑦 ≥ 2
*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 3 of 10
 16. For Visually Impaired:

If 𝑍 = 𝑎𝑥 + 𝑏𝑦 + 𝑐, where 𝑎, 𝑏, 𝑐 > 0, attains its maximum value at two of its

corner points (4,0) and (0,3) of the feasible region determined by the system of
linear inequalities, then

(A) 4𝑎 = 3𝑏 (B) 3𝑎 = 4𝑏 (C) 4𝑎 + 𝑐 = 3𝑏 (D) 3𝑎 + 𝑐 = 4𝑏

17. The feasible region of a linear programming problem is bounded but the 1
objective function attains its minimum value at more than one point. One of the
points is (5,0).

Then one of the other possible points at which the objective function attains its
minimum value is

(A) (2,9) (B) (6,6) (C) (4,7) (D) (0,0)

For Visually Impaired:

The graph of the inequality 3𝑥 + 5𝑦 < 10 is the

(A) Entire 𝑋𝑌 −plane

(B) Open Half plane that doesn’t contain origin
(C) Open Half plane that contains origin, but not the points of the line 3𝑥 +
5𝑦 = 10
(D) Half plane that contains origin and the points of the line 3𝑥 + 5𝑦 = 10

18. A person observed the first 4 digits of your 6-digit PIN. What is the probability 1
that the person can guess your PIN?

1 1 1
(A) (B) (C) (D) 1
81 100 90

*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 4 of 10
 ASSERTION-REASON BASED QUESTIONS

(Question numbers 19 and 20 are Assertion-Reason based questions

carrying 1 mark each. Two statements are given, one labelled Assertion (A)
and the other labelled Reason (R). Select the correct answer from the
options (A), (B), (C) and (D) as given below.)

(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.

19. √3
Assertion (A): Value of the expression sin−1 ( 2 ) + tan−1 1 − sec −1 (√2) is 4 .
𝜋 1
𝜋 𝜋
Reason (R): Principal value branch of sin−1 𝑥 is [− 2 , 2 ] and that of s𝑒𝑐 −1 𝑥
𝜋
is [0, 𝜋] − { 2 }.

20. Assertion(A): Given two non-zero vectors 𝑎⃗ and 𝑏⃗⃗ . If 𝑟⃗ is another non-zero 1
vector such that 𝑟⃗ × (𝑎⃗ + 𝑏⃗⃗) = ⃗⃗
0 . Then 𝑟⃗ is perpendicular to 𝑎⃗ × 𝑏⃗⃗ .
Reason (R): The vector (𝑎⃗ + 𝑏⃗⃗) is perpendicular to the plane of 𝑎⃗ and 𝑏⃗⃗

SECTION B
This section comprises of 5 very short answer (VSA) type questions of 2 marks each.

21A Evaluate tan (tan−1(−1) + 3 )

𝜋 2

OR

21B Find the domain of cos −1 (3𝑥 − 2)

22 𝜋 𝑥
If 𝑦 = log tan ( 4 + 2), then prove that
𝒅𝒚
− 𝐬𝐞𝐜 𝒙 = 𝟎 2
𝒅𝒙

23A (𝑥−3)
Find: ∫ (𝑥−1)3 𝑒 𝑥 𝑑𝑥 2

OR
23B Find out the area of shaded region in the enclosed figure.

*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 5 of 10
23 B For Visually Impaired:

Find out the area of the region enclosed by the curve 𝑦 2 = 𝑥 , 𝑥 = 3 and 𝑥-axis
in the first quadrant.

24. If 𝑓(𝑥 + 𝑦) = 𝑓(𝑥)𝑓(𝑦) for all 𝑥 , 𝑦 ∈ R and 𝑓(5)= 2, 𝑓 ′ (0) = 3, then using the 2
definition of derivatives, find 𝑓 ′ (5).

25. The two vectors 𝑖̂ + 𝑗̂ + 𝑘̂ and 3𝑖

̂ − 𝑗̂ + 3𝑘̂ represent the two sides 𝑂𝐴 and 𝑂𝐵, 2
respectively of a ∆𝑂𝐴𝐵, where 𝑂 is the origin. The point 𝑃 lies on 𝐴𝐵 such that
𝑂𝑃 is a median. Find the area of the parallelogram formed by the two adjacent
sides as 𝑂𝐴 and 𝑂𝑃.

SECTION C
This section comprises of 6 short answer (SA) type questions of 3 marks each.

26A. 𝑑𝑦 log 𝑥
If 𝑥 𝑦 = 𝑒 𝑥−𝑦 prove that 𝑑𝑥 = (log(𝑥𝑒))2 and hence find its value at 𝑥 = 𝑒. 3

OR
𝑑2 𝑦
26B. If 𝑥 = 𝑎(𝜃 − sin 𝜃 ), 𝑦 = 𝑎 (1 − cos 𝜃 ) find .
𝑑𝑥 2

27 A spherical ball of ice melts in such a way that the rate at which its volume 3
decreases at any instant is directly proportional to its surface area. Prove that
the radius of the ice ball decreases at a constant rate.

2
28A Sketch the graph 𝑦 = |𝑥 + 1| . Evaluate ∫−4|𝑥 + 1|𝑑𝑥. What does the value of this 3
integral represent on the graph?
OR
28B Using integration find the area of the region {(𝑥, 𝑦) ∶ 𝑥 2 − 4𝑦 ≤ 0, 𝑦 − 𝑥 ≤ 0}

For Visually Impaired:

2
28A Define the function 𝑦 = |𝑥 + 1| . Evaluate ∫−4|𝑥 + 1|𝑑𝑥. What does the value of
this integral represent?
OR
28B Using integration find the area enclosed within the curve: 25𝑥 2 + 16𝑦 2 = 400

29A Find the distance of the point (2, −1,3) from the line 3
𝑟⃗ = (2𝑖̂ − 𝑗̂ + 2𝑘̂) + 𝜇(3𝑖̂ + 6𝑗̂ + 2𝑘̂ )
measured parallel to the z-axis.
OR

29B Find the point of intersection of the line 𝑟⃗ = (3𝑖̂ + 𝑘̂) + 𝜇(𝑖̂ + 𝑗̂ + 𝑘̂) and the line
through (2, −1,1) parallel to the z-axis. How far is this point from the z-axis?

