CBSE Class 12 Mathematics Question Paper 2020 Set 65-B Solutions

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Senior School Certificate Examination-2020
Marking Scheme - MATHEMATICS
Subject Code: 041 Paper Code: 65 (B)
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1 P.T.O.

QUESTION PAPER CODE 65(B)
EXPECTED ANSWER/VALUE POINTS
SECTION – A
Question Numbers 1 to 20 carry 1 mark each.
Question Numbers 1 to 10 are multiple choice type questions.
Select the correct option.
Q.No. Marks
1. Total number of possible matrices of order 2 × 2 with each entry 2 or 3 is
(A) 4 (B) 8 (C) 16 (D) 32
Ans: (C) 16 1

2. The area of a triangle with vertices (–2, 0), (2, 0) and (0, k) is 4 sq. units.
The value of k is
(A) 4 (B) 2 (C) – 4 (D) 6
Ans: (B) 2 1

⎛1⎞
3. The value of the expression 2 cosec–12 + cos–1 ⎜ ⎟ is
2 ⎝ ⎠

π π 2π 2π
(A) (B) – (C) – (D)
3 3 3 3

2π
Ans: (D) 1
3

dx
4. ∫ 16 + 9x is equal to
2

1 3x 1 3x 1 –1 3x 1 9x
(A) tan –1 +c (B) tan –1 +c (C) tan + c (D) tan –1 +c
4 4 12 4 3 4 12 16

1 3x
Ans: (B) tan –1 +c 1
12 4

5. The area of the trinagle whose two sides are represented by the vectors
$i + k$ and 2i$ + $j + k$ is

3 3
(A) (B) 3 (C) 3 (D)
2 2

3
Ans: (A) 1
2

65(B) 2

6. If the direction cosines of a line are a, a, a, then

1 1
(A) a > 0 (B) a = 1 or a = –1 (C) 0 < a < 1 (D) a = or a = –
3 3

1 1
Ans: (D) a = or a = – 1
3 3

7. The plane 3x – 2y + 6z + 11 = 0, makes an angle sin–1 (α) with x-axis. The value of α
is

3 3 2 3
(A) – (B) (C) (D)
7 7 7 2

3
Ans: (B) 1
7

8. Let f : R → R be defined by f(x) = 5x – 3. Then, f–1(x) is given by

x+3 x x x
(A) (B) –5 (C) +3 (D) –3
5 3 5 5

x+3
Ans: (A) 1
5

9. If A and B be two events such that P(A) = 0⋅2, P(B) = 0⋅4 and P(A ∩ B) = 0⋅08, then
P(A | B) is

(A) 0⋅02 (B) 0.2 (C) 0⋅4 (D) 0⋅08
Ans: (B) 0.2 1

10. A bag contains 4 red and 3 black balls. If 2 balls are drawn from the bag at random
without replacement, then the probability of getting exactly one red ball is

1 2 4 3
(A) (B) (C) (D)
7 7 7 14

4
Ans: (C) 1
7

Fill in the blanks in questions numbers 11 to 15

11. Matrices A and B will be inverse of each other only if __________.
Ans: AB = BA = I 1

12. The function f(x) = |x – 3|, x ∈ R is not differentiable at x = __________.
Ans: 3 1

65(B) 3

⎛π ⎞
13. The function f(x) = loge (sin x) is strictly __________ on ⎜ , π ⎟ .
2 ⎝ ⎠

Ans: decreasing 1
OR

The approximate value of 26 , using differentials, up to 2 places of decimal is
__________.
Ans: 5.10 1
14. In an LPP, the linear function which has to be maximised or minimised is called a linear
_______ function.
Ans: Objective 1
uur
15. A vector of magnitude 14 in the direction of the vector a = – 2i$ + 3 $j + 6 k$ is
________.

