Class 11 Question Paper 2025 Maths

ANNUAL EXAMINATION

Question Paper
2025
NCERT BASED SYLLABUS

FOR CBSE AND STATE BOARD
FOLLOWING NCERT

KVS QUESTION PAPERS

Annual Exam 2025 Question Paper

SESSION ENDING EXAMINATION 2024-25
CLASS: XI ​ ​ ​ ​ ​ ​ ​ ​ ​ SUBJECT: MATHEMATICS
TIME: 3 HOURS​ ​ ​ ​ ​ ​ ​ ​ M. MARKS: 80
General Instructions:
(i) This Question Paper contains 38 questions. All questions are compulsory.
(ii) Question Paper is divided into five Sections – Section A, B, C, D and E.
(iii) In Section A – Questions no. 1 to18 are Multiple Choice Questions (MCQs) and Questions no.19 & 20 are
Assertion - Reason based questions of 1 mark each.
(iv) In Section B – Questions no.21 to 25 are Very Short Answer (VSA) type questions, carrying 2 marks each.
(v) In Section C – Questions no. 26 to 31 are Short Answer (SA) type questions, carrying 3 marks each.
(vi) In Section D – Questions no. 32 to 35 are Long Answer (LA) type questions, carrying 5 marks each.
(vii) In Section E– Questions no.36 to 38 are case study based questions, carrying 4 marks each.
(viii)There is no overall choice. However, an internal choice has been provided in some questions.
SECTION A
This section comprises multiple choice questions (MCQs) of 1 mark each.
1. Which of the following is a null set?
(a) {0} (b) { x : x is prime, 90 < x < 96 }
(c) {x: x ∈ N, 1 < x ≤ 2} (d) { x : x2 + 1 = 0, x ∈ R }
2. Let R be a relation in the set N given by R = {(a, b): a = b – 2, b > 4}. Then
(a) (8, 7) ∈ R (b) (6, 8) ∈ R (c) (3, 8) ∈ R (d) (2, 4) ∈ R
3. Let f : R → R given by f(x) = x2 then range of f is
(a) R (b) (-∞, 0) (c) [ 0, ∞) (d) (0, ∞)
4. If A = {3, {4, 5}, 6}, then which of the following statements is true?
(a) {4, 5} ⊆ A (b) {4, 5} ∈ A (c) 4 ∈ A (d) {3, 4, 5} ⊆ A

5. If 𝑓(𝑥) = 𝑥 −
3 1
3
𝑥
then 𝑓(𝑥) + 𝑓 ( ) is equal to
1
𝑥

(a) 2𝑥3 (b) 2 ( )
1
𝑥
3 (c) 0 (d) 1

6. The degree measure of the angle subtended at the centre of a circle of diameter 50 cm by an arc of length 11 cm
is
(a) 200 15' (b) 25015' (c) 25012' (d) 28012'
−1
7. If sin 𝑥 = 2
and 𝑥 lies in fourth quadrant then value of cos 2𝑥 is

−1 1
(a) - 2 (b) 2
(c) 2 (d) 2

Annual Exam 2025 Question Paper

8. Value of 𝑖( + ) is
37 1
67
𝑖

(a) 2i (b) i (c) – 1 (d) 0
9. Solution of the inequality |𝑥 + 1| > 4, 𝑥 ∈ R will be
(a) (3, ∞) (b) ( – ∞, 3) (c) (– 5, 3) (d) ( – ∞, – 5 ) ∪ (3, ∞)

10. If np3 = 20 x np2 then value of n is
(a) 18 (b) 20 (c) 22 (d) 24
11. In how many can the letters of the word ‘PENCIL’ be arranged so that E and N always come together?
(a) 120 ways (b) 240 ways (c) 720 ways (d) 360 ways
4
12. Fourth term in the expansion of 𝑥 + ( 2 3
𝑥 ) is
(a) x8 ​​ ​
(b) 54 x 2 ​
(c) 108 x ​ (d) 108/x
𝑎−𝑏
13. If a, b and c are in GP and r is the common ratio between them, then value of 𝑏−𝑐 is equal to