*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 6 of 10
 30. Solve graphically: 3
Maximise 𝑍 = 2𝑥 + 𝑦 subject to
𝑥 + 𝑦 ≤ 1200
𝑥 + 𝑦 ≥ 600
𝑥
𝑦≤ 2
𝑥 ≥ 0, 𝑦 ≥ 0 .

30 For Visually Impaired:

The objective function 𝑍 = 3𝑥 + 2𝑦 of a linear programming problem under

some constraints is to be maximized and minimized. The corner points of the
feasible region are 𝐴(600,0), 𝐵(1200,0), 𝐶(800, 400) and 𝐷(400, 200). Find the
point at which 𝑍 is maximum and the point at which 𝑍 is minimum. Also, find the
corresponding maximum and minimum values of 𝑍.)

31. Two students Mehul and Rashi are seeking admission in a college. The 3
probability that Mehul is selected is 0.4 and the probability of selection of exactly
one of the them is 0.5. Chances of selection of them is independent of each
other. Find the chances of selection of Rashi. Also find the probability of selection
of at least one of them.

SECTION D
This section comprises of 4 long answer (LA) type questions of 5 marks each

32. 3 −6 −1 1 −2 −1 5
For two matrices 𝐴 = [ 2 −5 −1] and 𝐵 = [0 −1 −1], find the product 𝐴𝐵
−2 4 1 2 0 3
and hence solve the system of equations:

3𝑥 − 6𝑦 − 𝑧 = 3
2𝑥 − 5𝑦 − 𝑧 + 2 = 0
−2𝑥 + 4𝑦 + 𝑧 = 5

33A Evaluate:
1 log(1+𝑥)
∫0 𝑑𝑥 5
1+𝑥 2

OR
33B (3 sin 𝜃−2) cos 𝜃
Find ∫ 5−𝑐𝑜𝑠2 𝜃−4 sin 𝜃 𝑑𝜃

34A 𝑑
Solve the differential equation: 𝑦 + 𝑑𝑥 ( 𝑥𝑦 ) = 𝑥 (sin 𝑥 + 𝑥) 5

OR

Find the particular solution of the differential equation:

34B
𝑥⁄ 𝑥⁄
2𝑦𝑒 𝑦 𝑑𝑥 +(𝑦−2𝑥𝑒 𝑦 ) 𝑑𝑦 = 0 given that 𝑦(0) = 1

*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 7 of 10
 35. The two lines
𝑥−1
= −𝑦 , 𝑧 + 1 = 0 and
−𝑥
=
𝑦+1
= 𝑧 + 2 intersect at a point 5
3 2 2
whose y-coordinate is 1. Find the co-ordinates of their point of intersection. Find
the vector equation of a line perpendicular to both the given lines and passing
through this point of intersection.

SECTION- E

This section comprises of 3 case-study/passage-based questions of 4 marks each with

subparts. The first two case study questions have three subparts (I), (II), (III) of marks 1,
1, 2 respectively. The third case study question has two subparts of 2 marks each

36. Case Study -1 4

A city’s traffic management department is planning to optimize traffic flow by
analyzing the connectivity between various traffic signals. The city has five major
spots labelled 𝐴, 𝐵, 𝐶, 𝐷, 𝑎𝑛𝑑 𝐸.

The department has collected the following data regarding one-way traffic flow
between spots:

1. Traffic flows from 𝐴 to 𝐵, 𝐴 to 𝐶, and 𝐴 to 𝐷.

2. Traffic flows from 𝐵 to 𝐶 and 𝐵 to 𝐸.
3. Traffic flows from 𝐶 to 𝐸.
4. Traffic flows from 𝐷 to 𝐸 and 𝐷 to 𝐶.

The department wants to represent and analyze this data using relations and
functions. Use the given data to answer the following questions:

I. Is the traffic flow reflexive? Justify. [1]

II. Is the traffic flow transitive? Justify. [1]
III A. Represent the relation describing the traffic flow as a set of ordered pairs.
Also state the domain and range of the relation.
OR
III B. Does the traffic flow represent a function? Justify your answer. [2]

*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 8 of 10
 37. Case Study -2 4

LED bulbs are energy-efficient because they use significantly less electricity than
traditional bulbs while producing the same amount of light. They convert more
energy into light rather than heat, reducing waste. Additionally, their long lifespan
means fewer replacements, saving resources and money over time.

A company manufactures a new type of energy-efficient LED bulb. The cost of

production and the revenue generated by selling x bulbs (in an hour) are
modelled as

𝐶(𝑥) = 0.5𝑥 2 − 10𝑥 + 150 and 𝑅(𝑥) = −0.3𝑥 2 + 20𝑥 respectively, where
𝐶(𝑥) and 𝑅(𝑥) are both in ₹.

To maximize the profit, the company needs to analyze these functions using
calculus. Use the given models to answer the following questions:

I. Derive the profit function 𝑃(𝑥) [1]

II. Find the critical points of 𝑃(𝑥). [1]
III A. Determine whether the critical points correspond to a maximum or a
minimum profit by using the second derivative test.

OR
III B. Identify the possible practical value of 𝑥 (i.e., the number of bulbs that can
realistically be produced and sold) that can maximize the profit, if the
resources available and the expenditure on machines allows to produce
minimum 10 but not more than 18 bulbs per hour. Also calculate the
maximum profit. [2]

*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 9 of 10
 38. Case Study -3 4

Excessive use of screens can result in vision problems, obesity, sleep disorders,
anxiety, low retention problems and can impede social and emotional
comprehension and expression. It is essential to be mindful of the amount of
time we spend on screens and to reduce our screen-time by taking regular
breaks, setting time limits, and engaging in non-screen-based activities.

In a class of students of the age group 14 to 17, the students were categorised
into three groups according to a feedback form filled by them. The first group
constituted of the students who spent more than 4 hours per day on the mobile
screen or the gaming screens, while the second group spent 2 to 4 hours /day
on the same activities. The third group spent less than 2 hours /day on the same.
The first group with the high screen time is 60% of all the students, whereas the
second group with moderate screen time is 30% and the third group with low
screen time is only 10% of the total number of students. It was observed that
80% students of first group faced severe anxiety and low retention issues, with
70% of second group, and 30% of third group having the same symptoms.

