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Class 12 - Mathematics
Sample Paper - 09 (2024-25)

Maximum Marks: 80
Time Allowed: : 3 hours

General Instructions:

i. This Question paper contains 38 questions. All questions are compulsory.

ii. This Question paper is divided into five Sections - A, B, C, D and E.
iii. In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs) and Questions no. 19 and 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 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.
ix. Use of calculators is not allowed.

Section A

1. Total number of possible matrices of order 3 × 3 with each entry 2 or 0 is

a) 27
b) 81
c) 9
d) 512

[ ] [ ]
1 2 1 0
2. Let A = 3 −5 and B = 0 2 and X be a matrix such that A = BX, then X is equal to

a)
1
2 [ −2
3
4
5 ]
[ ]
2 4
b)
3 −5

[ ]
1 2 4
c)
2 3 −5

[ ]
4 −2
d)
3 −5

3. The value of determinant

| x
x−1
x+1
x | is equal to

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a) 1
b) x2
c) x
d) 0

4. If y = tan-1 ( acos x − bsin x

bcos x + asin x ) then
dy
dx
= ?

a
a) b
b) 1
c) -1
−b
d)
a

→ →
5. If the projections of PQ on OX, OY, OZ are respectively 1, 2, 3 and 4, then the magnitude of PQ is

a) 13
b) 169
c) 19
d) 144

1 − y2

√
dy
6. The solution of the DE + is
dx 1 − x2

a) sin-1 y - sin-1 x = C
b) sin-1 y + sin-1 x = C
c) y + sin-1 y = sin-1 x + C
d) y - sin-1 y = - sin-1 x + C

7. Which of the following statements is correct?

a. Every LPP admits an optimal selection.

b. A LPP admits unique optimal solution.
c. If a LPP admits two optimal solutions it has an infinite solution.
d. The set of all feasible solutions of a LPP is not a convex set.
a) Option (d)
b) Option (a)
c) Option (b)
d) Option (c)

→
8. If b and c are the position vectors of the points B and C respectively, then the position vector of the point D such that BD
→
= 4BC, is

→
a) − 4(→
c − b)
→ →
b) 4( c − 3 b)

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→
c) 4(→
c − b)
→ →
d) 4 c + 3 b

π sec 5 x
9. ∫ 02 dx = ?
( sec 5 x + cosec 5x )
a) 1
π
b)
2
π
c) 4
d) 0

10. If A is a square matrix, then AA is a

a) none of these
b) skew-symmetric matrix
c) symmetric matrix
d) diagonal matrix

11. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

a) Minimum Z = – 13 at (4, 1)
b) Minimum Z = – 14 at (5, 0)
c) Minimum Z = – 12 at (4, 0)
d) Minimum Z = – 15 at (5, 1)

12. If the vectors 3î + λĵ + k̂ and 2î − ĵ + 8k̂ are perpendicular, then λ is equal to

a) 7
b) -14
c) 1/7
d) 14

13. The adjoint of matrix A = a ij = [ ] | |

p
r
q
s
is

| |
s −q
a)
r −p

| |
s −q
b)
−r p

c)
| | s
r −p
q

d) | | 0
0
0
q

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14. A bag X contains 2 white and 3 black balls and another bag Y contains 4 white and 2 black balls. One bag is selected at
random and a ball is drawn from it. Then, the probability of the chosen ball to be white is

2
a) 15
7
b)
15
8
c) 15
14
d) 15

dy
15. Solution of (x + 1) = 2xy is
dx

a) logy = {x + log|x|} + C
b) log | y | = 2(x − log | 1 + x | ) + c
c) log | y | = 2(x + log | 1 − x | ) + c
d) logy = {x − log|x|} + C

16. Vectors A and B are Collinear

a) if they are parallel to the same line irrespective of their magnitudes and directions.
b) if they are have equal magnitude
c) if the direction cosines of one are negatives of the other
d) if they are in the same line
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17. Lt
h→0 ( 3
h √8 + h
1
−
1
2h )
is equal to

1
a) − 48
1
b) 24
1
c) − 24
1
d)
48

18. The angle between the lines x = 1, y = 2 and y = - 1, z = 0 is

a) 0o
b) 30o
c) 60o
d) 90o

19. Assertion (A): f(x) = 2x3 - 9x2 + 12x - 3 is increasing outside the interval (1, 2).
Reason (R): f'(x) < 0 for x ∈ (1, 2).

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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.

20. Assertion (A): The f : R → R given by f(x) = [x] + x is one-one onto.

Reason (R): A function is said to be one-one and onto, if each element has unique image and range of f(x) is equal to
codomain of f(x).

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.
Section B

(
21. tan − 1 − √3 )
OR

( )
Find the value of sin − 1 sin
2π
3

22. Two sides of a triangle have lengths 'a' and 'b' and the angle between them is θ. What value of θ will maximize the area
of the triangle? Find the maximum area of the triangle also.

23. Find the points of local maxima or local minima and corresponding local maximum and Iocal minimum values of
the function. Also, find the points of inflection,if any: f(x) = xex.

OR

Find the set of values of 'a' for which f(x) = x + cos x + ax + b is increasing on R.

dx
24. Evaluate: ∫
√1 − e x

25. Find the maximum and minimum values, if any without using derivatives of the function f(x) = 16 x2 - 16x + 28 on R.

Section C

26. Evaluate : ∫ 10xlog(1 + 2x)dx

27. Two dice are thrown together and the total score is noted. The events E, F and G are a total of 4, a total of 9 or more,
and a total divisible by 5, respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are
independent.

