VIVEKANANDA EDUCATIONAL SOCIETY

QUARTERLY EXAMINATION [2024-25]

CLASS XII - MATHEMATICS

ANSWER KEY - SET 1

SECTION A

1. ( C ) not symmetric but transitive.

2. (b) 1
3. ( c ) −𝑒 𝑦−𝑥
4. ( d) 64
5. ( b) 𝜆 = 1
6. (c ) a linear function to be optimised.
7. ( b) 3X 5
𝜋
8. (a) 3
9. (d) f is neither one-one nor onto
3
10. (d) − 4
1
11. (d) r = 2√𝜋 units
12. (b) √1 − 𝑥 2
13. (b) 3
2
14. ( c) 3
15. (b) 12 ( any answer)
16. (d) tan x – cot x + c
1 3
17. (a) 3 𝑒 𝑥
18. (a) 𝐼
19. (a) Both A and R are true and R is the correct explanation of A
20. (d) A is false and R is true.

SECTION B
𝜋 2𝜋 𝜋 2𝜋
21. + − =
4 3 4 3
3𝑥
22. P=2 Q=𝑒
I.F. = 𝑒 ∫ 2𝑑𝑥 = 𝑒 2𝑥
General solution is
y𝑒 2𝑥 = ∫ 𝑒 2𝑥 𝑒 3𝑥 dx +c
= ∫ 𝑒 5𝑥 dx +c
𝑒 5𝑥
y𝑒 2𝑥 = +c
5
𝑒 3𝑥
y = + c𝑒 −2𝑥
5
2 2
23. AT = [ ]
9 6
4 11
A + AT = [ ]
11 12
0 7
A – AT=[ ]
−7 0
 1 1
A= [A + 𝐴𝑇 ] +2 [A − 𝐴𝑇 ]
2
11 7
2 9 2 0
2 2
[ ] = [11 ] + [ 7 ]
2 6 6 −2 0
2
(OR)
|𝐴| = −19
−2 −3
𝐴𝑑𝑗𝐴 = [ ]
−5 2
1 −2 −3
𝐴−1 = − 19 [ ]
−5 2
1
K = 19
1 𝑑𝑥
24. I =
2
∫
√ √𝑥 2 + 𝑥−1
3
2
1 𝑑𝑥
=
2
∫ 2 2
√ √(𝑥+3) −(5)
4 4
1 3 3
= log|𝑥 + 4 + √𝑥2 + 2 𝑥 − 1| + c
√2

(OR)
Put logx = t , x = 𝑒𝑡
1 1
I = ∫ ( + (− 2 )) 𝑒 𝑡 dt
𝑡 𝑡
𝑒𝑡
= + c
𝑡
𝑥
= +𝐶
𝑙𝑜𝑔𝑥
25. 𝑦𝑙𝑜𝑔𝑥 = 𝑥𝑙𝑜𝑔𝑦
𝑦 𝑑𝑦 𝑥 𝑑𝑦
+ logx 𝑑𝑥 = 𝑦 𝑑𝑥 + log y
𝑥
𝑑𝑦 𝑥 𝑦
(𝑙𝑜𝑔𝑥 − 𝑦) = log y− 𝑥
𝑑𝑥
𝑑𝑦 𝑦 𝑥𝑙𝑜𝑔𝑦−𝑦
= 𝑥 (𝑦𝑙𝑜𝑔𝑥−𝑥 )
𝑑𝑥

SECTION C
𝑥+1 𝐴 𝐵𝑥+𝑐
26. = +
𝑥(𝑥 2 +1) 𝑥 𝑥 2+1

x+1 = A((𝑥 2 + 1) + 𝐵𝑥(𝑥) + 𝑐𝑥

A=1 ; B = -1 ; C = 1
1 𝑥 1
I = ∫ 𝑑𝑥 - ∫ 2 dx + ∫ 2 𝑑𝑥
𝑥 𝑥 +1 𝑥 +1
1
= log|x| - log |𝑥 2 + 1| + 𝑡𝑎𝑛 −1 𝑥 + c
2

(OR)
1
I=∫ 𝑑𝑥 put 𝑒 𝑥 = t
𝑒 𝑥 −1

𝑒 𝑥 𝑑𝑥 = 𝑑𝑡
 1 𝐴 𝐵
= +
𝑡(𝑡−1) 𝑡 𝑡−1

A = -1 B = 1
1 1
I= ∫(− 𝑡 + 𝑡−1
) dt

= - log|t| +log|t-1| + c
𝑡−1
= log | | + C
𝑡
𝑒 𝑥 −1
= log| |+ c
𝑒𝑥
𝑠𝑖𝑛𝑦 𝑑𝑥
27. 𝑐𝑜𝑠𝑦
dy = − 1+2𝑒 −𝑥
𝑒𝑥
∫ 𝑡𝑎𝑛𝑦 𝑑𝑦 = − ∫ 𝑒 𝑥 +2 𝑑𝑥
log |𝑠𝑒𝑐𝑦| = − log|𝑒 𝑥 + 2| + log|c|
secy(𝑒 𝑥 + 2) = c
(OR)
𝑥
𝑥 𝑦
𝑑𝑥 ( −1)𝑒
𝑦
= 𝑥
𝑑𝑦
1+𝑒 𝑦