I. What is the total percentage of students who suffer from anxiety and low
retention issues in the class? [2]

II. A student is selected at random, and he is found to suffer from anxiety

and low retention issues. What is the probability that he/she spends
screen time more than 4 hours per day? [2]

*Please note that the assessment scheme of the Academic Session 2024-25 will continue in the current session i.e.
2025-26. Page 10 of 10
 MATHEMATICS – Code No. 041
MARKING SCHEME
CLASS – XII (2025-26)

SECTION-A (MCQs of 1 mark each)

Sol. N. Hint / Solution Marks

1 1
Clearly from the graph Domain is [− 2 , 2]
1 So graph is of the function sin−1(2𝑥) 1
Answer is (B) sin−1(2𝑥)
1 1
Domain is [− 3 , 3]
1
So the function is cos −1 (3𝑥) 1
(V.I.)
Answer is (C) cos−1 (3𝑥)

AB is defined so n=4
AC is defined so p=4
2 AB and AC are square matrices of same order 1
so 𝑚 × 3 = 𝑚 × 𝑞 ⇒ 𝑞 = 3 = 𝑚
Answer is (A) 𝑚 = 𝑞 = 3 𝑎𝑛𝑑 𝑛 = 𝑝 = 4

As A is skew symmetric
So 𝑝 = 0, 𝑞 = 2, 𝑟 = −3, 𝑡 = 4
𝑞+𝑡 6
3 So 𝑝+𝑟 = −3 = −2 1
Answer is (A) -2

|𝑎𝑑𝑗 𝐴| = 27 ⇒ |𝐴|3 = 27 = 33 ⇒ |𝐴| = 3

4 𝐴 (𝑎𝑑𝑗 𝐴) = |𝐴| 𝐼 = 3 𝐼 1
Answer is (C) 𝟑 𝑰
1
0 0
3 0 0 3
1
Inverse of the matrix [0 2 0] = 0 2 0
5 0 0 5 1 1
[0 0 5]
Answer is (B)

cos 67𝑜 sin 67𝑜

| | = cos 67𝑜 cos 23𝑜 − sin 67𝑜 sin 23𝑜 = cos(67𝑜 + 23𝑜 ) =
sin 23𝑜 cos 23𝑜
6 cos 90𝑜 = 0 1
Answer is (A) 0

𝑓(𝑥) is continuous at 𝑥 = 𝜋
⇒ lim−(𝑘𝑥 + 1) = lim+ 𝑐𝑜𝑠 𝑥 = 𝑓(𝜋)
𝑥→𝜋 𝑥→𝜋
⇒ lim [𝑘(𝜋 − ℎ) + 1] = lim 𝑐𝑜𝑠( 𝜋 + ℎ ) = 𝑘𝜋 + 1
7 ℎ→0 ℎ→0 1
−2
⇒ 𝑘𝜋 + 1 = −1 ⇒𝑘 = 𝜋
−𝟐
Answer is (D) 𝝅
Page 1 of 12
 𝑓(𝑥) = 𝑥 tan−1 𝑥
1
𝑓 ′ (𝑥) = 1. tan−1 𝑥 + 𝑥 . 1+𝑥 2
8 1 𝜋 1 1
𝑓 ′ (1) = 1 . tan−1 1 + 1+1 = +2
4
𝝅 𝟏
Answer is (B) 𝟒 + 𝟐

𝑓(𝑥) = 10 − 𝑥 − 2𝑥 2
⇒ 𝑓 ′ (𝑥) = −1 − 4𝑥
For increasing function 𝑓 ′ (𝑥) ≥ 0
⇒ − ( 1 + 4𝑥) ≥ 0
9 ⇒ ( 1 + 4𝑥) ≤ 0 1
⇒ 𝑥 ≤ − 1⁄4
1
⇒ 𝑥 ∈ (−∞, − 4]
𝟏
Answer is (A) (−∞, − ]
𝟒

𝑥𝑑𝑥 + 𝑦𝑑𝑦 = 0
⇒ ∫ 𝑥𝑑𝑥 = − ∫ 𝑦𝑑𝑦
𝑥2 𝑦2
⇒ =− +𝑘
2 2
10 2 2 1
⇒ 𝑥 + 𝑦 = 2𝑘
Solution is 𝑥 2 + 𝑦 2 = 2𝑘, 𝑘 being an arbitrary constant.
Answer is (C) Circles

𝑏 𝑏
𝐼 = ∫𝑎 𝑥 𝑓(𝑥)𝑑𝑥 = ∫𝑎 (𝑎 + 𝑏 − 𝑥)𝑓(𝑎 + 𝑏 − 𝑥)𝑑𝑥
𝑏
⇒ 𝐼 = ∫𝑎 (𝑎 + 𝑏 − 𝑥)𝑓(𝑥)𝑑𝑥 (given 𝑓(𝑎 + 𝑏 − 𝑥) = 𝑓(𝑥) )
𝑏 𝑏
⇒ 𝐼 = ∫𝑎 (𝑎 + 𝑏) 𝑓(𝑥)𝑑𝑥 − ∫𝑎 𝑥 𝑓(𝑥)𝑑𝑥
11 𝑏
⇒ 2 𝐼 = (𝑎 + 𝑏) ∫𝑎 𝑓(𝑥)𝑑𝑥 1
1 𝑏
⇒ 𝐼 = 2 (𝑎 + 𝑏) ∫𝑎 𝑓(𝑥)𝑑𝑥
𝒂+𝒃 𝒃
Answer is (D) 𝟐 ∫𝒂 𝒇(𝒙)𝒅𝒙

Let 𝐼 = ∫ 𝑥 3 𝑠𝑖𝑛4 (𝑥 4 ) cos(𝑥 4 ) 𝑑𝑥

1
Let sin(𝑥 4 ) = 𝑡 ⇒ 4𝑥 3 cos(𝑥 4 ) 𝑑𝑥 = 𝑑𝑡 ⇒ 𝑥 3 cos(𝑥 4 ) = 4 𝑑𝑡
1 1 1
Thus 𝐼 = ∫ 𝑡 4 ( 4 𝑑𝑡 ) = 20 𝑡 5 + C = 20 𝑠𝑖𝑛5 (𝑥 4 ) + C
12 1 1
⇒ 𝐼 = 20 𝑠𝑖𝑛5 (𝑥 4 ) + C = 𝑎 𝑠𝑖𝑛5 (𝑥 4 ) + C
1
So, 𝑎 = 20
𝟏
Answer is (B) 𝟐𝟎

The projection of the vector 𝑖̂ + 2𝑗̂ + 𝑘̂ on the line

1x1 +2x2+1x3 8
𝑟⃗ = (3𝑖̂ − 𝑗̂) + λ(𝑖̂ + 2𝑗̂ + 3𝑘̂ ) is = units
13 √12 2 2 +2 +3 √14 1
𝟖
Answer is (C) units
√𝟏𝟒

