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Class 12 - Mathematics
Sample Paper - 04 (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. If a matrix A = [1 2 3], then the matrix AA' (where A' is the transpose of A) is:

a) [14]
b) 14

[ ]
1 0 0
c) 0 2 0
0 0 3

[ ]
1 2 3
d) 2 3 1
3 1 2

2. Let A be a 3 × 3 matrix such that |adj A| = 64. Then |A| is equal to:

a) -8 only
b) 64
c) 8 only
d) 8 or -8

3. If A is skew symmetric matrix of order 3, then the value of |A| is:

a) 9
b) 3
c) 0
d) 27

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4.
| sin23 ∘
cos23 ∘
− sin67 ∘
cos67 ∘ | =?

√3
a) 2
b) sin 16o
c) 1
d) cos 16o

5. How many lines through the origin make equal angles with the coordinate axes ?

a) 8
b) 2
c) 4
d) 1

dy
6. The solution of the DE dx
= 2 x + y is

a) 2x + 2y = C
b) 2x - 2-y = C
c) 2x + 2-y = C
d) 2x + 2-y = C

7. A Linear Programming Problem is as follows:

Minimize Z = 2x + y
Subject to the constraints x ≥ 3, x ≤ 9, y ≥ 0
x - y ≥ 0, x + y ≤ 14
The feasible region has

a) 5 corner points including (0, 0) and (9, 5)

b) 5 corner points including (7, 7) and (3, 3)
c) 5 corner points including (3, 6) and (9, 5)
d) 5 corner points including (14, 0) and (9, 0)

→
8. Unit vector along PQ, where coordinates of P and Q respectively are (2, 1, -1) and (4, 4, -7), is

a) 2î + 3ĵ − 6k̂

2 iˆ 3 jˆ 6kˆ
b) 7
+ 7
− 7
c) − 2î − 3ĵ + 6k̂
− 2 iˆ 3 jˆ 6k̂
d) − +
7 7 7

d 3
9. If dx
f(x) = 2x + x
and f(1) = 1, then f(x) is

a) x2 + 3 log |x| – 4
b) x2 + 3 log |x| + 1

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3
c) 2 −
x2

d) x2 + 3 log |x|

[ ]
0 1 1
10. For the matrix X = 1 0 1 , (X2 - X) is:
1 1 0

a) 2I
b) 5I
c) I
d) 3I

11. In Corner point method for solving a linear programming problem the second step after finding the feasible region of the
linear programming problem and determining its corner points is

a) Evaluate the objective function Z = ax + by at the center point

b) Evaluate the objective function Z = ax + by at the mid points
c) Evaluate the objective function Z = ax + by at each corner point.
d) Evaluate the objective function Z = ax + by at the End points

12. The magnitude of the vector 6î − 2ĵ + 3k̂ is

a) 5
b) 12
c) 7
d) 1

[ ] [ ]
2 2 1 1
13. If A = and B = , then
4 0 2 0

a) |A| = 2|B|
b) |A| = |B|
c) |A| = -|B|
d) |A| = 22|B|

14. Two events E and F are independent. If P(E) = 0.3, P(E ∪ F)=0.5 then P(E|F) - P(F|E) equals

3
a) 35
2
b) 7
1
c) 7
1
d) 70

dy
15. The general solution of the differential equation = e x + y is
dx

a) ex + e-y = C
b) e-x + e-y = C

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c) e-x + ey = C
d) ex + ey = C

→ π → →
16. If the angle between →
a and b is and | →
a × b | = 3√3, then the value of →
a ⋅ b is
3

1
a) 3
1
b)
9
c) 9
d) 3
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{
x 2 − 2x − 3
x+1
, x≠ −1
17. Find the value of k for which the function (x) = is continuous at x = -1.
k, x= −1

a) -4
b) -3
c) 4
d) 2

→
(
18. Find the angle between the following pairs of lines: r = 2î − 5ĵ + k̂ +λ 3î + 2ĵ + 6k̂. ) and →
r = 7î − 6k̂ +

(
μ î + 2ĵ + 2k̂. ) ,λ, μ ∈ R

19
a) θ = cos − 1( 21 )

b) θ = sin − 1
( )
19
21

c) θ = cot − 1 ( )
19
21

d) θ = tan − 1 ( )
19
21

19. Assertion (A): The average rate of change of the function y = 15 - x2 between x = 2 and x = 3 is -5.
Reason (R): Average rate of change δ y = yat x = 3 - yat x = 2.

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): Let A and B be sets. Then, the function f : A × B → B × A such that f(a, b) = (b, a) is bijective.
Reason (R): A function f is said to be bijective, if it is both one-one and onto.

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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.
Section B

21. Simplify sec − 1

( ) 1
2
2x − 1
,0<x<
1

√2
.

OR

{ (
For the principal value, evaluate tan − 1 2cos 2sin − 1 2
1
)}.

22. Find the interval in function -2x3 - 9x2 - 12x + 1 is increasing or decreasing:

23. Prove that the function given by f(x) = log sinx is strictly increasing on 0, ( ) π
2
and strictly decreasing on ( )
π
2
,π

OR

Show that f(x) = (x – 1) ex + 1 is an increasing function for all x > 0.

x2 + x
24. Find: ∫ dx
x3 − x2 + x − 1

25. Prove that the function f given by f(x) = x2 - x + 1 is neither strictly increasing nor strictly decreasing on (-1, 1).

