Class XI Session 2024-25

Subject - Mathematics
Sample Question Paper - 7

Time Allowed: 3 hours Maximum Marks: 80

General Instructions:

1. This Question paper contains - five sections A, B, C, D and E. Each section is compulsory. However, there are

internal choices in some questions.

2. Section A has 18 MCQ’s and 02 Assertion-Reason based questions of 1 mark each.

3. Section B has 5 Very Short Answer (VSA)-type questions of 2 marks each.

4. Section C has 6 Short Answer (SA)-type questions of 3 marks each.

5. Section D has 4 Long Answer (LA)-type questions of 5 marks each.

6. Section E has 3 source based/case based/passage based/integrated units of assessment (4 marks each) with sub
parts.

Section A
1. sin 3x−sin x

cos x−cos 3x
is equal to [1]

a) cot 2x b) - cot 2x

c) - tan 2x d) tan 2x

2. Let f(x) = x3 . Then, dom (f) and range (f) are respectively [1]

a) R and R+ b) R+ and R

c) R+ and R+ d) R and R

3. A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean [1]
is

a) 7.6 b) 6.4

c) 8.6 d) 10.6
−−−−−−−−−−−−−−−−−−−−−−−
−−−−−−−−−−−−−−−−−−
4. If y =
−−−− −
√x + √x + √x + … + to∞ ∞ then
dy
= [1]
dx

a) 1

2y+1
b) 1

2y−1

−−−
c) d)
x x
√
y+1 y+1

5. The two lines ax + by = c and a′x + b′y = c′ are perpendicular if [1]

a) ab′ = ba′ b) aa′ + bb′ = 0

c) ab + a′b′ = 0 d) ab′ + ba′ = 0

6. A plane is parallel to yz-plane so it is perpendicular to: [1]

Page 1 of 17
 a) y-axis b) none of these

c) z-axis d) x-axis
5 5

7. If z = (
√3
+
i
) + (
√3
−
i
) , then [1]
2 2 2 2

a) Re(z)=0 b) Re(z) > 0, Imz) >

c) Re(z) > 0, Irs(z) < 0 d) Im(z)=0

8. 0! is always taken as [1]

a) 1 b) 2

c) ∞ d) 0
9. lim
sin x

x
= [1]
x→∞

a) 2 b) 1

c) ∞ d) 0
10. tan 15° = ? [1]

a) (√2+1)
b) (√3+1)

(√2−1) (√3−1)

c) (√3−1)
d) (√2−1)

(√3+1) (√2+1)

11. The set of all prime numbers is [1]

a) an infinite set b) a singleton set

c) a multi set d) a finite set

–
12. The integral part of ( √2 + 1)
6
is [1]

a) 98 b) 96

c) 99 d) 100
13. ∑
n

r=0
r
4 .
n
Cr is equal to [1]

a) 6n b) 5-n

c) 4n d) 5n

14. Solve the system of inequalities: [1]

x+7 2x+1
> 2, > 5
x−8 7x−1

a) (4, 8) b) (3, 6)

c) no solution d) (2, 5)
15. If A = {1, 2, 3, 4, 5, 6} then the number of proper subsets is [1]

a) 63 b) 36

c) 64 d) 25
16. cos 15° = ? [1]

a) b)
(√3+1) (√3+1)

√2 2√2

c) (√3−1)
d) (√3−1)

2√2 √2

dy
17. If y = sin x+cos x

sin x−cos x
, then dx
at x = 0 is equal to [1]

Page 2 of 17
 a) 0 b) -2

c) 1

2
d) Does not exist
18. The total number of 9 digit numbers which have all different digits is [1]

a) 9! b) 10 × 10!

c) 10! d) 9 × 9!
19. Assertion (A): Let A = {a, b} and B = {a, b, c}. Then, A ⊄ B. [1]
Reason (R): If A ⊂ B, then A ∪ B = B.

a) Both A and R are true and R is the correct b) Both A and R are true but R is not the
explanation of A. correct explanation of A.

c) A is true but R is false. d) A is false but R is true.

20. Assertion (A): A sequence is said to definite if it has finite no of terms. [1]
Reason (R): The sequence whose nth term if
n
2

n
if 2, 2, 8

3
, 4...

a) Both A and R are true and R is the correct b) Both A and R are true but R is not the
explanation of A. correct explanation of A.

c) A is true but R is false. d) A is false but R is true.