Page 2 of 12
 The distance of the point (a, b, c) from the y-axis is √𝑎2 + 𝑐 2
14 So, the distance is √32 + 52 = √34 units. 1
Answer is (B) √𝟑𝟒 units

(2𝑎⃗. 𝑖̂)𝑖̂ − (𝑏⃗⃗. 𝑗̂)𝑗̂ + (𝑐⃗. 𝑘̂)𝑘̂ = (2 × 3)𝑖̂ − (1)𝑗̂ + (2)𝑘̂

15 = 6𝑖̂ − 𝑗̂ + 2𝑘̂ = 𝑐⃗ 1
Answer is (D) 𝒄
⃗⃗

The points (1,0) and (0,2) satisfy the equation 2𝑥 + 𝑦 = 2

And shaded region shows that (0,0) doesn’t lie in the feasible solution region
16 So, the inequality is 2𝑥 + 𝑦 ≥ 2 1
Answer is (B) 𝟐𝒙 + 𝒚 ≥ 𝟐

(4,0) and (0,3) gives maximum value so

16 𝑍(4,0) = 𝑍(0,3) ⇒ 4𝑎 + 𝑐 = 3𝑏 + 𝑐 ⇒ 4𝑎 = 3𝑏 1
(V.I.) Answer is (A) 𝟒𝒂 = 𝟑𝒃
The student may read the point (2,9) from the line on the graph.
The student may find the equation 3𝑥 + 𝑦 = 15 joining (5,0) and (0,15) and
17 then verify the point (2,9) satisfies it. 1
Answer is (A) (2,9)

17 Answer is (C) Open Half plane that contains origin, but not the points of the
line 3𝑥 + 5𝑦 = 10 1
(V.I.)
𝟏
Answer is (B) 𝟏𝟎𝟎
The person knows the first 4 digits. So the person has to guess the remaining
18 two digits. 1
1 1 1
P (guessing the PIN )=1×1×1×1×10 × 10 = 100

√3 𝜋 𝜋 𝜋 𝜋 𝜋
sin−1 ( 2 ) + tan−1 1 − sec −1(√2) = 3 + 4 − 4 = ≠
3 4
So, A is false.
𝜋 𝜋 𝜋
19 Principal Value branch of sin−1 𝑥 is [− 2 , 2 ] and that of s𝑒𝑐 −1 𝑥 is [0, 𝜋] − {2 }. 1
So, R is true
Answer is (D)Assertion is false, but Reason is true

C. 𝑟⃗ × (𝑎⃗ + 𝑏⃗⃗) = ⃗0⃗ ⇒ 𝑟⃗ is parallel to (𝑎⃗ + 𝑏⃗⃗) and (𝑎⃗ + 𝑏⃗⃗) lies on the plane
of 𝑎⃗ and 𝑏⃗⃗ .
So, 𝑟⃗ is parallel to the plane of 𝑎⃗ and 𝑏⃗⃗ ⇒ 𝑟⃗ is perpendicular to (𝑎⃗ × 𝑏⃗⃗).
20 So, Assertion is true 1
But (𝑎⃗ + 𝑏⃗⃗) lies on the plane of 𝑎⃗ and 𝑏⃗⃗, so (𝑎⃗ + 𝑏⃗⃗) is not perpendicular to
the plane of 𝑎⃗ 𝑎𝑛𝑑 𝑏⃗⃗
Therefore, Reason is false.
Answer is (C) Assertion is true, but Reason is false

Page 3 of 12
 SECTION B
(VSA type questions of 2 marks each)

21A 𝜋
tan (tan−1(−1) + 3 ) = tan (− 4 + 3 )
𝜋 𝜋 ½
𝜋 𝜋
tan −tan
= 3
𝜋
4
𝜋 1
1+tan tan
3 4
√3−1
= 1+√3 𝑜𝑟 2 − √3 ½

OR OR

21B For domain, −1 ≤ 3𝑥 − 2 ≤ 1 ½

⇒ 1 ≤ 3𝑥 ≤ 3 ½
1 ½
⇒3≤𝑥≤1
1
So, domain of cos−1 (3𝑥 − 2) is [3 , 1] ½

22 𝑦 = log tan ( 4 + 2)
𝜋 𝑥

Differentiating with respect to 𝑥

𝑑𝑦 1 2 𝜋 𝑥 1
= 𝜋 𝑥 .𝑠𝑒𝑐 ( + ) .
½
𝑑𝑥 tan( + ) 4 2 2
4 2
𝜋 𝑥
cos( + ) 1 1
4 2
= 𝜋 𝑥 . 𝜋 𝑥 .2
sin( + ) 𝑐𝑜𝑠2 ( + )
4 2 4 2
1 1 1
= 𝜋 𝑥 𝜋 𝑥 = 𝜋 = cos 𝑥
2 sin( + )cos( + ))
4 2 4 2
sin( +𝑥)
2 1
𝑑𝑦
⟹ 𝑑𝑥 − sec 𝑥 = 0
½

23A (𝑥−3)𝑒 𝑥 (𝑥−1−2)𝑒 𝑥

∫ (𝑥−1)3
𝑑𝑥 = ∫ (𝑥−1)3
𝑑𝑥

1 2 1 𝑑 1
= ∫ ( (𝑥−1)2 − (𝑥−1)3 ) 𝑒 𝑥 𝑑𝑥 = ∫ ((𝑥−1)2 + 𝑑𝑥 ((𝑥−1)2 )) 𝑒 𝑥 𝑑𝑥 1
𝑒𝑥
= (𝑥−1)2
+c (as ∫(𝑓(𝑥) + 𝑓 ′ (𝑥))𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 𝑓(𝑥) + c) 1

OR OR
4 4
23B A = ∫0 𝑥 𝑑𝑦 = ∫0 √𝑦 𝑑𝑦
2 3⁄ 𝑦=4 16 1
=3×𝑦 2]
𝑦=0
= 3 sq. units
1
23B For Visually Impaired:
3 3
A = ∫0 𝑦 𝑑𝑥 = ∫0 √𝑥 𝑑𝑥 1
2 3⁄ 𝑥=3 1
=3×𝑥 2]
𝑥=0
= 2√3 sq. units

