CBSE Class 12 Mathematics Question Paper 2020 Set 65-2 Solutions

Strictly Confidential - (For Internal and Restricted Use Only)

Senior School Certificate Examination-2020
Marking Scheme - MATHEMATICS
Subject Code: 041 Paper Code: 65/2/1
General instructions:-
1. You are aware that evaluation is the most important process in the actual and correct assessment of the candidates. A
small mistake in evaluation may lead to serious problems which may affect the future of the candidates, education
system and teaching profession. To avoid mistakes, it is requested that before starting evaluation, you must read and
understand the spot evaluation guidelines carefully. Evaluation is a 10-12 days mission for all of us. Hence, it is
necessary that you put in your best efforts in this process.
2. Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done according to one's
own interpretation or any other consideration. Marking Scheme should be strictly adhered to and religiously followed.
However, while evaluating, answers which are based on latest information or knowledge and/or are innovative,
they may be assessed for their correctness otherwise and marks be awarded to them.
3. The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first day, to
ensure that evaluation has been carried out as per the instructions given in the Marking Scheme. The remaining
answer books meant for evaluation shall be given only after ensuring that there is no significant variation in the
marking of individual evaluators.
4. Evaluators will mark( √ ) wherever answer is correct. For wrong answer 'X"be marked. Evaluators will not put right
kind of mark while evaluating which gives an impression that answer is correct and no marks are awarded. This is
most common mistake which evaluators are committing.
5. If a question has parts, please award marks on the right-hand side for each part. Marks awarded for different parts of
the question should then be totaled up and written in the left-hand margin and encircled. This may be followed
strictly.
6. If a question does not have any parts, marks must be awarded in the left-hand margin and encircled. This may also be
followed strictly.
7. If a student has attempted an extra question, answer of the question deserving more marks should be retained and the
other answer scored out.
8. No marks to be deducted for the cumulative effect of an error. It should be penalized only once.
9. A full scale of marks 0 - 80 has to be used. Please do not hesitate to award full marks if the answer deserves
it.
10. Every examiner has to necessarily do evaluation work for full working hours i.e. 8 hours every day and evaluate 20
answer books per day in main subjects and 25 answer books per day in other subjects (Details are given in Spot
Guidelines).
11. Ensure that you do not make the following common types of errors committed by the Examiner in the past:-
• Leaving answer or part thereof unassessed in an answer book.
• Giving more marks for an answer than assigned to it.
• Wrong totaling of marks awarded on a reply
• Wrong transfer of marks from the inside pages of the answer book to the title page.
• Wrong question wise totaling on the title page.
• Wrong totaling of marks of the two columns on the title page.
• Wrong grand total.
• Marks in words and figures not tallying.
• Wrong transfer of marks from the answer book to online award list.
• Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly and clearly
indicated. It should merely be a line. Same is with the X for incorrect answer.)
• Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
12. While evaluating the answer books if the answer is found to be totally incorrect, it should be marked as cross (X) and
awarded zero (0)Marks.
13. Any unassessed portion, non-carrying over of marks to the title page, or totaling error detected by the candidate shall
damage the prestige of all the personnel engaged in the evaluation work as also of the Board. Hence, in order to
uphold the prestige of all concerned, it is again reiterated that the instructions be followed meticulously and judiciously.
14. The Examiners should acquaint themselves with the guidelines given in the Guidelines for spot Evaluation before
starting the actual evaluation.
15. Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title page, correctly
totaled and written in figures and words.
16. The Board permits candidates to obtain photocopy of the Answer Book on request in an RTI application and also
separately as a part of the re-evaluation process on payment of the processing charges.

1 P.T.O.

QUESTION PAPER CODE 65/2/1
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. The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1), (1, 1)} is
(A) symmetric and transitive, but not reflexive
(B) reflexive and symmetric, but not transitive
(C) symmetric, but neither reflexive nor transitive
(D) an equivalence relation
Ans: (C) symmetric, but neither reflexive nor transitive 1

3+ λ
2. tan–13 + tan–1 λ = tan–1 is valid for what values of λ?
1 − 3λ
⎛ 1 1⎞ 1
(A) λ ∈ ⎜ − , ⎟ (b) λ >
⎝ 3 3⎠ 3
1
(C) λ < (d) All real values of λ
3
1
Ans: (C) λ < 1
3
3. If A is a non-singular square matrix of order 3 such that A 2 = 3A, then
value of |A| is

