KENDRIYA VIDYALAYA SANGATHAN DEHRADUN REGION

BLUE PRINT PRE BOARD 1 EXAMINATION 2024 – 25

Class: XII Max. Marks: 80
Subject: Mathematics Time Allowed: 3 hours

Chapter Chapter Name TYPE OF QUESTIONS

No. MCQ A-R V.S.A S.A L.A CASE TOTAL
BASED/
CCT
1 Relation and Functions 1(1) 1(3) - 2(4)
2 Inverse Trigonometric - 1(4) 1(4)
Functions
3 Matrices 3(1) 3(3)
4 Determinants 1(1) 1(1) 1(5) 3(7)
5 Continuity and 2(1) 1(2) 1(3) - - 4(7)
Differentiability
6 Applications of 1(1) 1(2) - 1(4) 3(7)
Derivatives
7 Integrals 1(1) 1(2) 1(3) 1(5) 4(11)
8 Application of Integrals - 1(5) 1(5)
9 Differential Equations 2(1) 1(3) 3(5)
10 Vector Algebra 2(1) 2(2) 4(6)
11 Three Dimensional 2(1) 1(1) 1(5) 4(8)
Geometry
12 Linear Programming 2(1) 1(3) 3(5)
13 Probability 1(1) 1(3) 1(4) 3(8)
TOTAL 18(18) 2(2) 5(10) 6(18) 4(20) 3(12) 38(80)
 KENDRIYA VIDYALAYA SANGATHAN DEHRADUN REGION 2024-25
PRE BOARD I EXAMINATION
CLASS-XII
SUBJECT-MATHEMATICS(041)

TIME ALLOWED : 3 HOURS MAX. MARKS:80

General Instructions:
(i) All the questions are compulsory.

(ii) The question paper consists of 33 questions divided into 5 sections A, B, C, D

and E.
(iii) Section A comprises of 20 questions of 1 mark each. Section B comprises of 5
questions of 2 marks each. Section C comprises of 6 questions of 3 marks each.
Section D comprises of 4 questions of 5 marks each. Section E comprises of 2
questions of 4 marks each. In Section E has 3 sourse based/case based questions
of 4 marks each.

(iv) Use of calculators is not permitted.

Sr. SECTION A Marks

No. All questions are compulsory. In case of internal choices attempt any one.
1 The value of 𝑥 for which (𝑥 − 𝑥 2 ) is maximum is 1
3 1 1 1
(a) 4 (b) 2 (c) 3 (d) 4

2 3 𝑑𝑥 1
The value of ∫0 is
√9−𝑥 2
𝜋 𝜋 𝜋 𝜋
(a) 6 (b) 4 (c) 2 (d) 18

OR
(1+𝑥2 )
The value of ∫ 𝑥𝑒 𝑑𝑥 is
2 2 2
𝑒 (1+𝑥 ) 𝑒 (1+𝑥 ) 𝑒 (1+𝑥 )
(a) +c (b) +c (c) +c
2 3 4
2
𝑒 (1−𝑥 )
(d) +c
2

3 2 1
𝑑2𝑦
The difference of the order and the degree of the differential equation ( 2) +
𝑑𝑥
𝑑𝑦 3 4
(𝑑𝑥 ) + 𝑥 = 0 is :
(a) 1 (b) 2 (c) -1 (d) 0
OR
𝑑𝑦
Degree of the differential equation sin 𝑥 + cos (𝑑𝑥 ) = 𝑦 2 is
(a) 2 (b) 1 (c) not defined (d) 0

4 If 𝛼, 𝛽 𝑎𝑛𝑑 𝛾 are the angles which a half ray makes with the positive direction of 1
the axes, then 𝑠𝑖𝑛 2 𝛼 + 𝑠𝑖𝑛2 𝛽 + 𝑠𝑖𝑛2 𝛾 is equal to
(a) 1 (b) 2 (c) 0 (d) -1
5 1
Based on the given shaded region as the feasible region in the graph, at
which point(s) is the objective function Z = 3x + 9y maximum?

