Solution

ANNUAL EXAMINATION 2024-2025

Class 11 - Mathematics
Section A
1.
(c) {3, 5, 9}
Explanation:
The union of two sets A and B is the set of elements in A, or B, or both.
So smallest set A = {3, 5, 9}

2.
(c) A × B = {(1, a), (1, b), (2, a), (2, b)}
Explanation:
The set of all ordered pairs (a, b) such that a ∈ A and b ∈ B is called cartesian product of sets A and B.
∴ A × B = {(1, a), (1, b), (2, a), (2, b)}

3.
(c) ( 9π

10
)

Explanation:
c c c
∘ c ∘ π ∘ π 9π
180 = π ⇒ 1 = ( ) ⇒ 162 = ( × 162) = ( )
180 180 10

4. (a) 0
Explanation:
i109 + i114 + i119 + i124 = [1 + i4× 1 + (i4 )2× i2 + (i4)3× i3] = i109[1 + i + i2 + i3]
= i109× [1 + i - 1 - i] = i109× 0 = 0
5.
(d) x ∈ (10, ∞)
Explanation:
- 3x + 17 < - 13
⇒ - 3x + 17 - 17 < - 13 - 17
⇒ - 3x < - 30
−3x −30
⇒ >
−3 −3

⇒ x > 10
⇒ x ∈ (10, ∞)

6.
(d) 4536
Explanation:
To form a four-digit number with distinct digits we can use any four digits from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 without
repetition.
Since 0 cannot come as the first digit of a four-digit number( then it will be a three-digit number) the first place can be filled by
any of the 9 digits other than 0.
Now we have 9 more digits left and since repetition is not allowed the second place can be filled by any of these 9 digits.
Similarly the third and fourth can be filled by any of the 8 and 7 digits respectively.
Hence we get the four places can be filled together by 9 × 9 × 8 × 7 = 4536 different ways.

7.
(d) 1

1 / 10
 Explanation:
The pascal's triangle is given by

8. (a) y - 2 = 3 (x – 1)
Explanation:
Line is parallel to the line y = 3x - 1.
Therefore, slope of the line is ‘3’.
Also, line passes through the point (1,2).
Thus, equation of the line is: y - 2 = 3(x - 1)
9.
(d) 25π sq units
Explanation:
Given centre = (1,2)
−−−−−−−−−−−−−−−
Therefore radius = √(4 − 1) + (6 − 2) = 5 units 2 2

Area of the circle is πr = 25π sq units 2

10. (a) z = 0
Explanation:
For every point on xy-plane, z co-ordinate is always zero.
−−
11. (a) √50
Explanation:
Given points A(3, 4, 5) and the given O(0, 0, 0)
−−−−−−−−−−−−−−−−−−−−−− −
2 2
∴ √(3 − 0) + (4 − 0) + (5 − 0)
2
[using distance fomula]
−−−−−−−− − −−
= √9 + 16 + 25 = √50

12. (a) does not exists

Explanation:
|x|
R.H.S = lim
x
=
x

x
=1
+
x→0

|x| −x
and L.H.S = lim
x
=
x
= -1
−
x→0

13. (a) nan-1

Explanation:
n n
x −a
lim
x−a
x→a
n n
x −a
= lim
x−a
[∵ f(x) exists, lim f(x) = lim f(x)]
+ x→a +
x→a x→a
n n
(a+h ) −a
= lim a+h−a
h→0
n
h
[ (1+ ) −1]
a

= lim a n

h
h→0

= an lim [1 + n ⋅
n(n−1) h2
h

a
+
2
... + ... -1]
h→0 2! a

= an lim [
h(h−1)
n

a
+
h
+ ...]
h→0
2! a2

= an
n

= nan-1
14.
(c) 0
Explanation:

2 / 10
 Given that y = √−
x +
1

√x

dy 1 1
= −
dx 2√x 3/2
2x

dy 1 1
( ) = − = 0
dx
at x=1 2 2

15. (a) 1
Explanation:
f′ (x) = x cosx + sinx
So, f ( ) = cos
′ π

2
π

2
π

2
+ sin
π

2
= 1

16. (a) 1700

Explanation:
1700
17.
(b) dependent
Explanation:
S = [(H H H),(H H T),(H T H),(H T T),(T H H),(T H T),(T T H),(T T T)]
P(A) = P(2 heads) = 3

P(B) = P(last one is head) = 4

8
2 1
P (A ∩ B) = = ≠ P (A)P (B)
8 4

Thus, A and B are dependent.