Section C

2
|
26. Evaluate ∫ − 1 x 3 − x dx. |
1 1
27. Probabilities of solving a specific problem independently by A and Bare 2
and 3
respectively. If both try to solve problem
independently, then find the probability that

i. problem is solved.
ii. exactly one of them solves the problem.

π/4
28. Evaluate ∫ 0 log(1 + tanx)dx.

OR

4
Evaluate ∫ 0[ | x | + | x − 2 | + | x − 4 | ]dx.

29. Solve the initial value problem:

dy
dx
+ y cot x = 4x cosec x, y ()
π
2
=0

OR

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( )
Find the general solution of the differential equation 1 + y 2 + x − e tan ( −1 y
) dy
dx
= 0.

30. Minimize Z = 2x + 3y subject to the constraints x ≥ 0, y ≥ 0, x + 2y ≥ 1 and x+2y ≤ 10

OR

Solve the following LPP graphically:

Minimize Z = 3x + 5y
Subject to
- 2x + y ≤ 4
x+y ≥ 3
x-2y ≤ 2
x, y ≥ 0

{
π
asin 2 (x + 1), x≤0
31. Find the value of a for which the function f is defined as f(x) = tan x − sin x is continuous at x = 0.
, x>0
x3

Section D

32. Sketch the graph of y = Ix + 3I and evaluate the area under the curve y = Ix + 3I above X - axis and between x = - 6 to x =
0.

33. Let R be a relation on N × N, defined by (a, b) R (c, d) ⇔ a + d = b + c for all (a, b), (c, d) ∈ N × N. Show that R is an
equivalence relation.

OR

x−2
Let A = R - {3}, B = R - {1]. If f : A → B be defined by f(x) = ∀ x ∈ A. Then, show that f is bijective.
x−3

[ ]
1 1 1
34. For the matrix A = 1 2 − 3 , show that A3 - 6A2 + 5A + 11I = 0. Hence find A-1.
2 −1 3

35. Find the shortest distance between the given lines. →r = (î + 2ĵ − 4k̂)+ λ(2î + 3ĵ + 6k̂), →r = (3î + 3ĵ − 5k̂)
+ μ( − 2î + 3ĵ + 8k̂)

OR

x−1 y+1 z + 10
Find the perpendicular distance of the point (1, 0, 0) from the line = = . Also, find the coordinates of the
2 −3 8
foot of the perpendicular and the equation of the perpendicular.

Section E

36. Read the following text carefully and answer the questions that follow:
A shopkeeper sells three types of flower seeds A1, A2, A3. They are sold in the form of a mixture, where the proportions of

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these seeds are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively.

Based on the above information:

i. Calculate the probability that a randomly chosen seed will germinate. (1)
ii. Calculate the probability that the seed is of type A2, given that a randomly chosen seed germinates. (1)
iii. A die is throw and a card is selected at random from a deck of 52 playing cards. Then find the probability of getting an
even number on the die and a spade card. (2)
OR
If A and B are any two events such that P(A) + P(B) - P(A and B) = P(A), then find P(A|B). (2)

37. Read the following text carefully and answer the questions that follow:
Three friends Ganesh, Dinesh and Ramesh went for playing a Tug of war game. Team A, B, and C belong to Ganesh,
Dinesh and Ramesh respectively.
Teams A, B, C have attached a rope to a metal ring and is trying to pull the ring into their own area (team areas shown
below).
Team A pulls with F1 = 4î + 0ĵ KN
Team B → F2 = -2î + 4ĵ KN
Team C → F3 = -3î - 3ĵ KN

i. Which team will win the game? (1)

ii. What is the magnitude of the teams combine Force? (1)
iii. What is the magnitude of the force of Team B? (2)
OR
How many KN Force is applied by Team A? (2)

38. Read the following text carefully and answer the questions that follow:
The relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed by the following

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1
equation y = 4x - x2 where x is the number of days exposed to sunlight.
2

i. Find the rate of growth of the plant with respect to sunlight. (1)
ii. What is the number of days it will take for the plant to grow to the maximum height? (1)
iii. Verify that height of the plant is maximum after four days by second derivative test and find the maximum height of
plant. (2)
OR
What will be the height of the plant after 2 days? (2)
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Class 12 - Mathematics
Sample Paper - 04 (2024-25)

Solution

Section A
1. (a) [14]
Explanation:

[14]

2. (d) 8 or -8
Explanation:

8 or -8
Explanation

We know that |Adj A| = |A|n-1, n is the order of the matrix.

∵ 64 = |A|3-1
|A|2 = 64
|A| = ± 8

3. (c) 0
Explanation:

Determinant value of skew-symmetric matrix is always '0'.