28. Evaluate ∫ (5x + 3)√2x − 1dx

OR

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1
Evaluate: ∫ 4
dx
√ ( x − 1 )3 ( x + 2 )5
29. The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and
after 3 seconds it is 2 units, find the radius after time t.

OR

Find the particular solution of the differential equation x dx − y + x cosec

dy
()y
x
= 0.

30. Solve the following linear programming problem graphically:

Maximise: Z = x + 2y
subject to the constraints:
x + 2y ≥ 100,
2x - y ≤ 0,
2x + y ≤ 200,
x ≥ 0, y ≥ 0.

OR

Determine the maximum value of Z = 4x + 3y if the feasible region for an LPP is shown in Figure.

31. If y = tan − 1
( √1 + x 2 + √1 − x 2
√1 + x 2 − √1 − x 2 ), x2 ≤ 1, then find
dy
dx
.

Section D

32. Find the area bounded by the curves y = x and y = x3

33. Let A = {1, 2, 3} and R = {(a, b ):a, b ∈

| |
A and a 2 − b 2 ≤ 5. Write R as set of ordered pairs. Mention whether R is

i. reflexive
ii. symmetric
iii. transitive
Give reason in each case.

OR

x
Show that the function f : R → R defined by f(x) = 2 , ∀ x ∈ R, is neither one-one nor onto.
x +1

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[ ]
3 1
34. If A = , show that A2 - 5A + 7I = 0. Hence find A-1.
−1 2

x+5 y+3 z−6

35. Find the distance of a point (2, 4, –1) from the line = = .
1 4 −9

OR

x+1 y−1 z−9 x−3 y + 15 z−9

Find the length shortest distance between the lines: = = and = =
2 1 −3 2 −7 5

Section E

36. Read the following text carefully and answer the questions that follow:
In a bilateral cricket series between India and South Africa, the probability that India wins the first match is 0.6. If India
wins any match, then the probability that it wins the next match is 0.4, otherwise, the probability is 0.3. Also, it is given
that there is no tie in any match.

i. Find the probability that India won the second match, if India has already loose the first match. (1)
ii. Find the probability that India losing the third match, if India has already lost the first two matches. (1)
iii. Find the probability that India losing the first two matches. (2)
OR
Find the probability that India winning the first three matches. (2)

37. Read the following text carefully and answer the questions that follow:
Three slogans on chart papers are to be placed on a school bulletin board at the points A, B and C displaying A (Hub of
Learning), B (Creating a better world for tomorrow) and C (Education comes first). The coordinates of these points are
(1, 4, 2), (3, -3, -2) and (-2, 2, 6) respectively.

→ → → → → →
i. Let a, b and c be the position vectors of points A, B and C respectively, then find a + b + c . (1)
ii. What is the Area of △ABC. (1)

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→ → → → → →
iii. Suppose, if the given slogans are to be placed on a straight line, then find the value of | a × b + b × c + c × a | . (2)
OR
→ →
If a = 2î + 3ĵ + 6k̂, then find the unit vector in the direction of vector a. (2)

38. Read the following text carefully and answer the questions that follow:
An Apache helicopter of the enemy is flying along the curve given by y = x2 + 7. A soldier, placed at (3, 7) want to shoot
down the helicopter when it is nearest to him.

i. If P (x1, y1) be the position of a helicopter on curve y = x2 + 7, then find distance D from P to soldier place at (3, 7).
(1)
ii. Find the critical point such that distance is minimum. (1)
iii. Verify by second derivative test that distance is minimum at (1, 8). (2)
OR
Find the minimum distance between soldier and helicopter? (2)
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Class 12 - Mathematics
Sample Paper - 09 (2024-25)

Solution

Section A
1. (d) 512
Explanation:

Since each element aij can be filled in two ways (with either '2' or "0'), total number of possible matrices is 8 × 8 × 8 =
512

[ ]
1 2 4
2. (c) 2 3 −5
Explanation:

A = BX
B-1A = B-1BX
X = B-1A
Using Adjoint method of inverse

B −1 =
1
2 [ ] 2
0
0
1

X = B-1A

[ ][ ]
1 2 0 1 2
X=
2 0 1 3 −5

[ ]
1 2 4
X=
2 3 −5

3. (a) 1
Explanation:

| |
x x+1
We have, = x(x) - (x + 1)(x - 1)
x−1 x

= x2 − x2 − 1 ( )
= x2 − x2 + 1 = 1

4. (c) -1
Explanation:

Given that y = tan − 1 ( acos x − bsin x

bcos x + asin x )
Dividing by b cos x in numerator and denominator, we obtain

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( )
a
b
− tan x
y = tan − 1 a
1 + b tan x

a a
Let b
= tanα ⇒ α = tan − 1 b

They y = tan − 1 ( tan α − tan x

1 + tan αtan x )
tan A − tan B
Using tan (A - B) = 1 + tan Atan B
, we obtain
a
y = tan-1 tan (α − x) = α − x = tan − 1 b − x
Differentiating with respect to x, we
dy
= -1
dx

5. (a) 13
Explanation:

→
Let PQ = xî + yĵ + zk̂
Since OX = îOY = ĵOZ = k̂
→ →
PQ . OX
Projection of PQ on OX =
| OX |
x
12 =
1
x = 12
Similarly
→ →
PQ . OY
Projection of PQ on OY = | OY |
y
3= 1
y=3
Similarly
→ →
PQ . OZ
Projection of PQ on OZ =
| OZ |
z
4=
1
z=4
→
Hence PQ = 12î + 3ĵ + 4k̂
|PQ| = √122 + 32 + 42 = 13
6. (b) sin-1 y + sin-1 x = C
Explanation:

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1 − y2

√
dy
Given dx
+ =0
1 − x2

dy dx
− =
√ 1 − y2 √1 − x 2
On integrating on both sides, we obtain

− sin − 1y = sin − 1x + C As∫

( √
dx

1 − x2
= sin − 1x + C
)
sin-1 y + sin-1 x = C

7. (d) Option (c)

Explanation:

If a LPP admits two optimal solutions it has an infinite solution.

