KENDRIYA VIDYALAYA SANGATHAN, BHOPAL REGION

PRE-BOARD EXAM (2024-25)

CLASS- XII SUBJECT-MATHEMATICS (041)
Max.Marks:80 Time: 3 Hrs.
MARKING SCHEME
Q.No. Question Marks
Section - A (1X20=20)
1. (b) R is reflexive and transitive but not symmetric. 1
2. (c) 0 1
3. (d) 512 1
4. (a) 5 1
5. (c) 1 1
6. −1 1
(d) 2𝑎𝑡 2
7. (a) Continuous and differentiable at x=0 1
8. (c) 3 1
9. (𝑎) 𝑒 𝑥 𝑠𝑒𝑐𝑥 + 𝐶 1
10. (c) 𝑦 = 𝑐𝑥 1
11. 1 1
(b)±
√3
12. (a) (0,1) 1
13. (a) 2a = b 1

14. 3 1
(c) 13

15. 2𝜋 1
(a) 3

16. (d) 2𝑖̂ + 4𝑗̂ + 5𝑘̂ +μ (3𝑖̂ − 2𝑗̂ + 4𝑘̂) 1

17. 8 1
(c) 15
18. (d) 0 1

19. (d) (A) is false but (R) is true. 1

20. (b) Both (A) and (R) are true but (R) is not the correct explanation of(A). 1
Section - B (2X5=10)
21. √1+𝑥 2 −1
: 𝑡𝑎𝑛−1 [ 𝑥 ] , 𝑥 ≠ 0.

Put 𝑥 = 𝑡𝑎𝑛𝜃

−1
√1 + 𝑡𝑎𝑛𝜃 2 − 1
𝑡𝑎𝑛 [ ] 0.5
𝑡𝑎𝑛𝜃
𝑠𝑒𝑐𝜃 − 1
𝑡𝑎𝑛−1 [ ]
𝑡𝑎𝑛𝜃
1 − 𝑐𝑜𝑠𝜃 0.5
𝑡𝑎𝑛−1 [ ]
𝑠𝑖𝑛𝜃
𝜃
𝑡𝑎𝑛−1 [𝑡𝑎𝑛 ]
2 0.5
 𝜃 1 0.5
= = 2 𝑡𝑎𝑛−1 𝑥
2

22. 1 1 1 1
1
Area of triangle= 2 [𝑘 4 0] = ±4
0 0 2 0.5
−(𝑘 − 4) = ±4 0.5
𝑘 = 0 𝑎𝑛𝑑 8
OR
2 2 1 2 1 2 5 1
𝐴 =[ ].[ ]=[ ]
−2 3 −2 3 −10 7
2 5 −10 −5 8 0 0 0
𝐴2 − 5𝐴 + 8𝐼 = [ ]+[ ]+[ ]=[ ]=0 1
−10 7 +10 −15 0 8 0 0
23. 𝑉 = 𝑎3 differentiate w.r.t x
𝑑𝑉 𝑑𝑎 0.5
= 3𝑎2
𝑑𝑡 𝑑𝑡
𝑑𝑉
At 𝑑𝑡 = 9 𝑐𝑚3 /𝑠𝑒𝑐
𝑑𝑎
= 3/𝑎2 0.5
𝑑𝑡
𝑆 = 6𝑎2
𝑑𝑆 𝑑𝑎
= 12𝑎 0.5
𝑑𝑡 𝑑𝑡
𝑑𝑆
= 3.6 𝑐𝑚2 /𝑠𝑒𝑐 0.5
𝑑𝑡
24. (𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗)2 = (𝑎⃗)2 + ( 𝑏⃗⃗)2 + ( 𝑐⃗)2 + 2(𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ ) 1

1 0.5
𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ = − 2 (32 + 52 + 72 )

83 0.5
𝑎⃗. 𝑏⃗⃗ + 𝑏⃗⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ = − 2

OR
𝑎
⃗⃗ 6𝑖̂ − 2𝑗̂ + 3𝑘̂
𝑎̂ = =
|𝑎| √36 + 4 + 9 1

6𝑖̂ − 2𝑗̂ + 3𝑘̂

𝑎̂ = 0.5
7
60 20 30
vector in the direction of 𝑎
⃗⃗ magnitude 10 units = 𝑖̂ − 𝑗̂ + 𝑘̂ 0.5
7 7 7