4. (c) 1
Explanation:

By evaluating determinant we get

sin 223° + cos 223° = 1

5. (a) 8
Explanation:

since α = β = γ

1 1 1 1 1 1
cos 2α + cos 2β + cos 2γ = 1hence 3cos 2α = 1hence cosα = cosβ = cosγ = dcs of lines will be ( ± , ± , ± )
√3 √3 √3 √3 √3 √3

hence the lines will be 8

6. (d) 2x + 2-y = C
Explanation:

dy
Here, = 2x + y
dx
dy
dx
= 2 x2 y

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2-ydx = 2xdx
On integrating on both sides, we get
2 −y 2x
− log 2 + c 2 = log 2
+ c2
2x + 2-y = c3 log 2

2 x + 2-y = C

7. (b) 5 corner points including (7, 7) and (3, 3)

Explanation:

On plotting the constraints x = 3, x = 9, x = y and x + y = 14, we get the following graph. From the graph given below it,
clear that feasible region is ABCDEA, including corner points A(9, 0), B(3, 0), C(3, 3), D(7, 7) and E(9, 5).
Thus feasible region has 5 corner points including (7, 7) and (3, 3).

2 iˆ 3 jˆ 6k̂
8. (b) 7
+ 7
− 7
Explanation:

2 iˆ 3 jˆ 6kˆ
+ −
7 7 7

9. (d) x2 + 3 log |x|

Explanation:

x2 + 3 log |x|

10. (a) 2I
Explanation:

[ ][ ]
0 1 1 0 1 1
Here X2 = 1 0 1 1 0 1
1 1 0 1 1 0

[ ]
2 1 1
⇒ X2 = 1 2 1
1 1 2

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[ ][ ]
2 1 1 0 1 1
⇒ X2 - X 1 2 1 − 1 0 1
1 1 2 1 1 0

[ ]
2 0 0
= 0 2 0
0 0 2

[ ]
1 1 1
⇒ X2 - X = 2 1 1 1 = 2I
1 1 1

11. (c) Evaluate the objective function Z = ax + by at each corner point.

Explanation:

In Corner point method for solving a linear programming problem the second step after finding the feasible region of the
linear programming problem and determining its corner points is : To evaluate the objective function Z = ax + by at each
corner point.

12. (c) 7
Explanation:

13. (d) |A| = 22|B|

Explanation:

Let A = [ ]
2 2
4 0
and B = [ ]
1 1
2 0
Now, |A| = 0 - 8 = -8
and |B| = 0 - 2 = -2
Observe that |A| = 4(-2) = 22 |B|

1
14. (d) 70
Explanation:

Here P(E) = 0.3, P(E ∪ F)=0.5

Let P(F) = x
∴ P(E ∪ F) = P(E) + P(F) − P(E ∩ F)

= P(E) + P(F) − P(E) ⋅ P(F)

⇒ 0.5 = 0.3 + x - 0.3x

0.5 − 0.3 2
⇒ x= 0.7
= 7
= P(F)
P(E∩F) P(F∩E)
∴ P(E / F) − P(F / E) =
P(F)
− P(E)
P(E∩F) ⋅P(E) −P(F∩E) ⋅P(F)
=
P(E) ⋅P(F)

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P(E∩F) [P(E) −P(F) ]
= P(E∩F)
= P(E) − P(F)
3 2 21 − 20 1
= − = =
10 7 70 70

15. (a) ex + e-y = C

Explanation:

dy
We have, dx
= ex + y
dy
⇒
dx
= ex × ey
separating variables
⇒ e − ydy = e xdx
Integrating both sides
⇒ ∫ e − ydy = ∫ e xdx
⇒ − e − y = ex + c
⇒ ex + e − y = − c
Or,
e x + e − y = c (c is a constant)
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16. (d) 3
Explanation:

17. (a) -4
Explanation:

x 2 − 2x − 3
lim f(x) = lim
x+1
x→ −1 x→ −1
(x−3) (x+1)
lim = lim (x − 3)
x+1
x→ −1 x→ −1
= -1 - 3 = -4
For continuity at x = -1, we must have lim f(x) = f(-1)
x→ −1
⇒ -4 = k

19
18. (a) θ = cos − 1( )
21
Explanation:

→ → → →
→ →
If θ is the acute angle between r = a 1 + λb 1 and r = a 2 + λb 2 , then cosine of the angle between these two lines is given

| | || | |
→ →
b1 . b2
by : cosθ =
→ →
b1 b2

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→ →
Here, b 1 = 3î + 2ĵ + 6k̂, b 2 = î + 2ĵ + 2k̂
Then,

cosθ =
|| ( 3 iˆ + 2 jˆ + 6kˆ ) . ( iˆ + 2 jˆ + 2kˆ )

( 3 iˆ + 2 jˆ + 6kˆ ) | | ( i + 2 j + 2k ) |
ˆ ˆ ˆ |
cosθ =
| ||| 19

√49√9
=
19
21
⇒ θ = cos − 1
(| |)
19
21

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

Let y = f(x) = 15 - x2
If x changes from 2 to 3 then δ x = 3 - 2 = 1
Again f(3) = 15 - 9 = 16 and f(2) = 15 - 4 = 11
Therefore δ y = f(3) - f(2) = 6 - 11 = -5

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

Assertion: Here, f : A × B → B × A is defined as f(a, b) = (b, a).

Let (a1, b1), (a2, b2) ∈ A × B such that
f (a1, b1) = f(a2, b2)
⇒ (b1, a1) = (b2, a2)
⇒ b1 = b2 and a1 = a2
⇒ (a1, b1) = (a2, b2)
Therefore, f is one-one.
Now, let (b, a) ∈ B × A be any element.
Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a) [definition of f]
Therefore, f is onto. Hence, f is bijective.

