Model First Terminal Evaluation – 2023

MATHEMATICS
ANSWER KEY

1. (ii) one-one but not on to

(b) 𝑓 (𝑥) − 8𝑥 3 𝑔(𝑥) =
1
𝑥 3.
𝑔 ∘ 𝑓 (𝑥 ) = 𝑔(𝑓 (𝑥))
= 𝑔(8𝑥 3 )
1
= (8𝑥 3 )3
= 2𝑥

2. For any 𝑎 ∈ ℤ
|𝑎 − 𝑎| = 0 is even
Hence (𝑎, 𝑎) ∈ 𝑅.
∴ 𝑅 is reflexive.
→ if (𝑎, 𝑏) ∈ 𝑅
|𝑎 − 𝑏| is even
⇒ 𝑎 − 𝑏 is even
⇒ 𝑏 − 𝑎 is even
⇒ |𝑏 − 𝑎| is even.
⇒ (𝑏, 𝑎) ∈ 𝑅
∴ symmetric
→ if (𝑎, 𝑏) ∈ 𝑅 and (𝑏, 𝑐) ∈ 𝑅 |𝑎 − 𝑏| is even, |𝑏 − 𝑐| is even.
⇒ 𝑎 − 𝑏 is even, 𝑏 − 𝑐 is even
⇒ 𝑎 − 𝑏 + 𝑏 − 𝑐 is even
⇒ 𝑎 − 𝑐 is even
⇒ |𝑎 − 𝑐| is even
⇒ (𝑎, 𝑐) ∈ 𝑅.
∴ Transitive
Hence 𝑅 is an equivalence relation.
 1
cosec −1 (2) = sin−1 ( )
2
3. a) 𝜋
=
6

2𝜋 𝜋
sin−1 sin ( ) = sin−1 sin (𝜋 − )
3 3
−1 𝜋
b) = sin sin
3
𝜋
=
3

9 6 3
4. i) 3𝐴 = [ ]
12 −3 9

9 6 3 1 2 3
3𝐴 − 𝐵 =[ ]−[ ]
ii) 12 −3 9 −2 3 1
8 4 0
=[ ]
14 −6 8

5.
(1 + 1)2
𝑎11 = =2
2
(1 + 2)2 9
𝑎12 = =
2 2
2
(2 + 1) 9
𝑎21 = =
2 2
(2 + 2)2
𝑎22 = =8
2
(3 + 1)2
𝑎31 = =8
2
(3 + 2)2 25
𝑎32 = =
2 2
2 9/2
𝐴 = [9/2 8 ]
8 25/2
 5 0 1
1
0 = | 0 1 1|
2
𝑘 0 1
6. 1
0 = (5(1 − 0) − 0(0 − 𝑘) + 1(0 − 𝑘))
2
0=5+0−𝑘
𝑘 = ±5

𝑘𝑥 + 1, 𝑥 ⩽ 5
𝑓(𝑥) = {
3𝑥 − 5, 𝑥 > 5
lim𝑥→5− 𝑓(𝑥) = lim𝑥→5+ 𝑓(𝑥)
lim𝑥→5− 𝑘𝑥 + 1 = lim𝑥→5+ 3𝑥 − 5
7.
5𝑘 + 1 = 10
5𝑘 = 9
9
𝑘=
5

3(7 × 11 − 8 × 10) − 4(11 × 6 − 9 × 8) + 5(10 × 6 − 7 × 9)

8. = 3(−3) − 4(−6) + 5(−3)
= −9 + 24 − 15
=0

9. (a) {(1,1), (2,2), (3,3), (4,4), (1,2), (2,1), (2,3), (3,2)}

𝑓 (𝑥1 ) = 𝑓 (𝑥2 )
𝑥 𝑥
⇒ 1 = 2
(b) 𝑥1 +2 𝑥2 +2
⇒ 𝑥1 (𝑥2 + 2) = 𝑥2 (𝑥1 + 2)
𝑥1 𝑥2 + 2𝑥1 = 𝑥1 𝑥2 + 2𝑥2
2x1=2x2
x1 = x2

𝜋
10. a)
3
 b)
𝑥 𝑥 𝑥 𝑥
√1 + sin 𝑥 = √cos2 2 + sin2 2 + 2 sin 2 cos 2

𝑥 𝑥 2 𝑥 𝑥
= √(cos + sin ) = cos + sin
2 2 2 2
𝑥 𝑥 𝑥 𝑥
* √1 − sin 𝑥 = √cos2 + sin2 − 2 sin cos
2 2 2 2

