CBSE Class 12 Mathematics Question Paper 2020 Set 65-1 Solutions

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Senior School Certificate Examination-2020
Marking Scheme - Mathematics 65/1/1
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65/1/1 1

QUESTION PAPER CODE 65/1/1
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. If A is a square matrix of order 3 and |A| = 5, then the value of |2A’| is
(A) –10 (B) 10 (C) –40 (D) 40
Ans: (D) 40 1
2. If A is a square matrix such that A2 = A, then (I – A)3 + A is equal to
(A) I (B) 0 (C) I – A (D) I + A
Ans: (A) I 1

⎛ 3π ⎞
3. The principal value of tan −1 ⎜ tan ⎟
⎝ 5 ⎠

2π −2π 3π −3π
(A) (B) (C) (D)
5 5 5 5

2π
Ans: (B) − 1
5
r r
4. If the projection of a = ˆi − 2ˆj + 3kˆ on b = 2iˆ + λkˆ , is zero, then the value
of λ is
−2 −3
(A) 0 (B) 1 (C) (D)
3 2
2
Ans: (C) − 1
3
5. The vector equation of the line passing through the point (–1, 5, 4)
and perpendicular to the plane z = 0 is
r r
(A) r = −ˆi + 5ˆj + 4kˆ + λ (iˆ + ˆj) (B) r = −ˆi + 5ˆj + (4 + λ )kˆ
r
(C) r = ˆi − 5ˆj − 4kˆ + λkˆ (D) rr = λkˆ
r
Ans: (B) r = −ˆi + 5ˆj + (4 + λ )kˆ 1

6. The number of arbitrary constants in the particular solution of a
differential equation of second order is (are)
(A) 0 (B) 1 (C) 2 (D) 3
Ans: (A) 0 1

65/1/1 2

π
4

∫π sec xdx
2
7.
−
4
(A) –1 (B) 0 (C) 1 (D) 2
Ans: (D) 2 1
8. The length of the perpendicular drawn from the point (4, –7, 3) on the y-axis is
(A) 3 units (B) 4 units (C) 5 units (D) 7 units
Ans: (C) 5 units 1

1 1
9. If A and B are two independent events with P(A) = and P(B) = , then P(B′ | A) is
3 4
equal to
1 1 3
(A) (B) (C) (D) 1
4 3 4
3
Ans: (C) 1
4
10. The corner points of the feasible region determined by the system of linear inequalities
are (0, 0), (4,0), (2, 4) and (0, 5). If the maximum value of z = ax + by , where a, b > 0
occurs at both (2, 4) and (4,0), then
(A) a = 2b (B) 2a = b (C) a = b (D) 3a = b
Ans: (A) a = 2 b 1

Fill in the blanks in questions numbers 11 to 15
11. A relation R in a set A is called _____________, if (a1, a2) ∈ R implies
(a2, a1) ∈ R, for all a1, a2 ∈ A.
Ans: Symmetric 1

12. The greatest integer function defined by f (x) = [x], 0 < x < 2 is not
differentiable at x = __________.
Ans: 1
13. If A is a matrix of order 3 × 2, then the order of the matrix A′ is ________.
Ans: 2 × 3 1
OR
A square matrix A is said to be skew-symmetric, if _______
Ans: A = − A′ ( or, A′ = − A ) 1

14. The equation of the normal to the curve y 2 = 8x at the origin is _________
Ans: y = 0 1

65/1/1 3

OR
The radius of a circle is increasing at the uniform rate of 3 cm/s. At the
instant when the radius of the circle is 2 cm, its area increases at the rate
of ____________ cm2/s.
Ans: 12π 1
uuur
15. The position vectors of two point s A and B are OA = 2iˆ − ˆj − kˆ and
uuur
OB = 2iˆ − ˆj + 2kˆ , respectively. The position vector of a point P which divides
the line segment joining A and B in the ratio 2 : 1 is _______

Ans: 2 ˆi − ˆj + kˆ 1

Question numbers 16 to 20 are very short answer type questions

⎡ 2 0 0⎤
⎢ ⎥
16. If A = ⎢ −1 2 3⎥ , then find A (adj A).
⎢⎣ 3 3 5 ⎥⎦

Ans: A ⋅ adj(A) = |A| I 1/2

⎡2 0 0⎤
∴ A ⋅ adj (A) = 2I or ⎢ 0 2 0 ⎥ 1/2
⎢0 0 2⎥
⎣ ⎦

17. Find ∫ x 4 log xdx

x5 1 x5
Ans: ∫ x ·log x dx = log x· − ∫ · dx
4
1/2
5 x 5
x 5· log x x 5
= − +c 1/2
5 25
OR
2x
Find ∫ dx
3
x2 +1
Ans: Let, x2 + 1 = t ∴ 2x dx = dt 1/2
2x 1 3
∫ x + 1 dx = ∫ t dt = ∫ t
−1/3
dt = t 2/3 + c
3 2 3
2
3 2
( x + 1) + c
2/3
= 1/2
2

65/1/1 4

3
18. Evaluate ∫ | 2x − 1| dx .
1

3 3 3
⎡1 2⎤
Ans:
∫1 12x − 11dx = ∫1 (2x − 1) dx = ⎢⎣ 4 (2x − 1) ⎥⎦1 1/2

=6 1/2
19. Two cards are drawn at random and one-by-one without replacement from
a well-shuffled pack of 52 playing cards. Find the probability that one card
is red and the other is black.
26
C1 × 26 C1 26
Ans: 52
= 1/2+1/2
C2 51

dx
20. Find ∫ .
9 − 4x 2

dx dx
Ans: ∫ =∫ 1/2
9 − 4x 2
3 − (2x) 2
2

1 −1 ⎛ 2x ⎞
= sin ⎜ ⎟ + c 1/2
2 ⎝ 3 ⎠
SECTION-B
Question numbers 21 to 26 carry 2 marks each.

21. Prove that sin
−1
( 2x 1 − x ) = 2 cos x, 12 ≤ x ≤ 1
2 −1

Ans: Put x = cos θ ⇔ θ = cos–1x 1/2

L.H.S. = sin–1 2x 1 − x ( 2
)
1
= sin–1(2cos θ sin θ) = sin–1(sin 2θ) = 2θ = 2cos–1x = R.H.S. 1
2
OR
Consider a bijective function f : R + → (7, ∞) given by f (x) = 16x 2 + 24x + 7 ,
where R+ is the set of all positive real numbers. Find the inverse function of f.
Ans: Let y = f(x) = 16x2 + 24x + 7 = (4x + 3)2 – 2 1

−1 y+ 2 −3
⇒ f (y) = x = 1
4

65/1/1 5

2 d2 y
22. If x = at , y = 2at , then find .
dx 2

dy
dy 2a 1
Ans: = dt = = 1
dx dx 2at t
dt

d2 y 1 dt 1 1 1
2
=− 2· =− 2 · =− 1
dx t dx t 2at 2a t 3

23. Find the points on the curve y = x 3 − 3x 2 − 4x at which the tangent lines
are parallel to the line 4x + y − 3 = 0 .
dy
Ans: = – 4 ⇒ 3x2 – 6x – 4 = –4 1
dx
⇒ 3x(x – 2) = 0 ∴ x = 0 ; x = 2 1/2
Points on the curve are (0, 0), (2, –12) 1/2
r r
24. Find a unit vector perpendicular to each of the vectors a and b
r r
where a = 5iˆ + 6ˆj − 2kˆ and b = 7iˆ + 6ˆj + 2kˆ .

ˆi ˆj kˆ
r r
Ans: a × b = 5 6 −2 = 24iˆ − 24ˆj − 12kˆ 1
7 6 2

r r 2 2 1
Unit vector perpendicular to both a and b is ˆi − ˆj − kˆ 1
3 3 3
OR
Find the volume of the parallelopiped whose adjacent edges are represented by
r r r r r r
2a, − b and 3c , where a = ˆi − ˆj + 2kˆ , b = 3iˆ + 4ˆj − 5kˆ and c = 2iˆ − ˆj + 3kˆ

2 −2 4
Ans: Volume of the parallelopiped = −3 −4 5 1
6 −3 9

= |–24| = 24 1
25. Find the value of k so that the lines x = –y = kz and x – 2 = 2y + 1 = –z + 1
are perpendicular to each other.
1
y+
x y z x−2 2 = z −1
Ans: The lines, = = and = 1
1 −1 1 1 1 −1
k 2
1 1
are perpendicular ∴ 1 − − = 0 ⇒ k = 2 1
2 k

65/1/1 6

26. The probability of finding a green signal on a busy crossing X is 30%. What is the
probability of finding a green signal on X on two consecutive days out of three?