𝑎 1
(a) r (b) ar (c) 𝑟 (d) 𝑟

14. In which octant does the point (– 6, 5, –1) lie?
(a) ii octant (b) iii octant (c) vi octant (d) vii octant
15 What is the perpendicular distance of the point P (6, 7, 8) from xy-plane?
(a) 8 ​ ​ (b) 7 ​​ (c) 6 (d) None of these
16. Mean deviation about the median for the data 11, 3, 8, 7, 5, 14, 10, 2, 9 is given by
(a) 3 (b) 4 (c) 5 (d) 6
1 1 1
17. If E and F are events such that P(E) = 4 , P(F) = 2 and P (E and F) = 8 , what is the probability of event not E
and not F?
5 7 1 3
(a) 8 (b) 8 (c) 2 (d) 8

18. Three coins are tossed once, then probability of getting at least two heads is given by
2 1 3 1
(a) 3 (b) 2 (c) 8 (d) 3

ASSERTION – REASON BASED QUESTIONS
In the following questions 19 & 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, but (R) is false.
(d) (A) is false, but (R) is true.

Annual Exam 2025 Question Paper
19. Assertion (A): If A = {x : x is a multiple of 3} and B = {x : x is a multiple of 5} then
A ∩ B = { x : x is a multiple of 15}
Reason (R): A ∩ B is a set which contains common elements of set A and B.
4
20. Assertion (A): If tan θ = 3 , then sinθ is 5 but not
4 −4
5
.
Reason (R): If tan θ is positive θ lies either in first quadrant or third quadrant.
SECTION – B
This section comprises of Very Short Answer Type (VSA) questions, each of 2 marks.
21. Let U = { 1, 2, 3, 4, 5, 6}, A = { 2, 3} and B = { 3, 4,5}, then verify that ( A ∪ B)' = A' ∩ B'.
𝑎+𝑖𝑏
22. If x + iy = 𝑎−𝑖𝑏 , prove that x2 + y2 = 1.

23. Find the value of tan 12 .
13π

OR
Prove that: cos 6𝑥 = 32 cos6 𝑥 – 48 cos4 𝑥 + 18 cos2 𝑥 – 1.
24. Find the mean and variance of first 10 natural numbers.
25. A person has 2 parents, 4 grandparents, 8 great grandparents and so on. Find the number of his ancestors
during the 10 generations preceding his own.
OR
Which term of the G.P. 2, 8, 32……. up to n terms is 131072?

SECTION – C
The section comprises Short Answer (SA) type questions of 3 marks each.
4

) + (𝑎 − 𝑎 − 1) .
4

(
26.Evaluate 𝑎 +
2 2
𝑎 − 1
2 2

OR
2 4
Expand using Binomial Theorem 1 + ( 𝑥
2
− 𝑥) , 𝑥 ≠ 0.
27. Find the coordinates of the foci, the vertices, the length of major and minor axes, eccentricity and length of latus
rectum of the ellipse 9x2 + 4y2 = 36.
28. A manufacturer has 600 litres of 12% solution of acid. How many litres of a 30% acid solution must be added to it
so that acid content in the resulting mixture will be more than 15% but less than 18%?
OR
Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less
than 40.
29. If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1
(c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2)=0.

Annual Exam 2025 Question Paper
30. Calculate the mean, variance and standard deviation for the following distribution:
Class 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100
Frequency 3 7 12 15 8 3 2

31. Evaluate :
3 2
(i) ⎡⎢ 2
𝑥 −2𝑥 ⎤ 𝑎𝑥+𝑥 𝑐𝑜𝑠𝑥
⎥ (ii) 𝑏sin𝑠𝑖𝑛 𝑥
⎣ 𝑥 −5𝑥+6 ⎦
OR
Find the derivative of
4𝑥+5sin𝑠𝑖𝑛 𝑥
3𝑥+7cos𝑐𝑜𝑠 𝑥
.

