KENDRIYA VIDYALAYA SANGATHAN JAIPUR REGION

PRE BOARD EXAMINATION (2024 - 25)

SET No. 8
CLASS- XII SUBJECT: Mathematics (041)
Maximum Marks: 80 Time: 3 Hours
MARKING SCHEME

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

C D A D C B B C A C

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

B C B C A A D B C D

21. As sin-1(sin x) = x with x ∈ [-π/2, π/2] 1.

So, sin-1(sin -π/8) = – π/8 1.

22. Let f (x) = x4 - 62x2 + ax + 9

Therefore,On differentiating wrt x, we get

f' (x) = 4x3 - 124x + a

It is given that function f attains its maximum value on the interval [0, 2] at x =1.

Hence, f' (1) = 0 1

⇒ 4x3 - 124x + a = 0

⇒ 4 - 124 + a = 0

⇒ - 120 + a = 0

⇒ a = 120 Thus, the value of a = 120 1

dy m
23(a) . y=sin ( msin x )⇒ =cos ( m sin x ) .
−1 −1
dx √1−x2
dy
⇒ √ 1−x =mcos ( msin x )
2 −1
dx

( )
2
d y dy −2 x m
Again diff . w . r . t . x , √ 1−x =−msin ( msin x ) .
2 −1
+
2 √ 1−x √ 1−x 2
2
dx dx 2

1
2
2 d y dy
⇒ ( 1−x ) 2 −x =−m sin ( m sin x )=−m y
2 −1 2

dx dx
2
2 d y dy
⇒ ( 1−x ) 2 −x +m y=0
2
1
dx dx
OR 1

23.(b)√ 1+sinx = sin

√ 2 x
2
x x x x
+ cos2 + 2sin cos = cos + sin
2 2 2 2
x
2 ( ) 1
 2 x

√ 2 x x x
(x x
√ 1−sinx = sin 2 + cos 2 −2 sin 2 cos 2 = cos 2 −sin 2 )
y=tan−1
[ √√
1+ sinx+ √1−sinx
1+ sinx−√ 1−sinx] { }
−1
= tan cot
x
2

y=tan
−1
{( 2 2 )} 2 2 dx = 2
(
tan
π x
− ) π x dy −1
= −

24(a) .

24(b)

25.
Section C
26. f(x) = 3x4 -4x3 -12x2 +5

f’(x) = 12x3 -12x2 -24x = 12x ( x2 – x – 2 )

= 12x (x - 2)(x + 1) 1

Critical points are 0 , -1 , 2

i ) strictly increasing (-1,0)∪ (2,∞ ) 1

ii) strictly decreasing (-.∞ ,-1)∪ (0,2) 1

2 dy y tan x dy
⇒ + ( sec x ) y=tan x . sec x
2 2
27. Dividing both sides by cos x , + =
dx cos2 x cos 2 x dx
dy 2 2
Comparing with + Py=Q , we get P=sec x , Q=tan x . sec x
dx

∴ I . F .=e∫ =e∫
P dx sec2 xdx tan x
=e 1

Now, solution of Differential Equation is

y ( e tan x ) =∫ tan x . sec 2 x e tan x dx +c

2
¿ integrate RHS , put tan x=t ⇒ sec x dx=dt 1

∴ y ( e tan x ) =∫ t . et dt+ c now integrating by parts ,

y ( e tan x ) =t . e t−∫ 1. e t dt +c=t .e t −e t +c

⇒ y ( e tan x )=e t ( t−1 ) +c=e tan x ( tan x−1 ) +c

−tan x
⇒ y =tan x−1+c . e
28(a) 1
OR

28(b)

2
 1

29(a).

cos x dx
I =∫ ; p ut sin x=t ⇒ cos x dx=dt 1
(1+sin x )( 2+ sin x)

=∫
dt
(1+t )(2+t)
=∫ (
1 −1
¿
1+t 2+t
)dt =∫
dt
1+t
−∫
dt
2+t
¿ =log
1+t
2+t| |
+c 2

I =log |1+sin
2+sin x |
x
+c Putting t = sin x we obtain

29(b)
30

1.5

1.5

31.(a)

Sol. Given, refrigerator box contains 2 milk chocolates and 4 dark chocolates. 2 chocolates are drawn at
random, we have to find the probability distribution of milk chocolates. Let X denotes the probability
distribution, then X will have the value