→ →
8. (b) 4( c − 3 b)
Explanation:

→ →
Given, BD = 4BC
It means D divides the join of BC externally in the ratio 4 : 3.
∴ Position vector of D
→
→
4 c − 3b
=
4−3
→ →
4 c − 3b

π
9. (c)
4
Explanation:
π
sec 5 x
The integral is ∫ 02 dx
( sec 5 x + csc 5 x )
So out integral becomes,
1
sec 5 x cos 5 x
=
sec 5 x + csc 5 x 1
+
1
cos 5 x sin 5 x
sin 5 x
=
sin 5 x + cos 5 x
π sin 3 x
Here a = and f(x) =
2 sin 5 x + cos 5 x
5
cos x
f(a − x) =
sin x + cos 5 x
5

We know that,
a a
∴ ∫ 0f(x) = ∫ 0f(a − x) = 1 ...(let)
π
∴ 2I = ∫ 02 1dx
π
∴ 2I =
2

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π
∴ I=
2.2
π
= 4

10. (a) none of these

Explanation:

If A is a square matrix,

[ ]
1 2
Let A =
1 0

[ ][ ] [ ]
1 2 1 2 3 2
AA = × =
1 0 1 0 1 2
then AA is neither of the matrices given in the options of the question.

11. (c) Minimum Z = – 12 at (4, 0)

Explanation:

Objective function is Z = - 3x + 4 y ……………………(1).

The given constraints are : x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

The corner points obtained by constructing the line x+2y=8 and 3x +2y = 12 are (0,0),(0,2),(3,0) and (20/19,45/19)

Corner points Z = 5x + 3 y

O(0 , 0 ) 0

B(2,0) 10
C( 0 , 3 ) 9

D ( 20/19 , 45/19 ) 235/19 ……………….(Max.)

Here , Z = -12 is minimum at C ( 4 , 0 )

12. (d) 14
Explanation:

given vectors 3î + λĵ + k̂and2î − ĵ + 8k̂ are perpendicular to each other

⟹ (3î + λĵ + k̂). (2î − ĵ + 8k̂) = 0

⟹ 6-λ + 8 = 0 ⟹ λ = 14

13. (b) | s
−r
−q
p |
Explanation:

| |
s −q
−r p

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8
14. (c)
15
Explanation:

A white ball can be drawn in two mutually exclusive ways:

i. Selecting bag X and then drawing a white ball from it.

ii. Selecting bag Y and then drawing a white ball from it.

Let E1, E2 and A be three events as defined below:

E1 = Selecting bag X
E2 = Selecting bag Y
A = Drawing a white ball
We know that one bag is selected randomly.
1
∴ P(E1) =
2
1
P(E2) =
2

P
()E1
A
=
2
5

P
()E2
A
=
4
6
=
2
3

Using the law of total probability, we get

1 2 1 2
P(A) = × + ×
2 5 2 3
1 1
= +
5 3
3+5 8
= =
15 15

15. (b) log | y | = 2(x − log | 1 + x | ) + c

Explanation:

dy
(x + 1) = 2xy
dx

1 x
∫ y dy = 2∫ 1 + x dx

⇒
1
∫ dy = 2∫ 1 −
y ( 1+x
1
)
dx

log | y | = 2(x − log | 1 + x | ) + c

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16. (a) if they are parallel to the same line irrespective of their magnitudes and directions.
Explanation:

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Two vectors A and B are said to be collinear , if they are parallel to the same line irrespective of their magnitudes and
directions.

1
17. (a) − 48
Explanation:

lim
h→0 ( 3
h √8 + h
1
− 2h
1

)
{ }
1

( )
1 1 h −3 1
= lim h → 0 h 2
1+ 8
− 2

{ ( )( )
}
1 5
− −
1 1 1 h 3 3 h2
= lim h → 0 . 1− . + +... −1
h 2 3 8 2 64

= lim h → 0 2h − 24 +
1
{ h 5
18
.
h2
64
+...
}
= lim h → 0
1
2 { −
1
24
+
5
18 64
.
h
+... }
1
= −
48

18. (d) 90o

Explanation:

x = 1, y = 2 represents XY plane and y = - 1, z = 0 represents YZ plane. Since XY perpendicular to YZ. Hence angle is
90 degrees.

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

Assertion: We have, f(x) = 2x3 - 9x2 + 12x - 3

⇒ f'(x) = 6x2 - 18x + 12
For increasing function, f'(x) ≥ 0
∴ 6(x2 - 3x + 2) ≥ 0
⇒ 6(x - 2)(x - 1) ≥ 0
⇒ x ≤ 1 and x ≥ 2
∴ f(x) is increasing outside the interval (1, 2), therefore it is a true statement.
Reason: Now, f'(x) < 0

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⇒ 6(x - 2)(x - 1) < 0
⇒ 1<x<2
∴ Assertion and Reason are both true but Reason is not the correct explanation of Assertion.

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

Assertion: Since, greatest integer [x] gives only integer value.

But f(x) = [x] + x gives all real values and there is no repeated value of f(x) for any value of x.
Hence, f{x) is one-one and onto.