Section B
21. If A and B are any two non-empty sets, then prove that A × B = B× A ⇔ A = B [2]
OR
Find the domain of f(x) =
1

x+2
.
22. Evaluate: lim 1−cos 4θ

1−cos 6θ
. [2]
θ→0

23. Two dice are thrown simultaneously. Find the probability of getting a total of at least 10. [2]
OR
If A and B are two events associated with a random experiment such that P(A) = 0.25, P(B) = 0.4 and P(A or B) =
0.5, find the values of
i. P(A and B)
ii. P(A and B
¯
)
24. Is A = {x : x ∈ N, 1 < x ≤ 2} null set? [2]
25. Find the coordinates of the point which divides the join of A(-5, 11) and B(4, -7) in the ratio 2 : 7. [2]
Section C
n

[3]
2 [1⋅3⋅5.....(2n−1)]
26. Prove that : 2n
Cn =
n!
.
27. Find the point in yz-plane which is equidistant from the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2). [3]
−
28. Using binomial theorem, expand: ( √x + √y)
8
[3]
OR

In the expansion of (x + a)n, sums of odd and even terms are P and Q respectively, prove that

i. 2(P2 +Q2) = (x + a)2n + (x - a)2n

ii. P2 - Q2 = (x2 - a2)n

29. If f(x) = mx + c and f(0) = f'(0) = 1. What is value of f(2)? [3]
OR

Page 3 of 17
 Differentiate the function: 3x-5
30. The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term [3]
decreased by 1, the resulting numbers are in A.P. Find the numbers.
OR
The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.
39

10

31. Out of 25 members in a family, 12 like to take tea, 15 like to take coffee and 7 like to take coffee and tea both. [3]
How many like
i. at least one of the two drinks
ii. only tea but not coffee
iii. only coffee but not tea
iv. neither tea nor coffee
Section D
32. Find the mean and standard deviation for the following data: [5]

Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

Frequency 3 2 4 6 5 5 5 2 8 5

33. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the [5]
length of the latus rectum of the ellipse.
2
2
x y
+ = 1
100 400

OR
Find the lengths major and minor axes, coordinates of the vertices, coordinates of the foci, eccentricity, and length of
the latus rectum of the ellipse 9x2 + y2 = 36.
34. Solve the following system of linear inequalities. [5]
2(2x + 3) - 10 < 6(x - 2)
and
2x−3 4x
+ 6 ≥ 2+
4 3

35. If sin x =
√5
and x lies in the 2nd quadrant, find the values of cos , sin
x x
and tan .x
[5]
3 2 2 2

OR

Prove that: cos 36o cos 42o cos 60o cos 78o = 1

16
.
Section E
36. Read the following text carefully and answer the questions that follow: [4]
Representation of a Relation
A relation can be represented algebraically by roster form or by set-builder form and visually it can be
represented by an arrow diagram which are given below
i. Roster form In this form, we represent the relation by the set of all ordered pairs belongs to R.
ii. Set-builder form In this form, we represent the relation R from set A to set B as R = {(a, b): a ∈ A, b ∈ B
and the rule which relate the elements of A and B}.
iii. Arrow diagram To represent a relation by an arrow diagram, we draw arrows from first element to second
element of all ordered pairs belonging to relation R.
Questions:
i. If n(A) = 3 and B = {2, 3, 4, 6, 7, 8} then find the number of relations from A to B. (1)
ii. If A = {a, b} and B = {2, 3}, then find the number of relations from A to B. (1)

Page 4 of 17
 iii. If A = {a, b} and B = {2, 3}, write the relation in set-builder form. (2)
OR
Express of R = {(a, b): 2a + b = 5; a, b ∈ W} as the set of ordered pairs (in roster form). (2)
37. Read the following text carefully and answer the questions that follow: [4]
Two students Ankit and Vinod appeared in an examination. The probability that Ankit will qualify the
examination is 0.05 and that Vinod will qualify is 0.10. The probability that both will qualify is 0.02.
i. Find the probability that atleast one of them will qualify the exam. (1)
ii. Find the probability that atleast one of them will not qualify the exam. (1)
iii. Find the probability that both Ankit and Vinod will not qualify the exam. (2)
OR
Find the probability that only one of them will qualify the exam. (2)
38. Read the following text carefully and answer the questions that follow: [4]
The conjugate of a complex number z, is the complex number, obtained by changing the sign of imaginary part
of z. It is denoted by z̄ .
The modulus (or absolute value) of a complex number, z = a + ib is defined as the non-negative real number
−−−−−−
√a2 + b2 . It is denoted by |z|. i.e.
−−−−−−
2 2
|z| = √a + b

Multiplicative inverse of z is z̄

2
. It is also called reciprocal of z.
|z|

2
z z̄ = |z|

i. If f(z) = 7−z

2
, where z = 1 + 2i, then find |f(z)|. (1)
1−z

ii. Find the value of(z + 3)(z̄ + 3). (1)

iii. If (x - iy) (3 + 5i) is the conjugate of -6 - 24i, then find the value of x + y. (2)
OR
If z = 3 + 4i, then findz̄ . (2)

Page 5 of 17
 Solution
Section A
1. (a) cot 2x
(C+D) (C−D)
Explanation: Using sin C - sin D = 2 cos 2
sin
2
(C+D) (C−D)
and cos C - cos D = −2 sin 2
sin
2
, we get
4x 2x
2 cos( ) sin( )
sin 3x−sin x
= cot 2x.
2 2 cos 2x sin x
= =
cos x−cos 3x 4x 2x sin 2x sin x
2 sin( ) sin( )
2 2