Section B

21. Let x = cos θ

∴ θ = cos-1 x

sec-1
( ) (
1
2x 2 − 1
= sec-1
1
2cos 2 θ − 1 ) = sec-1
( ) 1
cos 2θ
= sec-1(sec 2θ)

= 2θ
= 2 cos-1 x

OR

1 π
We know that sin − 1 = .
2 6

∴
{ (
tan − 1 2cos 2sin − 1 2
1
)}
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{ ( )}
= tan − 1 2cos 2 ×
π
6

( ) ( )
= tan − 1 2cos
π
3
= tan − 1 2 ×
1
2
= tan − 11 =
π
4

22. It is given that function f(x) = -2x3 - 9x2 - 12x + 1

⇒ f '(x) = -6x2 - 18x - 12
⇒ f '(x) = -6(x2 + 3x + 2)
⇒ f '(x) = -6(x + 1)(x + 2)
If f '(x) = 0, then we get,
⇒ x = -1 and -2

So, the points x = -1 and x = -2 divide the real line into three disjoint intervals, ( − ∞, − 2), ( − 2, − 1) and ( − 1, ∞)
So, in intervals ( − ∞, − 2), ( − 1, ∞)
f '(x) = -6(x + 1)(x + 2) < 0
Therefore, the given function 'f ' is strictly decreasing for x < -2 and x > -1
Further, in interval (-2, -1)
f '(x) = -6(x + 1)(x + 2) > 0
Therefore, the given function (f) is strictly increasing for -2 < x < -1

1
23. f ′ (x) = . cosx
sin x
f ′ (x) = cotx

f ′ (x) > 0∀x ∈

( )
0,
π
2

and f ′ (x) < 0∀x

( )∈
π
2
,π

Hence, f(x) = log sinx is strictly increasing on 0,

( ) π
2
and decreasing on
( )
π
2
,π

OR

Given:- f(x) = (x – 1) ex + 1

((x − 1)e + 1)
d
⇒ f ′(x) = x
dx

= f'(x) = ex + (x – 1) ex
= f'(x) = ex(1+ x – 1)
= f'(x) = xex
as given
x>0
= ex > 0
= xex > 0
= f'(x) > 0
Hence, the condition for f(x) to be increasing
Thus, f(x) is increasing for all x > 0

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x2 + x
24. Let I = ∫ dx
x3 − x2 + x − 1
x2 + x A Bx + C
Now let = +
x−1 x2 + 1
(
( x − 1 ) x2 + 1 )
Getting A = 1, B = 0, C = 1
1 1
Therefore, I = ∫ x − 1 dx+∫ 2 dx
x +1
= log | x − 1 | +tan-1 x + C

25. Given: f(x) = x 2 − x + 1f(x) = x2 - x + 1

⇒ f'(x) = 2x - 1

f(x) is strictly increasing if f'(x) < 0

⇒ 2x - 1 > 0

1
⇒ x> 2

i.e., increasing on the interval ( ) 1

2
,1

f(x) is strictly decreasing if f'(x) < 0

⇒ 2x - 1 < 0

1
⇒ x< 2

i.e., decreasing on the interval − 1,

( ) 1
2

hence, f(x) is neither strictly increasing nor decreasing on the interval (-1, 1).

Section C

26. According to the question , I = ∫ − 1 x 3 − x dx

2
| |
We can observe that,

(x − x ), when − 1 < x < 0

3

|x − x | =
3

{ − (x − x ), when 0 ≤ x < 1
3

(x − x), when 1 ≤ x < 2

3

By Splitting the intervals , we get

0
| | | | | 1
|
I = ∫ − 1 x 3 − x dx + ∫ 0 x 3 − x dx + ∫ 1 x 3 − x dx
2

I = ∫ (x − x )dx + ∫ − (x − x )dx + ∫ (x − x )dx

0 3 1 3 2 3
∴
−1 0 1

= ∫ (x − x )dx − ∫ (x − x )dx + ∫ (x − x )dx

0 3 1 3 2 3
−1 0 1

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[ ] [ ] [ ]
0 1 2
x4 x2 x4 x2 x4 x2
= 4
− 2
− 4
− 2
+ 4
− 2
−1 0 1

= 0− [ ( )] [( ) ] [( ) ( )]
1
4
−
1
2
−
1
4
−
1
2
−0 +
16
4
−
4
2
−
1
4
−
1
2
1 1 1 1 1 1
= − + − + +4−2− +
4 2 4 2 4 2
3 3
= − 4 2
+ +2
−3+6+8
= 4
11
=
4
11
∴ I= sq units.
4

27. Let P(A) = Probability that A solves the problem

P(B) = Probability that B solves the problem

¯
P(A)= Probability that A does not solve the problem
¯
and P(B)= Probability that B does not solve the problem
According to the question, we have
1
P(A) =
2
¯ 1 1
then P(A) = 1 − P(A) = 1 − =
2 2
¯
[ ∵ P(A) + P(A) = 1]
1
and P(B) =
3
¯ 1 2
then P(B) = 1 − P(B) = 1 − 3
= 3

i. P (problem is solved)
¯ ¯
= P(A ∩ B) + P(A ∩ B) + P(A ∩ B)
¯ ¯
= P(A)P(B) + P(A) ⋅ P(B) + P(A) ⋅ P(B)
[ ∵ A and Bare independent events]