4 3 12
0, 1 and 2. Now probability distribution of milk chocolates P (X = 0) i.e., No milk chocolate = X =
6 5 30
2 4
P (X = 1) i.e., One milk chocolate = X X 2=16/30
6 5
2 1 2
P (X = 2) i.e., both milk chocolates= X = 2
6 5 30
Required probability distribution is
X 0 1 2

P(X) 12/30 16/30 2/30

Here most likely outcome is 16/30. Ie., one chocolates of each types

31(b) OR S = {1, 2, 3, 4, ......................12}

A = {2, 4, 6, 8, 10, 12}

B = {4, 5, 6, 7, 8, 9, 10, 11, 12}

and A ∩ B = {4, 6, 8, 10, 12}

6 1 9 3 5
 P( A)= = ;❑ P (B)= = ; P( A ∩ B)=
12 2 12 4 12
P( A ∩ B) 5 /12 5
P( A /B)= = =
P(B) 9 /12 9
SECTION D

32 Correct Figure 1

To find the area of the ellipse x2 + 9y2 = 36

2 2
x y
2
+ 2 + = 1 ...(i) 1
6 2
Clearly it is a horizontal ellipse as shown in
6
4
Required area is I=
3 ∫ √6 2−x 2 dx
0

4
= ¿¿ 2
3
=12 π 1

[ ][ ][ ]
1 −1 0 2 2 −4 6 0 0
33 : 2 3 4 −4 2 −4 = 0 6 0 = 6I,
0 1 2 2 −1 5 0 0 6

−1 1
i.e AB=6 I so . A = B
6

[ ]
2 2 −4
−1 1
So A = −4 2 −4
6
2 −1 5

System of equation can be written as AX=B and X= A-1B

[] [ ][ ] [ ] [ ]
x 2 2 −4 3 12 2
1 1
y = −4 2 −4 17 = −6 = −1
6 6
z 2 −1 5 7 24 4

Hence x=2, y=-1, z=4

dy 1
34. (a) = 2¿ . 2 2
dx x +1
¿> ¿ ( x2 + 1 )y1 = 2 ¿

2
( x2 + 1 )y2 + 2x.y1 = 2 2
x +1
( x2 +1 )2 y2 + 2x ( x2 +1 ) y1 = 2 1

OR

dy
34. (b) = (cosx) x ( x.logcosx ) ‘ + x sinx.(sinx.logx )’ 2
dx
dy −sinx sinx
= (cosx) x ( 1.logcosx + x ) + x sinx.(cosx.logx + ) 2
dx cosx x
dy sinx
= (cosx) x ( logcosx - xtanx) + x sinx.(cosx.logx + ) 1
dx x
x−2 y −2 z −3
35.(a) The co-ordinates of any point on the line = = = λ (let)
2 3 4
x= 2 λ+ 1 , y= 3 λ+ 2 z= 4 λ+ 3

Co-ordinates of the general point on the line be ( 2 λ+ 1 , 3 λ+ 2 , 4 λ+ 3)

x−4 y −1 z
The co-ordinates of any point on the line = = = μ (let)
5 2 1
x = 5 μ+ 4 , y=2 μ+1 , z=μ

Co-ordinates of the general point on the line are ( 5 μ+ 4 , 2 μ+1 , μ )

If both the lines intersect they must have a common point for some value of λ and μ

(2 λ+ 1 , 3 λ+ 2 , 4 λ+ 3) = (5 μ+ 4 , 2 μ+1 , μ )

2 λ+ 1 = 5 μ+ 4 ; 3 λ+ 2 = 2 μ+1 ; 4 λ+ 3 = μ

2 λ - 5 μ=3 ; 3 λ -2 μ=¿ -1 ; 4 λ - μ=¿-3

Getting λ = -1 and μ = -1 (solving first two equations) and it satisfy the third one .

Hence both the line will intersect and Point of Intersection is (-1,-1,-1).

OR

35(b)
36 . 1. 1

2.
 2

OR

Domain of f (x)=¿

Range of f (x)=¿

37
38

Let : Represent the event when many workers were not present for the job;

: Represent the event when all workers were present, and

E : Represent completing the construction work on time.

(i) The probability that all the workers are present for the job and work is completed on time

P(E∩E2) = P(E2) . P(E/E2) = 0.35x0.80 = 0.28

(ii) The probability that many workers are not present given that the construction work is completed on
time