𝑥 𝑑𝑥 𝑑𝑣
Put =v =v+y
𝑦 𝑑𝑦 𝑑𝑦

𝑑𝑣 𝑣𝑒 𝑣 −𝑒 𝑣
v + y𝑑𝑦 = 1+𝑒 𝑣
1+𝑒 𝑣 𝑑𝑦
∫ 𝑣+𝑒 𝑣 dv = − ∫ 𝑦
log| 𝑣 + 𝑒𝑣| = − log|y| + log|c|
𝑥
𝑥 𝑐
log| + 𝑒 | = log| | 𝑦
𝑦 𝑦
𝑥
𝑥 + 𝑦𝑒 = c 𝑦

𝑑𝑥
28. = 3a𝑠𝑖𝑛2 𝜃𝑐𝑜𝑠𝜃
𝑑𝜃
𝑑𝑦
= 𝑏 𝑐𝑜𝑠2𝜃𝑠𝑖𝑛𝜃[−3 − 4𝑐𝑜𝑠𝜃]
𝑑𝜃
𝑑𝑦 𝑏
= − 𝑐𝑜𝑡𝜃[3 + 4𝑐𝑜𝑠𝜃]
𝑑𝑥 3𝑎
𝑑2 𝑦 𝑏(12+6√2
=
𝑑𝑥 2 9𝑎2
 (OR)
x√1 + 𝑦 = − 𝑦 √1 + 𝑥
𝑥 (1 + 𝑦) = 𝑦 2 (1 + 𝑥)
2

𝑥2 + 𝑥2𝑦 = 𝑦 2 + 𝑦 2𝑥
𝑥2 + 𝑥2𝑦 − 𝑦 2 − 𝑦 2𝑥 = 0
(𝑥 + 𝑦)(𝑥 − 𝑦) + 𝑥𝑦(𝑥 − 𝑦) = 0
(𝑥 − 𝑦)[𝑥 + 𝑦 + 𝑥𝑦] = 0
𝑥
𝑦 = − 1+𝑥
𝑑𝑦 1
𝑑𝑥
=− (1+𝑥)2
′ (𝑥) 2
29. 𝑓 = −6𝑥 − 18𝑥 − 12
= −6(𝑥 2 + 3𝑥 + 2)
= −6(𝑥 + 2)(𝑥 + 1)
Put 𝑓 ′ (𝑥) = 0
𝑥 = −2 , 𝑥 = −1
In (−∞, −2) 𝑎𝑛𝑑 (−1, ∞) 𝑓 ′ (𝑥) < 0
=> f is strictly decreasing in (−∞, −2) 𝑎𝑛𝑑 (−1, ∞)
In (−2, −1) 𝑓 ′ (𝑥) > 0

=> f is strictly increasing in (-2,-1)

30.
1 1
Each line 2 mark and shading 2 mark & table and final answer 1 mark

31. Let 𝑥1 𝑎𝑛𝑑 𝑥2 ∈ 𝐴 such that f(𝑥1 ) = f(𝑥2 )

𝑥1 −1 𝑥2 −1
=
𝑥1 −2 𝑥2 −2
𝑥1 𝑥2 − 2𝑥1 − 𝑥2 + 2 = 𝑥1 𝑥2 − 2𝑥2 − 𝑥1 + 2
𝑥1 = 𝑥2
  f is one to one.
𝑥−1
Let y = 𝑥−2
𝑥𝑦 − 2𝑦 = 𝑥 – 1
𝑥(𝑦 − 1) = 2𝑦 − 1
2𝑦−1
x = 𝑦−1 ; y≠ 1
2𝑦−1
−1
𝑦−1
f(x) = 2𝑦−1
−2
𝑦−1

= y = > f is onto.
SECTION D
32. |A| = 67
−6 14 −15
Adj A = [ 17 5 9 ]
13 −8 −1
−6 14 −15
1
𝐴−1 67 [ 17 5 9 ]
13 −8 −1
The given equations can be written as
1 2 −3 𝑥 −4
[2 3 𝑦
2 ] [ ] = [ 14 ]
3 −3 −4 𝑧 −15
𝑇
𝐴 X=B
X = (𝐴−1 )𝑇 𝐵
−6 17 13 −4
1
= 67 [ 14 5 −8] [ 14 ]
−15 9 −1 −15