Ans: Probability of green signal on crossing X = 30 = 3 ⎫ 1
100 10 ⎪
⎪
⎬
3 7 ⎪
Probability of not a green signal on crossing X = 1 − =
10 10 ⎪⎭
Probability of a green signal on X on two concecutative days out of three
3 3 7 7 3 3 63
= × × + × × = 1
10 10 10 10 10 10 500

SECTION-C
Question numbers 27 to 32 carry 4 marks each.

27. Let N be the set of natural numbers and R be the relation on N × N defined by
(a, b) R (c, d) iff ad = bc for all a, b, c, d ∈ N. Show that R is an equivalence relation.
Ans: Reflexive: For any (a, b) ∈ N × N
a·b=b·a
∴ (a, b) R (a, b) thus R is reflexive 1
Symmetric: For (a, b), (c, d) ∈ N × N
(a, b) R (c, d) ⇒ a · d = b · c
⇒c·b=d·a
1
⇒ (c, d) R (a, b) ∴ R is symmetric 1
2
Transitive : For any (a, b), (c, d), (e, f), ∈ N × N
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a · d = b · c and c · f = d · e
⇒ a·d·c·f=b·c·d·e ⇒ a·f=b·e
1
∴ (a, b) R (e, f), ∴ R is transitive 1
2
∴ R is an equivalance Relation

2 dy
28. If y = e x cos x + (cos x) x , then find .
dx
Ans. Let u = (cosx)x ⇒ y = e x 2 ·cos x + u

dy 2 du 1
∴ = e x ·cos x (2x·cos x − x 2 ·sin x) + 1
dx dx 2
log u = log (cosx)x ⇒ log u = x . log(cosx)
Differentiate w.r.t. “x”

65/1/1 7

1 du du
= log(cosx) – x tan x ⇒ = (cosx)x {log(cos x) − x tan x} 2
u dx dx
Therefore,
dy
= e x 2 ·cos x (2x · cos x – x2 · sin x) + (cos x)x {log (cos x) − x tan x} 1/2
dx

29. Find ∫ sec3 xdx .

1
∫ sec xdx = ∫ sec x ·sec2 xdx = ∫ 1 + tan 2 x ·sec2 xdx
3
Ans. 1
2
(Put tan x = t ; sec2x dx = dt) 1/2

= ∫ 1 + t 2 dt

t 1 1
= 1 + t 2 + log t + 1 + t 2 + c 1
2 2 2
sec x·tan x 1
= + log tan x + sec x + c 1/2
2 2
30. Find the general solution of the differential equation ye y dx = (y3 + 2xe y )dy .

dy
Ans. y ·e y dx = ( y3 + 2xe y ) dy ⇒ y ·e y = y3 + 2xe y
dx
dx 2
∴ − x = y 2 ·e− y 1
dy y
1 1
−2 ∫ y dy −2log y
log
y2 1
I.F. (Integrating factor) = e =e =e = 1
y2
∴ Solution is
1 1
x · 2 = ∫ y 2 ·e − y · 2 dy + c = ∫ e− y dy + c 1
y y

⇒ x2 = − e− y + c or x = − y e + cy
2 −y 2
1
y

OR
Find the particular solution of the differential equation

dy ⎛y⎞ π
x = y − x tan ⎜ ⎟ , given that y = at x = 1.
dx ⎝x⎠ 4
Ans. The differential equation can be written as:
dy y y dy dv
= − tan , let y = v x ∴ =v+x 1
dx x x dx dx

65/1/1 8

dv 1
⇒ v+x = v − tan v ⇒ cot v dv = − dx
dx x
Integrate both sides
y c
log sin v = – log |x| + log c ⇒ log sin = log 2
x x
y π
⇒ x· sin = c , Put y = and x = 1
x 4
π 1
⇒ sin = c or c = 1/2
4 2
⎛y⎞ 1
∴ Particular solution is x·sin ⎜ ⎟ = 1/2
⎝x⎠ 2

31. A furniture trader deals in only two items – chairs and tables. He has `50,000 to
invest and a space to store at most 35 items. A chair costs him `1000 and a table
costs him `2000. The trader earns a profit of `150 and `250 on a chair and
table, respectively. Formulate the above problem as an LPP to maximise the
profit and solve it graphically.
Ans. Let No. of chairs = x, No. of tables = y
y Then L.P. P. is:
40 Maximize (Profit) : Z = 150x + 250y 1
x + 2y = 50
30 A (0, 25) Subject to : x + y ≤ 35 ⎫
⎪
20 1000x + 2000y ≤ 50000 ⇒ x + 2y ≤ 50 ⎬ 1
B (20, 15)
x,y≥0 ⎪
10 ⎭
C (35, 0) 1
x’ x Correct graph 1
0 10 20 30 40 50 2
y’
x + y = 35 Corner: Value of Z ⎫
⎪
A(0, 25) `6250 ⎪
B(20, 15) `6750 (Max) ⎪ 1/2
⎬
C(35, 0) `5250 ⎪
∴ Max (z) = `6750 ⎪
⎪
Number of chairs = 20, Tables = 15 ⎭
32. There are two bags, I and II. Bag I contains 3 red and 5 black balls and Bag II
contains 4 red and 3 black balls. One ball is transferred randomly from Bag I to
Bag II and then a ball is drawn randomly from Bag II. If the ball so drawn is
found to be black in colour, then find the probability that the transferred ball is
also black.
Ans. E1 = Event that the ball transfered from Bag I is Black ⎫
⎪
E2 = Event that the ball transfered from Bag I is Red ⎬ 1/2
⎪
A = Event that the ball drawn from Bag II is Black ⎭

5 3 ⎛A⎞ 4 1 ⎛A⎞ 3
P ( E1 ) = ; P ( E 2 ) = ; P ⎜ ⎟ = = ; P ⎜ ⎟ = 2
8 8 ⎝ E1 ⎠ 8 2 ⎝ E 2 ⎠ 8

65/1/1 9

Required Probability:

⎛A⎞
P ( E1 ) · P ⎜ ⎟ 5 1
·
⎛ E1 ⎞ ⎝ E1 ⎠ 8 2 20
P⎜ ⎟ = = = 1
⎝A⎠ ⎛A⎞ ⎛ A ⎞ 5 · 1 + 3 · 3 29 1
P ( E1 ) · P ⎜ ⎟ + P ( E 2 ) · P ⎜ ⎟ 2
⎝ E1 ⎠ ⎝ E2 ⎠ 8 2 8 8

OR
An urn contains 5 red, 2 white and 3 black balls. Three balls are drawn, one-
by-one, at random without replacement. Find the probability distribution of
the number of white balls. Also, find the mean and the variance of the number
of white balls drawn.
Ans. Let X = No. of white balls = 0, 1, 2
X: 0 1 2 1/2
8 7 6 7 8 7 2 7 2 1 8 1 1
P(X) : × × = 3× × × = 3× × × = 1
10 9 8 15 10 9 8 15 10 9 8 15 2

7 2
X · P(X) : 0 1/2
15 15
7 4
X2P(X) : 0
15 15
9 3
Mean = ∑ XP(X) = = 1/2
15 5
2
11 ⎡ 3 ⎤ 28
Variance = ∑ X P(x) − ⎡⎣ ∑ X P(X) ⎤⎦ = − ⎢ ⎥ =
2 2
1
15 ⎣ 5 ⎦ 75

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

⎡ 1 2 −3⎤
⎢ ⎥
33. If A = ⎢ 3 2 −2 ⎥ , then find A–1 and use it to solve the
⎢⎣ 2 −1 1 ⎥⎦
following system of the equations:
x + 2y − 3z = 6
3x + 2y − 2z = 3
2x − y + z = 2

⎡ 0 1 2⎤ 1 1⎡
0 1 2⎤ 1 1
Ans. A = 7 ; adj(A) = ⎢ −7 7 −7 ⎥ ; A−1 = adj A = ⎢−7 7 −7 ⎥ 1+1 +
⎢ −7 5 −4 ⎥ |A| 7 ⎢ −7 5 −4⎥ 2 2
⎣ ⎦ ⎣ ⎦

65/1/1 10

The system of equations in Matrix form can be written as :

⎡x ⎤ ⎡6⎤
A · X = B , where X = ⎢ y ⎥ ; B = ⎢ 3⎥ 1
⎢z⎥ ⎢ 2⎥
⎣ ⎦ ⎣ ⎦

⎡x⎤ 1 ⎡ 0 1 2⎤ ⎡6⎤
1⎡
7 ⎤ ⎡ 1⎤
X = A −1B ⇒ ⎢ y ⎥ = ⎢ −7 7 −7 ⎥ ⎢ 3 ⎥ = ⎢ −35⎥ = ⎢ −5⎥ 1
⎢ ⎥ 7 ⎢ −7 5 −4 ⎥ ⎢ 2 ⎥ 7 ⎢ −35⎥ ⎢ −5⎥
⎣z⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
∴ x = 1, y = –5 , z = –5 1
OR
Using properties of determinants, prove that