(A) –3 (B) 3 (C) 9 (D) 27
Ans: (D) 27 1

4. The function f : R → R given by f (x) = − | x − 1| is

(A) continuous as well as differentiable at x = 1
(B) not continuous but differentiable at x = 1
(C) continuous but not differentiable at x = 1

(D) neither continuous nor differentiable at x = 1
Ans: (C) continuous but not differentiable at x = 1 1

5. Let A = {1, 3, 5}. Then the number of equivalence relations in A
containing (1, 3) is
(A) 1 (B) 2 (C) 3 (D) 4
Ans: (B) 2 1

65/2/1 2

6. The interval in which the function f given by f (x) = x 2 e − x is strictly
increasing, is

(A) (–∞, ∞) (B) (–∞, 0) (C) (2, ∞) (D) (0, 2)
Ans: (D) (0, 2) 1
r r
7. If a = 4 and −3 ≤ λ ≤ 2, then λa lies in

(A) [0, 12] (B) [2, 3] (C) [8, 12] (D) [–12, 8]
Ans: (A) [0, 12] 1
^ ^
8. The vector 3 ^i − j + 2 k^ , 2 ^i + j + 3 k^ and ˆi + λˆj − kˆ are coplanar if value of λ is
(A) –2 (B) 0
(C) 2 (D) Any real number
Ans: (A) –2 1

9. The area of a triangle formed by vertices O, A and B, where
uuur uuur
OA = ˆi + 2ˆj + 3kˆ ^j and OB = −3iˆ − 2ˆj + kˆ is

(A) 3 5 sq. units (B) 5 5 sq. units
(C) 6 5 sq. units (D) 4 sq. units

Ans: (A) 3 5 sq. units 1

10. The coordinates of the foot of the perpendicular drawn from the point
(2, –3, 4) on the y-axis is
(A) (2, 3, 4) (B) (–2, –3, –4)
(C) (0, –3, 0) (D) (2, 0, 4)
Ans: (C) (0, –3, 0) 1

Fill in the blanks in questions numbers 11 to 15.
11. The range of the principal value branch of the function y = sec–1 x
is __________.
⎧π⎫
Ans: [ 0, π ] − ⎨ ⎬ 1
⎩2⎭
OR

−1 ⎛ 1 ⎞
The principal value of cos ⎜ − ⎟ is _____________.
⎝ 2⎠

2π
Ans: 1
3

65/2/1 3

⎡ 0 a 1⎤
12. Given a skew-symmetric matrix A = ⎢⎢ –1 b 1 ⎥⎥ , the value of
⎢⎣ –1 c 0 ⎥⎦
(a + b + c)2 is ______
Ans: 0 1

13. The distance between parallel planes 2x + y − 2z − 6 = 0 and
4x + 2y − 4z = 0 ______ units.

Ans: 2 1

OR

If P(1,0,–3) is the foot of the perpendicular from the origin to the plane,
then the cartesian equation of the plane is ________
Ans: x – 3z = 10 1
14. If the radius of the circle is increasing at the rate of 0.5 cm/s, then the rate
of increase of its circumference is ________
Ans: π cm/s 1

15. The corner points of the feasible region of an LPP are (0,0), (0,8), (2,7), (5,4)
and (6,0). The maximum profit P = 3x + 2y occurs at the point ________

Ans: (5, 4) 1

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

16. Differentiate sec 2 (x 2 ) with respect to x2.
Ans: Let x2 = u, differentiating sec2 u w.r.t u, putting u = x2 1/2
2sec 2 x 2 tan x 2 1/2
OR

2 dy
If y = f (x ) and f ′(x) = e x , then find .
dx

dy
Ans: = f ' ( x 2 ) 2x 1/2
dx
1/2
= 2xe x

⎧⎪ kx 2 + 5, if x ≤1
17. Find the value of k, so that the function f (x) = ⎨
⎪⎩ 2 , if x >1

is continuous at x = 1.
Ans: L.H.L. is k + 5 1/2
getting k = –3 1/2

65/2/1 4

π
2

∫π x cos xdx .
18. Evaluate: 2

−
2

Ans: Let f(x) = x cos2 x f(–x) = – f(x) or f is odd 1/2
π/2

∫ x cos xdx = 0
2
∴ 1/2
− π/2

dy
19. Find the general solution of the differential equation e y − x = 1.
dx
Ans: Given differential equation is eydy = ex dx 1/2

Integrating to get ey = ex + C 1/2

x −1 y + 4 z + 4
20. Find the coordinates of the point where the line = =
3 7 2
cuts the line xy-plane.
x −1 y + 4
Ans: Putting z = 0 in given equation gives = =2 1/2
3 7
Coordinates of required point are (7, 10, 0) 1/2

SECTION-B

Question numbers 21 to 26 carry 2 marks each.