2.
(d) R and R
Explanation: f(x) = x3
f(x) can assume any value, so domain of f(x) is R
The Range of the function can be positive or negative Real numbers, as the cube of any number depends on the sign of the
number, So Range of f(x) is R
3.
(c) 8.6
Explanation: First we arrange score in the ascending order
Then scores are : 34, 38, 42, 44, 46, 48, 54, 55, 63, 70
As there are 10 items in this data,
So, median will be the mean of fifth and sixth term
46+48
∴ Median = 2
= 47
Now, deviation from median for each value
di = 13, 9, 5, 3, 1, 1, 7, 8, 16, 23
∑ di 86
∴ Required Mean deviation = 10
=
10
= 8.6

4.
(b) 2y−1
1

−−−−−−
Explanation: y = √(x + y)
y2 = x + y
2yy' = 1 + y'
5.
(b) aa′ + bb′ = 0
′
−a −a
Explanation: We know that Slope of the line ax + by = c is b
, and the slope of the line a′x + b′y = c′ is ′
The lines are
b

perpendicular if tan θ = 5−x

3
(1)
′
−a −a ′ ′
= −1 or aa + bb = 0
b ′
b

6.
(d) x-axis
Explanation: Any plane parallel to yz-plane is perpendicular to x-axis.
7.
(d) Im(z)=0
Explanation: From given ,
2 3
√3 √3 2 √3
5 5 i 5 i
z = 2 C0 + C2 + C4
2 2 2 2 2

Since t2 = -1 and t4 = 1, will not contain any i and hence I m

(z) = 0

8. (a) 1
Explanation: We have nPr = n!

(n−r)!
.....(i)

Page 6 of 17
 Number of ways you can arrange n thing in n available spaces = n!
nP = n! ....(ii)
⇒ n

But from (i) we get nPn = n!

=
n!
.....(iii)
(n−n)! 0!

n!
Now from (ii) and (iii) we get 0!
= n! ⇒ 0! = 1

9.
(d) 0
Explanation: lim
sin x

x
x→∞

Let x = 1

x → ∞

∴ y → 0
1
sin
y
= lim
1
y→0
y

1
= lim y sin
y
y→0

1
= lim y × lim sin
y
y→0 y→0

1
= 0 × lim sin
y
y→0

=0
10.
( √3−1)
(c)
( √3+1)

1
(1− )
∘ ∘ √3 ( √3−1)
tan 45 −tan 30
Explanation: tan 15° = tan(45° - 30°) = 1+tan 45
∘
tan 30
∘ = =
1 ( √3+1)
(1+ )
√3

11. (a) an infinite set

Explanation: Set A = {2, 3, 5, 7,...} so it is infinite.
12.
(c) 99
Explanation: We have (1 + x)n = 1 + C (x) + C (x) + ....+ (x)n n
1
n
2
2

– – – – – – 5 – 6
Hence (√2 + 1) = 1 + C (√2) + C (√2) + C (√2) + C (√2) + C
6 6
1
6
2
2 6
3
3 6
4
4 6
5 (√2) + (√2)
– – – –
⇒ (√2 + 1) = 1 + 6(√2) + 15 × 2 + 20 × 2(√2) + 15 × 4 + 6 × 4(√2) + 8
6

–
= 99 + 70√2
–
Hence integral part of (√2 + 1) = 99 6

13.
(d) 5n
Explanation: ∑ n

r=0
r
4 .
n
Cr=4 0
⋅
n
C0 + 4
1 n
⋅ C1 + 4
2 n
⋅ C2 + ... + 4 n n
⋅ Cn

= 1 + 4. C + 4
n
1
2
.
n
C2 + .... + 4 n n
⋅ Cn

= (1 + 4)n = 5n
14.
(c) no solution
x+7
Explanation: x−8
> 2

x+7
⇒ − 2 > 0
x−8

x+7−2(x−8)
⇒
x−8
>0
x+7−2x+16
⇒
x−8
>0
(23−x) a
⇒
x−8
> 0 [∵ b
> 0 ⇒ (a > 0 and b > 0) or (a < 0 and b < 0)]
⇒ (23 - x > 0 and x - 8 > 0) or (23 - x < 0 and x - 8 < 0)
⇒ (x < 23 and x > 8) or (x > 23 and x < 8)
⇒ 8 < x < 23 [Since x > 23 and x < 8 is not possible]
⇒ xϵ (8, 23)
2x+1
Now 7x−1
>5

Page 7 of 17
 2x+1
⇒
7x−1
-5>0
2x+1−5(7x−1)
⇒
7x−1
>0
2x+1−35x+5
⇒
7x−1
>0
(6−33x) a
⇒
7x−1
> 0 [∵ b
> 0 ⇒ (a > 0 and b > 0) or (a < 0 and b < 0)]
⇒ (6 - 33x > 0 and 7x - 1 > 0) or (6 - 33x < 0 and 7x - 1 < 0)
⇒ (x < and x > ) or (x > and x < )
6 1 2 1

33 7 11 7
1 2
⇒ < x <
7 11

⇒ x ∈ (
1

7
,
11
2
) [Since x > 11
2
and x < 1

7
is not possible]
x+7 2x+1
Hence, the solution of the system x−8
> 2,
7x−1
> 5 will be (8, 23) ∩ ( 1