(b + c) 2 a2 bc
(c + a) 2 b2 ca = (a − b)(b − c)(c − a)(a + b + c)(a 2 + b 2 + c2 )
(a + b) 2 c2 ab

(b + c) 2 a2 bc
Ans. (c + a) 2 b2 ca
(a + b) 2 c2 ab

b2 + c2 a2 bc
2 2
= c2 + a 2 b2 ca (C1 → C1 – 2C3) 1
a +b c2 ab

a 2 + b2 + c2 a2 bc
= a 2 + b2 + c2 b2 ca (C1 → C1 + C2) 1
a 2 + b2 + c2 c2 ab

a 2 + b2 + c2 a2 bc
= 0 b2 − a 2 ca − bc (R2 → R2 – R1 , R3 → R3 – R1) 2
0 c2 − a 2 ab − bc

a 2 + b2 + c2 a2 bc
= (b − a)(c − a) 0 b + a −c 1
0 c + a −b

Expand along C1

= ( a 2 + b 2 + c 2 ) (b − a) (c − a) ( − b 2 − ab + c 2 + ac )

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

65/1/1 11

34. Using integration, find the area of the region bounded by the triangle whose
vertices are (2, –2), (4,5) and (6,2).
Ans. Let A(2, –2) ; B(4, 5) ; C(6, 2)
y
Equations of the lines
6 B(4, 5) 2 ⎤
5 AB : x = (y + 9)
7 ⎥
4 ⎥
2 ⎥ 1
2 C(6, 2) BC : x = − (y − 11) ⎥ 1
3 2
⎥
x’ x AC : x = y + 4 ⎥⎦
0 2 4 6
Correct graph 1/2
A(2, -2)
y’ 2 5 5
⎛ −2 ⎞ 2
ar(ΔABC) = ∫ (y + 4)dy + ⎜ ⎟ ∫ (y − 11)dy − ∫ (y + 9) dy 2
−2 ⎝ 3 ⎠2 −2
7
1 2 1 5 1 5 1
= ⎡⎣(y + 4) 2 ⎤⎦ − ⎡⎣(y − 11) 2 ⎤⎦ − ⎡⎣ (y + 9) 2 ⎤⎦ 1
2 −2 3 2 7 −2 2

= 16 + 15 – 21 = 10 1/2
35. Show that the height of the right circular cylinder of greatest volume which can
be inscribed in a right circular cone of height h and radius r is one-third of the
4
height of the cone, and the greatest volume of the cylinder is times the
9
volume of the cone.
Ans. Let H = Height of cylinder
A R = Radius of cylinder
(h - H) π 2
h Volume of cone = r h 1/2
R 3
D F
H V = Volume of cylinder = π R 2 H 1/2

E
B C
r h−H R r
ΔADF ~ Δ AEC ⇒ = ⇒ R = (h − H) 1
h r R
r2 π r2 3
∴ V = π · H · 2 (h − H) = 2 ( H − 2hH + Hh )
2 2 2
1
h h

π r2 h
V′(H) = 2 ( 3H − 4hH + h ) , V '(h) = 0 ⇒ H=
2 2
1+1
h 3

π r2 ⎛ h ⎞ π r2
V″(H) = 2 ( 6H − 4h ) ⎜
, V ″ H= ⎟= (−2h) < 0 1/2
h ⎝ 3 ⎠ h2

h 2r
∴ V is max iff H= and R =
3 3

65/1/1 12

Volume of cylinder 3πR 2 H 4r 2 h 1 4
= 2
= 3· · · 2 = 1/2
Volume of cone πr h 9 3 r h 9

36. Find the equation of the plane that contains the point A(2,1,–1) and is perpendicular to the
line of intersection of the planes 2x + y − z = 3 and x + 2y + z = 2 . Also find the angle
between the plane thus obtained and the y-axis.
Ans. Let equation of the required plane be:
1
a (x – 2) + b(y – 1) + c(z + 1) = 0 1
2
Also : 2a + b – c = 0
a + 2b + c = 0
a b c 1
Solving: = = = k ⇒ a = 3k, b = –3k, c = 3k 1
3 −3 3 2
∴ Equation of plane is : 3k(x – 2) –3k(y – 1) + 3k(z + 1) = 0
1
⇒ x–y+z=0 1
2
Let angle between y-axis and plane = θ

0 − 1+ 0 −1 ⎛ 1 ⎞ 1
then, sin θ = = ⇒ θ = sin −1 ⎜ ⎟ 1
1+1+1 3 ⎝ 3⎠ 2
OR
Find the distance of the point P(–2, –4, 7) from the point of intersection Q of the
r ˆ and the plane rr·(iˆ − ˆj + k)
line r = (3iˆ − 2ˆj + 6k)
ˆ + λ (2iˆ − ˆj + 2k) ˆ = 6 . Also write the
vector equation of the line PQ.
r
Ans. General point on line is: r = (3 + 2 λ) ˆi + ( − 2 − λ) ˆj + (6 + 2 λ) kˆ 1
For the point of intersection:

⎣ ⎦ ( )
⎡ (3 + 2 λ) ˆi + ( − 2 − λ) ˆj + (6 + 2 λ) kˆ ⎤ · ˆi − ˆj + kˆ = 6 1

⇒ 3 + 2 λ + 2 + λ + 6 + 2λ = 6 ⇒ λ = −1 1

∴ Q (iˆ − ˆj + 4 k)
ˆ = Q (1, − 1, 4) 1
r ˆ ˆ ˆ ˆ ˆ ( ˆ
PQ = 3 3 , equation of the line PQ : r = −2i − 4 j + 7k + μ 3i + 3 j − 3k ) 1+1

65/1/1 13

QUESTION PAPER CODE 65/1/2
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. If A is a 3 × 3 matrix and |A| = –2, then value of |A (adj A)| is
(A) –2 (B) 2 (C) –8 (D) 8
Ans: (C) –8 1
2. The number of arbitrary constants in the particular solution of a differential
equation of second order is (are)
(A) 0 (B) 1 (C) 2 (D) 3
Ans: (A) 0 1

⎛ 13π ⎞
3. The principal value of cos −1 ⎜ cos ⎟
⎝ 6 ⎠

13π π π π
(A) (B) (C) (D)
6 2 3 6
π
Ans: (D) 1
6
4. The corner points of the feasible region determined by the system of linear
inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If the maximum value of
z = ax + by, where a, b > 0 occurs at both (2, 4) and (4, 0), then
(A) a = 2b (B) 2a = b (C) a = b (D) 3a = b
Ans: (A) a = 2b 1
1 1
5. If A and B are two independent events with P(A) = and P(B) = ,
3 4
then P(B′ | A) is equal to
1 1 3
(A) (B) (C) (D) 1
4 3 4

3
Ans: (C) 1
4
6. If A is a square matrix such that A2 = A, then (I – A)3 + A is equal to
(A) I (B) 0 (C) I – A (D) I + A
Ans: (A) I 1

65/1/2 14

x
2
⎛1⎞
7. ∫ 1
2 sin ⎜ ⎟ dx, where x ≠ 0 is equal to
–
x x ⎝x⎠
2

(A) –2 (B) 0 (C) 1 (D) π
Ans: (B) 0 1
uur
8. The image of the point (2, –1, 5) in the plane r . $i = 0 is
(A) (–2, –1, 5) (B) (2, 1, –5) (C) (–2, 1, –5) (D) (2, 0, 0)
Ans: (A) (–2, –1, 5) 1
uur uur
9. If the projection of a = $i – 2 $j + 3k$ on b = 2 $i + λ k$ is zero ,
then the value of λ is
–2 –3
(A) 0 (B) 1 (C) (D)
3 2
–2
Ans: (C) 1
3
10. The vector equation of the line passing through the point (–1 , 5, 4)
and perpendicular to the plane z = 0 is
uur uur
(A) r = – $i + 5 $j + 4 k$ + λ (i$ + $j) (B) r = – $i + 5 $j + (4 + λ) k$
uur uur
(C) r = $i – 5 $j – 4 k$ + λ k$ (D) r = λ k$
uur
Ans: (B) r = – $i + 5 $j + (4 + λ) k$ 1

Fill in the blanks in questions numbers 11 to 15
uuur
11. The position vectors of two points A and B are OA = 2i$ – $j – k$ and
uuur
OB = 2i$ – $j + 2k$ , respectivley. The position vector of a point P which
divides the line segment joining A and B in the ratio 2 : 1 is __________.