Ans: −4 ˆi + 6 ˆj + 12 kˆ 1

Question numbers 16 to 20 are very short answer type questions.

bc 1 a(b + c)
ca 1 b(c + a)
16. Find the value of the determinant .
ab 1 c(a + b)

bc 1 1
Ans: Writing determinant as (ab + bc + ca) ca 1 1 1/2
ab 1 1
=0 1/2
17. Evaluate :
π4

∫ ( x + x cos x + tan x ) dx
3 5

–π 4

Ans: f(x) = x3 + x cos x + tan5x is odd 1/2

∴ Value of Integral = 0 1/2
18. Find :
6 cos x – 9 sin x
∫ 6 cos x + 4 sin x dx
3 2 cos x − 3 sin x
Ans: Given integral = 2 ∫ 3 cos x + 2 sin x dx 1/2

3
= log 3 cos x + 2 sin x + C 1/2
2

65(B) 4

OR

(x − 5)e x
Find :
∫ (x − 3)3
dx

⎡ 1 2 ⎤ x
Ans: Given integral = ∫ ⎢ − ⎥ e dx 1/2
2
⎣ (x − 3) (x − 3)3 ⎦

ex
= +C 1/2
(x − 3)2

19. Find :

∫ tan ( 3x + 5) dx m
2

(
Ans: Given integral = ∫ sec (3x + 5) − 1 dx
2
) 1/2

tan(3x+5)
= −x+C 1/2
3
20. Form the differential equation representing the family of curves
y = a cos (x + b), where a, b are arbitrary constants.
dy
Ans: = − a sin(x + b) 1/2
dx
d2 y d2 y
2
= − a cos(x + b) ⇒ =−y 1/2
dx dx 2
OR
Find the integrating factor for the solution of the differential equation
y dx – (x + 2y2) dy = 0.
Ans: Given differential equation can be written as
dx x
− =−y 1/2
dy y
1
∫ − y dy 1
Integrating factor is e = 1/2
y

SECTION-B

Question numbers 21 to 26 carry 2 marks.

21. Show at the relation R in the set of all positive real numbers defined by
R = {(a, b) : a ≤ b3} is neither symmetric nor transitive.
Ans: (1, 2) ∈ R but (2, 1) ∉ R ∴ R is not symmetriac 1
(9, 3) ∈ R, (3, 2) ∈ R but (9, 2) ∉ R ∴ R is not transitive 1

65(B) 5

OR
Prove that :

⎛ 12 ⎞ ⎛3⎞ ⎛ 56 ⎞
cos –1 ⎜ ⎟ + sin –1 ⎜ ⎟ = sin –1 ⎜ ⎟
⎝ 13 ⎠ ⎝5⎠ ⎝ 65 ⎠

−1 5 3
Ans: LHS = tan + tan −1 1/2+1/2
12 4

5 3
+
tan −1 12 4 = tan −1 56
= 5 3 33
1/2
1− ·
12 4

−1 56
= sin = RHS 1/2
65

dy
22. If (x2 + y2)2 = xy, then find .
dx
Ans: Differentiating both sides w.r.t. x

⎛ dy ⎞ dy
2(x 2 + y 2 ) ⎜ 2x+2y ⎟ = x +y 1+1/2
⎝ dx ⎠ dx

dy y − 4x 3 − 4xy 2
Getting dx = 2 1/2
4x y + 4y3 − x

23. The radius of a right circular cylinder is increasing at the rate of 2cm/s and its height is
decreasing at the rate of 8cm/s. Find the rate of change of its volume, when the radius is
3 cm and height is 6cm.
Ans: Let r be radius & h be the height

dr dh
= 2 cm/s = − 8cm/s 1/2
dt dt

dv ⎡ 2 dh dr ⎤
v = πr 2 h ⇒ = π ⎢r + 2hr ⎥ 1
dt ⎣ dt dt ⎦

dv
(at r = 3, h = 6) = 0 1/2
dt

24. $ $i – 3j$ – 5k$ and 3i$ – 4 $j – 4 k$
Show that the points with position vectors 2i$ – $j + k,
are the vertices of a right angled triangle.