SECTION – D

This section comprises Long Answer type (LA) questions of 5 marks each.
32. Let S be the sum, P the product and R the sum of reciprocal of n terms in a G.P. Prove that P2 Rn = Sn.
OR
The ratio of A.M. and G.M. of two positive numbers a and b is m : n. Show that

(
a:b= 𝑚 + 𝑚 −𝑛
2 2
)( : 𝑚 −
2
𝑚 −𝑛
2
)
33. Find the derivative of 𝑓(𝑥) = sin 𝑥 + cos 𝑥 from the first principles.

34. Find the domain and range of the function 𝑓= {(𝑥, 1− 𝑥
1
2 ): 𝑥∈𝑅, 𝑥≠±1 }.
35. Prove that:

cos 200 cos 400 cos 600 cos 800 = 16 .
1

OR

If tan 𝑥 = 4 and π < 𝑥 < 2 , find the values of sin 2 , cos 2 and tan 2 .
3 3π 𝑥 𝑥 𝑥

SECTION – E
This section comprises 3 source based/ case-based/ passage- based questions of 4 marks each.
36. A state cricket authority has to choose a team of 11 members. For this the authority asks 2 coaches of a
government academy to select the team members that have experience as well as the best performers in last 15
matches. They have to make a team of 11 cricketers amongst 15 possible candidates.

Annual Exam 2025 Question Paper

In how many ways can the final eleven be selected from 15 cricket players if:
(i) If one player must be included. (1)
(ii) one of them, who is in bad form, must be excluded. (1)
(iii) (a) Two of them being leg spinners, only one leg spinner must be included. (2)
OR
(b) If 6 players can bowl and team must include exactly 5 bowlers. (2)
37. Consider the triangle ABC with vertices A(1, 4), B(2, –3) and C(–1, –2) as shown in the given figure. AD is the
median and AM is the altitude through A.

Based on the above information, answer the following questions:
(i) Find the equation of BC. (1)
(ii) find length of altitude AM. (1)
(iii) (a) Find the equation of altitude AM. (2)
OR
(b) Find the equation of right bisector of side BC. (2)
38. From the employees of a company, 5 persons are selected to represent them in the managing committee of the
company. Particulars of five persons are as follows:
S. No. Name Sex Age in years
1 Harish M 30
2 Rohan M 33
3 Sheetal F 46
4 Alis F 28

Annual Exam 2025 Question Paper

5 Salim M 41

A person is selected at random from this group to act as a spokesperson. Based on the above information answer
the following questions:
(i) What is the probability that the spokesperson will be either female or over 35 years? (2)
(ii) What is the probability that the spokesperson will be neither male nor below 40 years? (2)

Annual
AnnualExam
Exam2025
2025Marking
QuestionScheme
Paper
MODEL ANSWER SHEET
SESSION ENDING EXAMINATION (2024 – 25)
CLASS : XI
SUBJECT : MATHEMATICS (041)
SECTION A
1 d 1
2 b 1
3 c 1
4 b 1
5 c 1
6 c 1
7 d 1
8 a 1
9 d 1
10 c 1
11 b 1
12 d 1
13 d 1
14 c 1
15 a 1
16 a 1
17 d 1
18 b 1
19 a 1
20 d 1
SECTION B
21 A ∪ B = {2, 3, 4, 5} 0.5
(A ∪ B)' = {1, 6} 0.5
A' = {1, 4, 5, 6}, B' = {1, 2, 6} 0.5
A' ∩ B' = {1, 6} 0.5
22 𝑥 + 𝑖𝑦 =
𝑎+𝑖𝑏 0.5
𝑎−𝑖𝑏
𝑎−𝑖𝑏
𝑥 – 𝑖𝑦 = 𝑎+𝑖𝑏 0.5
𝑎+𝑖𝑏 𝑎−𝑖𝑏
(𝑥 + 𝑖𝑦) (𝑥 – 𝑖𝑦) = 𝑎−𝑖𝑏 x 𝑎+𝑖𝑏
2 2 0.5
𝑎 +𝑏
𝑥2 + 𝑦2 = 2 2 = 1
𝑎 +𝑏