18. (a) A and B are mutually exclusive

Explanation:
We have, A = {2, 4, 6}, B = {1, 3, 5} and C = {4, 5, 6} since A ∩ B = ϕ, therefore A and B are mutually exclusive.
19.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:
Assertion:
(1 + x)n = n c0 + nc x + nc x
1 2
2
… + ncn x
n

Reason:
(1 + (-1))n = n c0 1
n
+ nc (1)
1
n−1
(−1)
1
+ nc (1)
2
n−2 2
(−1) +... + n n−n
cn (1)
n
(−1)

= n − n + n − n + ... (-1)nn
c8 c1 c2 c3 cn

Each term will cancel each other

∴ (1 + (-1))n = 0
Reason is also the but not the correct explanation of Assertion.

20. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
Two circles touch each other,
C1 C2 = r1 ± r2
−−−−−− −−−−− −−−−−
√a2 + b2 = √a2 − c ± √b2 − c

Squaring both side

−−− −−−− − −−−−−
a2 + b2 = (a2 - c) + (b2 - c) ± 2√(a 2 2
− c) (b − c)
−−−−−−−−−−− −
2c = 2 2
±2√(a − c)(b − c)
−−−−−−−−−−− −
c= 2 2
± √(a − c)(b − c)

Again Squaring both side

c2 = (a2 - c)(b2 - c)
c2 = a2b2 - b2c - a2c + c2
c(a2 + b2) = a2b2

3 / 10
 2 2
a +b 1
=
2 2 c
a b
1 1 1
+ =
2 2 c
b a

1 1 1
∴ = +
c 2 2
a b

Section B
21. x2 - 9 = 0 ⇒ (x + 3) (x - 3) = 0
⇒ x = -3 or x = 3

∴ A = {x : x ∈ N, x2 - 9 = 0}= {3} [∵ -3 ∉ N]
and B = {x : x ∈ Z, x2 - 9 = 0} = {-3, 3}.
Therefore, A ≠ B.
1 7 1 −4
22. [( 3
+
3
i) + (4 +
3
i)] − [
3
+ i]

1 4 7 1
= ( + 4 + ) + ( i + i − i)
3 3 3 3

1 4×3 4 7 i 3i
= ( + + ) + ( i + − )
3 3 3 3 3 3

1+12+4 7i+i −3i

= ( ) + ( )
3 3

17 5
= + i
3 3

23. Here, the total number of words will be equal to the number of ways of filling 4 vacant places by the 4 letters of the word ROSE.
i. When the repetition of the letters is not allowed
Here, vacant places are 4
First place can be filled by anyone of the 4 letters.
∴ Number of ways of filling first place = 4

Now, second place can be filled by anyone of the remaining 3 letters.

Number of ways of filling second place = 3
Similarly, third place can be filled by anyone of the remaining 2 letters and fourth place is filled by the last letter.
∴ Number of ways of filling third place = 2

and number of ways of filling fourth place = 1

Since each place can be filled after filling the previous place.
So, by FPM, the required number of ways
= 4 × 3 × 2 × 1 = 24
ii. When the repetition of the letters is allowed
Here, out of 4 vacant places, each place can be filled by 4 letters. So, each place can be filled in succession anyone of 4
different ways.
∴ the required number of ways

= 4 × 4 × 4 × 4

= 256 ways
Hence, a number of 4 letters word is 24 when repetition is not allowed and 256 when repetition is allowed.
24. Given points are A(x1, y1) = (-1, 1) and B(x2, y2) = (2, -4), then equation of line AB is
y2 − y1
y − y1 = (x − x1 )
x2 − x1

−4−1
⇒ y − 1 = (x + 1) [∵ x1 = −1, y1 = 1, x2 = 2, y2 = −4]
2+1

−5
⇒ y − 1 = (x + 1) ⇒ 3y − 3 = −5x − 5
3

⇒ 5x + 3y + 2 = 0

25.