Section B

(
21. Let tan − 1 − √3 = y )
⇒ tany = − √3
π
⇒ tany = − tan 3

⇒ tany = tan − 3 ( ) π

Since, the principal value branch of tan-1 is − 2 , [ ]

π π
2
.

π
Therefore, principal value of tan − 1 − √3 is − . ( ) 3

OR

sin − 1 sin( ) 2π
3

( )
= sin − 1 sin
3π − π
3

[ ( )]
= sin − 1 sin π −
π
3
π
= sin − 1sin
3
π
=
3

22. Let ABC be the given triangle with AB = a, BC = b and ∠ ABC = θ.

AD is perpendicular to BC.
∴ BD = a sinθ
Now,
1
Area of △ABC = × BC × AD
2
1
⇒ A = b × a sin θ ...(i)
2
dA 1
∵
dθ
= 2 ab cos θ
For maxima and minima,
dA
dθ
=0

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1
⇒ ab cosθ = 0
2
⇒ cosθ = 0
π
⇒ θ=
2
Now,
d 2A 1
= − ab sinθ
dθ 2 2
π d 2A 1
At θ = 2 , 2 = − 2
ab < 0
dθ
π
∵ θ= is point of local maxima
2
1 π 1
∵ Maximum area of △ = ab sin = ab.
2 2 2

23. We have, f(x) = xex

∵ f'(x) = ex + (x + 1)
f"(x) = ex (x + 1) + ex
= ex (x + 2)
For maxima and minima, we must have
f'(x) = 0
⇒ ex (x + 1) = 0
⇒ x = -1
1
Now, f"( -1) = e-1 = >0
e
x = -1 is point of local minima
Hence,
−1
local min value = f(-1) = .
e

OR

Here
f(x) = x + cos x + ax +b
⇒ f(x) = 1 - sin x + a
For F (x) to be increasing we must have f '(x) > 0
1 - sinx + a > 0
⇒ sinx < 1+a
We know that the maximum value of sin x is 1
⇒ 1 + a >1

⇒ a>0
⇒ a ∈ (0, ∞)

dx
24. Let I = ∫
√1 − e x
dx
Since − ∫
√ x2 ± a2
= log(x+ √x2 ± a2) + c , where c is the integrating constant
dx
∴ I=∫
√1 − e x

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dx
=∫
√e x ( e − x − 1 )
x
e − 2 dx
=∫
√e − x − 1
x
e − 2 dx
=∫
√( e− 2
x

) 2
− 12

1
Assuming e − ( x / 2 ) = a, − ( 2 )e − ( x / 2 ) dx = da i.e., e − ( x / 2 ) dx = -2da
x
e − 2 dx
∴ I=∫
√( e− 2
x

) 2
− 12

− 2da
=∫
√a 2 − 1 2
= − 2log | a + √a 2 − 1 | + c

| |
x
= − 2log e − 2 + √e − x − 1 + c, c being the integrating constant

25. Given: f(x) = 16 x2 - 16x +28 on R

= 16x2 -16x + 4 + 24
= (4x - 2)2 + 24
Now,
(4x - 2)2 ≥ 0 for all x ∈ R
2
⇒ (4x − 2) + 24 ≥ 24 for all x ∈ R

⇒ f(x) ≥ f () 1
2
1
Hence, the minimum value of f (x) is 24 at x =
2
Since f(x) can be made as large as possible by giving different values to x, Thus, maximum values do not exist.

Section C

26. Using Integration By parts

∫ fg = fg − ∫ fg ′
f ′ = x, g = log(2x + 1)
x2 2
f= 2
, g′ = 2x + 1
Therefore,
1
∫ 0xlog(1 + 2x)dx

[ ]
1
x 2log ( 1 + 2x ) 1 2x 2
= − ∫0 dx
2 2 ( 2x + 1 )
0
log ( 3 ) 1x 1 1
= 2
− ∫0 2 − 4
+ a 4 ( 2x + 1 ) dx

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[ ]
1
log ( 3 ) x2 x 1
= 2
− 4
− 4
+ 8 log | 2x + 1 |
0
log ( 3 ) 1
= 2
− 8 log(3)
3
= log e(3)
8
Hence the result.

27. Two dice are thrown together i.e., sample space (S) = 36 ⇒ n(S) = 36
E = A total of 4 = {(2, 2), (3, 1), (1, 3)}
⇒ n(E) = 3
F = A total of 9 or more
= {(3, 6), (6, 3), (4, 5), (4, 6), (5, 4), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
⇒ n(F) = 10

G = a total divisible by 5 = {(1, 4), (4, 1), (2, 3), (3, 2), (4, 6), (6, 4), (5, 5)}
⇒ n(G) = 7
Here, (E ∩ F) = ϕ and (E ∩ G) = ϕ
Also, (F ∩ G) = {(4, 6), (6, 4), (5, 5)}
⇒ n(F ∩ G) = 3 and (E ∩ F ∩ G) = ϕ
n(F) 3 1
∴ P(E) = = =
n(S) 36 12
n(F) 10 5
P(F) = = =
n(S) 36 18
n(G) 7
P(G) = n(S)
= 36
3 1
P(F ∩ G) = 36
= 12
5 7 35
And P(F) ⋅ P(G) = ⋅ =
18 36 648
Here, we see that P(F ∩ G) ≠ P(F) ⋅ P(G)
[since, only F and G have common events, so only F and G are used here]. Hence, there is no pair which is independent.