Ans: Let A (2iˆ − ˆj + k)
ˆ , B( ˆi − 3jˆ − 5kˆ ), C( 3iˆ − 4ˆj − 4kˆ )

uuur uuur
AB = − ˆi − 2jˆ − 6k,
ˆ AB = 41 1/2

65(B) 6

uuur uuur
AC = ˆi − 3jˆ − 5k,
ˆ AC = 35 1/2
uuur uuur
BC = 2iˆ − ˆj + k,
ˆ BC = 6 1/2

uuur 2 uuur 2 uuur 2
Clearly AC + BC = AB 1/2

∴ A, B, C are vertices of right angled triangle
OR
Using vectors, find the area of triangle ABC with vertices A(1, 1, 1) B(1, 2, 3)
and C(2, 3, 1).
uuur
Ans: AB = ˆj + 2kˆ 1/2
uuur
AC = ˆi + 2jˆ 1/2
uuur uuur
AB × AC = − 4iˆ + 2ˆj − kˆ 1/2

1
Required area = 21 1/2
2

25. Find the angle between the lines.

( ) + λ ( $i + 2 $j + 2 k$ ) and
ur
r = 2 $j – 3k$

( ) + μ ( 2i$ + 3 $j – 6 k$ ) .
ur
r = 2i$ + 6 $j + 3k$

Ans: Let θ be angle between lines

cos θ =
( ˆi + 2jˆ + 2 kˆ ) · ( 2iˆ + 3jˆ − 6 kˆ )
1
3· 7

−1 4
θ = cos 1
21

26. Two dice are thrown once. Given that two numbers appearing on the dice are different,
find the probability of the event ‘the sum of numbers on the dice is 6’.
Ans: A : Sum of numbers is 6
B : Number appearing on dice are different

A ∩ B = (1, 5) (5, 1) (2, 4) (4, 2) 1

4
⎛ ⎞
A P(A ∩ B) 2
P⎜ ⎟ = = 36 = 1/2+1/2
⎝B⎠ P (B) 30 15
36

65(B) 7

SECTION-C

Question numbers 27 to 32 carry 4 marks each.

27. Consider f : R + → [– 5, ∞] given by f(x) = 9x2 + 6x – 5. Show that f is invertible with
y + 6 –1
f–1(y) = , where R + is the set of all non-negative real numbers.
3

Ans: One-One
Let x1, x2 ∈ R + such that f (x1) = f (x2)

⇒ 9x12 + 6 x1 − 5 = 9x 22 + 6 x2 − 5

2
⇒ x1 = x2 or x1 + x2 = − rejected
3
∴ f is one-one 2
For range let f(x) = y
i.e. 9x2 + 6x – 5 = y

y + 6 −1
which gives x = ∈ R+ 1
3

if y + 6 –1 ≥ 0 i.e. y ≥ – 5

∴ Range = [–5, ∞) = codomain 1/2
f is onto

y + 6 −1
As f is one-one & onto ∴ f is invertible & f–1(y) = 1/2
3

dy
28. If y = (cos x)x, find .
dx
Ans: Taking log on both sides gives log y = x log cos x 1
differentiating w.r.t. x to get

1 dy
y dx = –x tan x + log cos x 2

dy
∴ = (cos x)x (–x tan x + log cos x) 1
dx

65(B) 8

OR

d2 y b4
If x = a cos θ and y = b sin θ, then prove that = – .
dx 2 a 2 y3

dx dy
Ans: = –a sin θ, = b cos θ 1/2+1/2
dθ dθ

dy b
∴ = − cot θ 1
dx a

d2 y b
+ cosec 2θ · −
1
2 = a a sin θ 1
dx

b ⎛ y⎞
= − ⎜ putting sin θ = ⎟ 1/2
a sin 3θ
2
⎝ b⎠

b4
= − 2 3 1/2
a y

29. Evaluate :

2
x dx
∫ ( x + 1)( x + 2 )
1

x 2 1
Ans: Writing (x+1)(x+2) as x + 2 − x + 1

2 2
dx dx
Given integral= 2 ∫ −∫
x + 2 1 x +1 1
1

= 2 log(x + 2) 12 − log(x + 1) 12 2

4 3 32
= 2 log − log or log 1
3 2 27

dy
30. Find the particular solution of the differential equation + y cot x = 2x + x2 cot x,
dx