7
,
2

11
) = ϕ

15. (a) 63
Explanation: 63
The no. of proper subsets = 2n - 1
Here n(A) = 6
In case of the proper subset, the set itself is excluded that's why the no. of the subset is 63. But if it is asked no. of improper or
just no. of subset then you may write 64
So no. of proper subsets = 63
16.
( √3+1)
(b)
2√2

Explanation: cos 15° = cos (45° - 30°) = cos 45°cos 30° + sin 45° sin 30°
1 √3 1 1 ( √3+1)
= ( × )+ ( × ) =
√2 2 √2 2 2√2

17.
(b) -2
sin x+cos x
Explanation: Given, y = sin x−cos x

dy (sinx−cosx)(cosx−sinx)−(sin x+cos x)(cos x+sin x)

=
dx 2
(sin x−cos x)
2 2
−(sin x−cos x ) −(sin x+cos x )
=
2
(sin x−cos x)

2 2 2 2
−[sin x+ cos x−2 sin x cos x+ sin x+ cos x+2 sin x cos x]

=
2
(sin x−cos x)

−2
=
2
(sin x−cos x)

dy −2 −2
∴ ( ) = = = −2
dx 2 2
at x=0 (sin 0−cos 0) (−1)

18.
(d) 9 × 9!
Explanation: We have to form 9 digit numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and we know that 0 can not be put on extremely
left place. Therefore, first place from the left can be filled in 9 ways.
Now repetition is not allowed. Therefore, the remaining 8 places can be filled in 9!
∴ The required number of ways = 9 × 9!

19.
(d) A is false but R is true.
Explanation: Assertion A = {a, b}, B = {a, b, c}
Since, all the elements of A are in B. So,
A⊂B
Reason ∵ A ⊂ B
⇒ A ∪ B = B

Hence, Assertion is false and Reason is true.

20.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation: Assertion is true.
Reason

Page 8 of 17
 n

Let tn = 2

Putting n = 1, 2, 7, x
8
t1 = 2, t2 = 2, t3 = 3
, t4 = x
8
80 the sequence is 2, 2, 3
,4
Reason is also correct but not the correct explanation for Assertion.
Section B
21. First, let A = B. Then, we have to prove that A × B = B × A
Now, A = B
⇒ A × B = A × A and B × A = A × A

⇒ A× B= B× A

Conversely, let A × B = B × A . Then, we have to prove that A = B.

Let x be an arbitrary element of A. Then,
x ∈ A

⇒ (x, b) ∈ A × B for all b∈B.

⇒ (x, b) ∈ B × A

⇒ x ∈ B

∴ A⊆ B

Let y be an arbitrary element of A. Then,

y ∈ B

⇒ (a, y) ∈ A × B for all a ∈ A

⇒ (a, y) ∈ B × A

⇒ y ∈ A

∴ B⊆ A

Hence, A = B.
Hence, A × B = B × A ⇔ A = B
OR
Here f (x) = 1

x+2

f(x) assume real values for all real values of x except for x + 2 = 0 i.e. x = - 2.
Thus domain of f (x) =R - {-2}.
1−cos(4θ)
22. We have:lim [ 1−cos 6θ
]
θ→0

{∵ 1 - cos A = 2 sin2 (
2

lim [
2 sin

2
2θ
]
A

2
}
)
θ→0 2 sin 3θ

⎡ 2 ⎤
2 (2θ)
sin 2θ
lim ⎢ × ⎥
2 sin 2 3θ
θ→0 (2θ) 2
⎣ ×(3θ ) ⎦
2
(3θ)

2 2
sin 2θ 3θ 4
lim [( ) × ( ) × ]
2θ sin 3θ 9
θ→0

4 sin x
= [∵ lim = 1]
9 x
X→0

23. We know that in a single throw of two dice, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space of the event and is given by
n(S) = 36.
Let E5 = event of getting a total of at least 10. Then,
E5 = event of getting a total of 10 or 11 or 12 = ((4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}.
⇒ n(E5) = 6
n( E5 )
∴ P(E3) = =
36
6
=
1

6
n(S)

OR
i. It is given that
: P(A) = 0.25, P(A or B) = 0.5 and P(B) = 0.4
To find : P(A and B)
Formula used : P(A or B) = P(A) + P(B) - P(A and B)
Substituting the value in the above formula we get,

Page 9 of 17
 0.5 = 0.25 + 0.4 - P(A and B)
0.5 = 0.65 - P(A and B)
P(A and B) = 0.65 - 0.5
P(A and B) = 0.15
ii. Given : P(A) = 0.25, P(A and B) = 0.15 ( from part (i))
To find : P(A and B¯
)
Formula used : P(A and B ¯
) = P(A) - P(A and B)
Substituting the value in the above formula we get,
P(A and B
¯
) = 0.25 - 0.15
P(A and B
¯
) = 0.10
P(A and B) = 0.10
¯

24. We know that,Natural numbers = 1, 2, 3, 4, 5, 6, 7,…

Natural number greater than 1 (1 < x) = 2, 3, 4, 5, ...
Natural number less than or equal to 2 (x ≤ 2) = 2
⇒ one element in this set

∴ It is not a null set.