Equation of the required circle is

4 / 10
 (x - 3)2 + (y - 4)2 = 42
or, x2 + y2 - 6x - 8y + 9 = 0
Section C
26. Here f (x) = x2
At x = 1.1
f(1.1) = (1.1)2 = 1.21
f (1) = (1)2 = 1
f (1.1)−f (1) 1.21−1 0.21
∴ = = = 2.1
(1.1−1) 0.1 0.1

27. (i) sin 75° = sin (45° + 30° )

= sin 45° cos 30° + cos 45° sin 30°
[ ∵ sin (A + B) = sin A cos B + sin B cos A ]
1 √3 1 1 √3 1 √3+1
= × + × = + =
√2 2 √2 2 2√2 2√2 2√2
∘ ∘
tan 45 −tan 30
(ii) tan 15 ∘
= tan(45
∘
− 30 ) =
∘

1+tan 45
∘
tan 30
∘

tan A−tan B
[∵ tan(A − B) = ]
1+tan A tan B

√3−1
1
1−
3 √3 √3−1 √3−1
= = = ×
1 √3+1
1+1× √3+1 √3−1
√3
√3

3+1−2√3 4−2√3 –
= = = 2 − √3
3−1 2
2x+5 2x+5
28. We have, x−1
>5⇒ x−1
-5>0
2x+5−5x+5 −3x+10
⇒
x−1
>0⇒ x−1
>0

(x-1)2 > 0 × (x-1)2

−3x+10
⇒ ×
(x−1)

⇒ (-3x + 10) (x - 1) > 0

⇒ (3x - 10) (x - 1) < 0 [Multiplying by - 1 on both sides to make coefficient of x positive]

Thus, product of (3x - 10) and (x - 1) will be negative.

Case I: If 3x - 10 > 0 and x - 1 < 0
⇒ 3x > 10 and x < 1 ⇒ x > and x < 1 10

So, this is impossible, since system of inequalities have no common solution.

Thus, there is no solution of given inequality in this case.
Case II: If 3x - 10 < 0 and x - 1 > 0
⇒ 3x < 10 and x > 1 ⇒ x < and x > 1 10

10 10
⇒ 1<x< 3
⇒ x ∈ (1, 3
)

10
Hence, the solution of given inequality is, 1 < x < 3

29. Using binomial theorem for the expansion of (2x - 3)6 we have
6 6 6 6 5 6 4 2 6 3 3
(2x − 3) = C0 (2x) + C1 (2x) (−3) + C2 (2x) (−3) + C3 (2x) (−3)

6 2 4 6 5 2 6 6
+ C4 (2x) (−3) + C5 2x (−3) + C6 (−3)

= 64x6 + 6.32x5 (-3) + 15.16x4.9 + 20.8x3 (-27) + 15.4x2.81 + 6.2x (-243) + 729
= 64x6 - 576x5 + 2160x4 - 4320x3 + 4860x2 - 2916x + 729

5 / 10
30. It is given that the vertex of the parabola is O(0, 0) and it is symmetric about the y-axis.

So, its equation is x2 = 4ay or x2 = -4ay.

Since the parabola passes through the point P(3, -4), so it lies in the 4th quadrant.
∴ it is a downward parabola. so equation of the parabola is x2 = -4ay.
Let Required equation of parabola is x2 = -4ay.
Since it passes through the point P(3, -4), we have
32 = -4 × a× (-4) ⇒ a = 9

16

So, the required equation is

x2 = -4 × 9

16
y
2 −9
⇒ x = y
4

⇒ 4x2 + 9y = 0
2

31. Given: lim 3

x −x−6

2
x→3 x −3x +x−3

Now, x2 - x-6
= x2 - x - 6
= x2 - 3x + 2x - 6
= x(x - 3)+2(x - 3)
= (x + 2)(x - 3) ...(i)
Dividing x3 - 3x2 + x - 3 by (x - 3), we get

Thus (x - 3) is a factor of x3 - 3x2 + x - 3 ...(ii)

