ANSWERKEY

SECTION A
MARKS
(Multiple Choice Questions) Each question carries 1 mark
1. A) I 1
2. C) −3 1
3. C) 0 1
4. (A) ±1 1
5. D) a= ∓√3 1
6. A) -7 1
7. B) Maximum value of Z is at Q. 1
8. B) 2√6 1
𝜋
9. (B) 2 1
10. C) ±3 1
11. (D) infinite 1
12. B) −4 1
13. (A) 16 1
14. (B) 62% 1
15. C) sec x 1
16. 7 1
A)
√14
17. 2 1
C) 𝜋
−2 1
18. (A) 3
ASSERTION-REASON BASED QUESTIONS
In the following questions, a statement of assertion (A) is followed by a
statement of
Reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
19. (B) Both A and R are true but R is not the correct explanation of A 1
20. (D) A is false but R is true. 1
SECTION B
This section comprises of very short answer type-questions (VSA) of 2 marks
each
2 𝑥−𝑠𝑖𝑛2 𝑥
21. 𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠
2 2
1
tan-1 (1+𝑠𝑖𝑛𝑥 )= tan-1 ( 𝑥 𝑥)
1+2𝑠𝑖𝑛 𝑐𝑜𝑠
2 2
𝜋 𝑥
= tan-1tan ( - )
4 2 1
𝜋 𝑥 𝜋 𝜋
=4 -2, 𝑥 ∈ (-2 , 2 )

(OR)
 −1 1 −𝜋 1
tan-1(√3)+cot-1(√3)+tan-1(sin( 2 ))
−𝜋 𝜋 −𝜋
= + 3+ 1
6 4
−𝜋
= 12
22.
𝑓(𝑥) = 2𝑥 3 − 9𝑥 2 + 12𝑥 + 15

𝑓 ′ (𝑥) = 6𝑥 2 − 18𝑥 + 12
𝑓 ′ (𝑥) = 0 ⇒ 6(𝑥 − 2)(𝑥 − 1) = 0 ⇒ 𝑥 = 2, 1 ½

Interval Sign of 𝑓′(𝑥) Remarks

(−∞, 1) >0 Strictly Increasing ½
(1,2) <0 Strictly decreasing
(2, ∞) >0 Strictly Increasing

So 𝑓(𝑥) is strictly decreasing in ( 1 , 2 ) and strictly increasing in (−∞, 1) ∪

(2, ∞).
At 𝑥 = 1, 2; 𝑓 ′ (𝑥) = 0
Therefore 𝑓(𝑥) is decreasing in [ 1 , 2 ] and strictly increasing in (−∞, 1] ∪ ½
[2, ∞)
23. 1
𝑓(𝑥) =𝑥 2 −3𝑥+2
½
For the maximum value of 𝑓(𝑥) , the value of the function 𝑥 2 − 3𝑥 + 2 will be
minimum.
½
𝑔(𝑥) = 𝑥 2 − 3𝑥 + 2 ⇒ 𝑔′ (𝑥) = 2𝑥 − 3
3
Now, 𝑔′ (𝑥) = 0 ⇒ 2𝑥 − 3 = 0 ⇒ 𝑥 = 2 ½
′′ (𝑥) 3
𝑔 = 2 > 0. Therefore 𝑔(𝑥) is minimum at 𝑥 = i.e. 𝑓(𝑥) is maximum at
2
3 1 ½
𝑥 = 2. Maximum value of 𝑓(𝑥) = 9 9 = −4
− +2
4 2

(OR) ½
P(x) =41+72x -18 x 2 ½
P’(x) = 72-36x
P’(x) =0 for x= 2 ½
P”(x) =-36 <0
So P(x) is maximum at x =2 ½
Maximum value of P(x) = P(2) = 41+ 72 x 2—18 x4=113
24. 𝑥 2 − 2𝑥=0 for x=0,2
3 2 3
∫1 |𝑥 2 − 2𝑥|𝑑𝑥=∫1 2𝑥 − 𝑥 2 𝑑𝑥+∫2 𝑥 2 − 2𝑥𝑑𝑥 1
=2 1