=
( ) ( ) ( )
2
1
2
×
1
2
3
1
+
1
2
4
×
1
3
2
+
1
2
×
1
3

= + + = =
6 6 6 6 3
2
Hence, probability that the problem is solved, is
3
ii. P (exactly one of them solve the problem)
= P (A solve but B do not solve) + P ( A do not solve but B solve)
¯ ¯
= P(A ∩ B) + P( A ∩ B)
¯ ¯
= P(A) P(B) + P(A)P(B)

=
( ) ( )
1
2
×
2
3
+
1
2
×
1
3
=
2
6
+
1
6
=
3
6
=
1
2

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π/4
28. According to the question, I = ∫ 0 log(1 + tanx)dx

∴
π/4
I = ∫ 0 log 1 + tan
[ ( )] π
4
−x dx

[ ∵ ∫ a0f(x)dx = ∫ a0f(a − x)dx ]

( )
π
tan 4 − tan x
π/4
= ∫ 0 log 1 + π
dx
1 + tan 4 tan x

[ ∵ tan(x − y) =
tan x − tan y
1 + tan xtan y ]
π/4
= ∫ 0 log 1 +
( 1 − tan x
1 + tan x ) dx

π/4
= ∫ 0 log
( 2
1 + tan x ) dx

π/4
= ∫ 0 (log2)dx − ∫ 0 log(1 + tanx)dx
π/4
[ ∵ log
m
n
= logm − logn ]
π/4
= ∫ 0 (log2)dx − I

⇒
π/4
2I = ∫ 0 log2dx = log2[x] 0
π/4
= log2 [ ]π
4
−0
π
∴ I= log2
8

OR

4
According to the question , I = ∫ 0[ | x | + | x − 2 | + | x − 4 | ]dx
For,
0< x <4, |x| = x
0 < x ≤ 2, | x − 2 | = − (x − 2)
2 ≤ x < 4, | x − 2 | = (x − 2)
0 < x < 4, | x − 4 | = − (x − 4)
4 2 4 4
∴ I = ∫ 0xdx + ∫ 0(2 − x)dx + ∫ 2(x − 2)dx + ∫ 0(4 − x)dx

[] [ ] [ ] [ ]
4 2 4 4
x2 x2 x2 x2
= + 2x − + − 2x + 4x −
2 2 2 2
0 0 2 0

= (8) + [(4 − 2) − 0] + [(8 − 8) − (2 − 4)] + 16 − [ 16

2 ]
= 8 + 2 + 2 + (16 − 8)
= 20
∴ I = 20 sq units.

29. Here,
dy
dx
+ y cot x = 4x cosec x, y ()π
2
=0

It is a linear differential equation. Comparing it with,

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dy
+ Py = Q
dx
P = cot x, Q = 4x cosec x
I.F. = e ∫ pdx
= e ∫ cot xdx
= elog sin x
= sin x
Solution of the equation is given by,
y × (I.F.) = ∫ Q × (I.F.) dx + c
y (sin x) = ∫ 4x cosec x × (sin x) dx + c
= ∫ 4x dx + c
x2
y sin x = 4 2 + c

= 2x2 + c
π
Put y = 0, x =
2
π2
0= 2
+c
2
π
c=−
2
π2
Now, y sin x = 2x2 −
2

OR

According to the question,

Given differential equation is

( 1 + y ) + (x − e ) dy
2 tan − 1 y
=0
dx

It can be rewritten as

(1 + y )
dx −1
2
dy
+ x − e tan y
=0
On dividing both sides by (1 + y 2), we get
dx 1 e tany − y
+ x=
dy 1 + y2
(1+y ) 2

dx
It is a linear differential equation of the form + Px = Q,
dy
tan − 1 y
1 e
here, P = and Q =
1 + y2 1 + y2
We know that,
Integrating factor IF = e ∫ Pdy
1
−1
= e ∫ 1 + y 2 = e tan y
∴ The general solution of linear differential equation is given by x × IF = ∫ (Q × IF)dy + C
−1
e tan y
tan − 1y −1
⇒ x×e =∫ × e tan y
dy + C
1 + y2
2tan − 1y
− 1y e
⇒ xe tan =∫ dy + C
1 + y2
put tan − 1y = t
1
⇒ dy = dt
1 + y2

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−1
xe tan y
= ∫ e 2tdt + C
e 2t
tan − 1 y
⇒ xe = 2
+C
−1
−1 e 2tan y
⇒ xe tan y
= 2
+C[ ∵ t = tan − 1y]

30. We have, Z = 2x + 3y subject to the constraints x ≥ 0, y ≥ 0, x + 2y ≥ 1 and x+2y ≤ 10

Now draw the line x + 2y = 1 and x + 2y = 10

and shaded region satisfied by above inequalities Here the feasible region is bounded.

The corner points are given as A(1, 0), B(10, 0), C(0, 5), and D 0, ( ) 1
2

The value of Z at A (1,0) = 2 ,B (10,0) = 20, at C (0,5) = 15 and D 0, ( ) 1

2
=
3
2

At corner points, the minimum value of Z is

3
2 ( )
this is the required solution which occurs at D 0,
1
2

OR

Converting the inequations into equations, we obtain the lines -2x + y = 4, x + y = 3 , x - 2y = 2, x = 0 and y = 0.
These lines are drawn on a suitable scale and the feasible region of the LPP is shaded in Fig.