28. Let I = ∫ (5x + 3)√2x − 1dx

Let 5x + 3 = λ(2x - 1) + μ
Comparing both sides, we get
2λ = 5 and -λ + μ = 3
5 −5
⇒ λ= 2
and 2
+μ=3
5 11
⇒ λ= 2
and μ = 2
∴ I = ∫ {λ(2x − 1) + μ}√2x − 1dx
= λ∫ (2x − 1)√2x − 1dx + μ∫ √2x − 1dx
3 1
= λ∫ (2x − 1) 2 dx + μ∫ (2x − 1) 2 dx
5 3
( 2x − 1 ) 2 ( 2x − 1 ) 2
=λ 5
+μ 3
2
×2 2
×2
5 3
( 2x − 1 ) 2 ( 2x − 1 ) 2
=λ 5
+μ 3
+c
5 3
5 ( 2x − 1 ) 2 11 ( 2x − 1 ) 2
= 2
× 5
+ 2
× 3
+c

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5
( 2x − 1 ) 2 11 3
= + (2x − 1) 2 + c
2 6

[ ]
1 3 11
= (2x − 1) 2 (2x − 1) + +c
2 3

[ ]
1 3 6x + 8
= (2x − 1) 2 +c
2 3

1 3 ( 3x + 4 )
= × (2x − 1) 2 × 2 +c
2 3
3 ( 3x + 4 )
= (2x − 1) 2 × +c
3
1 3
= (3x + 4)(2x − 1) 2 + c
3
1 3
∴ I= 3
(3x + 4)(2x − 1) 2 + c

OR

1
Let I = ∫ 4
dx. Then,
√ ( x − 1 )3 ( x + 2 )5

I=∫
1
dx = ∫
1
dx = ∫ ( )x−1
x+2
−3/4
×
1
( x + 2 )2
dx

√( ) √( )
4 4
x−1 3 x−1 3/4
8 2
x+2 (x+2) (x+2) x+2

x−1 3
Put, = t or, 1 − = t. Then,
x+2 x+2

(
d 1− x+2
3
) = dt ⇒
(x+2)
3
2 dx = dt ⇒
(x+2)
1
2 dx =
1
3
dt

x−1 1 1
Putting x+2
=t ⇒
2
dx = 3
dt, we obtain
(x+2)

I=
1
3
∫ t − 3 / 4dt =
4
3
t1 / 4 + C =
4
3 ( )
x−1
x+2
1/4
+C

29. Let A be the surface area of balloon, then,we have,

dA
∝ t
dt
dA
⇒ = λt
dt

(4πr ) = λt
d
2
⇒
dt
dr
⇒ 8πr = λt
dt
⇒ 8πrdr = λt
⇒ 8πrdr = λ∫ tdt
r2 λt 2
⇒ 8π 2 = 2
+c
λt 2
⇒ 4πr 2 = +c
2
Given r = 1 unit when t = 0, then,we have,
4π(1) 2 = 0 + c
⇒ 4π = c
Using it is equation (i),

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λt 2
4πr 2 = 2
+ 4π ....(2)
Also, given r = 2 units when t = 3 sec.
λ ( 3 )2
2
4π(2) = + 4π
2
9
⇒ 16π = 2
λ + 4π
1
⇒ r2 − 1 = 3
t2
1
⇒ r2 = 1 + t2
3

∴ r= √( 1 + 3 t2 .
1
)
OR

Given differential equation is

x
dy
dx
− y + xcosec () y
x
=0

On dividing both sides by x , we get

⇒
dy
dx
−
y
x
+ cosec ()
y
x
=0

⇒
dy
dx
=
y
x
− cosec ()
y
x

which is a homogeneous differential equation as

dy
dx
=F ()
y
x
.

put , y = vx
dy dv
⇒ =v+x
dx dx

v + x dx =
dv vx
x
− cosec ()
vx
x
dv
⇒ v+x = v − cosec v
dx
dv dv − dx
⇒ x = − cosec v ⇒ =
dx cosec v x
On integrating both sides, we get
dv dx
∫ cosec v = ∫ − x
dx
⇒ ∫ sinvdv = ∫ − x
⇒ − cosv = − log | x | + C

⇒
y
− cos = − log | x | + c
x [ put v =
y
x ]
y
⇒ cos = (log | x | − C)....(i)
x
Also, given that x = 1 and y = 0.
On putting above values in Eq. (i), we get
⇒ cos0 = log | 1 | − C

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⇒ 1=0−C ⇒ C= −1
y
∴ cos = log | x | + 1 [from Eq. (i)]
x
This is required solution of given differential equation.

30. Our problem is to mimmise and maximise the given objective function given as Z = x + 2 y .....(i)
Subject to the given constraints,
x + 2y ≥ 100 .......(ii)
2x - y ≤ 0 ...........(iii)
2x + y ≤ 200 ........(iv)
x ≥ 0, y ≥ 0 ........(v)
Table for line x + 2y = 100 is

x 0 100

y 50 0

So, the line x + 2y = 100 is passing through the points with coordinates (0, 50) and (100, 0).
On replacing the coordinates of the origin O (0, 0) in the inequality x + 2y ≥ 100, we get
2 × 0 + 0 ≥ 100
⇒ 0 ≥ 100(which is False)
So, the half plane for the inequality of the line ( ii) is away from the origin, which means that the point O( 0,0) does not
lie in the feasible region of the inequality of ( ii)
Table for the line ( iii) 2x - y = 0 is given as follows.

x 0 10
y 0 20

So, the line 2x - y = 0 is passing through the points (0, 0) and ( 10, 20).
On replacing the point (5, 0) in the inequality 2x - y ≤ 0, we get
2×5−0≤0
⇒ 10 ≤ 0 (which is False)
So, the half plane for the inequality of ( iii) is towards Y-axis.
Table of values for line 2x + y = 200 is given as follows.

x 0 100

y 200 0

So, the line 2x + y = 200 is passing through the points with coordinates (0, 200) and ( 100, 0).
On replacing O (0, 0) in the inequality 2 + y ≤ 200, we get
2 × 0 + 0 ≤ 200
⇒ 0 ≤ 200 (which is true)
So, the half plane for the inequality of the line ( iv) is towards the origin, which means that the point O ( 0,0) is a point in
the feasible region.
Aslo, x, y ≥ 0
So, the region lies in the I quadrant only.