25. Volume of a cylinder, V=πr2 h,

Differentiate both sides w.r .t ‘t’
 𝑑𝑉 𝑑𝑟 𝑑ℎ
=2𝜋𝑟ℎ 𝑑𝑡 +𝜋𝑟 2 𝑑𝑡
𝑑𝑡
𝑑𝑟 𝑑ℎ 1
Given that =3 m/s , =−4 m/s ,r=4 m and h=6 m.
𝑑𝑡 𝑑𝑡
𝑑𝑉
we get: 𝑑𝑡 =2π(4)(6)(3)+π(4)2(−4)
1
=144π−64π
=80π 𝑚3 /s
SECTION C
(This section comprises of short answer type questions (SA) of 3
marks each)
26.
3 cos 𝑥+2
∫ sin 𝑥+2 cos 𝑥+3 dx

Let (3𝑐𝑜𝑠𝑥 + 2) = A (𝑠𝑖𝑛𝑥 + 2 𝑐𝑜𝑠 𝑥 + 3) + B (𝑐𝑜𝑠𝑥 − 2𝑠𝑖𝑛𝑥 ) + C

1
6 3 −8
By comparing A = 5 B = 5 C= 5

6 𝑠𝑖𝑛𝑥+2 𝑐𝑜𝑠𝑥 +3 3 𝑐𝑜𝑠𝑥 −2𝑠𝑖𝑛𝑥 8 𝑑𝑥 1

I = 5 ∫ 𝑠𝑖𝑛𝑥 +2𝑐𝑜𝑠𝑥 +3 𝑑𝑥 + 5 ∫ 𝑠𝑖𝑛𝑥+2𝑐𝑜𝑠𝑥 +3 𝑑𝑥 − 5 ∫ 𝑠𝑖𝑛𝑥_2𝑐𝑜𝑠𝑥+3

𝑥
6𝑥 3 8 1+ 𝑡𝑎𝑛 1
I = 5 + 5 𝑙𝑜𝑔 {𝑠𝑖𝑛𝑥 + 2𝑐𝑜𝑠𝑥 + 3} − 5 tan−1( 2
) + K, where K is
2
integrating constant.
27.

60
P(H)= =0.6
100
40
P(E)=100=0.4
20
P(E∩H)=100=0.2

So, P(EυH)=0.8

P(𝐸̅ ∩ 𝐻
̅ )=1-P(EυH)=0.2
1
0.2 1
P(E/H)=0.6=3 1
0.2 1
P(H/E)=0.4=2 1
28.

Where C is integrating constant

(OR)
𝟑⁄
I = ∫−𝟏𝟐 |𝒙 𝒔𝒊𝒏𝝅𝒙| 𝒅𝒙
1
𝑥 𝑠𝑖𝑛𝜋 𝑥 𝑓𝑜𝑟 − 1 ≤ 𝑥 ≤ 1
f(x) = { 3
−𝑥 𝑠𝑖𝑛𝜋 𝑥 𝑓𝑜𝑟 1 ≤ 𝑥 ≤ 2
3⁄
Therefore ∫−12 |𝑥 𝑠𝑖𝑛𝜋𝑥| 𝑑𝑥 1
1 3⁄
= ∫−1 𝑥 𝑠𝑖𝑛𝜋𝑥 𝑑𝑥 + ∫1 2 − 𝑥 𝑠𝑖𝑛𝜋𝑥 𝑑𝑥