Page 4 of 12
24 Given 𝑓(𝑥 + 𝑦) = 𝑓(𝑥)𝑓(𝑦)
𝑓(0 + 5) = 𝑓(0)𝑓(5)
⟹ 𝑓(0) = 1 ½
′ (5) 𝑓 (5+ℎ)−𝑓(5) 𝑓 (5)𝑓(ℎ)−𝑓(5)
𝑓 = lim = lim [ ∵ 𝑓(𝑥 + 𝑦) = 𝑓(𝑥)𝑓(𝑦 ) ]
ℎ→0 ℎ ℎ→0 ℎ
2𝑓(ℎ)−2
= lim [ ∵ 𝑓(5) = 2] 1
ℎ→0 ℎ
𝑓(ℎ)−1 𝑓(ℎ)−𝑓(0)
= 2 lim = 2 lim = 2 𝑓 ′ (0)
ℎ→0 ℎ ℎ→0 ℎ
= 2 (3) [∵ 𝑓 ′ (0) = 3] ½
=6

25 1
𝑂𝑃 = 2 (4𝑖̂ + 4𝑘̂) = 2𝑖̂ + 2𝑘̂
The vector ⃗⃗⃗⃗⃗⃗ ½
Area of the parallelogram formed by the two adjacent sides as OA and OP
𝑖̂ 𝑗̂ 𝑘̂ ½
⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗
= |(𝑂𝐴 X 𝑂𝑃) | = ||1 1 1||
2 0 2
= |2𝑖̂ − 2𝑘̂ | ½
= 2√2 square units. ½

SECTION C
(SA type questions of 3 marks each)

26A 𝑥 𝑦 = 𝑒 𝑥−𝑦
Taking log of both sides
𝑦 log 𝑥 = (𝑥 − 𝑦) log 𝑒
𝑦 log 𝑥 + y = 𝑥 ( since log 𝑒 = 1)
𝑥
⟹ 𝑦 = 1+log 𝑥 1
Differentiating with respect to 𝑥
1
𝑑𝑦 (1+log 𝑥) .1− 𝑥 .
𝑥
= (1+log 𝑥)2
𝑑𝑥
log 𝑥
= (log 𝑒+log 𝑥)2
log 𝑥 1
= (log(𝑥 𝑒))2
𝑑𝑦 log 𝑒 1 1 1
Now ] = = (2𝑙𝑜𝑔𝑒)2 = 22 = (as log 𝑒 = 1) 1
𝑑𝑥 𝑥=𝑒 (log 𝑒 2 )2 4

Alternative Solution:

𝑥 𝑦 = 𝑒 𝑥−𝑦
Taking log of both sides
𝑦 log 𝑥 = (𝑥 − 𝑦) log 𝑒
𝑦 log 𝑥 + y = 𝑥 ( since log 𝑒 = 1)
Differentiating both sides w.r.t. x
𝑑𝑦 𝑦 𝑑𝑦
log 𝑥 𝑑𝑥 + 𝑥 + 𝑑𝑥 = 1
𝑑𝑦 𝑦
⇒ 𝑑𝑥 (1 + log 𝑥) = 1 − 𝑥
𝑥
𝑑𝑦 𝑥−𝑦 𝑥− 𝑥(1+log 𝑥)−𝑥 𝑥(1+log 𝑥−1) log 𝑥
1+log 𝑥
⇒ 𝑑𝑥 = 𝑥(1+log 𝑥) = 𝑥(1+log 𝑥) = = 𝑥(log 𝑒+log 𝑥)2 = (log(𝑥𝑒))2
𝑥(1+log 𝑥)2
𝑑𝑦 log 𝑒 1 1 1
Now ] = (log 𝑒 2 )2 = (2𝑙𝑜𝑔𝑒)2
= 22 = 4 (as log 𝑒 = 1)
𝑑𝑥 𝑥=𝑒

Page 5 of 12
 OR OR
𝑑𝑥 𝑑𝑦
26B = 𝑎(1 − cos 𝜃 ) , 𝑑𝜃 = 𝑎(0 + sin 𝜃), 1
𝑑𝜃
𝑑𝑦
𝑑𝑦 𝑎 sin 𝜃
⇒ 𝑑𝑥 = 𝑑𝜃
𝑑𝑥 = 𝑎(1−cos 𝜃 )
𝑑𝜃
𝜃
2 sin( ) cos( )
𝜃 1
𝜃
= 2
𝜃
2
= cot 2
2𝑠𝑖𝑛2 ( )
2
𝑑2 𝑦 1 𝜃 𝑑𝜃
⇒ 𝑑𝑥 2 = − 2 𝑐𝑜𝑠𝑒𝑐 2 ( 2) 𝑑𝑥
1 𝜃 1
= − 2𝑎 𝑐𝑜𝑠𝑒𝑐 2 ( 2) 𝜃
2𝑠𝑖𝑛2 ( )
2
1 𝜃 1
= − 4𝑎 𝑐𝑜𝑠𝑒𝑐 4 ( 2)

27 Let r be the radius of ice ball at time t .

4
𝑉 = 3 𝜋 𝑟 3 ………… (1)
½
S = 4𝜋𝑟 2 …………..(2)
𝑑𝑉
Given 𝑑𝑡 ∝ S
𝑑𝑉
⟹ 𝑑𝑡 = − k S (where k is some positive constant) ………(3) ½
Differentiating (1) w.r.t. t, we get
𝑑𝑉 4 2 𝑑𝑟
= 𝜋. (3 𝑟 ) 𝑑𝑡
𝑑𝑡 3 1
𝑑𝑉 𝑑𝑟
= 4𝜋𝑟 2 𝑑𝑡 ………. (4)
𝑑𝑡
𝑑𝑟
⟹ - k S = 4𝜋𝑟 2 𝑑𝑡 (from (3) and (4))
½
𝑑𝑟
⟹-kS=S (using (2))
𝑑𝑡
⟹ =−k
𝑑𝑟 ½
𝑑𝑡
⇒ radius of the ice-ball decreases at a constant rate

28A

2 −1 2
∫−4|𝑥 + 1|𝑑𝑥= ∫−4 (−𝑥 − 1) 𝑑𝑥 + ∫−1(𝑥 + 1) 𝑑𝑥 ½
(𝑥+1)2 −1 (𝑥+1)2 2
= − ] + ] ½
2 −4 2 −1
9 9
= − (0 − 2) + (2 − 0) = 9 ½
It represent the area of shaded region bounded by the curve 𝑦 = |𝑥 + 1|, ½
𝑥 − axis and the lines 𝑥 = −4 𝑎𝑛𝑑 𝑥 = 2