(a)Point B (b) Point C (c) Point D (d) every point on the line segment CD
6 The corner points of the bounded feasible region of an LPP are 1
O(0, 0), A(250, 0), B(200, 50) and C(0, 175). If the maximum value of the
objective function Z = 2ax+ by occurs at the points A(250, 0) and B(200, 50),
then the relation between a and b is:

(a 2a=b (b) 2a=3b (c) a=b (d) a=2b

7 1 1 1
If A and B are two independent events with P (A) = 3 and P (B) = 4 , then P (B ∣
A) is equal to
1 1 3
(a) 4 (b) 3 (c) 4 (d) 1
8 4−𝑥 2 1
The function 𝑓(𝑥) = 4𝑥−𝑥 3 is
(a) Discontinuous at only one point
(b) Discontinuous exactly at two point
(c) Discontinuous exactly at three point
(d) None of these
9 𝑑𝑦 1
The integrating factor of the differential equation(𝑥. log 𝑥) 𝑑𝑥 + 𝑦 = 2 log 𝑥 is
given by

(a) log(log 𝑥) (b)𝑒 𝑥 (c)log 𝑥 (d) 𝑥

10 ̂ ̂
If 𝜃 is the angle between the vectors 2𝑖̂ − 2𝑗̂ + 4𝑘 and 3𝑖̂ + 𝑗̂ + 2𝑘 , then sin 𝜃 = 1
2 √2 2
(a) 2/3 (b) (c) (d) √
√7 7 7
11 If 𝜃 is the angle between any two vectors 𝑎⃗⃗⃗ and 𝑏⃗, then |𝑎
⃗⃗⃗ . 𝑏⃗| = |𝑎
⃗⃗⃗ × 𝑏⃗| when 𝜃 1
is equal to
𝜋 𝜋
(a) 0 (b) 4 (c) 2 (d) 𝜋
12 Direction cosines of a line from P(1,-2,4) to Q(-1,1,-2) are 1
−2 3 −6
(a)−2,3, −6 (b) 2, −3,6 (c) 2,3,6 (d) 7 , 7 , 7
13 V is a matrix of order 3 such that |𝐴𝑑𝑗 𝑉| = 7. Which of these could be |𝑉|: 1
3
(a) 72 (b) 7 (c) √7 (d) √7
14 2 −1 −1 −8 1
The order of the matrix A such that [ 1 0 ] 𝐴 = [ 1 −2] is B
−3 4 9 22
(a) 3 × 2 (b) 2 × 2 (c)3 × 3 (d) 2 × 3
15 2𝑥 + 𝑦 4𝑥 7 7𝑦 − 13 1
If [ ]=[ ], then
5𝑥 − 7 4𝑥 𝑦 𝑥+6
(a) 𝑥 2 + 𝑦 2 = 25 (b) 𝑥 2 − 𝑦 2 = 5 (c) 𝑥 2 + 𝑦 2 = 13
(d) 𝑥 2 − 𝑦 2 = 9
16. 𝑑𝑦 1
If 3x+2y = sin y, then 𝑑𝑥 is :
3 sin 𝑦−1 2−sin 𝑦 2−cos 𝑦
(a) (b) (c) (d)
cos 𝑦−2 2 3 3

17. If the matrix A is both symmetric and skew-symmetric, then A is a 1

(a)diagonal matrix (b)zero matrix (c)square
matrix (d) an identity matrix
18. Q1. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), 1
(4,4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.

19. 1 3 𝜆+2 1
Assertion (A) : If matrix 𝐴 = [2 4 8 ] is singular then 𝜆 = 4.
3 5 10
Reason (R) : If A is a singular matrix, then |𝐴| = 0.

a. Both A and R are individually true and R is the correct explanation of A.

b. Both A and R are individually true and 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. 1 2 𝑑𝑦 2 1
Assertion: If 𝑦 = 4 𝑢 4 and 𝑢 = 3 𝑥 3 + 5, then 𝑑𝑥 = 3 𝑥 2 (2𝑥 3 + 15)3.
𝑑𝑦 𝑑𝑦 𝑑𝑢
Reason: If y is a function of 𝑢 and 𝑢 is a function of 𝑥, then 𝑑𝑥 = 𝑑𝑢 × 𝑑𝑥 .
a) Both A and R are true and R is the correct explanation of A.