𝑥 𝑥 2 𝑥 𝑥
= √(cos − sin ) = cos − sin
2 2 2 2
. 𝑥 𝑥 𝑥 𝑥
cos +sin +cos −sin
−1 2 2 2 2
⇒ cot [ 𝑥 𝑥 𝑥 𝑥 ]
cos +sin −(cos −sin )
2 2 2 2
𝑥
2 cos
−1 2
= cot [
2 sin 𝑥 ]
2
cot 𝑥
= cot −1 ( )
2
𝑥
=
2

1
11. a) 𝜋 − cos −1 ( )
2
𝜋 2𝜋
=𝜋− =
3 3

b)
cos 𝑥
tan−1 ( )
1−sin 𝑥
𝑥 𝑥
cos2 −sin2
tan−1 ( 𝑥
2
𝑥
2
𝑥 𝑥 )
cos2 +sin2 −2sin cos
2 2 2 2
𝑥 𝑥
cos +sin 𝑥
−1 2 2
= tan [𝑥 𝑥 ÷ cos ]
cos −sin 2
2 2
. 1+tan
𝑥
−1 2
= tan [ 𝑥 )
1−tan
2
𝜋 𝑥
= tan−1 (tan ( + )
4 2
𝜋 𝑥
= +
4 2
 3 1
12. We have, [ ]
−1 2

3 1 3 1 8 5
∴ 𝐴2 = 𝐴𝐴 = [ ][ ]=[ ]
−1 2 −1 2 −5 3
8 5 3 1
So, 𝐴2 − 5𝐴 + 7𝐼 = [ ] − 5[ ]+
−5 3 −1 2
1 0 8 − 15 + 7 5 − 5 + 0 0 0
7[ ]=[ ]=[ ]=0
0 1 −5 + 5 + 0 3 − 10 + 7 0 0

Now, 𝐴2 − 5𝐴 + 7𝐼 = 0
⇒ A−1 ( A2 − 5 A + 7I) = A−1 = 0 [Multiplying throughout by A−1 ]

⇒ 𝐴−1 𝐴2 − 5𝐴−1 𝐴 + 7𝐴−1 𝐼 = 𝑂

⇒ 𝐴 − 5𝐼 + 7𝐴−1 = 0
⇒ 7𝐴−1 = 5𝐼 − 𝐴
5 0 3 1 2 −1
7𝐴−1 = [ ]−[ ]=[ ]
0 5 −1 2 1 3
1 2 −1
⇒ 𝐴−1 = [ ]
7 1 3

13.
3𝑥 = 𝑥 + 4
2𝑥 = 4
𝑥 =2
3𝑦 = 6 + (𝑥 + 𝑦)
3𝑦 = 6 + 2 + 𝑦
3y = 8+y
2y = 4
Y=2
3w= 2w+3
W=3
3𝑧 = −1 + 𝑧 + 𝑤
3𝑧 = −1 + 𝑧 + 3
3𝑧 = 2 + 𝑧
2𝑧 = 2
𝑧 =1
14.
2 −1 −4
𝐴 = [−1 3 4]
1 −2 −3
1 1
. = (𝐴 + 𝐴′ ) + (𝐴 − 𝐴′ )
2 2
2 −1 1
𝐴′ = [−1 3 −2]
−4 4 −3
1 ′
(𝐴 + 𝐴 )=
2
1 4 −2 −3
[−2 6 2]
2
−3 2 −6
2 −1 −3/2
[ −1 3 1 ]
−3/2 1 −3
1 ′
(𝐴 − 𝐴 )=
2
0 0 −5
1/2 [0 0 6]
5 −6 0
2 −1 −3/2 0 0 −5/2
[ −1 3 1 ]+[ 0 0 3 ]
−3/2 1 −3 5/2 −3 0
2 −1 −4
= [−1 3 4 ]=4
1 −2 −3

1 −1 1
15. 𝐴 = [2 2 −3]
1 1 1
Cofactor matrix of 𝐴
5 −5 4
.= [+2 0 −2]
1 +5 4
5 2 1
Adj 𝐴 = [−5 0 5]
4 −2 4
.
|𝐴| = 1(2 + 3) + 1(2 + 3) + 1(2 − 2)
= 5 + 5 = 10 ≠ 0
 Non singular matrix

16. i) 𝑥 = 2𝑎𝑡, 𝑦 = 𝑎𝑡 2 .