0.5
23 tan =
13 π
12
= tan π + ( π
12 ) = tan π
12
0.5

Annual
AnnualExam
Exam2025
2025Marking
QuestionScheme
Paper

tan 12 = tan
π
( − )
π
3
π
4
π π
tan𝑡𝑎𝑛 3 − tan𝑡𝑎𝑛 4
= π π
1+ 𝑡𝑎𝑛 3 tan𝑡𝑎𝑛 4

3−1
1
= =2– 3
1+ 3

OR 0.5
Cos 6x = cos 2.3x
= 2 cos2 3x – 1
= 2 (4 cos3 x – 3 cos x)2 – 1 0.5
6 4 2
= 2 (16 cos x – 24 cos x + 9 cos x) – 1 0.5
6 4 2
= 32 cos x – 48 cos x + 18 cos x – 1 0.5
0.5
24 Mean =
1+2+3+4+5+6+7+8+9+10
= 5.5 0.5
10

2
2
( )
σ = 𝑁 ∑ 𝑥 − 𝑥 = 10 x 2 x (20.25 + 12.25 + 6.25 + 2.25 + 0.25)
𝑖
1 1 1
0.5
1
= 10 x 82.50 = 8.25

25 a= 2, r =2, n = 10 0.5
𝑛
Sn = (
𝑎 𝑟 −1 )
𝑟−1
0.5
2 2 −1
S10 = 2−1 ( 𝑛 ) = 2 (1024 – 1) = 2046
1
OR
a=2, r = 4, an = 131072
𝑛−1 0.5
an = a 𝑟
131072 = 2 (4)n – 1
0.5
65536 = (4)n – 1
48 = (4)n – 1
1
n=9
SECTION C
26 ( x + y)4 + ( x – y)4 = 2 ( 4C0 x4 + 4C2 x2y2 + 4C4 y4) 1
4 4

( 𝑎 +
2 2
𝑎 − 1 + 𝑎 − ) ( 2 2
𝑎 − 1 )
2 4 1
= 2 [ C0 𝑎 4
( ) 2 4 4
+ C2 𝑎 ( ) 2 2
( 2
𝑎 − 1 ) + C44
( 2
)
𝑎 − 1 ]

8 4 1
= 2 [(𝑎) + 6 (𝑎) (𝑎2 – 1) + (𝑎2 – 1)2]
= 2 𝑎8 + 12 𝑎6 – 10 𝑎4 – 4 𝑎2 + 2
1
OR

Annual
AnnualExam
Exam2025
2025Marking
QuestionScheme
Paper

2 4 4 𝑥 4 𝑥 3 2 𝑥 2
⎡1 +
⎣
𝑥
2
− ⎤ = C0
𝑥⎦ ( 1+ 2 ) − 4C1 1 + ( 2 )
𝑥 (
+ 4C2 1 + 2) x
2 2 4 2 3 4 2 4
( ) 𝑥
– C3 (1 + ) ( ) 𝑥
2 𝑥
+ C4 ( ) 𝑥 1

( ) – 4 (1 + 3 + 3 ) +
2 3 4 2 3
𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 2
= 1+ 42 + 6 4
+ 4 + 8 16 2 4
+ 8 𝑥
1
6 (1 + 2 + ) – 4 (1 + ) +
2
𝑥 𝑥 4 𝑥 8 16
2 4 2 2 3 4
𝑥 𝑥 𝑥
2 3 4
16 32 816 𝑥 𝑥
= 4 – 3 + 2 + 𝑥 𝑥
– 5 – 4𝑥 + 2 + 2 + 16
𝑥 𝑥 𝑥