Page 6 of 12
 OR

28B

4 4 𝑥2
Required Area = ∫0 𝑥 𝑑𝑥 − ∫0 𝑑𝑥 1
4
4
𝑥2 1 ½
= ] − 12 [𝑥 3 ]40
2 0
1 1 16 8 ½
= 2 (16 − 0) − 12 (64 − 0) = 8 − = 3 sq. units
3

For Visually Impaired:

−𝑥 − 1, 𝑥 < −1
𝑦 = |𝑥 + 1| = 𝑓(𝑥) = {
𝑥 + 1, 𝑥 ≥ −1 1
2 −1 2
∫−4|𝑥 + 1|𝑑𝑥= ∫−4 (−𝑥 − 1) 𝑑𝑥 + ∫−1(𝑥 + 1) 𝑑𝑥
(𝑥+1)2 −1 (𝑥+1)2 2
= − ] + ] 1
2 −4 2 −1
9 9
= − (0 − ) + ( − 0) = 9
2 2
It represent the area of shaded region bounded by the curve 𝑦 = |𝑥 + 1|, 1
𝑥 − axis and the lines 𝑥 = −4 𝑎𝑛𝑑 𝑥 = 2

OR

Page 7 of 12
 𝑥2 𝑦2 𝑥2 𝑦2 5
25𝑥 2 + 16𝑦 2 = 400 ⇒ 16 + 25 = 1 ⇒ 42 + 52 = 1 ⇒ 𝑦 = 4 √42 − 𝑥 2
45
Required Area = 4 ∫0 4 √42 − 𝑥 2 𝑑𝑥 1
4
𝑥√42 −𝑥 2 42 𝑥
= 5[ + sin−1 (4)] 1
2 2
0
−1 (1)
= 5[0 + 8 sin − 0]
𝜋
= 40 × 2 = 20𝜋 sq. units 1

29A The line through (2, −1,3) parallel to the z-axis is given by
𝑟⃗ = (2𝑖̂ − 𝑗̂ + 3𝑘̂ ) + λ(𝑘̂) 1
Any point on this line is P (2, −1, 3 + 𝜆 ) ½
Any point on the given line 𝑟⃗ = (2𝑖̂ − 𝑗̂ + 2𝑘̂) + 𝜇(3𝑖̂ + 6𝑗̂ + 2𝑘̂) is
Q (2 + 3𝜇, −1 + 6𝜇, 2 + 2𝜇)
½
For the intersection point
½
Q (2 + 3𝜇, −1 + 6𝜇, 2 + 2𝜇) = P (2, -1, 3+ λ ) ⇒ 2 = 2 + 3 𝜇 ⇒ 𝜇 = 0
½
The point of intersection is (2, −1,2)
The distance from (2, −1,3) to (2, −1,2) is clearly 1 unit.

Alternative Solution:
Any point on the line through (2, −1, 3) parallel to the z-axis is (2, −1, 𝜆)
Any point on the given line is (2 + 3𝜇, −1 + 6𝜇, 2 + 2𝜇) 1
Therefore, 2 = 2 + 3 𝜇 ⇒ 𝜇 = 0 1
The point of intersection is (2, −1,2) ½
The distance from (2, −1,3) to (2, −1,2) is clearly 1 unit. ½
OR
29B
The line through (2,−1,1)parallel to the z-axis is 𝑟⃗ = (2𝑖̂ − 𝑗̂ + 𝑘̂ ) + λ(𝑘̂) 1
Any point on this line is P (2, −1, 1+ λ )
Any point on the given line is A (3+ 𝜇, 𝜇, 1+ 𝜇 )
A (3+ 𝜇, 𝜇, 1+ 𝜇 ) = P (2, −1, 1+ λ ) ⇒ 𝜇 = −1 1
The point of intersection is (2, −1,0) ½
The distance of (2, −1, 0) from the z-axis is √22 + (−1)2 = √5 units. ½
30 Sketching the graph 1
1
2

Page 8 of 12
 Corner points A(600,0), B(1200,0), C(800,400), D(400,200)
Values of Z: 𝑍𝐴 = 1200, 𝑍𝐵 = 2400, 𝑍𝐶 = 2000, 𝑍𝐷 = 1000 ½
Maximum 𝑍 = 2400 when 𝑥 = 1200 and 𝑦 = 0 ½
½
For Visually Impaired:
At Corner points A(600,0), B(1200,0), C(800,400), D(400,200)
30 Values of Z are 𝑍𝐴 = 1800, 𝑍𝐵 = 3600, 𝑍𝐶 = 3200, 𝑍𝐷 = 1600 1
Maximum Value of Z = 3600 at B(1200,0) 1
And Minimum Value of Z= 1600 at D(400,200) 1