25. Let P(x, y) be the point that divides the join of A(-5, 11) and B(4, -7) in the ratio 2 : 7
We know that: If m1 : m2 is the ratio in which the join of two points is divided by another point (x, y), then
m1 x2 + m2 x1
x=
m1 + m2
m1 y + m2 y
2 1
y=
m1 + m2

Here, x1 = -5, x2 = 4, y1 = 11 and y2 = -7

Substituting,we get
2×4+7×−5
x =
2+7
8−35
x =
9
−27
x =
9

⇒ x = -3
2×−7+7×11
y =
2+7
−14+77
y =
9
63
y =
9

⇒ y=8
Thus, the coordinates of the point which divided the join of A(-5, 11) and B(4, -7) in the ratio 2 : 7 is (-3, 8).
Section C
n
2 [1⋅35.....(2n−1)]
26. 2n
Cn =
n!
(2n)!
=
n!n!

(2n)(2n−1)(2n−2)(2n−3).....

4⋅3⋅2⋅1
=
n!n!
n
[2 ....4⋅2][(2n−1)....3⋅1]
=
n!n!
n
2 [1⋅2⋅....n][1⋅3⋅5.....⋅(2n−1)]
=
n!n!
n
2 ×n![1⋅3⋅5....⋅(2n−1)]
=
n!n!
n
2 [1⋅3⋅5....⋅(2n−1)]
=
n!

27. The general point on yz plane is D(0, y, z).

Consider this point is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).
∴ AD = BD
−−−−−−−−−−−−−−−−−−−−−− − −−−−−−−−−−−−−−−−−−−−−− −
√(0 − 3)2 + (y − 2)2 + (z + 1)2 = √(0 − 1)2 + (y + 1)2 + (z − 0)2

Squaring both sides,

(0 - 3)2 + (y - 2)2 + (z + 1)2 = (0 - 1)2 + (y + 1)2 + (z - 0)2
9 + y2 - 4y + 4 + z2 + 2z + 1 = 1 + y2 + 2y + 1+ z2
-6y + 2z + 12 = 0 ….(1)
Also, AD = CD
−−−−−−−−−−−−−−−−−−−−−− − −−−−−−−−−−−−−−−−−−−−−− −
√(0 − 3)2 + (y − 2)2 + (z + 1)2 = √(0 − 2)2 + (y − 1)2 + (z − 2)2

Page 10 of 17
 Squaring both sides,
(0 - 3)2 + (y - 2)2 + (z + 1)2 = (0 - 2)2 + (y - 1)2 + (z - 2)2
9 + y2 - 4y + 4 + z2 + 2z + 1 = 4 + y2 - 2y + 1+ z2 - 4z + 4
-2y + 6z + 5 = 0 ….(2)
By solving equation (1) and (2) we get
−3
y= 31

16
z= 16
31 −3
The point which is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2) is ( 16
, 16
).
−
28. We hand to find value of (√x + √y)
8

Formula used: n
Cr =
n!

(n−r)!(r)!

(a + b)n = C a + C n
0
n n
1a
n−1
b +
n
C2 a
n−2 2
b +…… +
n
Cn−1 ab
n−1
+
n
Cn b
n

We have, (√−
x + √y)
8

1 1

We can write √−
x as x 2 and √y as y 2

8
1 1

Now, we have to solve for (x 2 + y 2 )

8−0 8−1 1 8−2 2 8−3 3

2 1 2 1 1 1 1
8
= [8C0 (x 2
) ] + [8C1 (x 2
) (y 2
) ] + [8C2 (x 2
) (y 2
) ] + [ C3 (x 2
) (y 2
) ]

8−4 8−5 5 8−6 6

1 4 1 2 1 1
8 1 8 8
+ [ C4 (x 2 ) ( ) ] + [ C5 (x 2 ) (y 2 ) ]+ [ C6 (x 2 ) (y 2 ) ]
y2

8−7 7 8
1 1 1

+ [8C7 (x 2 ) (y 2 ) ] + [8C8 (y 2 ) ]

5 2 1 6 2
8! 8! 8!
= [ (x 2 )] + [ (x 2 ) (y 2 )] + [ (x 2 ) (y 2 )]
0!(8−0)! 1!(8−1)! 2!(8−2)!

5 3 4 4
8! 8!
+[ (x 2 ) (y 2 )] + [ (x 2 ) (y 2 )]
3!(8−3)! 4!(8−4)!

2 5 2 1 5
8! 8! 6 8! 2 8!
+[ (x 2 ) (y 2 )] + [ (x 2 )( )] + [ (x 2 )( )] + [ (y 2 )]
2 2
5!(8−5)! 6!(8−6)! y 7!(8−7)! y 8!(8−8)!