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

c) A is true but R is false.
d) A is false but R is true.
 Section B
𝜋
21. Find the values of k so that the function f is continuous at x = − 6 , where 2
√3𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥 𝜋
𝜋 , 𝑥 ≠ −6
𝑥+ 6
𝑓(𝑥) = {
𝜋
𝑘, 𝑥 = − 6

22. 2
The radius r of a right circular cylinder is decreasing at the rate of
7 3 cm/min.
and its height h is increasing at the rate of 2 cm/min. When r = 7 cm and h
= 2 cm, find the rate of change of the volume of cylinder.
OR
Find the absolute maximum value and the absolute minimum value of the
1 9
function: 𝑓(𝑥) = 4𝑥 − 2 𝑥 2 , 𝑥 ∈ [−2, 2]
23. 𝑑𝑥 2
Evaluate ∫ .
𝑥 2 +4𝑥+8
24. . If 𝜃 is the angle between two vectors i − 2j + 3k and 3i − 2j + k, find sin 𝜃. 2

25. 2
If 𝑎 is a unit vector, then find |𝑥 | if (𝑥 − 𝑎)(𝑥 + 𝑎) = 8

Section C
26. Show that the relation R defined by (a, b) R (c, d) ⇒ a + d = b + c 3
on the set N × N is an equivalence relation.

OR
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b|
is even}, is an equivalence relation.
27. 3
𝑑2𝑦
If x = a (cos 2t + 2t sin 2t) and = a (sin 2t − 2t cos 2t), then find 𝑑𝑥 2.

28. Solve the following differential equation : 3

𝑑𝑦
4 + 8𝑦 = 5𝑒 −3𝑥
𝑑𝑥
29. 3
Solve the following LPP graphically:
Minimize Z = 5x +10y
Subject to constraints: x + 2y ≤ 120, x + y ≥ 60, x – 2y≥ 0 y, x≥0, y ≥0

30. 1 3
Evaluate I= ∫ 1+𝑐𝑜𝑡𝑥 𝑑𝑥

31. 3
If A and B are two independent events such that P(A’∩ 𝐵)= 5/12 and P(A ∩ 𝐵′)=
1/6. Then find P(A) - P(B)
 Section D

32. 1 −1 0 2 2 −4 5
Determine the product [3 3 4] [−4 2 −4], and use it to solve the
0 1 2 2 −1 5
system of equations:
𝑥 − 𝑦 = 3, 2𝑥 + 3𝑦 + 4𝑧 = 17, 𝑦 + 2𝑧 = 7.

33. 𝟐𝒙 5
Find : I = ∫ (𝒙𝟐+𝟏)(𝒙𝟐+𝟑) 𝑑𝑥

OR
𝜋
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
4
Find I =∫0 𝑑𝑥
16+9𝑠𝑖𝑛2𝑥

34. Find the area of region bounded by the parabola 4𝑥 2 = 𝑦 and the line y= 5
8x + 12.
OR
Using integration, find the area of the region bounded by the triangle whose
vertices are A(2,5), B(4, 7) and C(6,2).

35. Find the shortest distance between the lines 𝑟 = (3𝑖̂ + 2𝑗̂ − 4𝑘̂) + 𝜆(𝑖̂ + 2𝑗̂ + 2𝑘̂) 5
and 𝑟 = (5𝑖̂ − 2𝑗̂) + 𝜇(3𝑖̂ + 2𝑗̂ + 6𝑘̂) .
Iflines intersect, find their point of intersection.
Section E
(Attempt any four parts out of five)
36. A gardener wants to construct a rectangular bed of garden in a circular patch of 4
land. He takes the maximum perimeter of the rectangular region as possible.
(Refer to the images given below for calculations)

(i) Find the perimeter of rectangle in terms of any one side and radius of
circle.
(ii) Find critical points to maximize the perimeter of rectangle?
(iii) Check for maximum or minimum value of perimeter at critical point.
OR
If a rectangle of the maximum perimeter which can be inscribed in a circle of
radius 10 cm is square, then the perimeter of region.
37. 4
Two men on either side of a temple of 30 meters high observe its top at the angle
of elevation 𝛼 and𝛽 respectively (as shown in the figure below). The distance
 between the two men is 40√3 meters and the distance between the first person A
and the temple is 30√3 meters.