π
(x ≠ 0), given that y = 0, when x = .
2

Ans: Integrating factor is e ∫ cot x dx = sin x 1

(
Solution is y · sin x = ∫ x cos x + 2x sin x dx + C
2
) 1

65(B) 9

y sin x = x2 sin x – ∫ 2x sin x dx + ∫ 2x sin x dx+ C

y sin x = x2 sin x + C 1

π π2
Putting y = 0, x = to get C = − 1/2
2 4

π2
∴ Particular solution is y sin x = x2 sin x − 1/2
4

31. A furniture firm manufactures chairs and tables, each requiring the use of three machines
A, B and C. Production of one chair requires 2 hours on machine A, 1 hour on machine
B and 1 hour on machine C. Each table requires 1 hour each on machines A and B and 3
hours on machine C. The profit obtained by sellling one chair is ` 300; while by selling
one table, the profit is ` 600. The total time available per week on machine A is 70 hours,
on machine B is 40 hours and on machine C is 90 hours. Formulate an LPP to determine
the number of chairs and tables the firm should make per week in order to get maximum
profit.
Ans: Let number of chairs be x and table be y.
Maximize profit Z = 300x + 600 y 1
Subject to constraints
2x + y ≤ 70 ⎫
⎪
x + y ≤ 40 ⎪⎪ 3
⎬
x + 3y ≤ 90 ⎪
⎪
x, y ≥ 0 ⎪⎭
32. A bag contains 2 white, 3 red and 4 blue balls. Two balls are drawn one-by-one without
replacement from the bag. Find the probability distribution of the number of red balls.
Also, find the mean of the number of red balls.
Ans: Let X denotes number of Red balls
X 0 1 2 1

15 5 18 1 3 1
P(X) or or or 2
36 12 36 2 36 12

18 6
XP(X) 0 1/2
36 36

24 2
Mean = ∑ XP(X) = or 1/2
36 3

65(B) 10

OR
A card from a pack of 52 playing cards is lost. From the remaining cards of the pack, two
cards are drawn (one-by-one without replacement) and both are found to be diamonds. Find
the probability of the lost card being a diamond card.
Ans: Let E1 : Lost card is diamond

E2 : Lost card is not a diamond
A : Two cards drawn are diamonds
1 3
P(E1) = P(E2) = 1
4 4

12 13
C2 C2
P(A/E1) = 51 P(A/E2) = 51 1
C2 C2

P ( E1 ) P ( A/E1 )
P(E1 / A) = P E P A/E + P E P A/E
( 1) ( 1) ( 2 ) ( 2 )

1 12 C 2
×
4 51 C 2
= 1
1 12 C 2 3 13 C2
× + ×
4 51 C 2 4 51 C2

11
= 1
50

SECTION-D
Question numbers 33 to 36 carry 6 marks each.

⎡ 2 –1 1⎤
⎢ ⎥
33. If A = ⎢ –1 2 –1⎥ , show that A3 – 6A2 + 9A – 4I = O and hence find A–1.
⎣⎢ 1 –1 2 ⎥⎦

⎡ 6 –5 5 ⎤
⎢ ⎥
Ans: A2 = ⎢ –5 6 –5⎥ 1
⎢⎣ 5 –5 6 ⎥⎦

⎡ 22 –21 21⎤
⎢ 22 –21⎥⎥
A3 = –21
⎢ 1
⎢⎣ 21 –21 22 ⎥⎦

65(B) 11

∴ A3 – 6A2 + 9A – 4I

⎡ 22 –21 21⎤ ⎡ 36 –30 30 ⎤ ⎡ 18 –9 9⎤ ⎡4 0 0⎤
⎢ ⎥ ⎢ ⎥ ⎢ 18 –9 ⎥⎥ − ⎢⎢ 0 4 0 ⎥⎥
= ⎢ –21 22 –21⎥ − ⎢ –30 36 –30 ⎥ + ⎢ –9
⎢⎣ 21 –21 22 ⎥⎦ ⎢⎣ 30 –30 36 ⎥⎦ ⎢⎣ 9 –9 18 ⎥⎦ ⎢⎣ 0 0 4 ⎥⎦