Ans: 2 $i – $j + kˆ 1

12. The equation of the normal to the curve y2 = 8x at the origin is __________.
Ans: y = 0 1
OR
The radius of a circle is increasing at the uniform rate of 3 cm/sec. At the
instant when the radius of the circle is 2 cm, its area increases at the rate
of __________ cm2/s.
Ans: 12 π 1

65/1/2 15

13. On applying elementary column operation C 2 → C 2 – 3C1 in the matriax

⎡ 4 − 2⎤ ⎡ 2 1 ⎤ ⎡1 −1 ⎤
equation ⎢ 5 = , the RHS (Right Hand Side) of the
⎣ 3 ⎥⎦ ⎢⎣ −3 4 ⎥⎦ ⎢⎣ 2 0 ⎥⎦

equation becomes __________.
⎡ 2 1 ⎤ ⎡1 − 4⎤
Ans: ⎢ −3 4 ⎥ ⎢ 2 − 6 ⎥ 1
⎣ ⎦⎣ ⎦

OR
A square matrix A is said to be symmetric if _______
Ans: A = A′ 1
14. A relation R in a set A is called ________, if (a1, a2) ∈ R implies (a2, a1) ∈ R,
for all a1, a2 ∈ A.
Ans: Symmetric 1
15. The greatest integer fucntion defined by f(x) = [x], 0 < x < 2 is not
differentiable at x = ______.
Ans: 1 1

Question numbers 16 to 20 are very short answer type questions

16. If A is non-singular square matrix of order 3 and A2 = 2A, then find
the value of | A |.
Ans: |A|2 = 8|A| 1/2

⇒ |A| = 8 1/2

17. Two cards are drawn at random and one-by-one without replacement from a
wll-shuffled pack of 52 playing cards. Find the probability that one card is red
and the other is black.
26
C1 × 26 C1 26
Ans: 52
= 1/2+1/2
C2 51

3

18. Evaluate ∫ | 2x – 1| dx .
1

3 3 3
⎡1 ⎤
Ans: ∫ 2x − 1 dx = ∫ (2x − 1)dx = ⎢ (2x − 1) 2 ⎥ 1/2
1 1 ⎣4 ⎦1
=6 1/2

65/1/2 16

dx
19. Find : ∫ 9 – 4x 2

dx dx
Ans: ∫ =∫ 1/2
9 − 4x 2 32 − (2x)2
1 ⎛ 2x ⎞
= sin −1 ⎜ ⎟ + C 1/2
2 ⎝ 3 ⎠
Find: ∫ x log x dx .
4
20.

x5 1 x5
∫ x · log xdx = log x· − ∫ · dx
Ans: 4 1/2
5 x 5
x 5·log x x 5
= − +c 1/2
5 25
OR
2x
Find: ∫ 3 dx .
x2 + 1

Ans: Let, x 2 + 1 = t
∴ 2xdx = dt 1/2

2x 1 −1/3 3 2/3
∫ 3 x 2 + 1 dx = ∫ 3 t dt = ∫ t dt =
2
t +c

(
3 2
)
2/3
= x +1 +c 1/2
2
SECTION-B
Question numbers 21 to 26 carry 2 marks each.
r r
21. Find a unit vector perpendicular to each of the vectors a and b
r r
where a = 5iˆ + 6ˆj − 2kˆ and b = 7iˆ + 6ˆj + 2kˆ .

ˆi ˆj kˆ
r r
Ans: a × b = 5 6 −2 = 24iˆ − 24ˆj − 12kˆ 1
7 6 2

r r 2 2 1
Unit vector perpendicular to both a and b is ˆi − ˆj − kˆ 1
3 3 3
OR
Find the volume of the parallelopiped whose adjacent edges are represented by
r r r r r r
2a, − b and 3c , where a = ˆi − ˆj + 2kˆ , b = 3iˆ + 4ˆj − 5kˆ and c = 2iˆ − ˆj + 3kˆ

65/1/2 17

2 −2 4
Ans: Volume of the parallelopiped = −3 −4 5 1
6 −3 9

= |–24| = 24 1

22. Examine the applicability of Rolle’s theorem for the function f (x) = sin 2x
in [0, π]. Hence find the points where the tangent is parallel to x-axis.
Ans: As, sine function and polynomial function are everywhere continuous
and differentiable.
∴ (i) f(x) = sin 2x is continuous on [0, π] ⎫
⎪
(ii) f(x) = sin 2x is differentiable on (0, π) ⎬
(iii) f(0) = 0 = f(π) ⎪
⎭ 1
∴ Rolle’s Theorem is applicable for f(x) = sin2x

Solving, f ′(x) = 0 or 2 cos 2x = 0 ⇒ cos 2x = 0
⎫
π 3π π 3π ⎪
∴ 2x = , ⇒ x = , ⎬ 1/2
2 2 4 4 ⎪
⎭
⎛ π ⎞ ⎛ 3π ⎞
The points where the tangent is parallel to x-axis are: ⎜ ,1⎟ ; ⎜ , −1⎟ 1/2
⎝4 ⎠ ⎝ 4 ⎠

23. Find the values of x for which the fucntion f(x) = 2 + 3x – x3 is decreasing.
Ans: f(x) is decreasing iff f ′(x) ≤ 0 1
⇔ 3 − 3x 2 ≤ 0 or x 2 ≥ 1
⇔ x ≤ −1 or x ≥ 1 1
24. The probability of finding a green signal on a busy crossing X is 30%. What is the
probability of finding a green signal on X on two consecutive days out of three?

Ans: Probability of green signal on crossing X = 30 = 3 ⎫ 1
100 10 ⎪
⎪
3 7 ⎬
Probability of not a green signal on crossing X = 1 − = ⎪
10 10 ⎪⎭
Probability of a green signal on X on two concecutative days out of three

3 3 7 7 3 3 63
= × × + × × = 1
10 10 10 10 10 10 500

65/1/2 18

25. Prove that sin
−1
( 2x 1 − x ) = 2 cos x, 12 ≤ x ≤ 1
2 −1

Ans: Put x = cos θ ⇔ θ = cos–1x 1/2

(
L.H.S. = sin–1 2x 1 − x
2
)
1
= sin–1(2cos θ sin θ) = sin–1(sin 2θ) = 2θ = 2cos–1x = R.H.S. 1
2
OR
Consider a bijective function f : R + → (7, ∞) given by f (x) = 16x 2 + 24x + 7 ,
where R+ is the set of all positive real numbers. Find the inverse function of f.
Ans: Let y = f(x) = 16x2 + 24x + 7 = (4x + 3)2 – 2 1
y+ 2 −3
⇒ f −1 (y) = x = 1
4
26. Find the value of k so that the lines x = –y = kz and x – 2 = 2y + 1 = –z + 1
are perpendicular to each other.
1
y+
x y z x−2 2 = z −1
Ans: The lines, = = and = 1
1 −1 1 1 1 −1
k 2
1 1
are perpendicular ∴ 1 − − = 0 ⇒ k = 2 1
2 k

SECTION-C

Question numbers 27 to 32 carry 4 marks each.

27. A furniture trader deals in only two items – chairs and tables. He has `50,000 to
invest and a space to store at most 35 items. A chair costs him `1000 and a table
costs him `2000. The trader earns a profit of `150 and `250 on a chair and
table, respectively. Formulate the above problem as an LPP to maximise the
profit and solve it graphically.
Ans. Let No. of chairs = x, No. of tables = y
y
40 Then L.P. P. is:
x + 2y = 50
30 A (0, 25) Maximize (Profit) : Z = 150x + 250y 1
20
B (20, 15) Subject to: x + y ≤ 35 ⎫
10 ⎪
1000x + 2000y ≤ 50000 ⇒ x+2y ≤ 50 ⎬
C (35, 0) 1
x’ x x,y ≥ 0 ⎪
0 10 20 30 40 50 ⎭
y’
x + y = 35
1
Correct graph 1
2

65/1/2 19

Corner: Value of Z
⎫
A(0, 25) `6250 ⎪
B(20, 15) `6750 (Max) ⎪
⎪
C(35, 0) `5250 ⎬ 1/2
⎪
∴ Max (Z) = `6750 ⎪
⎪
Number of chairs = 20, Tables = 15 ⎭

d2 y π
28. If x = a sec3θ, y = a tan3θ, then find 2
at θ = .
dx 4

dy dy / dθ 3a tan 2 θ sec2 θ
Ans. = = = sin θ 2
dx dx / dθ 3a sec3 θ tan θ

d2 y dθ 1 1
2
= cos θ· = 4
1
dx dx 3a sec θ·tan θ 2

d2 y 1
2
=
dx θ= π 12a 1/2
4

2x + 1
29. Find: ∫ dx .
3 + 2x – x 2

2x + 1 2 − 2x 1
Ans. ∫ 3 + 2x − x 2
dx = − ∫
3 + 2x − x 2
dx + 3∫
22 − (x − 1) 2
dx
2