𝑑𝑦 𝑑𝑦∣𝑑𝑡 2𝑎𝑡
= =
,𝑑𝑥 𝑑𝑥∣𝑑𝑡 2𝑎
=𝑡

ii)
𝑦 = tan−1 𝑥
𝑑𝑦 1
=
𝑑𝑥 1 + 𝑥 2
𝑑2 𝑦 −1
2
= × 2𝑥
𝑑𝑥 ( 1 + 𝑥 2 )2
(1 + 𝑥 2 )𝑦2 + 2𝑥𝑦1 =
−2𝑥 2𝑥
(1 + 𝑥 2 ) × 2 2
+ 2
(1+𝑥 ) 1+𝑥
−2𝑥 2𝑥
= +
1+𝑥 2 1+𝑥 2
=0

𝜋
We have sin−1 𝑥 + cos −1 𝑥 =
2
−1 −1 −1 −1 𝜋 −1
17. a) ∴ cos ( ) + sin ( ) = tan−1 (1) + cos −1 ( ) +
2 2 2 2
−1 𝜋 𝜋 3𝜋
sin−1 ( ) = + =
2 4 2 4
b)
𝑥 = tan 𝜃 , 𝜃 = tan−1 𝑥
√1 + 𝑥 2 − 1 √1 + tan2 𝜃 − 1)
tan−1 ( ) = tan−1 (
𝑥 tan 𝜃
sec 𝜃 − 1
= tan−1 ( )
tan 𝜃
1
−1
= tan−1 [ cos 𝜃 ]
sin 𝜃/cos 𝜃
1 − cos 𝜃
= tan−1 ( )
𝑠𝑖𝑛 𝜃
𝜃
2sin
= tan−1 [ 2 ]
𝜃 𝜃
2sin cos
2 2

sin 𝜃/2
= tan−1 ( )
cos 𝜃/2
= tan−1 (tan 𝜃/2)
𝜃 tan−1 𝑥
= =
2 2

1 2 5
𝐴=[ 2 3 1]
18. −1 1 1
−1 1
𝐴 = Adj 𝐴
|𝐴|

|𝐴| = 1(3 − 1) − 2(2 + 1) + 5(2 + 3)

= 2 − 6 + 25
= 21
2 −3 5
Cofactor matrix of 𝐴 = [ +3 6 −3]
−12 +9 −1
 2 3 −12
Adjoint = [−3 6 9 ]
5 −3 −1
−1
1 2 3 −12
𝐴 = [−3 6 9 ]
21
5 −3 −1
2/21 1/7 −4/7
= [−1/7 2/7 3/7 ]
5/21 −1/7 −1/21

3 −2 3 𝑥 8
19. Let 𝐴 = [2 1 −1] 𝑋 = [𝑦] 𝐵 = [1]
4 −3 2 𝑧 4
Then the system of equations can be written as AX =B
3 −2 3
|𝐴| = |2 1 −1| = −17 ≠ 0
4 −3 2
−1 −5 −1
adj 𝐴 = [ −8 −6 9 ]
−10 1 7
−1 −5 −1
1 1
𝐴−1 = |𝐴| adj 𝐴 = [ −8 −6 9 ]
−17
−10 1 7
∴ Unique solution is given by
−1 −5 −1 8
−1 1
𝑥=𝐴 𝐵= [ −8 −6 9 ] [1]
−17
−10 1 7 4
−17 1
. 1
= − [−34] = [2]
17
−51 3
∴x=1 y=2 z=3

20. i)
𝑦 = 𝑥𝑥
log 𝑦 = 𝑥log 𝑥
1 𝑑𝑦 1
= 𝑥 × + 1 ⋅ log 𝑥
𝑦 𝑑𝑥 𝑥
= 1 + log 𝑥
𝑑𝑦
= 𝑦(1 + log 𝑥)
𝑑𝑥
= 𝑥 𝑥 (1 + log 𝑥)
ii)

𝑥 2 + 𝑥𝑦 + 𝑦 2 = 100
𝑑𝑦 𝑑𝑦
2𝑥 + 𝑥 ⋅ + 𝑦 + 2𝑦 =0
𝑑𝑥 𝑑𝑥
. 𝑑𝑦
(2𝑥 + 𝑦) + (𝑥 + 2𝑦) =0
𝑑𝑥
𝑑𝑦 −(2𝑥+𝑦)
=
𝑑𝑥 𝑥+2𝑦

iii)
𝑥 = 𝑎(𝜃 − sin 𝜃)
𝑦 = 𝑎(1 + cos 𝜃)
𝑑𝑦 𝑎(1−cos 𝜃)
. =
𝑑𝑥 𝑎(−sin 𝜃)
cos 𝜃−1
=
sin 𝜃