25. 𝐴 𝑃(𝐴 ∩ 𝐵) 1
𝑃( ) = =
𝐵 𝑃(𝐵) 4

1 0.5
𝑃(𝐴 ∩ 𝐵) =
12
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

3 0.5
𝑃(𝐴 ∪ 𝐵) =
4
1
𝑃(𝐴′ ∩ 𝐵 ′ ) = 𝑃(𝐴 ∪ 𝐵)′ = 1 − 𝑃(𝐴𝑈𝐵) = 1
4
 Section - C (3X6=18)
26.
2 2 2 3 1
∫ (𝑥 − 1)𝑑𝑥 + ∫ (𝑥 − 1)𝑑𝑥 + ∫ −(𝑥 − 2)𝑑𝑥 + ∫ (𝑥 − 2)𝑑𝑥
1 1 1 2
2 3
∫ 1 𝑑𝑥 + ∫ (2𝑥 − 3)𝑑𝑥 1
1 2

=6 1

OR
𝜋 𝑥 𝑠𝑖𝑛𝑥
Evaluate : ∫0 dx
1+𝑐𝑜𝑠2 𝑥
𝜋 (𝜋−𝑥) 𝑠𝑖𝑛(𝜋−𝑥)
I= ∫0 1+𝑐𝑜𝑠2 (𝜋−𝑥) dx 1
𝜋
𝑠𝑖𝑛𝑥
2𝐼 = ∫ 2
0 1 + 𝑐𝑜𝑠 𝑥
Put cosx=t 1
𝜋2 1
𝐼=
4

27. Find the intervals in which the function 𝑓(𝑥) = 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 𝑜𝑛 [0,2𝜋]
𝜋
𝑓 ′ (𝑥) = 𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥 = √2sin (𝑥 + )
4
𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥 = 0

𝑡𝑎𝑛𝑥 = −1
3𝜋 7𝜋 1
𝑥 = −1 ⟹ 𝑥 = ,
4 4
3𝜋 7𝜋
𝑓(𝑥)𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 {0 < 𝑥 < } ∪ { < 𝑥 < 2𝜋} 1
4 4
3𝜋 7𝜋 1
𝑓(𝑥)𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 { <𝑥< }
4 4
28. 𝑑𝑦 2𝑥𝑦 1
+ (1+x2) = (1+𝑥 2 )2
𝑑𝑥
𝑑𝑦
+ 𝑃(𝑥)𝑦 = 𝑄(𝑥) 0.5
𝑑𝑥

I.F=𝑒 ∫ 𝑃(𝑥)𝑑𝑥
2𝑥
∫ 𝑑𝑥
I.F=𝑒 (1+x2) = (1 + x 2 )
1
y× 𝐼𝐹 = ∫ 𝑄(𝑥) × 𝐼𝐹𝑑𝑥 + 𝐶
1
𝑦 × (1 + x 2 ) = ∫ 𝑑𝑥 + 𝐶
(1 + x 2 )
𝑦(1 + x 2 ) = 𝑡𝑎𝑛−1 𝑥 + 𝑐 1
y = 0 when x = 1
𝜋
𝑦(1 + x 2 ) = 𝑡𝑎𝑛−1 𝑥 − 0.5
4
29. Mmaximise Z = 3x + 9y
subject to x+3y ≤ 60, x + y ≥ 10 , x≤ y , & x ≥ 0, y ≥0
for correct feasible region 1
𝐴𝑡(5,5) 𝑍 = 60

𝐴𝑡(15,15), 𝑍 = 180 1

𝐴𝑡(0,20) , 𝑍 = 180

𝐴𝑡(0,10) , 𝑍 = 90,
1
𝑍 ℎ𝑎𝑠 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑥 + 3𝑦 = 60

30.

0.5

0.5

0.5
0.5

OR
𝑥 2 +𝑥 2 +1
: ∫ 𝑥 2 (𝑥 2 +1) 𝑑𝑥
𝑥2 𝑥 2 +1
=∫ 𝑥 2(𝑥 2 +1) 𝑑𝑥 + ∫ 𝑥 2 (𝑥 2 +1) 𝑑𝑥
1
1 1
∫ 2 𝑑𝑥 + ∫ 2 𝑑𝑥
(𝑥 + 1) 𝑥 1
1
𝑡𝑎𝑛−1 𝑥 − + 𝐶
𝑥 1

31. lines 2𝑥 = 3𝑦 = −𝑧 𝑎𝑛𝑑 6𝑥 = −𝑦 = −4𝑧.