7 1 5
4 3 2
= [1 (x )] + [8 (x 2 ) (y 2 )] + [28 (x ) (y)] + [56 (x 2 )( )]
2
y

3 5 1 2
2 2 2 3 4
+ [70 (x ) (y )] + [56 (x 2
) (y 2
)] + [28 (x ) (y )] + [8 (x 2
) (y 2
)] + [1 (y )]

OR
Here (x + a)n =n C 0x
n n
+ C1 x
n−1
a+ C2 x
n n−2 2
a +. . . +
n
Cn a
n

= P + Q . . . (i)
where P = C x n
0
n n
+ C3 x
n−3 3
a +. . . .

n n−1 n n−3 3
Q= C1 x a+ C3 x a +. . . .

nn
Also (x − a) = C x − C x a+ C x
n n
0
n n n−1 n
2
n−2 2
a +. . . + (−1) Cn a
n
. . . (ii)
=P-Q
(i) Squaring and adding (i) and (ii) we have
(x + a)2n + (x - a)2n = (P + Q)2 + (P - Q)2
= P2 + Q2 + 2PQ + P2 + Q2 - 2PQ
= 2P2 + 2Q2 = 2(P2 +Q2)
(ii) Multiplying(i) and (ii) we have
(x + a)n (x - a)n = (P + Q)(P - Q)
(x2 - a2)n = P2 - Q2
29. We have,
f(x) = mx + c ....(i)
Differentiating with respect to x, we get
f'(x) = m.1 + 0
⇒ f'(x) = m ....(ii)

Put, x = 0 in (i) and (ii), we get

f(0) = c and f'(0) = m

Page 11 of 17
 ⇒ 1 = c and 1 = m [ ∴ f(0) = f'(0) = 1]
Put the values of m and c in f(x) = mx + c, we get f(x) = x + 1.
∴ f(2) = 2 + 1 = 3. [Put x = 2 in f(x) = x +1]

OR
We have,
d n n−1
x = nx
dx

Therefore,
d

dx
3x
−5
= 3(-5)x-5-1
= -15x-6
a
30. Let three number in G.P. are r
, a, ar
Here,
a
× a × ar = 729
r

3
⇒ a = 729

⇒ a=9
From the given conditions we can write ,
a a
( × a) + (a × ar) + ( × ar) = 819
r r
81
⇒ + 81r + 81 = 819
r
9
⇒ + 9r + 9 = 91
r

⇒ 9 + 9r2 + 9r = 91r
⇒ 9r2 - 82r + 9 = 0
⇒ 9r2 - 81r - r + 9 = 0
⇒ 9r(r - 9) -1(r - 9) = 0

r = 9, 1

So, required G.P. are

81, 9, 1, ....
or, 1, 9, 81, ....
OR
Let a

r
, a, ar be first three terms of the given G.P.
a

r
+ a + ar =
39

10
...(i)
(
a

r
)(a)(ar) = 1 ...(ii)
From (ii) we obtain a3 = 1 ⇒ a = 1 (considering real roots only)
Substituting a = 1 in equation (i), we obtain
1 39
+ 1 + r=
r 10

2 39
⇒ 1 + r+ r = r
10

⇒ 10 + 10r + 10r2 - 39r = 0

⇒ 10r2 - 29r + 10 = 0
⇒ 10r2 - 25r - 4r + 10 = 0
⇒ 5r(2r - 5)-2(2r - 5) = 0
⇒ (5r - 2)(2r - 5) = 0

⇒ r= or 2

5
5

corresponding terms of the G.P

i. when r = 2

5
5 2
⇒
2
, 1, 5
5
ii. when r = 2

⇒
2

5
, 1, 5

31. Given that, n(T) = 12

n(C) = 15
n(T ∩ C) = 7
i. n(T ∪ C) = n(T) + n(C) - n(T ∩ C)
= 12 + 15 - 7

Page 12 of 17
 n(T ∪ C) = 20
20 members like at least one of the two drinks.
ii. Only tea but not coffee
= n(T) - n(T ∩ C)
= 12 - 7
=5
iii. Only coffee but not tea
= n(C) - n(T ∩ C)
= 15 - 7
=8
iv. Neither tea nor coffee
= n(U) - n(T ∪ C)
= 25 - 20
=5
Section D
32. We make the table from the given data:

Class marks Mid value (xi) di = xi -a = xi - 45 fi fidi d

i
2
fid 2
i

0-10 5 -40 3 -120 1600 4800

10-20 15 -30 2 -60 900 1800

20-30 25 -20 4 -80 400 1600

30-40 35 -10 6 -60 100 600

40-50 45 0 5 0 0 0

50-60 55 10 5 50 100 500

60-70 65 20 5 100 400 2000

70-80 75 30 2 60 900 1800

80-90 85 40 8 320 1600 12800

90-100 95 50 5 250 2500 12500

∑ fi = 45 ∑ fidi = 460 ∑ fid = 38400

2
i

Let a = 45.
∑ fi di
∴ Mean = a + ∑f
i
460
= 45 + 45

= 45 + 10.22 = 55.22
−−−−−−−−−−−−−−−
2 2
∑ fi d ∑ fi di
∴ Standard deviation = √ i
− ( )
∑ fi ∑ fi

−−−−−−−−−−−−−
=√ 38400

45
− (10.22)
2

−−−−−−−−−−−−−
= √853.33 − 104.45
−−−−−
= √748.88
= 27.36
2 2
y
33. The equation of given ellipse is x