⎛ x −1 ⎞
= −2 3 + 2x − x 2 + 3sin −1 ⎜ ⎟+c 2
⎝ 2 ⎠

30. There are two bags, I and II. Bag I contains 3 red and 5 black balls and Bag II
contains 4 red and 3 black balls. One ball is transferred randomly from Bag I to
Bag II and then a ball is drawn randomly from Bag II. If the ball so drawn is
found to be black in colour, then find the probability that the transferred ball is
also black.
Ans. E1 = Event that the ball transfered from Bag I is Black
⎫
E2 = Event that the ball transfered from Bag I is Red ⎪⎬ 1/2
A = Event that the ball drawn from Bag II is Black ⎪
⎭
5 3 ⎛A⎞ 4 1 ⎛A⎞ 3
P ( E1 ) = ; P ( E 2 ) = ; P ⎜ ⎟ = = ; P ⎜ ⎟ = 2
8 8 ⎝ E1 ⎠ 8 2 ⎝ E 2 ⎠ 8

65/1/2 20

Required Probability:

⎛A⎞
P ( E1 ) · P ⎜ ⎟ 5 1
·
⎛ E1 ⎞ ⎝ E1 ⎠ 8 2 20
P⎜ ⎟ = = = 1
⎝A⎠ ⎛A⎞ ⎛ A ⎞ 5 · 1 + 3 · 3 29 1
P ( E1 ) · P ⎜ ⎟ + P ( E 2 ) · P ⎜ ⎟ 2
⎝ E1 ⎠ ⎝ E2 ⎠ 8 2 8 8

OR
An urn contains 5 red, 2 white and 3 black balls. Three balls are drawn, one-
by-one, at random without replacement. Find the probability distribution of
the number of white balls. Also, find the mean and the variance of the number
of white balls drawn.
Ans. Let X = No. of white balls = 0, 1, 2
X: 0 1 2 1/2
8 7 6 7 8 7 2 7 2 1 8 1 1
P(X) : × × = 3× × × = 3× × × = 1
10 9 8 15 10 9 8 15 10 9 8 15 2

7 2
X · P(X) : 0 1/2
15 15
7 4
X2P(X) : 0
15 15
9 3
Mean = ∑ XP(X) = = 1/2
15 5
2
11 ⎡ 3 ⎤ 28
Variance = ∑ X P(x) − ⎡⎣ ∑ X P(X) ⎤⎦ = − ⎢ ⎥ =
2 2
1
15 ⎣ 5 ⎦ 75

31. Find the general solution of the differential equation ye y dx = (y3 + 2xe y )dy .

dy
Ans. y ·e y dx = ( y3 + 2xe y ) dy ⇒ y ·e y = y3 + 2xe y
dx
dx 2
∴ − x = y 2 ·e− y 1
dy y
1 1
−2 ∫ y dy −2log y
log
y2 1
I.F. (Integrating factor) = e =e =e = 1
y2
∴ Solution is
1 1
x · 2 = ∫ y 2 ·e − y · 2 dy + c = ∫ e− y dy + c 1
y y

x 2 −y
= − e− y + c or x = − y e + cy
2
⇒ 2
1
y

65/1/2 21

OR
Find the particular solution of the differential equation

dy ⎛y⎞ π
x = y − x tan ⎜ ⎟ , given that y = at x = 1.
dx ⎝x⎠ 4
Ans. The differential equation can be written as:
dy y y dy dv
= − tan , let y = v x ∴ =v+x 1
dx x x dx dx
dv 1
⇒ v+x = v − tan v ⇒ cot v dv = − dx
dx x
Integrate both sides
y c
log sin v = – log |x| + log c ⇒ log sin = log 2
x x
y π
⇒ x· sin = c , Put y = and x = 1
x 4
π 1
⇒ sin = c or c = 1/2
4 2
⎛y⎞ 1
∴ Particular solution is x·sin ⎜ ⎟ = 1/2
⎝x⎠ 2

32. Let N be the set of natural numbers and R be the relation on N × N defined by
(a, b) R (c, d) iff ad = bc for all a, b, c, d ∈ N. Show that R is an equivalence relation.
Ans: Reflexive: For any (a, b) ∈ N × N
a·b=b·a
∴ (a, b) R (a, b) thus R is reflexive 1
Symmetric: For (a, b), (c, d) ∈ N × N
(a, b) R (c, d) ⇒ a · d = b · c
⇒c·b=d·a
1
⇒ (c, d) R (a, b) ∴ R is symmetric 1
2
Transitive : For any (a, b), (c, d), (e, f), ∈ N × N
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a · d = b · c and c · f = d · e
⇒ a·d·c·f=b·c·d·e ⇒ a·f=b·e
1
∴ (a, b) R (e, f), ∴ R is transitive 1
2
∴ R is an equivalance Relation

65/1/2 22

SECTION-D
Question numbers 33 to 36 carry 6 marks.
33. Show that the height of the right circular cylinder of greatest volume which can
be inscribed in a right circular cone of height h and radius r is one-third of the
4
height of the cone, and the greatest volume of the cylinder is times the
9
volume of the cone.
Ans. Let H = Height of cylinder
A
R = Radius of cylinder
(h - H)
h π 2
R Volume of cone = r h 1/2
D F 3
H
V = Volume of cylinder = π R 2 H 1/2
E
B C
r
h−H R r
ΔADF ~ Δ AEC ⇒ = ⇒ R = (h − H) 1
h r R
r2 π r2 3
∴V = π ·H ·
h2
(h − H) 2
=
h2
( H − 2hH2 + Hh 2 ) 1

π r2 h
V′(H) = 2 ( 3H − 4hH + h ) , V '(h) = 0 ⇒ H=
2 2
1+1
h 3

π r2 ⎛ h ⎞ π r2
V″(H) = 2 ( 6H − 4h ) , V″ ⎜ H= ⎟ = 2 (−2h) < 0 1/2
h ⎝ 3⎠ h

h 2r
∴ V is max iff H= and R =
3 3

Volume of cylinder 3πR 2 H 4r 2 h 1 4
= 2
= 3· · · 2 = 1/2
Volume of cone πr h 9 3 r h 9

34. Using integration, find the area of the region {(x, y) : x2 + y2 ≤ 9, x + y ≥ 3}
Ans. Correct graph. 2
Required area
x+y=3
3 3 3
= ∫ 9 − x 2 dx − ∫ (3 − x)dx 2
0 0
–3 3 3 3
x 9 ⎛ x ⎞⎤ 1 ⎤ 1
O = 9 − x 2 + sin −1 ⎜ ⎟ ⎥ + (3 − x) 2 ⎥ 1
2 2 ⎝ 3 ⎠⎦0 2 ⎦0 2

x2 + y2 = 9 9π 9 9
–3 = − or ( π − 2) 1/2
4 2 4

65/1/2 23

35. Find the equation of the plane that contains the point A(2,1,–1) and is perpendicular to the
line of intersection of the planes 2x + y − z = 3 and x + 2y + z = 2 . Also find the angle
between the plane thus obtained and the y-axis.
Ans. Let equation of the required plane be:
1
a (x – 2) + b(y – 1) + c(z + 1) = 0 1
2
Also : 2a + b – c = 0
a + 2b + c = 0
a b c 1
Solving: = = = k ⇒ a = 3k, b = –3k, c = 3k 1
3 −3 3 2
∴ Equation of plane is : 3k(x – 2) –3k(y – 1) + 3k(z + 1) = 0
1
⇒ x–y+z=0 1
2
Let angle between y-axis and plane = θ

0 − 1+ 0 −1 ⎛ 1 ⎞ 1
then, sin θ = = ⇒ θ = sin −1 ⎜ ⎟ 1
1+1+1 3 ⎝ 3⎠ 2
OR
Find the distance of the point P(–2, –4, 7) from the point of intersection Q of the
r ˆ and the plane rr·(iˆ − ˆj + k)
line r = (3iˆ − 2ˆj + 6k)
ˆ + λ (2iˆ − ˆj + 2k) ˆ = 6 . Also write the
vector equation of the line PQ.
r
Ans. General point on line is: r = (3 + 2 λ) ˆi + ( − 2 − λ) ˆj + (6 + 2 λ) kˆ 1
For the point of intersection:

⎣ ⎦ ( )
⎡(3 + 2 λ) ˆi + ( − 2 − λ) ˆj + (6 + 2 λ) kˆ ⎤ · ˆi − ˆj + kˆ = 6 1

⇒ 3 + 2 λ + 2 + λ + 6 + 2λ = 6 ⇒ λ = −1 1

∴ Q (iˆ − ˆj + 4 k)
ˆ = Q (1, − 1, 4) 1
r ˆ ˆ ˆ ˆ ˆ ( ˆ
PQ = 3 3 , equation of the line PQ : r = −2i − 4 j + 7k + μ 3i + 3 j − 3k ) 1+1

⎡ 1 2 −3⎤
⎢ ⎥
36. If A = ⎢ 3 2 −2 ⎥ , then find A–1 and use it to solve the
⎢⎣ 2 −1 1 ⎥⎦
following system of the equations:
x + 2y − 3z = 6
3x + 2y − 2z = 3
2x − y + z = 2

65/1/2 24

⎡ 0 1 2⎤ 1 1 ⎡⎢
0 1 2⎤ 1 1
Ans. ⎢ ⎥ −1
A = 7 ; adj(A) = −7 7 −7 ; A = adj A = −7 7 −7 ⎥ 1+1 +
⎢ −7 5 −4 ⎥ |A| 7 ⎢ −7 5 −4⎥ 2 2
⎣ ⎦ ⎣ ⎦
The system of equations in Matrix form can be written as :

⎡x ⎤ ⎡6⎤
A · X = B , where X = ⎢ y ⎥ ; B = ⎢ 3⎥ 1
⎢z⎥ ⎢ 2⎥
⎣ ⎦ ⎣ ⎦

⎡x⎤ 1 ⎡ 0 1 2⎤ ⎡6⎤
1 ⎡⎢
7 ⎤ ⎡ 1⎤
X = A B ⇒ y = −7 7 −7 3 = −35⎥ = ⎢ −5⎥
⎢ ⎥
−1
⎢ ⎥ ⎢ ⎥
1
⎢ ⎥ 7 ⎢ −7 5 −4 ⎥ ⎢ 2 ⎥ 7 ⎢ −35⎥ ⎢ −5⎥
⎣z⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
∴ x = 1, y = –5 , z = –5 1
OR

Using properties of determinants, prove that

(b + c) 2 a2 bc
(c + a) 2 b2 ca = (a − b)(b − c)(c − a)(a + b + c)(a 2 + b 2 + c2 )
(a + b) 2 c2 ab

(b + c) 2 a2 bc
Ans. (c + a) 2 b2 ca
(a + b) 2 c2 ab

b2 + c2 a2 bc
2 2
= c2 + a 2 b2 ca (C1 → C1 – 2C3) 1
a +b c2 ab

a 2 + b2 + c2 a2 bc
= a 2 + b2 + c2 b2 ca (C1 → C1 + C2) 1
a 2 + b2 + c2 c2 ab

a 2 + b2 + c2 a2 bc
= 0 b2 − a 2 ca − bc (R2 → R2 – R1 , R3 → R3 – R1) 2
0 c2 − a 2 ab − bc

65/1/2 25

a 2 + b 2 + c2 a2 bc
= (b − a)(c − a) 0 b + a −c 1
0 c + a −b

Expand along C1

= ( a + b + c ) (b − a) (c − a) ( − b − ab + c + ac )
2 2 2 2 2

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

65/1/2 26

QUESTION PAPER CODE 65/1/3
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
⎡ 2 −1 3 ⎤
⎢ 0 7 ⎥ is not invertible for
1. The matrix ⎢ λ ⎥
⎣ −1 1 4 ⎦
(A) λ = –1 (B) λ = 0 (C) λ = 1 (D) λ ∈ R – {1}
Ans: (C) λ = 1 1
2. The number of arbitrary constants in the particular solution of a differential
equation of second order is (are)
(A) 0 (B) 1 (C) 2 (D) 3
Ans: (A) 0 1
⎛ 7π ⎞
The value of tan–1 ⎜ tan is
6 ⎟⎠
3.
⎝

π π π 7π
(A) (B) (C) (D)
6 2 3 6
π
Ans: (A) 1
6
4. The corner points of the feasible retgion determined by the system of linear
inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If the maximum value of
z = ax + by, where a, b > 0 occurs at both (2, 4) and (4, 0), then
(A) a = 2b (B) 2a = b (C) a = b (D) 3a = b
Ans: (A) a = 2b 1

1 1
5. If A and B are two independent events with P(A) = and P(B) = ,
3 4
then P(B’ | A) is equal to

1 1 3
(A) (B) (C) (D) 1
4 3 4

3
Ans: (C) 1
4
6. If A is a square matrix such that A2 = A, then (I – A)3 + A is equal to
(A) I (B) 0 (C) I – A (D) I + A
Ans: (A) I 1

65/1/3 27

e
log x
7. ∫1
x dx, is equal to

e2 1
(A) (B) 1 (C) (D) – ∞
2 2
1
Ans: (C) 1
2
8. A point P lies on the line segment joining the points (–1, 3, 2) and (5, 0, 6).
If x-coordinate of P is 2, then its z-coordinate is
3
(A) –1 (B) 4 (C) (D) 8
2
Ans: (B) 4 1
uur uur
9. If the projection of a = $i – 2 $j + 3k$ on b = 2 $i + λ k$ is zero , then the value of λ is

–2 –3
(A) 0 (B) 1 (C) (D)
3 2
–2
Ans: (C) 1
3
10. The vector equation of the line passing through the point (–1 , 5, 4) and perpendicular to
the plane z = 0 is
uur uur
(A) r = – $i + 5 $j + 4 k$ + λ (i$ + $j) (B) r = – $i + 5 $j + (4 + λ) k$
uur uur
(C) r = $i – 5 $j – 4 k$ + λ k$ (D) r = λ k$
uur
Ans: (B) r = – $i + 5 $j + (4 + λ) k$ 1

Fill in the blanks in question numbers 11 to 15
uuur
11. The position vectors of two points A and B are OA = 2i$ – $j – k$ and
uuur
OB = 2i$ – $j + 2k$ , respectivley. The position vector of a point P which
divides the line segment joining A and B in the ratio 2 : 1 is __________.

Ans: 2 $i – $j + kˆ 1

12. The equation of the normal to the curve y2 = 8x at the origin is __________.
Ans: y = 0 1
OR
The radius of a circle is increasing at the uniform rate of 3 cm/sec. At the
instant when the radius of the circle is 2 cm, its area increases at the rate
of __________ cm2/s.
Ans: 12 π 1

65/1/3 28

13. If A is a square matrix of order 3 and Aij is the cofactor of the element aij,
then value of a21 A11 + a22 A12 + a23 A13 is __________.
Ans: 0 1
OR
If the matrix A is both symmetric and skew symmetric, then A is a __________.
Ans: Zero matrix 1

14. A relation R in a set A is called ________, if (a1, a2) ∈ R implies (a2, a1) ∈ R,
for all a1, a2 ∈ A.
Ans: Symmetric 1
15. The greatest integer fucntion defined by f(x) = [x], 0 < x < 2 is not
differentiable at x = ______.
Ans: 1 1

Question numbers 16 to 20 are very short answer type questions

16. If A is a square matrix of order 3 and |A| = 2, then find the value of | –AA’ |.

2
Ans: −AA′ = − A 1/2

= –4 1/2

17. Two cards are drawn at random and one-by-one without replacement from a
well-shuffled pack of 52 playing cards. Find the probability that one card is red
and the other is black.
26
C1 × 26 C1 26
Ans: 52
= 1/2+1/2
C2 51

3

18. Evaluate: ∫ | 2x – 1| dx .
1

3 3 3
⎡1 2⎤
Ans: ∫ 2x − 1 dx = ∫ (2x − 1)dx = ⎢ 4 (2x − 1) ⎥ 1/2
1 1 ⎣ ⎦1
=6 1/2

dx
19. Find : ∫ 9 – 4x 2

dx dx
Ans: ∫ =∫ 1/2
9 − 4x 2 32 − (2x)2
1 ⎛ 2x ⎞
= sin −1 ⎜ ⎟ + C 1/2
2 ⎝ 3 ⎠

65/1/3 29

Find: ∫ x log x dx .
4
20.

x5 1 x5
∫ x · log xdx = log x· − ∫ · dx
Ans: 4 1/2
5 x 5
x 5·log x x 5
= − +c 1/2
5 25
OR
2x
Find: ∫ 3 dx .
x2 + 1

Ans: Let, x 2 + 1 = t
∴ 2xdx = dt 1/2

2x 1 −1/3 3 2/3
∫ 3 x 2 + 1 dx = ∫ 3 t dt = ∫ t dt =
2
t +c

(
3 2
)
2/3
= x +1 +c 1/2
2
SECTION-B
Question numbers 21 to 26 carry 2 marks each.
r r
21. Find a unit vector perpendicular to each of the vectors a and b
r r
where a = 5iˆ + 6ˆj − 2kˆ and b = 7iˆ + 6ˆj + 2kˆ .