Now, give a value, say 15 equal to (lcm of 3 and 5) to Z to obtain the line 3x + 5y = 15. This line meets the coordinate axes
at P1 (5, 0) and Q1(0, 3). Join these points by a dotted line. Move this line parallel to itself in the decreasing direction
towards the origin so that it passes through only one point of the feasible region. Clearly, P3Q3 is such a line passing
through the vertex P of the feasible region.

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The coordinates of P are obtained by solving the lines x - 2y = 2 and x + y = 3.
8 1
Solving these equations, we get x = and y = .
3 3
8 1
Put, x = and y = in Z = 3x + 5y, we get
3 3
8 1 29
Z=3 × 3
=5 × 3
= 3
29 8 1
Hence, the minimum value of Z is 3
at x = 3 , y = 3 .

{
π
asin (x + 1), x≤0
2
31. According to the question,we are given f(x) = tan x − sin x is continuous at x = 0, and we have to find value of
, x>0
x3

a.
Therefore, LHL = RHL = f(0)....(i)
π
Now, LHL = lim asin 2 (x + 1)
x→0−
π
⇒ LHL = lim asin 2 ( − h + 1)
h→0
π
= asin = a..........(ii)
2
π
Also,f(0) = asin 2 = a.......(iii)
Now, we need to evaluate RHL at x = a [since LHL = f(0) = a and from this,it is not possible to find the value of a]
tan x − sin x
Now, RHL = lim
x3
x→0+
tan h − sin h
⇒ RHL = lim
h3
h→0
sin h
cos h
− sin h
= lim
h3
h→0
sin h − sin hcos h
= lim
h 3cos h
h→0
sin h ( 1 − cos h )
= lim
h 3cos h
h→0
sin h 1 − cos h 1
= lim ⋅ lim ⋅ lim
h h2 cos h
h→0 h→0 h→0
1 − cos h
= 1 × lim ×1
h→0 h2
1 − cos h
= lim
h2
h→0
h

[ ]
2sin 2 2 x
= lim ∵ 1 − cosx = 2sin 2
h2 2
h→0
h
2 × sin 2 2
= lim
h→0
( ) h2
4
×4

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()
h 2

[ ]
1 sin 2 1 sin x
= × lim = ×1 ∵ lim =1
2 h 2 x
h→0 2
x→0

1
∴ RHL = ..........(iv)
2
Therefore, on substituting the values from Eqs. (ii), (iii) and (iv) to Eq. (i), we get,
1 1
a= =a ⇒ a=
2 2

Section D

32. First, we sketch the graph of y = | x + 3 |

{
x + 3, if x+3≥0
∴ y = |x + 3| =
− (x + 3), if x+3<0

⇒ y = |x + 3| = { x + 3,
− x − 3,
if
if
x≥ −3
x< −3
So, we have y = x + 3 for x ≥ - 3 and y = -x - 3 for x < -3
A sketch of y = | x + 3 | is shown below:

y = x + 3 is the straight line which cuts X and Y-axes at (-3, 0 and (0, 3), respectively.
∴ y = x + 3 for x ≥ -3 represents the part of the line which lies on the right side of x = -3.
Similarly,y = − x − 3, x < -3 represents the part of line y = − x − 3, which lies on left side of x = -3
Clearly, required area = Area of region ABPA + Area of region PCOP
−3 0
= ∫ − 6 ( − x − 3)dx + ∫ − 3 (x + 3)dx

[ ] [ ]
−3 0
x2 x2
= − − 3x + + 3x
2 2
−6 −3

= [( ) −
9
2
+ 9 − ( − 18 + 18) + 0 − ] [ ( )] 9
2
−9

= − ( ) 9
2
−
9
2
+ (9 + 9)

= 18 - 9
= 9 sq. units

33. Here R is a relation on N × N , defined by (a, b) R (c, d) ⇔ a + d = b + c for all (a, b), (c, d) ∈ N×N
We shall show that R satisfies the following properties

i. Reflexivity:
We know that a + b = b + a for all a, b ∈ N.

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∴ (a, b) R (a, b) for all (a, b) ∈ (N × N)
So, R is reflexive.
ii. Symmetry:
Let (a, b) R (c, d). Then,
(a, b) R (c, d) ⇒ a+d=b+c
⇒ c+b=d+a
⇒ (c, d) R (a, b).
∴ (a, b) R (c, d) ⇒ (c, d) R (a, b) for all (a, b), (c, d) ∈ N×N
This shows that R is symmetric.
iii. Transitivity:
Let (a, b) R (c, d) and (c, d) R (e, f). Then,
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a + d = b + c and c + f = d + e
⇒ a+d+c+f=b+c+d+e
⇒ a+f=b+e
⇒ (a, b ) R (e, f).
Thus, (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f)
This shows that R is transitive.
∴ R is reflexive, symmetric and transitive
Hence, R is an equivalence relation on N × N

OR

Given that, A = R - {3}, B = R - {1}.

x−2
f : A → B is defined by f(x) = ∀ x ∈ A
x−3
For injectivity
x1 − 2 x2 − 2
Let f(x 1) = f(x 2) ⇒ =
x1 − 3 x2 − 3
⇒ (x1 - 2)(x2 - 3) = (x2 - 2)(x1 - 3)
⇒ x1x2 - 3x1 - 2x2 + 6 = x1x2 - 3x2 - 2x1 + 6
⇒ -3x1 - 2x2 = -3x2 - 2x1
⇒ -x1 = -x2 ⇒ x1 = x2
So, f(x) is an injective function
For surjectivity
x−2
Let y = x−3
⇒ x - 2 = xy - 3y
2 − 3y
⇒ x(1 - y) = 2 - 3y ⇒ x=
1−y
3y − 2
⇒ x= ∈ A, ∀ y ∈ B [codomain]
y−1
So, f(x) is surjective function.
Hence, f(x) is a bijective function.