2 − 1
= 𝜋 − [𝜋 2 − 𝜋 ] = 𝜋 + 𝜋 2
3 1 1

𝑥 𝑥
29.
𝑦𝑒 𝑦 𝑑𝑥 = (𝑥𝑒 𝑦 + 𝑦 2 ) 𝑑𝑦 ½
𝑥
(𝑥𝑒𝑦 +𝑦2 )
𝑑𝑥
⟹ 𝑑𝑦 = 𝑥
𝑦𝑒𝑦
𝑑𝑥 𝑑𝑣
Let 𝑥 = 𝑣𝑦, ∴ 𝑑𝑦 = 𝑣 + 𝑦 𝑑𝑦 1
𝑑𝑣 (𝑥𝑒𝑣 +𝑦2 ) 𝑑𝑣 (𝑣𝑦× 𝑒𝑣 +𝑦2 )
Therefore ⟹ 𝑣 + 𝑦 𝑑𝑦 = ⟹ 𝑣 + 𝑦 𝑑𝑦 =
𝑦𝑒𝑣 𝑦𝑒𝑣
𝑑𝑣 (𝑣 + 𝑦) 𝑑𝑣𝑒𝑣 (𝑣 + 𝑦) 𝑒𝑣
𝑣 𝑒𝑣 + 𝑦 − 𝑣𝑒𝑣
⟹𝑣+𝑦 = ⟹ 𝑦 = − 𝑣 = 1
𝑑𝑦 𝑒𝑣 𝑑𝑦 𝑒𝑣 𝑒𝑣
𝑑𝑣 𝑦 𝑑𝑣
⟹ 𝑦 𝑑𝑦 = 𝑒𝑣 ⟹ 𝑒𝑣 𝑑𝑦 = 1 ⟹ 𝑒𝑣 𝑑𝑣 = 𝑑𝑦 ½
Integrate, we get ∫ 𝑒 𝑑𝑣 = ∫ 𝑑𝑦 𝑣
𝑥
⟹ 𝑒𝑣 = 𝑦 + 𝑐 ⟹ 𝑒𝑦 = 𝑦 + 𝑐 , where c is integrating constant.
 (OR)
𝑑𝑦
− 3𝑦 = 𝑠𝑖𝑛2𝑥
𝑑𝑥
Since this is a linear differential equation, ½

IF (Integrating factor) = 𝑒 −3𝑥

½
Solution of D.E is

𝑦. 𝑒 −3𝑥 = ∫ 𝑠𝑖𝑛2𝑥 𝑒 −3𝑥 𝑑𝑥 + 𝐶, where C is integrating constant

Integrating the RHS by parts,

1½
1
∫ 𝑠𝑖𝑛2𝑥 𝑒 −3𝑥 𝑑𝑥 = −13 𝑒 −3𝑥 (3𝑠𝑖𝑛2𝑥 + 2𝑐𝑜𝑠2𝑥) + 𝐶

Hence solution becomes,

1 ½
𝑦. 𝑒 −3𝑥 = − 13 𝑒 −3𝑥 (3𝑠𝑖𝑛2𝑥 + 2𝑐𝑜𝑠2𝑥) + 𝐶
3𝑠𝑖𝑛2𝑥+2𝑐𝑜𝑠2𝑥
So, solution is 𝑦 = − + 𝐶 𝑒 3𝑥
13

30. 1

The feasible region is the region EAQPA.

The vertices are E(15,0) , A(40,0) , Q(4,18) and P( 6,12) 1
At E (15,0) Z= 300 At Q (4,18) Z= 260
At A(40, 0) Z = 800 At P( 6, 12) Z = 240
Hence Min Z 240 at x = 6 and y=12
(OR)
 1

Feasible region is not bounded so −300 may or may not be smallest value.
Let −50𝑥 + 20𝑦 < −300 => −5𝑥 + 2𝑦 < −30
It has common point. So Z has no minimum value subject to given constrains.
Z has a maximum value 100 at (0,5)
𝑛
31. Given y = {𝑥 + √𝑥 2 + 1}
𝑑𝑦 𝑛−1 1
= 𝑛 [𝑥 + √(𝑥 2 + 1)] {1 + . 2𝑥}
𝑑𝑥 2√𝑥 2 + 1
𝑛−1 √1+𝑥 2 +𝑥
= 𝑛[𝑥 + √(𝑥 2 + 1)] { √𝑥 2 +1
}
𝑛𝑦 1½
= √1+𝑥2
𝑑𝑦 2
(1 + 𝑥 ( ) = 𝑛2 𝑦 2
2)
𝑑𝑥
Differentiating again w.r.t.x, we get
2
𝑑𝑦 1 𝑑2 𝑦 𝑑𝑦 2 𝑑𝑦 1½
(1 + 𝑥 )2 ( ) 2
+ ( ) 2𝑥 = 𝑛2 2𝑦1
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑦
Dividing both sides by 2𝑑𝑥 , we get
𝑑2 𝑦 𝑑𝑦
(1 + 𝑥 2 ) + 𝑥 𝑑𝑥 = 𝑛2 𝑦
𝑑𝑥 2
SECTION D
(This section comprises of long answer-type questions (LA) of 5 marks
each)
32.
2