Substituting the values of Eqn. (i) and (ii) in the given expression
(x+2)(x−3)
= lim 2
x→3 (x +1)(x−3)

x+2 3+2 5
= 2
=
9+1
=
10
x +1

= 1

Section D
32. i. n(A × A) = 9
⇒ n(A) ⊂ n(A) = 9 ⇒ n(A) = 3
(-1,0) ∈ A × A ⇒ -1 ∈ A, 0 ∈ A
(0,1) ∈ A × A ⇒ 0 ∈ A, 1 ∈ A
⇒ -1, 0, 1 ∈ A

Also, n(A) = 3 ⇒ A = (-1, 0, 1)

Hence, A = {-1, 0, 1}

6 / 10
 Also, A × A = {-1, 0, 1} × {-1, 0, 1}
= {(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)}
Hence, the remaining elements of A × A are
(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0) and (1, 1).
ii. Given, (A × B) = 6 and (A × B) = {(1, 3), (2, 5), (3, 3)}
We know that Cartesian product of set A = {a, b} & B = {c, d} is A × B = {(a, c), (a, d), (b, c), (b, d)}
Therefore, A = {1, 2, 3} & B = {3, 5}
⇒ A × B = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)}

Thus, remaining elements are A × B = {(1, 5), (2, 3), (3, 5)}
iii. If the set A has 3 elements and set B has 4 elements, then the number of elements in A × B = 12
OR
Clearly, A is the set of all first entries in ordered pairs in A × B and B is the set of all second entries in ordered pairs in A × B
∴ A = {a, b} and B = {1, 2, 3}

33. i. We have, profit = Revenue - Cost

= (60x + 2000) - (20x + 4000)
= 40x - 2000
To earn some profit, 40x - 2000 > 0.
⇒ x > 50

Hence, the manufacturer must sell more than 50 items to realise some profit.
ii. 12x + 7 < -11
12x < -11 - 7
12x < -18
−3
x< 2

iii. 5x - 8 > 40
5x > 40 + 8
5x > 48
x> 48

OR

34. i. From the given figure, it is clear that the coordinates of points A and B are (8, 10, 0) and (-6, 4, 0), respectively.
ii. From the given figure, it is clear that the coordinates of points C and D are (15, -20, 0) and (0, 0, 30), respectively.
iii. From the given figure, it is clear that the coordinates of points A and D are (8, 10, 0) and (0, 0, 30), respectively.
Thus, the equation of line passing through (8, 10, 0) and (0, 0, 30) is:
x−0 y−0 z−30

8−0
= 10−0
= 0−30

x y z−30
⇒
8
= 10
= −30

x y z−30
⇒
4
= 5
= −15
y
⇒
x

4
= 5
= 30−z

15

OR
Here, Coordinates of O = (0, 0, 0)
Coordinates of A = (8, 10, 0)
Coordinates of B = (-6, 4, 0)
Coordinates of C = (15, -20, 0)
−− −−−−− −−−
Also, OA = ∣∣√8 2 2
+ 10 ∣∣ = √164 units
−− −−−−
OB = ∣√ 2
∣
2∣
6 + 4 ∣

−−−−−− −−
= |√36 + 16| = √52 units
−−−−−−−−
and OC = ∣∣√15 + 20 ∣∣ 2 2

−−−−−−−−
= |√225 + 400|
−−−
= √625 = 25 units
∴ The sum of the distances OA, OB, and OC
−−− −−
= √164 + √52 + 25
−− −−
= (2√41 + 2√13 + 25) units

7 / 10
 Section E
35. Given that:
3x2 + 2y2 = 18
After Divide by 18 to both the sides, we get
2 2
3 y

18
2
x +
18
2
y
2
= 1 ⇒
x

6
+
9
= 1 ... (i)
Now, above equation is of the form,
2 2
y
x

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

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

a2 = 9 and b2 = 6 ⇒ a = √9 and b = √6 ⇒ a = 3 and b = √6
– – –

i. Length or major axes

∴ Length of major axes = 2a = 2 × 3 = 6 units

ii. Length or minor axes

– –
∴ Length or minor axes =2b = 2 × √6 = 2√6

iii. Coordinates of the vertices

∴ Coordinates of 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 = 9 - 6 ⇒ c2 = 3 ⇒ c = √3 ... (iii)
–