[ ]
1 1 1
34. Given: A = 1 2 −3
2 −1 3

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[ ][ ]
1 1 1 1 1 1
∴ A2 = 1 2 −3 1 2 −3
2 −1 3 2 −1 3

[ ]
1+1+2 1+2−1 1−3+3
⇒ A = 2 1+2−6 1+4+3 1−6−9
2−1+6 2−2−3 2+3+9

[ ]
4 2 1
= −3 8 − 14
7 −3 14

[ ][ ]
4 2 1 1 1 1
Now A 3 = A 2A = −3 8 − 14 1 2 −3
7 −3 14 2 −1 3

[ ]
4+2+2 4+4+1 4−6+3
= − 3 + 8 − 28 − 3 + 16 + 14 − 3 − 24 − 42
7 − 3 + 28 7 − 6 − 14 7 + 9 + 42

[ ]
8 7 1
= − 23 27 − 69
32 − 13 58

L.H.S. = A 3 − 6A 2 + 5A + 11I

[ ][ ][ ] [ ]
8 7 1 4 2 1 1 1 1 1 0 0
= − 23 27 − 69 − 6 −3 8 − 14 +5 1 2 −3 + 11 0 1 0
32 − 13 58 7 −3 14 2 −1 3 0 0 1

[ ][ ][ ][ ]
8 7 1 24 12 6 5 5 5 11 0 0
= − 23 27 − 69 − − 18 48 − 84 + 5 10 − 15 + 0 11 0
32 − 13 58 42 − 18 84 10 −5 15 0 0 11

[ ][ ]
8 − 24 + 5 7 − 12 + 5 1−6+5 11 0 0
= − 23 + 18 + 5 27 − 48 + 10 − 69 + 84 − 15 + 0 11 0
32 − 42 + 10 − 13 + 18 − 5 58 + 84 + 15 0 0 11

[ ][ ][ ]
− 11 0 0 11 0 0 0 0 0
= 0 − 11 0 + 0 11 0 = 0 0 0 = 0= R.H.S.
0 0 − 11 0 0 11 0 0 0

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Now, to find A − 1, multiplying A 3 − 6A 2 + 5A + 11I = 0 by A − 1

⇒ A 3A − 1 − 6A 2A − 1 + 5AA − 1 + 11I. A − 1 = 0.A − 1

⇒ A 2 − 6A + 5I + 11A − 1 = 0

⇒ 11A − 1 = 6A − 5I − A 2

[ ] [ ][ ]
1 1 1 1 0 0 4 2 1
⇒ 11A − 1 = 6 1 2 −3 −5 0 1 0 − −3 8 − 14
2 −1 3 0 0 1 7 −3 14

[ ][ ][ ]
6 6 6 5 0 0 4 2 1
⇒ 11A − 1 = 6 12 − 18 − 0 5 0 − −3 8 − 14
12 −6 18 0 0 5 7 −3 14

[ ]
6−5−4 6−2 6−1
−1 6+3 12 − 5 − 8 − 18 + 14
⇒ 11A =
12 − 7 −6 + 3 18 − 5 − 14

[ ]
−3 4 5
⇒ 11A − 1 = 9 −1 −4
5 −3 −1

[ ]
−3 4 5
1
⇒ A −1 = 9 −1 −4
11
5 −3 −1

35. Given
→
r = (î + 2ĵ − 4k̂)+ λ(2î + 3ĵ + 6k̂)
→
r = (3î + 3ĵ − 5k̂)+ μ( − 2î + 3ĵ + 8k̂)
Here, we have
→
a 1 = ı̂ + 2ȷ̂ − 4k̂
→
b 1 = 2ı̂ + 3ȷ̂ + 6k̂
→
a 2 = 3î + 3ĵ − 5k̂
→
b 2 = − 2ı̂ + 3ȷ̂ + 8k̂
Thus,

| |
→ → î ĵ k̂
b1 × b2 = 2 3 6
−2 3 8

= î(24 − 18) − ĵ(16+ 12) + k̂(6 − 6)