67
1
=67 [134]
201
𝑥 = 1; 𝑦 = 2; 𝑧 = 3

33. For Reflexivity: Let a ∈ 𝐴 , |a-a|= 0 is a multiple of 3 is true for all a ∈ 𝐴

(a,a) ∈ 𝑅 => R is reflexive.
For symmetric : Let a , b∈ 𝐴 such that (a,b) ∈ 𝑅
|a-b| is a multiple of 3
|b-a | is also a multiple of 3
(b,a) ∈ 𝑅 => R is symmetric.
 For transitivity: Let a,b and c∈ 𝐴 such that (a,b) ∈ 𝑅 and (b,c) ∈ 𝑅
|𝑎 − 𝑏| = 3m and |𝑏 − 𝑐| = 3n where m,n ∈ 𝑁
𝑎 − 𝑏 = ± 3𝑚 and 𝑏 − 𝑐 = ±3𝑛
( 𝑎 − 𝑏) + (𝑏 − 𝑐) = ±3(𝑚 + 𝑛)
𝑎 − 𝑐 = ±3(𝑚 + 𝑛)
|𝑎 − 𝑐| = 3𝑑 => is a multiple of 3
(𝑎, 𝑐) ∈ 𝑅 => R is transitive.
Elements related to 5 = {2,5,8,11}
(OR)
For reflexivity: Let (a,b)∈ 𝐴 (a,b) R (a,b) as ab = ba (multiplication is
commutative) is true for all (a,b)∈ 𝐴 => R is reflexive.
For symmetric : Let (𝑎, 𝑏) 𝑎𝑛𝑑 (𝑐, 𝑑) ∈ 𝐴 such that (a,b) R (c,d)
ad = bc
bc = ad
cb = da => (c,d) R (a,b)
R is symmetric
For transitivity: Let (𝑎, 𝑏) (𝑐, 𝑑)𝑎𝑛𝑑(𝑒, 𝑓) ∈ 𝐴 such that
(a,b) R (c,d) and (c,d) R (e,f)
ad = bc and cf = de
adcf = bcde
af = be => (a,b)R(e,f)
R is transitive.
Hence R is an equivalence relation.
5
5 (𝑥−2)2 9
34. I1 = ∫2 (𝑥 − 2)𝑑𝑥 = [ ] =
2 2 2

3 5
3 5 (𝑥−3)2 (𝑥−3)2
I2 = − ∫2 (𝑥 − 3)𝑑𝑥 + ∫3 (𝑥 − 3)𝑑𝑥 = − [ ] +[ ]
2 2 2 3

1 4 5
=2 +2 = 2

5
5 (𝑥−5)2 9
I3 = − ∫2 (𝑥 − 5)𝑑𝑥 = − [ ] =
2 2 2
 9 5 9 23
I= + +2 =
2 2 2

35.

2
−
Ar(ACB) = |∫−13(3𝑥 + 2)𝑑𝑥 |

2
(3𝑥+2)2 − 3 1 1
=|( ) | |− 6| = 6
6 −1

1 (3𝑥+2)2 1 25
Ar(AED) = ∫− 2(3𝑥 + 2)𝑑𝑥 = ( )2 =
3 6 −3 6

1 25 26 13
Required area = + = =
6 6 6 3

(OR)
 SECTION E
36. i) Price = (300 −3x) ; Quantity = (80 + x)
(ii) R(x) =(300 −3x) (80 + x)
(iii) 𝑅 ′ (𝑥) = 60 − 6𝑥
X = 10
(OR)
R(5) – R(2) =(300−15)(80+5) – (300−6)(80+2)
= 285 X 85 − 294 X 82
= 24225 – 24108 = Rs.117
37. (i) Solving 2𝑥 + 5𝑦 = 100 and 8𝑥 + 5𝑦 = 200
50 40
Point of intersection is ( , )
3 3

50 40
(ii) ( 3 , 3 ) , (0,20) , (25,0) and (0,0)

(iii)
Corner points Z = 6x - 9y
(0,0) 0
(0,20) −180
50 40
( , )
3 3 −20
(25,0) 150
𝑍𝑚𝑖𝑛 = −180
(OR)
Corner points Z = 6x + 3y
(0,0) 0
(0,20) 60
50 40
( , )
3 3 140
 (25,0) 150

𝑍𝑚𝑎𝑥 = 150

10

38. A = [𝑥 35000 − 𝑥 ] ; B= [100

8 ]
100

10
100
(i) Total interest = [15000 20000] [ 8 ] = 1500 + 1600
100

= 𝑅𝑠. 3100
10
100
(ii) [𝑥 35000 − 𝑥] [ 8 ] = 3200
100

10x +280000 – 8x = 320000

2x = 40000
X = 20000
Amount invested in two bonds are Rs.20000 and Rs.15000
-------------*********-------------