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On solving equations 2x - y = 0 and x + 2y = 100, we get the point of intersection as B(20, 40).
Again, solving the equations 2x - y = 0 and 2x + y = 200, we get C(50, 100).
∴ Feasible region is ABCDA, which is a bounded feasible region.

The coordinates of the corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100) and D(0, 200).
The values of Z at corner points are given below:

Corner points Z = x + 2y

A(0, 50) Z = 0 + 2 × 50 = 100

B(20, 40) Z = 20 + 2 × 40 = 100

C(50, 100) Z = 50 + 2 × 100 = 250

D(0, 200) Z = 0 + 2 × 200 = 400

The maximum value of Z is 400 at D(0, 200) and the minimum value of Z is 100 at all the points on the line segment
joining A(0, 50) and B(20, 40).

OR

We have to determine the maximum value of Z = 4x + 3y if the feasible region for an LPP is shown in Fig.

The feasible region is bounded. Therefore, maximum of Z must occur at the corner point of the feasible region (Fig.
12.1).

Corner Point Value of Z

O, (0, 0) 4(0) + 3(0) = 0

A(25, 0) 4(25) + 3(0) = 100

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B(16, 16) 4(16) + 3(16) = 112 (maximum)

C(0, 24) 4(0) + 3(24) = 72

Hence, the maximum value of Z is 112.

31. Given, y= tan − 1

( √1 + x 2 + √1 − x 2
√1 + x 2 − √1 − x 2 )
Put x2 = sinθ ⇒ θ = sin − 1x 2

∴ y = tan − 1
( √1 + sin θ + √1 − sin θ
√1 + sin θ − √1 − sin θ )

( √ √
)
θ θ θ θ θ θ θ θ
cos 2 2 + sin 2 2 + 2sin 2 cos 2 + cos 2 2 + sin 2 2 − 2sin 2 cos 2

= tan − 1
√ √
θ θ θ θ θ θ θ θ
cos 2 2 + sin 2 2 + 2sin 2 cos 2 − cos 2 2 + sin 2 2 − 2sin 2 cos 2

[ √( ) √( )
]
θ θ 2 θ θ 2
cos 2 + sin 2 + cos 2 − sin 2

= tan − 1

√( ) √( )
θ θ 2 θ θ 2
cos 2 + sin 2 − cos 2 − sin 2

[ ( ) ( )
]
θ θ θ θ
cos 2 + sin 2 + cos 2 − sin 2
−1
= tan
( θ
cos 2 + sin 2
θ
) (
−
θ
cos 2 − sin 2
θ
)

( )
θ
2cos 2
−1
= tan θ
2sin 2

= tan − 1 cot
( ) θ
2

= tan − 1 tan [ ( )] π
2
−
θ
2
π θ
= −
2 2
π 1
⇒ y= 2
− 2 sin − 1x 2
Therefore, on differentiating both sides w.r.t x, we get,

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dy 1 1
dx
= − 2
(2x)
√1 − ( x 2 ) 2
−x
=
√1 − x 4
Section D

32. The given curves are ,

y = x ...(i)
and y = x3 ...(ii)
The sketch of the curve y = x3 is shown in Fig. Clearly, y =x is a line passing through the origin and making an angle of
45° with x-axis. The shaded portion shown in Fig. is the region bounded by the curves y = x and y = x3. Solving y = x
and y = x3 simultaneously, we find that the two curves intersect at O (0, 0), A (1,1) and B (- 1, - 1).
When we slice the shaded region into vertical strips, we observe that the vertical strips change their character at O.

Therefore, the required area is given by,

Required area = Area BCOB + Area ODAO
Area BCOB: Each vertical strip in this region has its lower end on y = x and the upper end on y = x3. Therefore, the

| |
approximating rectangle shown in this region has length = y 4 − y 3 , width = dx and area = y 4 − y 3 dx. Since the | |
approximating rectangle can move from x = - 1 to x = 0.

∴
| |
Area BCOB = ∫ 00 y 4 − y 3 dx = ∫ 0− 1 − y 4 − y 3 dx [ ( ) ∵ y4 < y3 ∴ y4 - y3 < 0]

(
= ∫ 0− 1 − x − x 3 dx [ ) ∵ R (x, y3) and S (x, y4) lie on (ii) and (i) respectively ∴ y3 = x3 and y4 = x]

[ ]
0
=
0
∫ −1 (x − x )dx =
3
x4
4
−
x2
2
−1
=0− ( )
1
4
−
1
2
=
1
4
sq. units

Area ODAO: Each vertical strip in this region has its two ends on (ii) and (i) respectively. So, the approximating

| |
rectangle shown in this region has length = y 2 − y 1 , width = dx and therefore,we have,

Area ODAO = ∫ 10 |y 2 − y 1 |dx = ∫ 10 (y 2 − y 1 )dx [ y2 > y1 ∵ ∴ y2 - y1 > 0]

1
(
= ∫ 0 x − x 3 dx [ ) ∵ P (x, y1) and Q (x, y2) lie on (ii) and (i) respectively ∴ y1 = x3 and y2 = x]

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[ ]
1
x2 x4 1 1 1
= 2
− 4
= 2
− 4
= 4
sq. units
0
1 1 1
∴ Required area = Area BCOB + Area ODAO = + = sq. units
4 4 2

33. Given that

Let A = {1, 2, 3} and R = {(a, b) : a, b ∈

|
A and a 2 − b 2 ≤ 5 |
| |
Put a = 1, b = 1 1 2 − 1 2 ≤ 5, (1, 1) is an ordered pair.