⎡ −3 2 ⎤ ⎡1 0 ⎤ 2
21. If A = ⎢ ⎥ and I = ⎢ ⎥ , find scalar k so that A + I = kA .
⎣ 1 − 1⎦ ⎣ 0 1 ⎦

⎡ 11 −8⎤
Ans: A2 = ⎢ −4 3 ⎥ 1
⎣ ⎦

⎡12 −8⎤ ⎡ −3k 2k ⎤
A2 + I = kA ⇒ ⎢ −4 4 ⎥ = ⎢ k − k ⎥⎦ 1/2
⎣ ⎦ ⎣
k = –4 1/2

65/2/1 5

sec x − 1 ⎛π⎞
22. If f (x) = , find f ′ ⎜ ⎟ .
sec x + 1 ⎝3⎠

1
−1
cos x x
Ans: f(x) = = tan 1
1 2
+1
cos x

1 x
f ’(x) = sec 2 1/2
2 2

⎛π⎞ 2
f′⎜ ⎟ = 1/2
⎝3⎠ 3
OR

Find f ′ ( x ) if f ( x ) = (tan x) tan x .
Ans: Taking log on both sides. log f(x) = tan x log tan x 1/2

f '(x)
differentiating to get = sec2 x + sec2 x log tan x 1
f (x)
Thus, f ’(x) = (tan x)tan x · sec2 x(1 + log tan x) 1/2

tan 3 x
23. Find ∫ dx .
cos3 x

sin 3 x
Ans: Given Integral is I = ∫ dx 1/2
cos6 x
Put cos x = t
sin x dx = – d t 1/2
⎡ −1 1 ⎤
= ∫ ⎢ 6 + 4 ⎥ dt
⎣t t ⎦

t −5 t −3
= − +c 1/2
5 3
1 1 sec5 x sec3 x
= 5(cos x)5 − 3(cos x)3 + c or − +c 1/2
5 3
r
24. Find a vector r equally inclined to the three axes and whose magnitude
is 3 3 units.
r
Ans: Let the vector r = aiˆ + ajˆ + akˆ 1/2

∴ 3a 2 = 3 3 1

required vector is 3iˆ + 3jˆ + 3kˆ or − 3iˆ − 3jˆ − 3kˆ 1/2

65/2/1 6

OR
r r r r
Find the angle between unit vectors a and b so that 3 a – b is also a
unit vector.
r r r r2
Ans: Using 3 a – b = 1 i.e. 3 a – b = 1 1/2

r r 3
& getting a · b = 1
2

π
getting angle or 30° 1/2
6
r
25. Find the points of intersection of the line r = 2iˆ − ˆj + 2kˆ + λ (3iˆ + 4ˆj + 2k)
ˆ
r
and the plane r·(iˆ − ˆj + k)
ˆ = 5.

Ans: Any point on line is ( 2 + 3 λ, −1 + 4 λ, 2 + 2 λ ) 1
Putting in equation of plane to get λ = 0 1/2
Thus point of intersection is (2, –1, 2) 1/2

26. A purse contains 3 silver and 6 copper coins and a second purse contains
4 silver and 3 copper coins. If a coin is drawn at random from one of the
two purses, find the probability that it is a silver coin.
Ans: E1 : coin is drawn from purse 1.
E2 : coin is drawn from purse 2. 1/2
A : Silver coin is drawn
1 1 ⎛A⎞ 3 ⎛A⎞ 4
P ( E1 ) = , P ( E2 ) = , P ⎜ ⎟ = , P ⎜ ⎟ = 1
2 2 ⎝ E1 ⎠ 9 ⎝ E 2 ⎠ 7

1 3 1 4
P (A) = × + ×
2 9 2 7
19
= 1/2
42
SECTION-C

Question numbers 27 to 32 carry 4 marks each.