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→ →
b 1 × b 2 = 6î − 28ĵ + 0k̂
→ →
∴ |b × b |
1 2 = 6 2 + ( − 28) 2 + 0 2
√
= √36 + 784 + 9
= √820
→ →
a 2 − a 1 = (3 − 1)î+ (3 − 2)ȷ̂ + ( − 5 + 4)k̂
→ →
∴ a − a
2 1 = 2 ı̂ + ȷ̂ − k̂
Now, we have
→ → → →
(b 1 × b 2) ⋅ ( a 2 − a 1 ) = (6ı̂ − 28ȷ̂ + 0k̂) ⋅ (2ı̂ + ȷ̂ − k̂)
= (6 × 2) + (( − 28) × 1) + (0 × ( − 1))
= 12 - 28 + 0
= -16
Thus,the shortest distance between the given lines is

| |
→ → → →
( b1 × b2 ) ⋅ ( a2 − a1 )
d= → →
| b1 × b2 |

⇒ d=

16
| |
− 16

√820

∴ d= units
√820

OR

Suppose the point (1, 0, 0) be P and the point through which the line passes be Q(1,-1,-10). The line is parallel to the
→
vector b = 2î − 3ĵ + 8k̂
Now,
→
PQ = 0î − ĵ − 10k̂

| |
→ î ĵ k̂
→
∴ b × PQ = 2 −3 8
0 −1 − 10

= 38î + 20ĵ − 2k̂

→
→
⇒ | b × PQ | =
√382 + 202 + 22
= √1444 + 400 + 4
= √1848
→
→
| b × PQ |
d= →
|b|

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√1848
=
√77
= √24
= 2√ 6
Suppose L be the foot of the perpendicular drawn from the point P(1,0,0) to the given line-

The coordinates of a general point on the line

x−1 y+1 z + 10
2
= −3
= 8
are given by
x−1 y+1 z + 10
= = =λ
2 −3 8
⇒ x = 2λ + 1
y = − 3λ − 1
z = 8λ − 10
Suppose the coordinates of L be
(2λ + 1, − 3λ − 1, 8λ − 10)
Since,The direction ratios of PL are proportional to,
2λ + 1 − 1, − 3λ − 1 − 0, 8λ − 10 − 0, i.e., 2λ, − 3λ − 1, 8λ − 10
Since,The direction ratios of the given line are proportional to 2, -3, 8, but PL is perpendicular to the given line.
∴ 2(2λ) − 3( − 3λ − 1) + 8(8λ − 10) = 0
⇒ λ = 1 Substituting λ = 1 in (2λ + 1, − 3λ − 1, 8λ − 10) we get the coordinates of L as (3, -4, -2). Equation of the line
PL is given by
x−1 y−0 z−0
= =
3−1 −4−0 −2−0
x−1 y z
= = =
1 −2 −1
→
⇒ r = î + λ(î − 2ĵ − k̂)

Section E

36. i.

4 4 2
Here, P(E1) = , P(E2) = , P(E3) =
10 10 10

P
() () ()
A
E1
=
45
100
,P
A
E2
=
60
100
,P
A
E3
=
35
100

4
() ()
P(A) = P(E1)

45 4
⋅ P

60
A
E1

2
+ P(E2)

35
⋅ P
A
E2
+ P(E3) ⋅ P
()
A
E3

= × + × + ×
10 100 10 100 10 100

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180 240 70
= + +
1000 1000 100
490
= = 4.9
1000

ii. Required probability = P

()
E2
A

=
P E2( ) ⋅ P
( ) A
E2

P(A)
4 60
×
10 100
= 490
1000
240 24
= =
490 49

iii. Let,
E1 = Event for getting an even number on die and
E2 = Event that a spade card is selected
3
∴ P(E1) = 6
1
= 2
13 1
and P(E2) = 52
= 4
Then, P(E1 ∩ E2) = P(E1) ⋅ P(E2)
1 1 1
= 2, 4
= 8
OR
P(A) + P(B) - P(A and B) = P(A)
⇒ P(A) + P(B) - P(A ∩ B) = P(A)
⇒ P(B) - P(A ∩ B) = 0
⇒ P(A ∩ B) = P(B)
P(A∩B)
∴ P(A|B) = P(B)
P(B)
=
P(B)
=1

37. i. Force applied by team A

= √4 2 + 0 2
=4N
Force applied by team B
= √( − 2) 2 + 42
= √4 + 16
= √20
= 2√5 N
Force applied by team C
= √( − 3) 2 + ( − 3) 2

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= √9 + 9 = √18 = 3√2
Hence, the force applied by team B is maximum.
So, Team 'B' will win.

ii. Sum of force applied by team A, B and C

= (4 + (-2) + (-3))î + (0 + 4 + (-3))ĵ
= − î + ĵ
Magnitude of team combine force
= √( − 1) 2 + (1) 2
= √2N
iii. Force applied by team B
= √( − 2) 2 + 42
= √4 + 16
= √20
= 2√5 N
OR
Force applied by team A
= √4 2 + 0 2
=4N

dy
38. i. The rate of growth =
dx
1
d ( 4x − 2 x2 )
=
dx
=4-x
ii. For the height to be maximum or minimum
dy
dx
=0

dy
(
d 4x − 2 x 2
1
) 1
⇒
dx
= dx
=4− 2
⋅ 2x = 0
dy
=4-x=0
dx
⇒ x=4
∴ Number of required days = 4
dy
iii. =4-x
dx
d 2y
⇒ = -1 < 0
dx 2
⇒ Function attains maximum value at x = 4
We have
1
y = 4x - x 2
2
1
∴ when x = 4 the height of the plant will be maximum which is y = 4 × 4 − × (4) 2 = 16 - 8 = 8 cm
2
OR
1
We have, y = 4x - x 2
2
∴ When x = 4 the height of the plant will be maximum which is

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1
y=4×4− × (4) 2
2
= 8 - 2 = 6 cm
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