=0 2
A3 – 6A2 + 9A – 4I = 0 ⇒ A2 – 6A + 9I – 4A–1 = 0
1⎡ 2
⇒ A–1 = A − 6A + 9I ⎤⎦ 1
4⎣

3 1 −1⎤
1 ⎡⎢
∴ A–1 = 1 3 1⎥ 1
4 ⎢ −1 1 3 ⎥
⎣ ⎦

OR
Using the properties of determinants,

3a −a+b −a+c
prove that − b+a 3b − b+c = 3(a + b + c) (ab + bc + ca)
−c+a −c+b 3c

a+b+c −a+b −a+c
Ans: Applying C1 → C1 + C2 + C3 LHS = a+b+c 3b −b+c 2
a+b+c −c+b 3c

1 −a+b −a+c
= (a + b + c) 1 3b − b+c 1
1 −c+b 3c

R2 → R2 – R1, R3 → R3 – R1

1 −a+b −a+c
= (a + b + c) 0 2b+a a − b 2
0 a −c 2c+a

Expanding to get

= (a + b + c) (3ab + 3bc + 3ca) 1

= 3(a + b + c) (ab + bc + ca) = RHS

65(B) 12

34. Find the absolute maximum and absolute minimum values of a fucntion f given by
f(x) = 12x4/3 – 6x1/3, x ∈ [–1 , 1].

1/3 2
Ans: f ′(x) = 16x − −2/3 2
x

1
f ′(x) = 0 ⇒ x = 1
8

1
Critical points are –1, 0, ,1 1
8

⎛1⎞ 9
f (– 1) = 18, f(0) = 0, f ⎜ ⎟ = − , f(1) = 6 1
8 ⎝ ⎠ 4
absolute maximum value is 18 1/2
9
absolute minimum value is − 1/2
4

35. Using integration, find the area of the region bounded by the lines
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0.

1
Ans: Point of intersection of given lines are (2, 0), (1, 2) and (4, 3) 1
2

1 4 2 3 4
Required area =
3 ∫1
(x + 5)dx − 2∫ (2 − x)dx − ∫ ( x − 2)dx
1 2 2
2

(x + 5) 2 4 2(2 − x) 2 2 3 4 1
= + − (x − 2) 2 1
6 1 2 1 4 2 2

15 7
= −1− 3 = 1
2 2

36. Prove that the line through A (0, –1, –1) and B(4, 5, 1) intersects the line throught
C(3, 9, 4) and D(– 4, 4, 4).
Ans: dr’s of AB : 4, 6, 2 1
dr’s of CD : –7, –5, 0 1

x 2 − x1 y 2 − y1 z 2 − z1 3 − 0 9 +1 4 +1
Consider a1 b1 c1 = 4 6 2 3
a2 b2 c2 −7 −5 0

= 30 – 140 + 110 = 0
∴ lines intersect each other 1

65(B) 13

OR
Find the coordinates of the foot of perpendicular and the perpendicular distance from the
point P(4, 3, 2) to the plane x + 2y + 3z = 2. Also, find the image of P in the plane.
Ans: Let foot of perpendicular be A

x −4 y−3 z−2
Equation of PA is = = =λ 1
1 2 3

Coordinate of A (λ + 4, 2λ + 3, 3λ + 2) 1
Putting in equation of plane we get λ = –1 1
Coordinate of A (3, 1, –1) 1

distance AP = (4 − 3) 2 + (3 − 1) 2 + (2 + 1) 2

= 14 1

For coordinate of image P′(x, y, z)
x+4 y+3 z+2
= 3, = 1, = −1
2 2 2

to get P′(2, –1, –4) 1

65(B) 14