27 2
𝑦
2
Equation of ellipse in standard form: 4 + 9 = 1
𝑥

2 2
𝑦 0.5
2 +
𝑥
Comparing this equation with 2 = 1, we get b = 2, a = 3
𝑏 𝑎
0.5
2 2
c= 𝑎 − 𝑏 = 9 − 4= 5
foci are (0, 5) and (0, − 5) 0.5

vertices are (0, 3) and (0, –3)
length of major axis is 6 units and length of minor axis is 4 units 0.5
0.5
5
e= 𝑎 = 3
𝑐
0.5
2
8
latus rectum = 2 𝑎 = 3
𝑏

28 Let x litres of 30% acid solution is required to be added, so total mixture =(x + 600)
litre
30% of x + 12% of 600 > 15% of (x + 600) ………………………. (1) 0.5
30% of x + 12% of 600 < 18% of (x + 600) ……………………….. (2) 0.5
Solving (1) we get x > 120 1
Solving (2) we get x < 300 1
i.e. 120 < x < 300
OR
Let the consecutive odd numbers be x and x+2 0.5
Given: x > 10 , X+ 2 > 10 ………………. (1) 0.5
x + (x + 2) < 40 ………………… (2) 0.5
solving (2) we get 2x + 2 < 40
i.e. x < 19 1
so, 10 < x < 19 0.5
So required possible pairs will be (11,13), (13,15), (15, 17), (17,19)
29 y = m1x + c1 ………………. (1)
y = m2x + c2 ……………….. (2)
y = m3x + c3 ………………... (3)

Annual
AnnualExam
Exam2025
2025Marking
QuestionScheme
Paper
𝑐 −𝑐 𝑚2𝑐1− 𝑚1𝑐2 1
solving (1) and (2) we get x = 𝑚1− 𝑚2 , y = 𝑚2− 𝑚1
2 1

substituting these values in (3)
𝑚2𝑐1− 𝑚1𝑐2 𝑐1− 𝑐2
𝑚2− 𝑚1
= m3 𝑚 − 𝑚 + c3 1
2 1

𝑚1 𝑐2 – 𝑚2𝑐1 + 𝑚2𝑐3 – 𝑚1𝑐3 + 𝑚3𝑐1 – 𝑚3𝑐2 = 0
1
𝑚1 (𝑐2 – 𝑐3) + 𝑚2 (𝑐3 – 𝑐1) + 𝑚3 (𝑐1 − 𝑐2) = 0

30 1.5
class frequency Mid 𝑓𝑖 𝑥𝑖 2
fi (𝑥𝑖 − 𝑥)2 Marks
(𝑥 − 𝑥)
point
𝑖

For
30 – 40 3 35 105 729 2187 Correc
40 – 50 7 45 315 289 2023 t
50 – 60 12 55 660 49 588 Table
60 – 70 15 65 975 9 135
70 – 80 8 75 600 169 1352
80 – 90 3 85 255 529 1587
90 – 100 2 95 190 1089 2178
50 3100 10050

3100
Mean ( 𝑥 ) = 𝑁 ∑ 𝑓 𝑥 = = 62
1
𝑖 𝑖 50 0.5

2
2
Variance ( σ ) = 𝑁 ∑ 𝑓
1
𝑖 (
𝑥𝑖 − 𝑥 ) 0.5
= 50 x 10050 = 201
1 0.5

Standard deviation (σ) = 201 = 14.18
31 3 2
𝑥 (𝑥−2)
2
0.5
(i) ⎡⎢ 2
𝑥 −2𝑥 ⎤
⎥ = ⎡⎢ (𝑥−2)(𝑥−3) ⎤⎥
⎣ 𝑥 −5𝑥+6 ⎦ ⎣ ⎦
2
2
2 1
= ⎡⎢ (𝑥−3) ⎤⎥ = 2−3 = – 4
𝑥
⎣ ⎦
𝑎𝑥
𝑎𝑥+𝑥 𝑐𝑜𝑠𝑥
(ii) 𝑏sin𝑠𝑖𝑛 𝑥 = 𝑏sin𝑠𝑖𝑛 𝑥 + 𝑏sin𝑠𝑖𝑛
𝑥𝑐𝑜𝑠 𝑥
𝑥 0.5
𝑎 1 1 cos𝑐𝑜𝑠 𝑥
= 𝑏 sin𝑠𝑖𝑛 𝑥 + 𝑏 sin𝑠𝑖𝑛 𝑥
𝑥 𝑥 0.5
1 𝑎+1 0.5
= 𝑏 + 𝑏 =
𝑎
𝑏
OR
4𝑥+5sin𝑠𝑖𝑛 𝑥
𝑦 = 3𝑥+7cos𝑐𝑜𝑠 𝑥