100
+
400
= 1

Now 400 > 100 ⇒ a2 = 400 and b2 = 100

2
y 2

So the equation of ellipse in standard form is 2

+
x

2
= 1
a b

∴ a2 = 400 ⇒ a = 20 and b2 = 100 ⇒ b = 10

−−−−−−
We know that c = √a 2
− b
2

−−−−−− −− −−− –
∴ c = √400 − 100 = √300 = 10√3
–
∴ Coordinates of foci are (0, ±c) i.e. (0, ±10√3)
Coordinates of vertices are (0, ±a) i.e. (0, ±20)

Page 13 of 17
 Length of major axis = 2 a = 2 × 20 = 40
Length of minor axis = 2 b = 2 × 10 = 20
10√3 √3
Eccentricity (e) = c

a
=
20
=
2
2
2b 2×100
Length of latus rectum = a
=
20
= 10

OR
Given that:
9x2 + y2 = 36.
After divide by 36 to both the sides, we get
2 2
y
9

36
2
x +
1

36
y
2
= 1 ⇒
x

4
+
36
= 1 ... (i)
Now, the above equation is of the form,
2 2
y
x

2
+
2
= 1 ... (ii)
b a

Comparing Eq. (i) and (ii), we get

a2 = 36 and b2 = 4 ⇒ a = √36 and b = √4 ⇒ and a = 6 and b = 2
−− –

i. Length of major axes

∴ Length of major axes = 2a = 2 × 6 = 12 units

ii. Length of major axes

∴ Length of major axes = 2b = 2 × 2 =4 units

iii. Coordinates of the Vertices

∴ Coordinates of the Vertices = (0, a) and (0, -a) = (0, 6) and (0, -6)

iv. Coordinates of the foci

As we know that, Coordinates of foci = (0, ±c) where c2 = a2 - b2
Now
c2 = 36 - 4 ⇒ c2 = 32 ⇒ c = √32 ... (iii)
−−

−−
Coordinates of foci = (0, ±√32)
∴

v. Eccentricity
√32
As we know that, Eccentricity = c

a
⇒ e=
6
[from (iii)]
vi. Length of the Latus Rectum
2
2 2×(2)
As we know that, Length of Latus Rectum = 2b

a
=
6
=
8

6
4
=
3

34. The given system of linear inequalities is

2(2x + 3) -10 < 6(x - 2) ....(i)
2x−3 4x
and 4
+ 6 ≥ 2 +
3
...(ii)
From inequality (i), we get
2(2x + 3) - 10 < 6(x - 2)
⇒ 4x + 6 - 10 < 6x - 12

⇒ 4x - 4 < 6x - 12

⇒ 4x - 4 + 4 < 6x - 12 + 4 [adding 4 on both sides]

⇒ 4x < 6x - 8

⇒ 4x - 6x < 6x - 8 - 6x [subtracting 6x from both sides]

⇒ -2x < -8
⇒ 2x > 8 [dividing both sides by - 1 and then inequality sign will change]

⇒ > [dividing both sides by 2]

2x

2
8

∴ ...(iii)
x > 4

Thus, any value of x greater than 4 satisfies the inequality.

∴ Solution set is x ∈ (4, ∞)

The representation of solution of inequality (i) is

From inequality (ii), we get

Page 14 of 17
 2x−3 4x 2x−3+24 6+4x
+ 6 ≥ 2 + ⇒ ≥
4 3 4 3
2x+21 6+4x
⇒ ≥ ⇒ 3(2x + 21) ≥ 4(6 + 4x)
4 3

⇒ 6x + 63 ≥ 24 + 16x

⇒ −10x ≥ −39 ⇒ 10x ≤ 39

10x 39
⇒ ≤
10 10

⇒ ...(iv)
x ≤ 3.9

Thus, any value of x less than or equal to 3.9 satisfies the inequality.
∴ Solution set is x ∈ (−∞, 3.9] .
Its representation on number line is

From Eqs. (iii) and (iv), it is clear, that there is no common value of x, which satisfies both inequalities (iii) and (iv).
Hence, the given system of inequalities has no solution.
35. We know,
sin2 x + cos2 x = 1
cos2 x = 1 - sin2 x
−−−−−
cos x = ±√1 − 5

9
=± 2

Since, x ∈ ( π

2
, π)

cos x will be negative in second quadrant

therefore. cos x = − 2

We know,
cos 2x = 2 cos2 x - 1
−−−−−
x 1−cos x
cos 2
=±√ 2
−−−−−−
−2
1+( ) −
−
√
Now, cos = ±√
x 3 1
= ±
2 2 6