Based on the above information, answer the following:

(i) Find 𝛼, ∠𝐶𝐴𝐵.
(ii) Find 𝛽, ∠𝐵𝐶𝐴.
𝜋
(iii) Find whether ∠𝐴𝐵𝐶 = 6
OR
Find domain and range of cos −1 𝑥.
38. Read the following passage and answer the questions given below.
4

There are two antiaircraft guns, named as A and B. The probabilities

that the shell fired from them hits an airplane are 0.3 and 0.2
respectively. Both of them fired one shell at an airplane at the same
time.
(i) What is the probability that the shell fired from exactly one of them hit the
plane?
(ii) If it is known that the shell fired from exactly one of them hit
the plane, then what is the probability that it was fired from B?
 KENDRIYA VIDYALAYA SANGATHAN DEHRADUN REGION 2024-25
PRE BOARD I EXAMINATION
CLASS-XII
SUBJECT-MATHEMATICS

Marking scheme
1. 1 1
(b) 2
2
2. 𝜋 𝑒 (1+𝑥 ) 1
(c) 2 OR (c) 2 +c
3. (d) 0 OR (c) not defined 1
4. (b) 2 1
5. (d) every point on the line segment CD 1
6. (a) 2a=b 1
7. (a) 1 1
4
8. (c) Discontinuous exactly at three point 1
9. (c)log 𝑥 1
10. (b) 2 1
√7
11. (b) 𝜋 1
4
12. (d) −2 , 3 , −6 1
7 7 7
13. (c) √7 1
14. (b) 2 × 2 1
15. (c) 𝑥 2 + 𝑦 2 = 13 1
3
16. (a) cos 𝑦−2 1
17. (b)zero matrix 1
18. (b)reflexive, transitive but not symmetric 1
19. (a) 1
20. (d) 1
𝜋
21. Since f(x) is cont. at 𝑥 = − 6 , hence
𝜋 1
L.HL.=R.H.L. =f( − 6 )
Solving for k , get k = 2 1

22. dv/dt = -3cm/min, dh/dt= 2cm/min ½

V = 𝜋r2h/3 ½
Dv/dt = 14𝜋/3 cm3/min 1

OR
f’(x) =4-x , x=4
f(4)=8 , f(-2)= -10, f(9/2)= 7.8
23. 𝑑𝑥 𝑑𝑥 1
∫ 2 =∫
𝑥 + 4𝑥 + 8 (𝑥 + 2)2 + 4 1
1 𝑥+2
= 2 tan−1 ( 2 ) + 𝐶
⃗⃗⃗ 𝑏⃗ = |𝑎.
24. To write 𝑎. ⃗⃗⃗ ||𝑏⃗ | sin 𝜃 𝑛̂ 1
2√6
To find value of sin 𝜃 = 1
7
25. To write |𝑎| = 1 1
To find |𝑥 | = 3 1
26. To show that 1
R is reflexive 1
R is symmetric 1
R is transitve
27. To find 𝑑𝑥 and 𝑑𝑦 1
𝑑𝑡 𝑑𝑡
𝑑𝑦 1
𝑑𝑦 𝑑𝑡
To write = 𝑑𝑥 = tan2t
𝑑𝑥
𝑑𝑡
𝑑2𝑦 1 1
2
= 𝑠𝑒𝑐 3 2𝑡
𝑑𝑥 2𝑡𝑎
28. To find I.F. =e2x 1
−5 1
y = 4 e-3x +c e-2x
29. Plotting three lines correctly 1
Shading the feasible region 1
Minimize Z = 300 at (60, 0) 1
30. I= ∫ 1 𝑑𝑥
1+𝑐𝑜𝑡𝑥
1
=∫ 𝑐𝑜𝑠𝑥 𝑑𝑥
1+
𝑠𝑖𝑛𝑥
𝑠𝑖𝑛𝑥
=∫ 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 𝑑𝑥
1 2𝑠𝑖𝑛𝑥
= 2 ∫ 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 𝑑𝑥 1
1 (𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥)+𝑠𝑖𝑛𝑥− 𝑐𝑜𝑠𝑥)
= 2∫ 𝑑𝑥
𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
1 1 𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥
= 2 ∫ 1𝑑𝑥 + ∫ 𝑑𝑥
2 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
1 1 𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥 1
= 2 𝑥 + 2 ∫ 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 𝑑𝑥
Let sinx + cosx = t ⇒ (cosx - sinx)dx = dt
1 1 −1
= 2 𝑥 + 2 ∫ 𝑡 𝑑𝑡
1 1
= 2𝑥 - log |𝑡| + c
2
1 1
= 2𝑥 - log |𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥| + c 1
2