1
2 2
Required area =∫0 √4 − 𝑥 2 𝑑𝑥 − ∫02(2 − 𝑥)𝑑𝑥

1
= 1
=0+2 sin-11-0-4+2
=𝜋 − 2 sq.units

33. R is reflexive:
(𝑎, 𝑎)𝜖𝑅 ⇒ 𝑎 − 𝑎 + √3 = √3𝜖𝑆
Hence R is reflexive. 1
R is Symmetric:
√3 & 1𝜖 𝑅 ⇒ (√3, 1)𝜖𝑅 ⇒ √3 − 1 + √3 =2√3 𝜖𝑆
2
(1, √3) ⇒ 1 − √3 + √3 = 1 ∉ 𝑆 Hence ⇒ (√3, 1)𝜖𝑅 𝑏𝑢𝑡 (√3, 1) ∉ 𝑅
Hence R is not symmetric.
R is Transitive
(1, √2)𝜖𝑅 ; (√2, , √3)𝜖𝑅
i.e, 1 − √2 + √3 𝜖 𝑆 2
√2 − √3 + √3 = √2𝜖 𝑆
But (1, √3) ⇒ 1 − √3 + √3 = 1 ∉ 𝑆
Hence R is not transitive.
(OR)
1
One – One
Case 1 When 𝑥1 , 𝑥2 are even numbers
f (𝑥1 ) = f (𝑥2 ) =>𝑥1 + 1 = 𝑥2 + 1 =>𝑥1 = 𝑥2
So function is one -one.
1
Case 2 When 𝑥1 , 𝑥2 are odd numbers
f (𝑥1 ) = f (𝑥2 ) =>𝑥1 − 1 = 𝑥2 - 1 =>𝑥1 = 𝑥2
So function is one -one.
1
Case 3 When 𝑥1 is odd, 𝑥2 is even number
Then 𝑥1 ≠ 𝑥2 . Also in this case f (𝑥1 ) is even and f (𝑥2 ) is odd
So f (𝑥1 ) ≠ f (𝑥2 )
i.e. 𝑥1 ≠ 𝑥2 => So f (𝑥1 ) ≠ f (𝑥2 )
so function is one- one 2
Onto
For every even number ‘y’ of codomain and odd number y +1 in domain and
for every odd number y od codomain there exists even number y – 1 in
domain.
 Hence function is onto function.
34.
1 −1 2
A= [0 2 −3]
3 −2 4
−2 0 1
B=[ 9 2 −3]
6 1 −2 2
1 −1 2 −2 0 1
AB=[0 2 −3] [ 9 2 −3]
3 −2 4 6 1 −2
1 0 0
= [0 1 0]
0 0 1
−2 0 1
A-1= [ 9 2 −3] 1
6 1 −2
The system of equations can be written in the form AX=C where
1 −1 2 𝑥 1
A= [0 2 −3],X=[𝑦] and C=[1]
3 −2 4 𝑧 2

Then X= (A)-1C
𝑥 −2 0 1 1 0
[𝑦] =[ 9 2 −3] [1] =[5]
𝑧 2
6 1 −2 2 3
Thus, 𝑥 = 0, 𝑦 = 5, 𝑧 = 3
35. Equation of line passing through (1,2, −4) can be taken as
𝑥−1 𝑦−2 𝑧+4 1
= = …(i)
𝑎 𝑏 𝑐
𝑥−8 𝑦+19 𝑧−10
Given lines are = = …………(ii) 1
3 −16 7
𝑥−15 𝑦−29 𝑧−5
= = …………(iii)
3 8 −5
Since (i)⊥ (ii) 1
so 3𝑎 − 16𝑏 + 7𝑐 = 0 …….. (iv)
Since (i)⊥ (iii)
So 3𝑎 + 8𝑏 − 5𝑐 = 0. ……….. (v)
From (iv) and (v) , we get 1
𝑎 𝑏 𝑐
= =
80−56 21+15 24+48
𝑎 𝑏 𝑐 𝑎 𝑏 𝑐 1
 = = or = = =𝜆 ∴ 𝑎 = 2𝜆, 𝑏 = 3𝜆, 𝑐 = 6𝜆
24 36 72 2 3 6
Putting these values in equation (i), we get
𝑥−1 𝑦−2 𝑧+4 𝑥−1 𝑦−2 𝑧+4
Equations of required lines are = =  = =
2𝜆 3𝜆 6𝜆 2 3 6
i.e. 𝑟⃗ = 𝑖̂ + 2𝑗̂ − 4𝑘̂ + 𝜆(2𝑖̂ + 3𝑗̂ + 6𝑘̂ )
OR
x 1 y  3 z  2
  =  (i)
3 2 1
½
 x y7 z7 ½
  =  (ii)
1 3 2
Any point on the line (i) is (-3   1,2  3,   2)
Any point on the line (ii) is (  ,3  7,2  7)
1
For intersection of lines (i) and (ii)
 3  1    1