–
∴ Coordinates of foci = (0, ± √3)

v. Eccentricity
c √3 1
As we know that, Eccentricity = a
⇒ e=
3
= [from (iii)]
√3

vi. Length of the Latus Rectum

2
2 2×( √6)
As we know that, Length of Latus Rectum = 2b

a
=
3
2×6
= = 4
3

3 2
x +3x −9x−2
36. We have to find the value of lim 3
x→2 x −x−6

We have
3 2
x +3x −9x−2
lim
3
x→2 x −x−6

Divide x3 + 3x2 - 9x - 2 by x3 - x - 6

3 2 2
x +3x −9x−2 3x −8x+4
⇒ lim
3
= lim 1 + lim 3
x→2 x −x−6 x→2 x→2 x −x−6

2
3x −2x−6x+4
= 1 + lim 3
x→2 x −x−6

2
3x −2x−6x+4
= 1 + lim
x→2 x3 −x−6

3 2 (3x−2)(x−2)
x +3x −9x−2
⇒ lim
3
= 1 + lim 3
x→2 x −x−6 x→2 x −x−6

Divide x3 - x - 6 by x - 2

8 / 10
 3 2 (3x−2)(x−2)
x +3x −9x−2
⇒ lim
3
= 1 + lim 2
x→2 x −x−6 x→2 (x−2)(x +2x+3)

(3x−2)
= 1 + lim 2
x→2 (x +2x+3)

3×2−2
=1+ 2
2 +2×2+3

=1 + 4

11
15
= 11

37. We have to find the derivative of f(x) = cos x

f (x+h)−f (x)
Derivative of a function f(x) is given by f’(x) = lim = h
{where h is a very small positive number}
h→0

cos x f (x+h)−f (x)

∴ Derivative of f(x) = x
is given as f’(x) = lim = h
h→0
c os(x +h) c os x
−
x +h x

⇒ f (x) = lim
h
h→0

x c os(x +h)−(x +h) c os x

x (x +h) x cos(x+h)−(x+h) cos x

⇒ f (x) = lim
h
= lim
h→0 h→0 h(x)(x+h)

Using the algebra of limits we have:

x cos(x+h)−(x+h) cos x
1
⇒ f (x) = lim × lim
h
h→0 h→0 x(x+h)

x cos(x+h)−(x+h) cos x 1
⇒ f (x) = lim ×
h x(x+o)
h→0

x cos(x+h)−(x+h) cos x
1
⇒ f (x) = lim
x2 h→0
h

1 x cos(x+h)−x cos x−h cos x

⇒ f (x) = lim
2 h
x h→0

Using the algebra of limits, we have:

1 −h cos x x cos(x+h)−x cos x
⇒ f (x) = {lim + lim }
2 h h
x h→0 h→0

1 x(cos(x+h)−cos x)
⇒ f (x) = {− lim cos x + lim }
2 h
x h→0 h→0

Using the algebra of limits we have:

cos x 1 cos(x+h)−cos x
∴ f’(x) = − 2
+
x
lim
h
x h→0

We can’t evaluate the limits at this stage only as on putting value it will take 0

0
form. So, we need to do little modifications.
(A+B) (A−B)
Use: cos A – cos B = – 2 sin ( 2
) sin ( 2
)
2x +h h
−2 sin( ) sin( )
cos x 1
f’(x) = −
2 2
∴ + lim
2 x h
x h→0
h h
sin(x+ ) sin( )

f’(x) = − cos x 1 2 2
⇒ + lim
x2 x h
h→0
2

Using algebra of limits:

h
sin( )

⇒ f’(x) = − cos x

2
+
1

x
lim
h
2
× lim sin(x +
h

2
)
x h→0 h→0
2

By using the formula we get : lim sin x

x
= 1
x→0

9 / 10
 ⇒ f’(x) =− cos x

2
+
1

x
lim sin(x +
h

2
)
x h→0

Put the value of h to evaluate the limit:

cos x cos x sin x
∴ f’(x) =− 2
+
1

x
× sin (x + 0) = − 2
−
x
x x

Hence,
Derivative of f(x) = (cos x)/x is − cos x

2
−
sin x

x
x

38. 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 +
∑ fi

= 45 + 460

45

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

−−−−−−−−−−−−−
38400
=√ 45
− (10.22)
2

−−−−−− −−−−−−−
= √853.33 − 104.45
−−−−−
= √748.88
= 27.36

10 / 10