Annual
AnnualExam
Exam2025
2025Marking
QuestionScheme
Paper
𝑑
(3𝑥+7cos𝑐𝑜𝑠 𝑥) 𝑑𝑥 (4𝑥+5sin𝑠𝑖𝑛 𝑥)−(4𝑥+5sin𝑠𝑖𝑛 𝑥) 𝑑𝑥 (3𝑥+7cos𝑐𝑜𝑠 𝑥)
𝑑
1
𝑑𝑦
𝑑𝑥
= 2
(3𝑥+7cos𝑐𝑜𝑠 𝑥) )

=
(3𝑥+7cos𝑐𝑜𝑠 𝑥) (4+5cos𝑐𝑜𝑠 𝑥)−(4𝑥+5sin𝑠𝑖𝑛 𝑥) (3−7sin𝑠𝑖𝑛 𝑥) 1
2
(3𝑥+7cos𝑐𝑜𝑠 𝑥) )
15𝑥cos𝑐𝑜𝑠 𝑥+28𝑥sin𝑠𝑖𝑛 𝑥+28cos𝑐𝑜𝑠 𝑥−15sin𝑠𝑖𝑛 𝑥+35
= 2 1
(3𝑥+7cos𝑐𝑜𝑠 𝑥) )

SECTION D
32 Let the given GP be a, ar, ar2, …….., arn – 1
𝑛
0.5
S = (1−𝑟 )
𝑎 1−𝑟

P = a x ar x ar2 x ……….x arn – 1
= an r1+2+3+………….+ (n – 1)
(𝑛−1)𝑛

= an 𝑟 2

P2 = a2n r(n – 1)n 1.5
1 1
R = 𝑎 + 𝑎𝑟 + …………….. +
1
𝑛−1
𝑎𝑟

=
1
𝑎 ( −1) = 𝑟 (1−𝑟 ) =
1
𝑟
𝑛
𝑛
(1−𝑟𝑛) 1.5
1
𝑎 𝑛 𝑛−1
𝑟
−1 (1−𝑟)𝑟 𝑎(1−𝑟)𝑟
𝑛
𝑆 𝑎 1−𝑟
= 1−𝑟 ( ) x 𝑎(1−𝑟)𝑟 𝑛−1 𝑛−1
= 𝑎2 𝑟
1
𝑛
𝑅 (1−𝑟 )
𝑆 𝑛 = 2n 𝑟𝑛(𝑛−1)
( )
𝑅
𝑎 = P2 0.5

P2 Rn = Sn
OR
0.5
𝑎+𝑏
A.M. = 2
, G.M. = 𝑎𝑏 0.5
𝑚
So,
𝑎+𝑏
= 𝑛
2 𝑎𝑏

Applying componendo and dividendo rule
𝑚+𝑛
𝑎+𝑏+2 𝑎𝑏
= 𝑚−𝑛 1
𝑎+𝑏−2 𝑎𝑏
2 0.5
( 𝑎+ 𝑏) = 𝑚+𝑛
2 𝑚−𝑛
( 𝑎− 𝑏 )
( 𝑎+ 𝑏) = 𝑚+𝑛
( 𝑎− 𝑏) 𝑚−𝑛 0.5
Again applying componendo and dividendo rule
( 𝑎+ 𝑏)+( 𝑎− 𝑏) = 𝑚+𝑛 + 𝑚−𝑛
( 𝑎+ 𝑏)−( 𝑎− 𝑏) 𝑚+𝑛 − 𝑚−𝑛