2  3  3  7
  2  2  7  1

Solving first two equations simultaneously 1

  1,   2
These values substituted in third equation, it holds true
Therefore the given lines intersect and their point of intersection is ( 2,1,-3)

SECTION E
This section comprises of 3 case study/passage -based questions of 4
marks each with three sub-parts (i),(ii),(iii) of marks 1,1,2 respectively.
An internal choice has been provided in the 2marks questions.

36. Case study 1

PROBABILITY
Let E1 and E2 be the events that the person selected is actually having
COVID and the ‘person selected is actually not having COVID respectively
Let A be the Event that the person s test is diagnosed as positive .
90 1
i)P(tested Covid / has Covid)= P(A/𝐸1 )= 90% = 100 = 0.9
ii) P(tested Covid positive/ does not has covid)= P( A/E2) = 1% = 0.01
A A 1
iii) P(𝐸1 ) = 0.001 , P(𝐸2 ) = 0.999, P (𝐸 ) = 0.9 and P ( 2 ) = 0.01
1
𝑃(𝐸1 )𝑃(𝐴/𝐸1 ) 0.001×0.9 0.0009 2
P(E1/A) = (𝐸1 )𝑃(𝐴/𝐸1 )+ (𝐸2 )𝑃(𝐴/𝐸2 )
= 0.001×0.9+0.999×0.01 = 0.0009+0.0049
0.0009
= 0.01089 = 0.083
OR
𝐴 𝐴 2
iii)P(A) =(𝐸1 )𝑃 ( ) + (𝐸2 )𝑃 ( ) = 0.001 × 0.9 + 0.999 × 0.01
𝐸1 𝐸2
= 0.01089

37. Case study 2

i) 𝐴⃗ = 15 𝜄̂ + 0 𝑗̂ + 0𝑘̂. ⃗⃗ = 0 𝜄̂ + 8𝑗̂ + 6 𝑘̂ .
𝐵 1

𝐴⃗ .𝐵
⃗⃗
ii) Projection= ⃗⃗ 1
|𝐵|

15.0 + 0.0+0.6
= =0
√0+64+36
iii) The dot product is 𝐹⃗ . 𝑁
⃗⃗ = 455 X 0 −780 𝑋 (−90) + 130 𝑋 120 =
85, 800 watts. 2
|𝐹⃗ | = 910 |𝑁 ⃗⃗| = 150
 𝐹⃗ .𝑁
⃗⃗
cos 𝜃 = |𝐹⃗| |𝑁 ⃗⃗|

85800
= 910 𝑋 150 = 0.629

𝜃 = cos −1(0.629)

OR

⃗⃗ = 𝐴⃗ 𝑋 𝐵
𝑁 ⃗⃗
𝜄̂ 𝑗̂ 𝜅̂
= [15 0 0] = 15 (+6 𝑗̂ − 8𝑘̂.)= −90 𝑗̂ + 120𝑘̂.
0 8 6
2
So components 0, −90 𝑎𝑛𝑑 120
1 6 1
𝐹⃗ = 910 (2 𝑖̂ − 7 𝑗̂ + 7 𝑘̂) = 455𝑖̂ − 780𝑗̂ + 130𝑘̂

1 6 1
𝐹⃗ = 910 (2 𝑖̂ − 7 𝑗̂ + 7 𝑘̂) = 455𝑖̂ − 780𝑗̂ + 130𝑘̂

38. Case Study- 3

2
2