Put a = 1, b = 2 | 1 2
− 2 | ≤ 5, (1, 2) is an ordered pair.
2

Put a = 1, b = 3 | 1 2
− 3 | > 5, (1, 3) is not an ordered pair.
2

Put a = 2, b = 1 | 2 2
− 1 | ≤ 5, (2, 1) is an ordered pair.
2

Put a = 2, b = 2 | 2 2 − 2 | ≤ 5, (2, 2) is an ordered pair.

2

Put a = 2, b = 3 | 2 2 − 3 | ≤ 5, (2, 3) is an ordered pair.

2

Put a = 3, b = 1 | 3 2
− 1 | > 5, (3, 1) is not an ordered pair.
2

Put a = 3, b = 2 | 3 2
− 2 | ≤ 5, (3, 2) is an ordered pair.
2

Put a = 3, b = 3 | 3 2
− 3 | ≤ 5, (3, 3) is an ordered pair.
2

R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3)}

i. For (a, a) ∈ R

|a 2
|
− a 2 = 0 ≤ 5. Thus, it is reflexive.
ii. Let (a, b) ∈ R

(a, b) ∈
|
R, a 2 − b 2 ≤ 5 |
|b 2
− a2 ≤ 5 |
(b, a) ∈ R
Hence, it is symmetric
iii. Put a = 1, b = 2, c = 3

|1 2
| − 22 ≤ 5

|2 −3 |≤5
2 2

But |1 − 3 | > 5
2 2

Thus, it is not transitive

OR

For x1, x2 ∈ R, consider

f(x1) = f(x2)

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x1 x2
⇒ =
x 21 + 1 x 22 + 1

⇒ x 1x 22 + x 1 = x 2x 21 + x 2
⇒ x1x2(x2 - x1) = x2 - x1
⇒ x1 = x2 or x1x2 = 1
1
We note that there are point, x1 and x2 with x1 ≠ x2 and f(x1) = f(x2) for instance, if we take x1 = 2 and x 2 = , then we
2
2 2 1
have f(x 1) and f(x 2) = but 2 ≠ . Hence f is not one-one. Also, f is not onto for if so then for 1 ∈ R∃x ∈ R such
5 5 2
x
that f(x) = 1 which gives = 1. But there is no such x in the domain R, since the equation x2 - x + 1 = 0 does not
x2 + 1
give any real value of x.

[ ]
3 1
34. Given: A =
−1 2

[ ][ ]
3 1 3 1
∴ A 2 = A. A =
−1 2 −1 2

[ ] [ ]
9−1 3+2 8 5
= =
−3 − 2 −1 + 4 −5 3

L.H.S = A2 - 5A + 7I = A2 - 5A + 7I2

=
[ ] [
8
−5
5
3
−5
−1
3 1
2 ] [ ]+7
1
0
0
1

[ ][ ][ ]
8 5 15 5 7 0
= − +
−5 3 −5 10 0 7

[ ][ ]
8 − 15 5−5 7 0
= +
−5 + 5 3 − 10 0 7

[ ][ ]
−7 0 7 0
= +
0 −7 0 7

= [ −7 + 7
0+0
0+0
−7 + 7 ][ ]
+
0
0
0
0
=0

= R.H.S.

⇒ A 2 − 5A + 7I 2 = 0 …(i)

To find: A − 1, multiplying eq. (i) by A − 1.

⇒ A 2A − 1 − 5A. A − 1 + 7I 2A − 1 = 0.A − 1

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⇒ A − 5I 2 + 7A − 1 = 0

⇒ 7A − 1 = − A + 5I 2

[ ] [ ]
3 1 1 0
= +5
−1 2 0 1

= [ ][ ] [ ]
3
−1
1
2
+
5
0
0
5
=
2
1
−1
3

⇒ A −1 =
1
7 [ ]
2
1
−1
3

x+5 y+3 z−6

35. We have equation of the line as 1
= 4
= −9
= λ.
⇒ x = λ − 5, y = 4λ − 3, z = 6 − 9λ
Let the coordinates of L be (λ − 5, 4λ − 3, 6 − 9λ), then Dr’s of PL are (λ − 7, 4λ − 7, 7 − 9λ).
Also, the direction ratios of given line are proportional to 1, 4, -9.
Since, P L is perpendicular to the given line.
∴(λ − 7) ⋅ 1 + (4λ − 7) ⋅ 4 + (7 − 9λ) ⋅ ( − 9) = 0
⇒ λ − 7 + 16λ − 28 + 81λ − 63 = 0

⇒ 98λ = 98 ⇒ λ = 1
So, the coordinates of L are (-4, 1, -3).
∴ Required distance, PL = √( − 4 − 2) 2 + (1 − 4) 2 + ( − 3 + 1) 2
= √36 + 9 + 4 = 7 units
OR