𝑎 𝑚+𝑛 + 𝑚−𝑛
=
𝑏 𝑚+𝑛 − 𝑚−𝑛

1

Annual
AnnualExam
Exam2025
2025Marking
QuestionScheme
Paper
2
𝑎
= ( 𝑚+𝑛+ 𝑚−𝑛)
𝑏 2
( 𝑚+𝑛− 𝑚−𝑛)
2 2 1
𝑎 𝑚+𝑛+𝑚−𝑛 + 2 (𝑚+𝑛)(𝑚−𝑛) 𝑚 + 𝑚 −𝑛
𝑏
= =
𝑚+𝑛+𝑚−𝑛 − 2 (𝑚+𝑛)(𝑚−𝑛) 2
𝑚− 𝑚 −𝑛
2

33 𝑓(𝑥) = sin 𝑥 + cos 𝑥
𝑓΄(𝑥) =
𝑓(𝑥+ℎ)−𝑓(𝑥) 0.5
ℎ
0.5
𝑠𝑖𝑛(𝑥+ℎ)+cos𝑐𝑜𝑠 (𝑥+ℎ) −sin𝑠𝑖𝑛 𝑥−cos𝑐𝑜𝑠 𝑥
= ℎ

=
2𝑐𝑜𝑠 ( 𝑥+ℎ+𝑥
2 ) 𝑠𝑖𝑛( 𝑥+ℎ−𝑥
2 )−2𝑠𝑖𝑛( 𝑥+ℎ+𝑥
2 )sin𝑠𝑖𝑛 ( 𝑥+ℎ−𝑥
2 )
ℎ 2

=
ℎ 2𝑥+ℎ
2sin𝑠𝑖𝑛 2 ⎡𝑐𝑜𝑠 2
⎣ ( )−sin𝑠𝑖𝑛 ( 2𝑥+ℎ
2 ) ⎤⎦
ℎ
1
ℎ
sin𝑠𝑖𝑛 2
= ℎ
2
. lim .⎡𝑐𝑜𝑠
ℎ→0 ⎣
( 2𝑥+ℎ
2 ) − sin 𝑠𝑖𝑛 ( 2𝑥+ℎ
2 ) ⎤⎦ 0.5
= 1 x (cos 𝑥 – sin 𝑥) = cos 𝑥 – sin 𝑥 0.5
34 𝑓(𝑥) =
1
2 ,x∈R
1−𝑥

𝑓(𝑥) is defined for all real numbers except 1 – x2 = 0
1
Or x2 – 1 = 0
i.e. x = ± 1
so Domain of 𝑓= R – {-1,1}
1
let 𝑓(𝑥) = 𝑦
1
2 =𝑦
1−𝑥
1
𝑥2 = 1 - 𝑦

𝑦−1
𝑥=± 𝑦 1.5
𝑥 will take real values , if
𝑦−1
𝑦
≥0

y< 0 or y ≥ 1
1.5
Range of 𝑓= (-∞, 0) ∪ [1, ∞)
35 L.H.S.= cos 200 cos 400 cos 600 cos 800
1
= 2 x cos 600 cos 400 ( 2 cos 800 cos 200) 1

1 1
= 2 x 2 x cos 400 [ cos (800 + 200) + cos (800 - 200)]
1
1
= 4 x cos 400 [cos 1000 + cos 600]
1 1
= 4
cos 1000 cos 400 + 4
cos 600 cos 400
1

Annual
AnnualExam
Exam2025
2025Marking
QuestionScheme
Paper
1 1
= 8 x 2 cos 1000 cos 400 + 8 cos 400
1 1
1
= 8 [cos 1400 + cos 600 ] + 8 cos 400
1 1 1
= 8
cos 1400 + 16 + 8 cos(1800 -1400) 0.5
1 1 1
= 8 cos 1400 + 16
− 8 cos 1400
0.5
1
= 16 = R.H.S.