ˆi ˆj kˆ
r r
Ans: a × b = 5 6 −2 = 24iˆ − 24ˆj − 12kˆ 1
7 6 2

r r 2 2 1
Unit vector perpendicular to both a and b is ˆi − ˆj − kˆ 1
3 3 3
OR
Find the volume of the parallelopiped whose adjacent edges are represented by
r r r r r r
2a, − b and 3c , where a = ˆi − ˆj + 2kˆ , b = 3iˆ + 4ˆj − 5kˆ and c = 2iˆ − ˆj + 3kˆ

2 −2 4
Ans: Volume of the parallelopiped = −3 −4 5 1
6 −3 9

= |–24| = 24 1

65/1/3 30

⎛ π2 ⎞
22. If f(x) = tan x , then find f′ ⎜ 16 ⎟ .
⎝ ⎠

Ans: f ′(x) = sec2 x 1
4 x tan( x)
⎛π⎞ 2 2
f ′⎜ ⎟ = =
⎝ 16 ⎠ 4 ⋅ π π 1
4
23. Using differentials, find the approximate value of 25 ⋅ 3 up to two places
of decimals.
Ans: Let y = f (x) = x , Let x = 25, x + Δx = 25.3, Δx = 0.3 1

dy 1
Δy ~ ·Δx = (0.3) = 0.03 1/2
dx x =25 2 25

25.3 = f (25) + Δy = 5 + 0.03 = 5.03 (approx.) 1/2

24. The probability of finding a green signal on a busy crossing X is 30%. What is the
probability of finding a green signal on X on two consecutive days out of three?
⎫
Ans: Probability of green signal on crossing X = 30 = 3 ⎪ 1
100 10 ⎪
⎬
3 7 ⎪
Probability of not a green signal on crossing X = 1 − = ⎪
10 10 ⎭
Probability of a green signal on X on two concecutative days out of three
3 3 7 7 3 3 63
= × × + × × = 1
10 10 10 10 10 10 500

25. Prove that sin
−1
( 2x 1 − x ) = 2 cos x, 12 ≤ x ≤ 1
2 −1

Ans: Put x = cos θ ⇔ θ = cos–1x 1/2

(
L.H.S. = sin–1 2x 1 − x
2
)
1
= sin–1(2cos θ sin θ) = sin–1(sin 2θ) = 2θ = 2cos–1x = R.H.S. 1
2

65/1/3 31

OR
Consider a bijective function f : R + → (7, ∞) given by f (x) = 16x 2 + 24x + 7 ,
where R+ is the set of all positive real numbers. Find the inverse function of f.
Ans: Let y = f(x) = 16x2 + 24x + 7 = (4x + 3)2 – 2 1

−1 y+ 2 −3
⇒ f (y) = x = 1
4
26. Find the value of k so that the lines x = –y = kz and x – 2 = 2y + 1 = –z + 1
are perpendicular to each other.
1
y+
x y z x−2 2 = z −1
Ans: The lines, = = and = 1
1 −1 1 1 1 −1
k 2
1 1
are perpendicular ∴ 1 − − = 0 ⇒ k = 2 1
2 k

SECTION-C
Question numbers 27 to 32 carry 4 marks each.

27. A furniture trader deals in only two items – chairs and tables. He has `50,000 to invest
and a space to store at most 35 items. A chair costs him `1000 and a table costs him
`2000. The trader earns a profit of `150 and `250 on a chair and table, respectively.
Formulate the above problem as an LPP to maximise the profit and solve it graphically.
Ans. Let No. of chairs = x, No. of tables = y
Then L.P. P. is:
y
40 Maximize (Profit) : Z = 150x + 250y 1
x + 2y = 50 Subject to : x + y ≤ 35
30 A (0, 25) ⎫
⎪
20 1000x + 2000y ≤ 50000 ⇒ x + 2y ≤ 50 ⎬ 1
B (20, 15) ⎪
x,y≥0 ⎭
10
C (35, 0) 1
x’ x Correct graph 1
0 10 20 30 40 50 2
y’
x + y = 35 Corner: Value of Z ⎫
A(0, 25) `6250 ⎪
⎪
B(20, 15) `6750 (Max) ⎪ 1/2
⎬
C(35, 0) `5250 ⎪
⎪
∴ Max (Z) = `6750 ⎪
⎭
Number of chairs = 20, Tables = 15

65/1/3 32

2
28. If x = a (θ – sin θ), y = a (1 – cos θ), a > 0, then find d y at θ = π .
dx 2 3

dy dx
Ans. = a sin θ , = a(1 − cos θ) 1/2+1/2
dθ dθ

θ θ
2sin cos
dy sin θ 2 2 = cot θ
= =
dx 1 − cos θ θ 2 1
2sin 2
2

θ
cosec2
d2 y 1 θ dθ 2
= − · cosec2 · = − 1
dx 2
2 2 dx 2a(1 − cos θ)

d2 y ⎤ 1 4 4
2 ⎥ π
=− × =−
dx ⎦ θ= 2 ⎛ 1⎞ a
3
a ⎜1 − ⎟ 1
⎝ 2⎠
3

Evaluate ∫ e dx as limit of the sums.
x
29.
2

Ans. Let f (x) = e x , a = 1, b = 3, nh = 2 , ....... 1

f (1) + f (1 + h) + f (1 + 2h) + ..... + f (1 + (n − 1)h)

e(enh − 1)
= e + e1+ h + e1+ 2h + ..... + e1+(n −1)h = 2
eh − 1
3
e(e nh − 1)
∫
x
e dx = lim h· h
= e(e 2 − 1) or e3 − e 1
1
h →0 e −1

30. There are two bags, I and II. Bag I contains 3 red and 5 black balls and Bag II
contains 4 red and 3 black balls. One ball is transferred randomly from Bag I to
Bag II and then a ball is drawn randomly from Bag II. If the ball so drawn is
found to be black in colour, then find the probability that the transferred ball is
also black.
Ans. E1 = Event that the ball transfered from Bag I is Black ⎫
⎪
E2 = Event that the ball transfered from Bag I is Red ⎬ 1/2
⎪
A = Event that the ball drawn from Bag II is Black ⎭

5 3 ⎛A⎞ 4 1 ⎛A⎞ 3
P ( E1 ) = ; P ( E 2 ) = ; P ⎜ ⎟ = = ; P ⎜ ⎟ = 2
8 8 ⎝ E1 ⎠ 8 2 ⎝ E 2 ⎠ 8

65/1/3 33

Required Probability:

⎛A⎞
P ( E1 ) · P ⎜ ⎟ 5 1
·
⎛ E1 ⎞ ⎝ E1 ⎠ 8 2 20
P⎜ ⎟ = = = 1
⎝A⎠ ⎛A⎞ ⎛ A ⎞ 5 · 1 + 3 · 3 29 1
P ( E1 ) · P ⎜ ⎟ + P ( E 2 ) · P ⎜ ⎟ 2
⎝ E1 ⎠ ⎝ E2 ⎠ 8 2 8 8

OR
An urn contains 5 red, 2 white and 3 black balls. Three balls are drawn, one-
by-one, at random without replacement. Find the probability distribution of
the number of white balls. Also, find the mean and the variance of the number
of white balls drawn.
Ans. Let X = No. of white balls = 0, 1, 2
X: 0 1 2 1/2
8 7 6 7 8 7 2 7 2 1 8 1 1
P(X) : × × = 3× × × = 3× × × = 1
10 9 8 15 10 9 8 15 10 9 8 15 2

7 2
X · P(X) : 0 1/2
15 15
7 4
X2P(X) : 0
15 15
9 3
Mean = ∑ XP(X) = = 1/2
15 5
2
11 ⎡ 3 ⎤ 28
Variance = ∑ X P(x) − ⎡⎣ ∑ X P(X) ⎤⎦ = − ⎢ ⎥ =
2 2
1
15 ⎣ 5 ⎦ 75
31. Find the general solution of the differential equation ye y dx = (y3 + 2xe y )dy .