Here,it is given equations of lines

x+1 y−1 z−9
L1 : 2
= 1
= −3
x−3 y + 15 z−9
L2 : 2
= −7
= 5
Direction ratios of L1 and L2 are (2, 1, -3) are (2, -7, 5) respectively.
Suppose, general point on line L1 is P = (x1, y1, z1)
x1 = 2s - 1, y1 = s + 1, z1 = -3s + 9
and let general point on line L2 is Q = (x2, y2, z2)
x2 = 2t + 3, y2 = -7t - 15, z2 = 5t + 9 then ,we get
→
∴ ( ) (
PQ = x 2 − x 1 ı̂ + y 2 − y 1 ȷ̂+ z 2 − z 1 k̂ ) ( )
= (5t + 9 − 2s + 1)î + ( − 7t − 15− s − 1)ȷ̂ + (5t + 9 + 3s − 9)k̂
→
∴ PQ = (5t − 2s + 10) î + ( − 7t − s − 16) ȷ̂ + (5t + 3s)k̂

→
Direction ratios of PQ are ((5t - 2s + 10), (-7t - s - 16), (5t + 3s))
PQ will be the shortest distance if it is perpendicular to both the given lines
Thus, by the condition of perpendicularity,
2(5t - 2s + 10) + 1(-7t - s - 16) - 3(5t - 3s) = 0 and

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2(5t - 2s + 10) -7(-7t - s - 16) + 5(5t + 3s) = 0
⇒ 10t - 4s + 20 - 7t - s - 16 - 15t - 9s = 0 and

10t - 4s + 20 + 49t + 7s + 112 + 25t + 15s = 0

⇒ -12t - 14s = -4 and 84t + 18s = -132
Solving above two equations, we obtain
t = -2 and s = 2
thus,
P = (3, 3, 3) and Q = (-1, -1, -1)
Now, distance between points P and Q is
d= √(3 + 1) 2 + (3 + 1) 2 + (3 + 1) 2
= √(4) 2 + (4) 2 + (4) 2
= √16 + 16 + 16
= √48
= 4√3
Thus, the shortest distance between two given lines is
d = 4√3 units
Now, equation of line passing through points P and Q is
x − x1 y − y1 z − z1
x1 − x2
= y1 − y2
= z1 − z2
x−3 y−3 z−3
∴
3+1
= 3+1
= 3+1
x−3 y−3 z−3
∴
4
= 4
= 4
∴ x-3=y-3=z-3
⇒ x = y = z

Therefore, equation of line of shortest distance between two given lines is x = y = z

Section E
36. i. It is given that if India loose any match, then the probability that it wins the next match is 0.3.
∴ Required probability = 0.3

ii. It is given that, if India loose any match, then the probability that it wins the next match is 0.3.
∴ Required probability = 1 - 0.3 = 0.7

iii. Required probability = P(India losing first match) ⋅ P(India losing second match when India has already lost first
match) = 0.4 × 0.7 = 0.28
OR
Required probability = P(India winning first match) ⋅ P(India winning second match if India has already won first
match) ⋅ P(India winning third match if India has already won first two matches) = 0.6 × 0.4 × 0.4 = 0.096

→ → →
37. i. a = î + 4ĵ + 2k̂, b = 3î − 3ĵ − 2k̂ and c = − 2î + 2ĵ + 6k̂
→ → →
∴ a + b + c = 2î + 3ĵ + 6k̂

ii. We have, A(1, 4, 2), B(3, -3, -2) and C(-2, 2, 6)

→ → →
→ → →
Now, AB = b − a = 2î − 7ĵ − 4k̂ and AC = c − a = − 3î − 2ĵ + 4k̂

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| |
→ → î ĵ k̂
∴ AB × AC = 2 −7 −4
−3 −2 4

= î(-28 - 8) - ĵ(8 - 12) + k̂(-4 - 21) = -36î +4ĵ - 25k̂

→ →
√
Now, | AB × AC | = ( − 36) 2 + 4 2 + ( − 25) 2
= √1296 + 16 + 625 = √1937
1 → → 1
∴ Area of △ABC = 2
| AB × AC | = 2 √1937 sq. units
iii. If the given points lie on the straight line, then the points will be collinear and so area of △ABC = 0
→ → → → → → → →
⇒ | a × b + b × c + cˉ × a | = 0 [ ∵ If a, b, c are the position vectors of the three vertices A, B and C of △ABC,
1 → → → → → →
then area of triangle = 2
|a × b + b × c + c × a |]
OR
→
Here, a = 2î + 3ĵ + 6k̂
→
|a| = √4 + 9 + 36 = √49 = 7
2 iˆ + 3 jˆ + 6k̂
Now unit vector â =
7
2 3 6
â = î + ĵ + k̂
7 7 7

38. i. P(x1, y1) is on the curve y = x2 + 7 ⇒ y 1 = x 21 + 7

Distance from p(x 1, x 21 + 7) and (3, 7)

D= √ (x − 3 ) + (x + 7 − 7 )
1
2 2
1
2

⇒
√ (x − 3 ) + (x )
1
2 2 2
1

√x
4
⇒ D= 1 + x 21 − 6x 1 + 9

√x
4
ii. D = 1 + x 21 − 6x 1 + 9

D' = x 41 + x 21 − 6x 1 + 9
dD ′
dx
= 4x 31 + 2x 1 − 6 = 0
dD ′ 3
dx
= 2x 1 + x 1 − 3 = 0

⇒
(
(x1 - 1) 2x 1 2 + 2x 1 + 3 = 0 )
2
x1 = 1 and 2x 1 + 2x1 + 3 = 0 gives no real roots
The critical point is (1, 8).
dD ′
iii. = 4x 31 + 2x 1 − 6
dx
d 2D ′ 2
= 12x 1 + 2
dx 2

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d 2D ′
dx 2 ] x1 = 1
= 12 + 2 = 14 > 0

Hence distance is minimum at (1, 8).

OR

√x
4
D= 1 + x 21 − 6x 1 + 9

D= √1 + 1 − 6 + 9 = √5 units
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