Since, x ∈ ( π

2
, π) ⇒
x

2
∈ (
π

4
,
π

2
)

cos x

2
will be positive in 1st quadrant.
x 1
So, cos 2
=
√6

We know,
cos 2x = 2 cos2 x - 1
cos x = 1 - 2 sin2 x

2
... [∵ cos x = − ] 2

−
2

3
= 1 - 2 sin2 x

2 sin2 x 2
= +12

sin2
2+3
2 x

2
= 3

sin2 = x 5

2 6
−
−
sin2
x 5

2
= ±√ 6

π x π π
Since, x ∈ ( 2
, π) ⇒
2
∈ (
4
,
2
)

x
sin 2
will be positive in 1st quadrant
So,
−
−
sin x

2
=√ 5

We know,
x
sin

tan x

2
= 2

x
cos
2

5
√
6

tan x

2
= 1

√6

Page 15 of 17
 –
tan =√5 x

2
−
−
x x x 5 –
Hence, values of cos , sin , tan 2 2 2
are 1
,√ 6
and √5
√6

OR
We have to prove that cos 36o cos 42o cos 60o cos 78o = 1

16
.
LHS = cos 36o cos 42o cos 60o cos 78o
By regrouping the LHS and multiplying and dividing by 2 we get,
= cos 36o cos 60o (2 cos 78o cos 42o)
1

But 2 cos A cos B = cos (A + B) + cos (A - B)

Then the above equation becomes,
= cos 36o cos 60o (cos(78o + 42o) + cos(48o - 42o))
1

= cos 36o cos 60o (cos(120o) + cos(36o))

1

= cos 36o cos 60o (cos(180o - 60o) + cos(36o))

1

But cos(90o - θ) = sinθ and cos(180o - θ)= -cos(θ).

Then the above equation becomes,
= cos 36o cos 60o (-cos(60o) + cos(36o))
1

Now, cos(36o) =
√5+1

cos(60o)
1
= 2

Substituting the corresponding values, we get

√5+1 √5+1
= 1

2
(
4
)(
1

2
) (−
1

2
+
4
)

√5+1 √5+1−2
=( 16
)(
4
)

2 2
( √5) − 1
=( 16×4
)

5−1
=( 64
)

16

LHS = RHS
Hence proved.
Section E

36. i. Number of relations = 2mn

= 23× 6 = 218
ii. Number of relations = 2mn
= 22× 2 = 24 = 16
iii. R = {(x, y): x ∈ P, y ∈ Q and x is the square of y}
OR
Here, W denotes the set of whole numbers.
We have 2a + b = 5 where a, b ∈ W
∴ a = 0 ⇒ b = 5

⇒ a = 1 ⇒ b = 5 − 2 = 3

and a = 2 ⇒ b = 1
For a > 3, the values of b given by the above relation are not whole numbers.
∴ A = {(0, 5), (1, 3), (2, 1)}

37. i. Let E1 and E2 denotes the events that Ankit and Vinod will respectively qualify the exam.
P (E1 ∪ E2 ) = P(E1) + P(E2) − P(E1 ∩ E2)
= 0.05 + 0.10 - 0.02 = 0.13
ii. Let E1 and E2 denotes the events that Ankit and Vinod will respectively qualify the exam.
Probability of atleast one of them does not qualify
′ ′ ′
= P (E ∪ E ) = P ((E1 ∩ E2 ) )
1 2

= 1 - p(E1 ∩ E2) = 1 - 0.02 = 0.98

Page 16 of 17
 iii. Let E1 and E2 denotes the events that Ankit and Vinod will respectively qualify the exam.
′ ′ ′
= P (E ∩ E ) = P ((E1 ∪ E2 ) )
1 2

= 1 - P(E1 ∪ E2) = 1 - 0.13 = 0.87

OR
Let E1 and E2 denotes the events that Ankit and Vinod will respectively qualify the exam.
The probability that Vinod will not qualify the exam.
Probability that only one of them will qualify the exam = P((E1 - E2) ∪ (E2 - E1))
= P(E1 - E2) + P(E2 - E1)
= P(E1 ∪ E2) - P(E1 ∩ E2)
= 0.13 - 0.02 = 0.11
38. i. Let z = 1 + 2i
−−−− –
⇒ |z| = √1 + 4 = √5
7−z 7−1−2i
Now, f(z) = =
1−z 2 1−(1+2i)
2

6−2i 6−2i
= 1−1−4i2−4i
=
4−4i

(3−i)(2+2i)
=
(2−2i)(2+2i)
2
6−2i+6i−2i 6+4i+2
= 2
=
4+4
4−4i
8+4i
= 8
= 1 +
1

2
i

f(z) = 1 + 1

2
−−−−− −−−
4+1 √5 |z|
∴ |f(z)| = √1 + 1

4
= √
4
=
2
=
2

ii. Given that: (z + 3)(z̄ + 3)

Let z = x + yi
So (z + 3)(z̄ + 3) = (x + yi + 3)(x - yi + 3)
= [(x + 3) + yi][(x + 3) - yi]
= (x + 3)2 - y2i2
= (x + 3)2 + y2
= |x + 3 + iy|2
= |z + 3|2
iii. The conjugate of -6 - 24i is -6 + 24i.
It is given that -6 + 24i = (x – iy) (3 + 5i)
-6 + 24i = 3x + 5xi - 3iy -5yi2
-6 + 24i = (3x + 5y) + i(5x - 3y)
Comparing the real and imaginary parts,
3x + 5y = -6
5x - 3y = 24
Solving these two equations we get x = 3 and y = -3.
Therefore, x = 3 and y = -3
Then x + y = 3 - 3 = 0
OR
z = 3 + 4i
⇒ z̄ = 3 - 4i

Page 17 of 17