31. A and B are two independent events

P(A’∩ 𝐵)= 5/12
P(A’)P(B)= 5/12
[1 – P(A)]P(B) =5/12 1

P(A ∩ 𝐵′)= 1/6.

P(A)P(B’)= 1/6
P(A)[1 – P(B)] =1/6 1

P(A) - P(B) = 1/30

1

32. 1− 1 0 2 2 −4
𝐴𝐵 = [ 3 3 4] [−4 2 −4]
0 1 2 2 −1 5 1
6 0 0
= [0 6 0]
0 0 6
 1 0 0
=6 [0 1 0] 1
0 0 1
AB = 6I
1
A-1 = 6 𝐵 1
2 2 −4
1
= = [−4 2 −4]
6
2 −1 5 1
Writing given equation in matrix form
1 −1 0 𝑥 3
[3 3 4] [𝑦] = [17]
0 1 2 𝑧 7
A X = C
1
X = A-1 C = 6 𝐵𝐶 1
𝑥 2 2 −4 3
1
𝑦
[ ] = 6 [−4 2 −4] [17]
𝑧 2 −1 5 7
12
1
= 6 [−6]
24
2
= [−1]
4

33. I = ∫ 𝟐𝒙
𝑑𝑥
(𝒙𝟐 +𝟏)(𝒙𝟐 +𝟑)
Let, x2 = t
2xdx = dt,
𝟏
I = ∫ (𝒕+𝟏)(𝒕+𝟑) 𝑑𝑡
𝑑𝑡 1
= ∫ (𝑡+1)(𝑡+3)
1 𝑨 𝑩
Let, (𝑡+1)(𝑡+3) = 𝒕+𝟏 + 𝒕+𝟑
1
1 = A(t+3) + B(t+1)
Equating the coefficients of t and the constant term, we get
A+B = 0
3A + B = 1
On solving these equations, we get
1
A = 1/2 , B = -1/2
1 𝟏/𝟐 −𝟏/𝟐
= + 𝒕+𝟑
(𝑡+1)(𝑡+3) 𝒕+𝟏
𝑑𝑡 1/2 −1/2
∫ (𝑡+1)(𝑡+3) = ∫ 𝑡+1 𝑑𝑡 +∫ 𝑡+3
𝑑𝑡
1 1
= log|t+1| - log|t+3| + c
2 2
1 𝑡+1 1
= log| |+c
2 𝑡+3
1 𝒙𝟐 +𝟏
= 2log|𝒙𝟐+𝟑| + c

OR 1

𝜋
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
I=∫04 𝑑𝑥
16+9𝑠𝑖𝑛2𝑥

Let sinx – cosx = t ⇒ 1 – sin2x = t2 1

(sinx + cosx)dx = dt
 𝜋
When x = 0 , t = -1 and when x = 4 , t = 0
𝜋
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥 1
I = ∫04 𝑑𝑥
16+9𝑠𝑖𝑛2𝑥