31 Let the events be:

A: Mehul is selected
B: Rashi is selected
Then according to the question,
A and B are independent events and
𝑃(𝐴) = 0.4, 𝑃(𝐴 ∩ 𝐵̅ ) + 𝑃(𝐵 ∩ 𝐴̅) = 0.5 1
Let 𝑃(𝐵) = 𝑥
Then 𝑃(𝐴 ∩ 𝐵̅ ) + 𝑃(𝐵 ∩ 𝐴̅) = 0.5
⇒ 𝑃(𝐴)𝑃(𝐵̅ ) + 𝑃(𝐵)𝑃(𝐴̅) = 0.5
⇒ 0.4(1 − 𝑥) + 𝑥(1 − 0.4) = 0.5
⇒ 0.4 − 0.4𝑥 + 0.6𝑥 = 0.5
⇒ 0.2𝑥 = 0.5 − 0.4 = 0.1
0.1 1
⇒ 𝑥 = 0.2 = 2 = 0.5 1
So, probability of selection of Rashi = 0.5
Probability of selection of at least one of them = 1 − 𝑃(𝐴̅ ∩ 𝐵̅ )
= 1 − 𝑃(𝐴̅)𝑃(𝐵̅ )
= 1 − 0.6 × 0.5
= 1 − 0.3 = 0.7 1
SECTION D
(LA type questions of 5 marks each)
32 3 −6 −1 1 −2 −1 1 0 0
𝐴𝐵 = [ 2 −5 −1] [0 −1 −1] = [0 1 0] = 𝐼 1
−2 4 1 2 0 3 0 0 1
−1 −1
So, 𝐴 = 𝐵 𝑎𝑛𝑑 𝐵 = 𝐴
½
Given system of equations is
3𝑥 − 6𝑦 − 𝑧 = 3, 2𝑥 − 5𝑦 − 𝑧 + 2 = 0, − 2𝑥 + 4𝑦 + 𝑧 = 5
In matrix form it can be written as: 𝐴𝑋 = 𝐶,
𝑥 ½
3
where 𝑋 = [ ] 𝑎𝑛𝑑 𝐶 = [−2]
𝑦
𝑧 5 ½
Here |𝐴| = −3 − 0 + 2 = −1 ≠ 0 ½
So, the system is consistent and has unique solution given by the
expression 𝑋 = 𝐴−1 𝐶 = 𝐵𝐶 ½
1 −2 −1 3 𝑥 2
⇒ 𝑋 = [0 −1 −1] [−2] ⇒ [𝑦] = [−3] 1
2 0 3 5 𝑧 21
½
Thus 𝑥 = 2, 𝑦 = −3, 𝑧 = 21
Page 9 of 12
33A Let 𝑥 = tan 𝜃 ⇒ 𝑑𝑥 = 𝑠𝑒𝑐 2 𝜃 𝑑𝜃 ½
𝜋
log(1+𝑡𝑎𝑛𝜃)
𝐼= ∫ 4
0 1+𝑡𝑎𝑛2 𝜃
. 𝑠𝑒𝑐 2 𝜃 𝑑𝜃
𝜋 𝜋
𝜋
𝐼 = ∫04 𝑙𝑜𝑔 (1 + 𝑡𝑎𝑛𝜃)𝑑𝜃 = ∫0 𝑙𝑜𝑔 [1 + tan (4 − 𝜃)] 𝑑𝜃
4
1
𝜋
1−tan 𝜃
=∫0 log [1 + 1+tan 𝜃] 𝑑𝜃
4
1
𝜋
1+tan 𝜃+1−tan 𝜃
=∫0 log [
4 ] d𝜃
1+tan 𝜃
𝜋
2
=∫04 log [1+tan 𝜃 ] 𝑑𝜃
𝜋 𝜋
=∫0 log 2 𝑑𝜃 − ∫0 log[1 + 𝑡𝑎𝑛𝜃] 𝑑𝜃
4 4

𝜋 1
4
= log 2 × 𝑥]0 − 𝐼
𝜋 1
⇒ 2𝐼 = log 2
4
𝜋
⇒ 𝐼= log 2
8 ½

33B OR OR

(3 sin 𝜃−2) cos 𝜃 (3 𝑠𝑖𝑛 𝜃−2) 𝑐𝑜𝑠 𝜃

I = ∫ 5−𝑐𝑜𝑠2 𝜃−4 sin 𝜃 𝑑𝜃 = ∫ 5−(1−𝑠𝑖𝑛2 𝜃)−4 𝑠𝑖𝑛 𝜃 𝑑𝜃 ½
𝐿𝑒𝑡 𝑠𝑖𝑛 𝜃 = 𝑡 ⟹ 𝑐𝑜𝑠 𝜃 𝑑𝜃 = 𝑑𝑡
(3𝑡−2)
𝐼 =∫ 5−(1−𝑡 2 )−4𝑡 𝑑𝑡
(3𝑡−2)
1
3𝑡−2
=∫ 𝑡 2 −4𝑡+4 𝑑𝑡 =∫ (𝑡−2)2 𝑑𝑡
3𝑡−2 𝐴 𝐵
Let (𝑡−2)2
= (𝑡−2) + (𝑡−2)2
3𝑡 − 2 = 𝐴(𝑡 − 2) + 𝐵
Comparing the coefficients of t and constant terms on both sides
½
𝐴 = 3 , − 2𝐴 + 𝐵 = − 2 , 𝐵 = 4
(3 sin 𝜃−2) cos 𝜃 3 4
+½
∫ 5−𝑐𝑜𝑠2 𝜃−4 sin 𝜃 𝑑𝜃 =∫ 𝑡−2 𝑑𝑡 + ∫ (𝑡−2)2 dt
4 1+1
= 3log|𝑡 − 2| − +𝐶
𝑡−2
4 ½
= 3 log|𝑠𝑖𝑛 𝜃 − 2| − sin 𝜃−2 + C

34A 𝑑
𝑦 + 𝑑𝑥 ( 𝑥𝑦 ) = 𝑥 (sin 𝑥 + 𝑥)
𝑑𝑦
⟹ 𝑦 + (𝑥 𝑑𝑥 + 𝑦 ) = 𝑥 (sin 𝑥 + 𝑥)
𝑑𝑦
1
⟹ 2𝑦 + 𝑥 𝑑𝑥 = 𝑥 (sin 𝑥 + 𝑥)
𝑑𝑦 2𝑦
⟹ 𝑑𝑥 + = (sin 𝑥 + 𝑥)
𝑥
𝑑𝑦
This a linear differential equation of the form 𝑑𝑥 + 𝑃𝑦 = 𝑄
2
P= ,Q=(sin 𝑥 + 𝑥)
𝑥
2
1
∫ 𝑥 dx 2 𝑙𝑜𝑔 𝑥 𝑙𝑜𝑔 𝑥 2 2
I.F =𝑒 = 𝑒 =𝑒 =𝑥
Solution will be 𝑦 . I.F =∫ 𝑄. 𝐼𝐹 𝑑𝑥 1
𝑦𝑥 2 = ∫(𝑠𝑖𝑛 𝑥 + 𝑥) 𝑥 2 dx
𝑦𝑥 2 =∫ 𝑠𝑖𝑛𝑥. 𝑥 2 𝑑𝑥 + ∫ 𝑥 3 𝑑𝑥

