VIVEKANANDA EDUCATIONAL SOCIETY

QUARTERLY EXAMINATION [2024 -25]

CLASS XII - MATHEMATICS

ANSWER KEY - SET 2

SECTION A

1. ( c ) −𝑒 𝑦−𝑥
2. ( d) 64
3. ANY ANSWER
4. ( C) not symmetric but transitive
5. ( b) 3X 5
6. (d) f is neither one-one nor onto
1
7. d) r = 2√𝜋 units
2𝜋
8. ( c ) 9
𝑒𝑥
9. (c) 𝑥
1
10. (c )
4
11. (b) 3
12. Any answer
13. (c ) a linear function to be optimised.
14. (d) - tan x – cot x + c
1 3
15. (a) 𝑒𝑥
3
16. (a) 𝐼
17. ( b) 𝜆 = 1
256
18. ( c) 3
19. (d) A is false and R is true.
20. (a) Both A and R are true and R is the correct explanation of A

SECTION B

21. P = 2 Q = 𝑒 3𝑥
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 =
5
+ c𝑒 −2𝑥
 𝜋 𝜋 𝜋 𝜋
22. − 6 + 3 − 4 = − 12

23. 𝑦𝑙𝑜𝑔𝑥 = 𝑥𝑙𝑜𝑔𝑦

𝑦 𝑑𝑦 𝑥 𝑑𝑦
+ logx = + log y
𝑥 𝑑𝑥 𝑦 𝑑𝑥

𝑑𝑦 𝑥 𝑦
(𝑙𝑜𝑔𝑥 − 𝑦) = log y−
𝑑𝑥 𝑥

𝑑𝑦 𝑦 𝑥𝑙𝑜𝑔𝑦−𝑦
= ( )
𝑑𝑥 𝑥 𝑦𝑙𝑜𝑔𝑥−𝑥

1 𝑑𝑥
24. 𝐼 = ∫
√2 √4−2𝑥−𝑥 2
1 𝑑𝑥
= ∫
√2 √(√5) −(𝑥+1)2
2

1 𝑥+1
= 𝑠𝑖𝑛 −1 +c
√2 √5
(OR)
𝑡
Put logx = t , x = 𝑒
I = ∫(𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡)𝑒 𝑡 dt
=𝑒 𝑡 𝑠𝑖𝑛𝑡 + c
= 𝑥𝑠𝑖𝑛(𝑙𝑜𝑔𝑥) + 𝐶
2 2
25. AT = [ ]
9 6
4 11
A + AT = [ ]
11 12
0 7
A – AT=[ ]
−7 0
1 1
A = [A + 𝐴𝑇 ] + [A − 𝐴𝑇 ]
2 2
11 7
2 9 2 0
2 2
[ ] = [11 ] + [ 7 ]
2 6 6 − 0
2 2
(OR)
|𝐴| = −19
−2 −3
𝐴𝑑𝑗𝐴 = [ ]
−5 2
1 −2 −3
𝐴−1 = − 19 [ ]
−5 2
1
K = 19
 SECTION – C
1 1
26. Each line 2 mark and shading 2 mark & table and final answer 1
mark

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

𝑥1 −1 𝑥2 −1 1
= -----------------
𝑥1 −2 𝑥2 −2 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.
28. 𝑓 ′ (𝑥) = 3𝑥 2 + 4𝑥

= 𝑥(3𝑥 + 4)

Put 𝑓 ′ (𝑥) = 0
4
𝑥 = 0 , 𝑥 = −3
4
In (−∞, − 3 ) 𝑎𝑛𝑑 (0, ∞) 𝑓 ′ (𝑥) > 0

4
=> f is strictly increasing in (−∞, − 3 ) 𝑎𝑛𝑑 (0, ∞)
 4
In (− 3 , 0) 𝑓 ′ (𝑥) < 0

4
=> f is strictly decreasing in(− , 0)
3

𝑑𝑥 1 𝑑𝑦
29. = a(1- cos𝜃) ---- , = −𝑎𝑠𝑖𝑛𝜃
𝑑𝜃 2 𝑑𝜃

𝜃 𝜃
𝑑𝑦 2𝑠𝑖𝑛 2 𝑐𝑜𝑠 2 𝜃
= 𝜃 = - cot
𝑑𝑥 2𝑠𝑖𝑛2 2 2
𝑑2 𝑦 𝜃 𝑑𝜃
= 𝑐𝑜𝑠𝑒𝑐 2 2 𝑑𝑥
𝑑𝑥 2
1 𝜃
= 4𝑎 𝑐𝑜𝑠𝑒𝑐 4 2
(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
=− (1+𝑥)2
𝑑𝑥
𝑠𝑖𝑛𝑦 𝑑𝑥
30. dy = − 1+2𝑒 −𝑥
𝑐𝑜𝑠𝑦

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 + y𝑑𝑦
𝑦 𝑑𝑦

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

𝑥
𝑥 + 𝑦𝑒 = c 𝑦

𝑥+1 𝐴 𝐵𝑥+𝑐
31. 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 = ∫ 𝑥 𝑑𝑥 - ∫ 𝑥 2 +1 dx + ∫ 𝑥 2 +1 𝑑𝑥

1
= log|x| - log |𝑥 2 + 1| + 𝑡𝑎𝑛−1 𝑥 + c
2

(OR)
1
I = ∫ 𝑒 𝑥 −1 𝑑𝑥 put 𝑒 𝑥 = t

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 | 𝑡
| + C

𝑒 𝑥 −1
= log| |+ c
𝑒𝑥

SECTION D

32. |A| = −12

6 −6 −6
Adj A = [−3 3 −3]
4 0 −4
 6 −6 −6
−1 1
𝐴 − 12 [−3 3 −3]
4 0 −4
The given equations can be written as
1 2 −3 𝑥 1
[2 0 −3] [𝑦] = [2]
1 2 0 𝑧 3
AX = B
X = 𝐴−1 𝐵
6 −6 −6 1
1
= − 12 [−3 3 −3] [2]
4 0 −4 3

−24
1
= − 12 [ −6 ]
−8
1 2
𝑥 = 2; 𝑦 = ; 𝑧 = 3
2
33.

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)

𝜋
𝑑𝑥
34. I = ∫𝜋3 1+√𝑡𝑎𝑛𝑥 ________(𝑖)
6

𝜋
3 𝑑𝑥
I=∫ 𝜋
𝜋 𝜋
6 1+√𝑡𝑎𝑛( + −𝑥)
6 3

𝜋
3 𝑑𝑥
I=∫ 𝜋
𝜋
6 1+√𝑡𝑎𝑛( −𝑥)
2
 𝜋
𝑑𝑥
I = ∫𝜋3 1+√𝑐𝑜𝑡𝑥
6

𝜋
3 √𝑡𝑎𝑛𝑥
=∫ 𝜋 𝑑𝑥 -------(ii)
1+√𝑡𝑎𝑛𝑥
6

Adding (i) and (ii) we get

𝜋 𝜋
3 3
2I = ∫ 𝑑𝑥
𝜋 = [𝑥] 𝜋
6 6

𝜋 𝜋 𝜋
= [3 − 6 ] = 6

𝜋
I = 12

35. 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.

SECTION - E
10

36. 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

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

38. (i) Price = (300 −3x) ; Quantity = (80 + x)

(ii) R(x) =(300 −3x) (80 + x)

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.
-------------*********-------------