dy
Ans. y ·e y dx = ( y3 + 2xe y ) dy ⇒ y ·e y = y3 + 2xe y
dx
dx 2
∴ − x = y 2 ·e− y 1
dy y
1 1
−2 ∫ y dy log
y2 1
I.F. (Integrating factor) = e = e −2log y = e = 1
y2
∴ Solution is
1 1
x · 2 = ∫ y 2 ·e − y · 2 dy + c = ∫ e − y dy + c 1
y y

x x = − y 2 e − y + cy 2
⇒ = − e −y
+ c or 1
y2

65/1/3 34

OR
Find the particular solution of the differential equation

dy ⎛y⎞ π
x = y − x tan ⎜ ⎟ , given that y = at x = 1.
dx ⎝x⎠ 4
Ans. The differential equation can be written as:
dy y y dy dv
= − tan , let y = v x ∴ =v+x 1
dx x x dx dx
dv 1
⇒ v+x = v − tan v ⇒ cot v dv = − dx
dx x
Integrate both sides
y c
log sin v = – log |x| + log c ⇒ log sin = log 2
x x
y π
⇒ x· sin = c , Put y = and x = 1
x 4
π 1
⇒ sin = c or c = 1/2
4 2
⎛y⎞ 1
∴ Particular solution is x·sin ⎜ ⎟ = 1/2
⎝x⎠ 2

32. Let N be the set of natural numbers and R be the relation on N × N defined by
(a, b) R (c, d) iff ad = bc for all a, b, c, d ∈ N. Show that R is an equivalence relation.
Ans: Reflexive: For any (a, b) ∈ N × N
a·b=b·a
∴ (a, b) R (a, b) thus R is reflexive 1
Symmetric: For (a, b), (c, d) ∈ N × N
(a, b) R (c, d) ⇒ a · d = b · c
⇒c·b=d·a
1
⇒ (c, d) R (a, b) ∴ R is symmetric 1
2
Transitive : For any (a, b), (c, d), (e, f), ∈ N × N
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a · d = b · c and c · f = d · e
⇒ a·d·c·f=b·c·d·e ⇒ a·f=b·e
1
∴ (a, b) R (e, f), ∴ R is transitive 1
2
∴ R is an equivalance Relation

65/1/3 35

SECTION-D

Question numbers 33 to 36 carry 6 marks each.

33. Show that the height of the right circular cylinder of greatest volume which can
be inscribed in a right circular cone of height h and radius r is one-third of the
4
height of the cone, and the greatest volume of the cylinder is times the
9
volume of the cone.
Ans. Let H = Height of cylinder
A R = Radius of cylinder
(h - H)
h π 2
Volume of cone = r h 1/2
R 3
D F
H V = Volume of cylinder = π R 2 H 1/2
E
h−H R r
B
r
C ΔADF ~ Δ AEC ⇒ = ⇒ R = (h − H) 1
h r R
r2 π r2 3
2 (
∴V = π·H· 2
(h − H) 2
= H − 2hH 2 + Hh 2 ) 1
h h
π r2 h
2 (
V′(H) = 3H 2 − 4hH + h 2 ) , V '(h) = 0 ⇒ H= 1+1
h 3

π r2 ⎛ h ⎞ π r2
V″(H) = 2 ( 6H − 4h ) , V ″ ⎜ H= ⎟= (−2h) < 0 1/2
h ⎝ 3 ⎠ h2
h 2r
∴ V is max iff H= and R =
3 3

Volume of cylinder 3πR 2 H 4r 2 h 1 4
= 2
= 3· · · 2 = 1/2
Volume of cone πr h 9 3 r h 9

34. Using integration, find the area of the region enclosed by the parabola y = 3x2
and the line 3x – y + 6 = 0.
Ans. y
Points of intersection x = –1, 2 1
12
Correct Graph 1
10 Required area
8 2 2

6
= ∫ 3(x + 2)dx − 3 ∫ x 2 dx 2
−1 −1
4 3 2 2 3 2 1
= ⎡⎣ (x + 2) ⎤⎦ −1 − ⎡⎣ x ⎤⎦ −1 1
2 2 2
x
–4 –2 –1 2 4 3 27
y′ = × 15 − 9 = 1/2
2 2

65/1/3 36

35. Find the equation of the plane that contains the point A(2,1,–1) and is perpendicular to the
line of intersection of the planes 2x + y − z = 3 and x + 2y + z = 2 . Also find the angle
between the plane thus obtained and the y-axis.
Ans. Let equation of the required plane be:
1
a (x – 2) + b(y – 1) + c(z + 1) = 0 1
2
Also : 2a + b – c = 0
a + 2b + c = 0
a b c 1
Solving: = = = k ⇒ a = 3k, b = –3k, c = 3k 1
3 −3 3 2
∴ Equation of plane is : 3k(x – 2) –3k(y – 1) + 3k(z + 1) = 0
1
⇒ x–y+z=0 1
2
Let angle between y-axis and plane = θ

0 − 1+ 0 −1 ⎛ 1 ⎞ 1
then, sin θ = = ⇒ θ = sin −1 ⎜ ⎟ 1
1+1+1 3 ⎝ 3⎠ 2

OR
Find the distance of the point P(–2, –4, 7) from the point of intersection Q of the
r ˆ and the plane rr·(iˆ − ˆj + k)
line r = (3iˆ − 2ˆj + 6k)
ˆ + λ (2iˆ − ˆj + 2k) ˆ = 6 . Also write the
vector equation of the line PQ.
r
Ans. General point on line is: r = (3 + 2 λ) ˆi + ( − 2 − λ) ˆj + (6 + 2 λ) kˆ 1
For the point of intersection:

⎣ ⎦ ( )
⎡(3 + 2 λ) ˆi + ( − 2 − λ) ˆj + (6 + 2 λ) kˆ ⎤ · ˆi − ˆj + kˆ = 6 1

⇒ 3 + 2 λ + 2 + λ + 6 + 2λ = 6 ⇒ λ = −1 1

∴ Q (iˆ − ˆj + 4 k)
ˆ = Q (1, − 1, 4) 1
r ˆ ˆ ˆ ˆ ˆ ( ˆ
PQ = 3 3 , equation of the line PQ : r = −2i − 4 j + 7k + μ 3i + 3 j − 3k ) 1+1

⎡ 1 2 −3⎤
⎢ ⎥
36. If A = ⎢ 3 2 −2 ⎥ , then find A–1 and use it to solve the
⎢⎣ 2 −1 1 ⎥⎦
following system of the equations:
x + 2y − 3z = 6
3x + 2y − 2z = 3
2x − y + z = 2

65/1/3 37

⎡ 0 1 2⎤ 1 1 ⎡⎢
0 1 2⎤ 1 1
Ans. ⎢ ⎥ −1
A = 7 ; adj(A) = −7 7 −7 ; A = adj A = −7 7 −7 ⎥ 1+1 +
⎢ −7 5 −4 ⎥ |A| 7 ⎢ −7 5 −4⎥ 2 2
⎣ ⎦ ⎣ ⎦
The system of equations in Matrix form can be written as :

⎡x ⎤ ⎡6⎤
A · X = B , where X = ⎢ y ⎥ ; B = ⎢ 3⎥ 1
⎢z⎥ ⎢ 2⎥
⎣ ⎦ ⎣ ⎦

⎡x⎤ 1 ⎡ 0 1 2⎤ ⎡6⎤
1 ⎡⎢
7 ⎤ ⎡ 1⎤
X = A B ⇒ y = −7 7 −7 3 = −35⎥ = ⎢ −5⎥
⎢ ⎥
−1
⎢ ⎥ ⎢ ⎥
1
⎢ ⎥ 7 ⎢ −7 5 −4 ⎥ ⎢ 2 ⎥ 7 ⎢ −35⎥ ⎢ −5⎥
⎣z⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
∴ x = 1, y = –5 , z = –5 1
OR
Using properties of determinants, prove that

(b + c) 2 a2 bc
(c + a) 2 b2 ca = (a − b)(b − c)(c − a)(a + b + c)(a 2 + b 2 + c2 )
(a + b) 2 c2 ab

(b + c) 2 a2 bc
Ans. (c + a) 2 b2 ca
(a + b) 2 c2 ab

b2 + c2 a2 bc
2 2
= c2 + a 2 b2 ca (C1 → C1 – 2C3) 1
a +b c2 ab

a 2 + b2 + c2 a2 bc
= a 2 + b2 + c2 b2 ca (C1 → C1 + C2) 1
a 2 + b2 + c2 c2 ab

a 2 + b2 + c2 a2 bc
= 0 b − a2
2
ca − bc (R2 → R2 – R1 , R3 → R3 – R1) 2
0 c2 − a 2 ab − bc

a 2 + b2 + c2 a2 bc
= (b − a)(c − a) 0 b + a −c 1
0 c + a −b

65/1/3 38

Expand along C1

= ( a 2 + b 2 + c 2 ) (b − a) (c − a) ( − b 2 − ab + c 2 + ac )

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

65/1/3 39