Page 10 of 12
 𝑥4 1
⟹ 𝑦𝑥 2 = −𝑥 2 cos 𝑥 + 2 ∫ 𝑥 cos 𝑥 𝑑𝑥 + +C
4
𝑥4
⟹ 𝑦𝑥 2 = −𝑥 2 𝑐𝑜𝑠 𝑥 +2 (𝑥 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠 𝑥) + 4 + C
Which is the required solution 1
OR
𝑥⁄ 𝑥⁄
2𝑦 𝑒 𝑦 𝑑𝑥 + ( 𝑦 − 2𝑥 𝑒 𝑦 ) 𝑑𝑦 = 0
34B 𝑥 𝑥 𝑦
𝑥
𝑑𝑥 2𝑥 𝑒 ⁄𝑦 −𝑦 2 𝑒 −1
𝑦
⟹ = 𝑥 = 𝑥
𝑑𝑦 2𝑦 𝑒 ⁄𝑦 2 𝑒𝑦 1
It is a homogeneous differential equation.
𝑑𝑥 𝑑𝑣
Let 𝑥 = 𝑣𝑦 ⟹ 𝑑𝑦 = 𝑣 +𝑦 𝑑𝑦 1
𝑑𝑣 2𝑣𝑒 𝑣 −1
𝑣 +𝑦 𝑑𝑦 = 2𝑒 𝑣
𝑑𝑣 2𝑣𝑒 𝑣 −1 2𝑣𝑒 𝑣 −1−2𝑣𝑒 𝑣
⟹ 𝑦 𝑑𝑦 = –𝑣=
2𝑒 𝑣 2𝑒 𝑣
𝑑𝑣 −1
⟹ 𝑦 =
𝑑𝑦 2𝑒 𝑣
𝑣 𝑑𝑦
⟹ 2𝑒 𝑑𝑣 = − 𝑦 1
𝑣 𝑑𝑦
∫ 2𝑒 𝑑𝑣 = − ∫ 𝑦
𝑣
⟹ 2𝑒 = − log |𝑦| + C
𝑥
⟹ 2𝑒 𝑦 +" 𝑙𝑜𝑔|𝑦| = C 1
When 𝑥 = 0 , 𝑦 = 1 , C = 2
𝑥
Required solution 2 𝑒 𝑦 + log|𝑦| = 2
1
35 Let
𝑥−1
=
𝑦−0
=
𝑧+1
=𝜆 ⇒ Any point on it is (3 𝜆 +1, − 𝜆, −1) ½
3 −1 0
1
For the point where 𝑦 = 1 ⇒ 𝜆 = −1
½
⇒ The point is (−2, 1 , −1)
1
The directions of the two lines are 𝑚 ⃗⃗⃗ = 3𝑖̂ − 𝑗̂
½
and 𝑛⃗⃗ = −2𝑖̂ + 2𝑗̂ + 𝑘̂
𝑖̂ 𝑗̂ 𝑘̂
⃗⃗⃗ × 𝑛⃗⃗ = | 3 −1 0| = −𝑖̂ − 3𝑗̂ + 4𝑘̂
𝑚 1
−2 2 1
The equation of the required line is
𝑟⃗ = (−2𝑖̂ + 𝑗̂ − 𝑘̂) + 𝜇(−𝑖̂ − 3𝑗̂ + 4𝑘̂) ½

Alternative Solution:
𝑥−1 𝑦−0 𝑧+1
Let = = = 𝜆 ⇒ Any point on it is (3 𝜆 + 1, − 𝜆, −1) ½
3 −1 0
For the point where 𝑦 = 1 ⇒ 𝜆 = −1 1
⇒ The point is (−2, 1, −1) ½
Let the direction ratios of the required line be a, b, c
Then 3𝑎 − 𝑏 = 0
And −2𝑎 + 2𝑏 + 𝑐 = 0 1
𝑎 −𝑏 𝑐 𝑎 𝑏 𝑐 1
Solving we get = = 4 ⇒ −1 = −3 = 4
−1 3
𝑥+2 𝑦−1 𝑧+1
The required line is = = =𝜇 ½
−1 −3 4
In vector form 𝑟⃗ = (−2𝑖̂ + 𝑗̂ − 𝑘̂) + 𝜇(−𝑖̂ − 3𝑗̂ + 4𝑘̂) ½

Page 11 of 12
 SECTION- E
(3 case-study/passage-based questions of 4 marks each)
36 I. Traffic flow is not reflexive as (𝐴, 𝐴) ∉ 𝑅 (or no major spot is connected
with itself) 1
II. Traffic flow is not transitive as (𝐴, 𝐵) ∈ 𝑅 𝑎𝑛𝑑 (𝐵, 𝐸) ∈ 𝑅, 𝑏𝑢𝑡 (𝐴, 𝐸) ∉ 𝑅 1
III A. 𝑅 = {(𝐴, 𝐵), (𝐴, 𝐶), (𝐴, 𝐷), (𝐵, 𝐶), (𝐵, 𝐸), (𝐶, 𝐸), (𝐷, 𝐸), (𝐷, 𝐶)} 1
Domain = {𝐴, 𝐵, 𝐶, 𝐷} ½+
Range = {𝐵, 𝐶, 𝐷, 𝐸} ½
OR
III B. No, the traffic flow doesn’t represent a function as A has three images. 1+1

37 I. 𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = −0.3𝑥 2 + 20𝑥 − (0.5𝑥 2 − 10𝑥 + 150)

= −0.8𝑥 2 + 30𝑥 − 150 1
II. For critical points 𝑃′ (𝑥) = 0 ⇒ −1.6𝑥 + 30 = 0
30 300 1
⇒ 𝑥 = 1.6 = = 18.75
16
III A. Now 𝑃′′ (𝑥) = −1.6
1
In particular 𝑃′′ (18.75) = −1.6 < 0
1
So, critical value 𝑥 = 18.75 corresponds to a maximum profit.

OR
III B. As 𝑥 is the number of bulbs, so practically 18 bulbs correspond to a
maximum profit. 1
Maximum profit is 𝑃(18) = −0.8 × 182 + 30 × 18 − 150
= −259.2 + 540 − 150 1
= 540 − 409.2 = ₹130.80
38 Let the events be
E1: the student is in the first group (time spent on screen is more than 4 hours)
E2: the student is in the second group (time spent on screen is 2 to 4 hours)
E3: the student is in third group (time spent on screen is less than 2 hours)
A: the event of the student showing symptoms of anxiety and low retention
60 30 10
P(E1) = 100 P(E2) = 100 and P(E3) = 100
80 70 30
P(A|E1) = 100 P(A|E2) = 100 and P(A|E3) = 100

I. P(A)= P(E1) x P(A|E1) + P(E2) x P(A|E2) + P(E3) x P(A|E3)

60 80 30 70 10 30 72
= 100 x 100 + 100 x 100 + 100 x 100 = 100 = 72% 2

𝑃(𝐸1 ∩𝐴)
II. P(E1|A) = 𝑃(𝐴)
60 80
( x
100 100
) 48 2 2
= 72 = 72 = 3
( )
100

Page 12 of 12