𝑥−0 𝑦−0 𝑧−0 𝑥−0 𝑦−0 𝑧−0
= = 𝑎𝑛𝑑 = =
1 1 −1 1 −1 −4 1
2 3 6
1 1 0.5
⃗⃗⃗⃗
𝑏1 = 𝑖̂ + 𝑗̂ − 𝑘̂
2 3
1 1 0.5
⃗⃗⃗⃗⃗
𝑏2 = 𝑖̂ + 𝑗̂ − 𝑘̂
6 4 1
the angle between the lines 90°
Section - D (5X4=20)
32. Evaluate the product AB, were
1 −1 0 4 4 −8
A = [2 3 4] and B = [−8 4 −8]
0 1 2 4 −2 10
and use product AB to solve the system of linear equations:

x − y = 4, 2x + 3y + 4z = 0, y + 2z = 2
1 −1 0 4 4 −8 12 0 0
𝐴𝐵 = [2 3 4] . = [−8 4 −8] = [ 0 12 0 ] = 12 𝐼 2
0 1 2 4 −2 10 0 0 12
 𝐴 = 12𝐵 −1 𝑂𝑅 𝐵 = 12𝐴−1

x − y = 4, 2x + 3y + 4z = 0, y + 2z = 2

Matrix representation of the system of equation

1
1 −1 0 𝑥 4
[2 3 4] . [𝑦] = [0]
0 1 2 𝑧 2
𝐴×𝑋 =𝐶

𝑋 = 𝐶 × 𝐴−1 1

𝐵 0.5
𝑋=𝐶×
12
𝑥 0 0.5
[𝑦] = [−4] 0.5
𝑧 3 0.5

33.

1
33 or

0.5

1.5

1.5

0.5

34. Find the shortest distance between lines

𝑥−4 𝑦+3 𝑧+1 𝑥−1 𝑦+1 𝑧+10
= 4 = 5 𝑎𝑛𝑑 4 = −3 = 5
3
𝑥−4 𝑦−(−3) 𝑧−(−1) 𝑥−1 𝑦−(−1) 𝑧−(−10) 0.5
= = 𝑎𝑛𝑑 = =
3 4 5 4 −3 5

⃗⃗⃗⃗⃗1 = 4𝑖̂ − 3𝑗̂ − 𝑘̂

𝑎 0.5
𝑎2 = 𝑖̂ − 𝑗̂ − 10𝑘̂
⃗⃗⃗⃗⃗
⃗⃗⃗⃗
𝑏1 = 3𝑖̂ + 4𝑗̂ + 5𝑘̂
⃗⃗⃗⃗⃗ 0.5
𝑏2 = 4𝑖̂ − 3𝑗̂ + 5𝑘̂
⃗⃗⃗⃗
𝑏1 × ⃗⃗⃗⃗⃗
𝑏2 = 35𝑖̂ + 5𝑗̂ − 25𝑘̂ 1
|𝑏⃗⃗⃗⃗1 × ⃗⃗⃗⃗⃗
𝑏2 | = 125√3
Formula of shortest distance 2
 Distance=120/125√3 0.5
OR

35. For correct graph 2

To identify correct area and formula 1
13
For finding correct area sq. units 2
3

Section - E (4X3=12)
36 (i) probability that an airplane reached its destination late. 2

(ii) If the airplane reached its destination late, find the probability that it was due to moderate 2
turbulence

37
2
(i) Show that the function: 𝑓: 𝑅 → {𝑥 ∈ 𝑅: −1 ≤ 𝑥 ≤ 1} defined by
𝑥
𝑓(𝑥) = 1+|𝑥| , 𝑥 ∈ 𝑅 is one -one.

(ii) Show that the function: 𝑓: 𝑅 → {𝑥 ∈ 𝑅: −1 ≤ 𝑥 ≤ 1} defined by 2

𝑥
𝑓(𝑥) = 1+|𝑥| , 𝑥 ∈ 𝑅 is onto.

(i) Write the expression for the volume (V) in terms of 𝑥. 1

𝑑𝑉 1
(ii) Find 𝑥 for which 𝑑𝑥 = 0.

(iii) What will be the maximum volume of the box? 2

OR
What is the outer surface area of the box of maximum volume?