OR
9 25
Sec2𝑥 = 1 + tan2𝑥 = 1 + 16 = 16
16
Cos2 𝑥 = 25
1
4
Cos 𝑥 = - 5 (𝑥 lies in third quadrant) 0.5
π 𝑥
As π < 𝑥 < 2 , so 2 < 2 < 4 ( 2 lies in second quadrant)
3π 𝑥 3π

9
2sin2 2 = 1 – cos x = 1 + 5 = 5
𝑥 4

𝑥 9
sin2 2 = 10
1.5
𝑥
𝑥
sin 2 =
3
, ( 2 lies in second quadrant, so sin 2𝑥 > 0)
10

1
2cos2 2 = 1 + cos x = 1 - 5 = 5
𝑥 4

𝑥 1
cos2 2 = 10
𝑥 𝑥 1.5
𝑐𝑜𝑠 2 = −1 , ( 2 lies in second quadrant, so cos 2𝑥 < 0)
10
3
𝑠𝑖𝑛 2
𝑥 0.5
𝑡𝑎𝑛
𝑥
2
= 𝑥 = 10
−1 = -3
𝑐𝑜𝑠 2
10

SECTION E
36 (i) one player must be included so 10 players will be selected from 14 players
No of ways = 14C10 = 1001 1
(ii) one player must be excluded from total no of players, so 11 players will be
selected from 14 players 1
14
No of ways = C11 = 336
(iii) (a) no of ways = 2C1 x 13C10 = 572 2
OR
6 9
(b) no of ways = C5 x C6 = 504 2
37 (i) equation of BC:
1
y + 3 = −3 (x – 2)
1

Annual
AnnualExam
Exam2025
2025Marking
QuestionScheme
Paper

x + 3y – 7 = 0
| 1.1+3.4−7 | 6 3 10 1
(ii) length of altitude AD = | |= =
| 1+9 | 10 5
1
(iii) (a) slope of AM = 3, (as AD is perpendicular to BC)
Equation of AM:
y- 4 = 3 (x – 1)
1
3x – y + 1 = 0
OR
1
1 −5
(b) coordinates of point D are ( 2 , 2 ), slope = 3

Equation of right bisector: 1
3x – y – 4 = 0
38 (i) Let A be the event that spokesperson is a female and B over 35 years.
2 2
P(A) = 5 , P(B) = 5 , P(A∩B) = 5
1 1

3
P(A∪B) = P(A) + P(B) - P(A∩B) = 5
1
(ii) Let E be the event that spokesperson is a male and F over 40 years.
3 2 1
P(E) = 5 , P(F) = 5 , P(E∩F) = 5
4 1
P(E∪ F) = P(E) + P(F) - P(E∩F) = 5
1
1
P(E´∩F´) = P(E∪F)´ = 1 – P(E∪F) = 5

QUESTION PAPERS
GET PREVIOUS YEAR QUESTION PAPERS FOR ALL CLASSES FOR CBSE, ISCSE, ISC AND ALL ALL
STATE BOARDS HERE AT

Andhra Pradesh Board Question Papers
BOARD
WISE Assam Board Question Papers
Bihar Board Question Papers
Chhattisgarh Board Question Papers
Goa Board Question Papers
Gujarat Board Question Papers
Haryana Board Question Papers
Himachal Pradesh Board Question Papers
J&K State Board Question Papers
Jharkhand Board Question Papers
Karnataka Board Question Papers
Kerala Board Question Papers
Madhya Pradesh Board Question Papers
Maharashtra State Board Question Papers
Manipur Board Question Papers
Meghalaya Board Question Papers
Mizoram Board Question Papers
Nagaland Board Question Papers
Orissa Board Question Papers
Punjab Board Question Papers
Rajasthan Board Question Papers
Tamil Nadu Board Question Papers
Telangana State Board Question Papers
Tripura Board Question Papers
Uttarakhand Board Question Papers
Uttar Pradesh (UP) Board Question Papers
West Bengal (WB) Board Question Papers