0 1
=∫−1 𝑑𝑡
16+9(1−𝑡 2) 1
0 1
=∫−1 𝑑𝑡
25− 9𝑡 2

1
=15 𝑙𝑜𝑔2
2
34. Given parabola 4𝑥 2 = 𝑦 and the line y = 8x + 12

For getting common points, we solve both equations

4𝑥 2 = 8𝑥 + 12
4𝑥 2 − 8𝑥 − 12 = 0
𝑥 2 − 2𝑥 − 3 = 0
(x-3)(x+1)=0
x=3,-1
Hence common points are (-1, 4) and (3, 36).
Requared area = area of shaded region.
3 3 3
=∫−1 𝑦𝑑𝑥 = ∫−1(8𝑥 + 12)𝑑𝑥 − ∫−1 4𝑥 2 𝑑𝑥
3
=∫−1[8𝑥 + 12 − 4𝑥 2 ]𝑑𝑥
3
= 4∫−1[2𝑥 + 3 − 𝑥 2 ]𝑑𝑥
5
=4[9+ ] = 128/3 sq. units
3

OR

Using integration, find the area of the region bounded by the triangle
whose vertices are A(2,5), B(4, 7) and C(6,2).
Sol.: Equation of line passing through two points (x1, y1) and (x2, y2) is
given by
𝑦2−𝑦1
y-y1 = ( x2−x1 )(𝑥 − 𝑥1)

Hence, equation of line AB is y = x +3

−5𝑥
equation of line BC is y = 2 + 17
−3𝑥 13
equation of line AC is y = = +
4 2
 Requared area = area of shaded region.
4 6 −5𝑥 6 −3𝑥 13
=∫2 (𝑥 + 3)𝑑𝑥 + ∫4 ( 2 + 17) 𝑑𝑥 − ∫2 ( 4 + 2 ) 𝑑𝑥
= 12 + 9 – 14= 7 sq. units.

35. 𝑎
⃗⃗⃗⃗2 − ⃗⃗⃗⃗
𝑎1=2𝑖 ̂ − 4𝑗̂ + 4𝑘̂
⃗⃗⃗
𝑏1 × ⃗⃗⃗⃗𝑏2 = 8𝑖̂ − 4𝑘̂ 1
(𝑏 ⃗⃗⃗1 × ⃗⃗⃗⃗
𝑏2 ).( ⃗⃗⃗⃗
𝑎2 − ⃗⃗⃗⃗
𝑎1 ) = 16-16=0
Hence lines are intersecting and the S.D. between lines is 0 2
For point of intersection
𝑟 = (3𝑖̂ + 2𝑗̂ − 4𝑘̂ ) + 𝜆(𝑖̂ + 2𝑗̂ + 2𝑘̂ ) = (5𝑖̂ − 2𝑗̂) + 𝜇(3𝑖̂ + 2𝑗̂ + 6𝑘̂)
Get here 𝜆= - 4, 𝜇= - 2
𝑟 = (3𝑖̂ + 2𝑗̂ − 4𝑘̂ ) + (−4)(𝑖̂ + 2𝑗̂ + 2𝑘̂) =(−𝑖̂ − 6𝑗̂ − 12𝑘̂) 1
point of intersection is (-1, -6 , -12)
1
36. (i) Perimeter =2(𝑥 + √4𝑎2− 𝑥2) 1
(ii) 𝑥 = √2𝑎 2
(iii) To show that 𝑥 = √2𝑎 is point of maxima 1

37. 1 1
(i) 𝛼 = sin−1 (2)
1
(ii) 𝛽 = tan−1 √3 2
𝜋
(iii) ∠𝐴𝐵𝐶 ≠ 6
𝑂𝑅
Domain=[-1 , 1], Range=[0 , 𝜋]
38. (i). P(E) = 0.38 2
𝐸 7 2
(ii) 𝑃 ( 𝐸3) = 19