27. Check whether the relation R in the set N of natural numbers given by
R = {(a, b) : a is divisor of b}
is reflexive, symmetric or transitive. Also determine whether R is an
equivalence relation.
Ans: For reflexive
Let a ∈ N clearly a divides a ∴ (a, a) ∈ R
∴ R is reflexive 1

65/2/1 7

For symmetric
(1, 2) ∈ R but (2, 1) ∉ R 1
∴ R is not symmetric
For transitive
Let (a, b) , (b, c) ∈ R
∴ a divides b and b divides c
⇒ a divides c ∴ (a, c) ∈ R 1
R is transitive
As R is not symmetric ∴ It is not an equivalence relation 1

OR

−1 1 2 1 −1 ⎛ 4 ⎞
Prove that tan + tan −1 = sin ⎜ ⎟ .
4 9 2 ⎝5⎠

⎛ 1 2 ⎞
−1
⎜ 4+9 ⎟ −1 1
Ans: LHS = tan ⎜ ⎟ = tan 2
1 2
⎜ 1− · ⎟ 2
⎝ 4 9⎠

1 1
= · 2 tan −1 1
2 2

⎛ 1⎞
⎜ 2× ⎟
1
sin −1 ⎜ 2 = 1 sin −1 ⎛ 4 ⎞
= 2 1⎟ ⎜ ⎟
⎝5⎠ 1
⎜ 1+ ⎟ 2
⎝ 4⎠
= RHS

⎛y⎞ dy x + y
28. If tan −1 ⎜ ⎟ = log x 2 + y 2 , prove that = .
⎝x⎠ dx x − y
Ans: Differentiating both sides w.r.t. x to get

dy dy
x −y 2x + 2y
1 dx 1 dx 1 1
· = · 1 +1
y2 x2 x + y 2 x + y2
2 2 2 2 2
1+ 2
x

dy dy 1
Simplyfying we get x −y=x+y
dx dx 2
dy x+y 1
getting =
dx x − y 2
OR
d2 y
If y = e
a cos −1 x
(
, −1 < x < 1 , then show that 1 − x 2 ) dx 2 − x dx
dy 2
−a y = 0

65/2/1 8

-l
dy −ae a cos x
Ans: = 1
dx 1− x2
dy -l 1
1− x2 = −ae a cos x
dx 2
Differentiating again & getting
-l
d2 y x dy a 2 e a cos x
1− x 2
− · = 2
dx 2 1 − x 2 dx 1− x2
2
d y dy
(1 − x ) dx
1
2
−x −a y = 0
2
2

dx 2

x3 + 1
29. Find ∫ 3 dx
x −x

x3 − x + x + 1
Ans: Writing Integral as I = ∫ dx 1
x3 − x
1
= ∫ 1dx + ∫ dx 1
x(x − 1)

⎛ 1 1⎞
= x + ∫⎜ − ⎟ dx 1
⎝ x-1 x⎠

= x + log x − 1 − log x + c 1
30. Solve the following differential equation:

(1 + e ) dy + e
y/x y/x ⎛ y⎞
⎜ 1 − ⎟ dx = 0 (x ≠ 0)
⎝ x⎠

dy e y/ x (y / x − 1)
Ans: Writing given differential equation as = 1
dx 1 + e y/ x
dy dv 1
Putting y = vx & =v+x
dx dx 2

dv − ( e +v )
v

to get x = 1
dx 1 + ev

ev + 1 dx
⇒ ∫ ev + v dv = − ∫ x
log e v + v = – log | x | + log c 1

y c 1
e y/x + = or x e y/x + y = c
x x 2

65/2/1 9

31. Find the shortest distance between the lines
r
r = 2iˆ − ˆj + kˆ + λ (3iˆ − 2ˆj + 5k)
ˆ
r
r = 3iˆ + 2ˆj − 4kˆ + μ (4iˆ − ˆj + 3k)
ˆ
r r
Ans: a 2 − a1 = ˆi + 3jˆ − 5 kˆ 1

ˆi ˆj kˆ
r r
b1 × b 2 = 3 −2 5 = − ˆi + 11jˆ + 5kˆ
2
4 −1 3

r r
( a 2 − a1 ) · ( b1 × b2 )
r r
Using, shortest distance = r r
b1 × b 2

7 1
= units or units 1
147 3

32. A cotton industry manufactures pedestal lamps and wooden shades. Both the
products require machine time as well as craftsman time in the making. The
number of hour(s) required for producing 1 unit of each and the corresponding
profit is given in the following table

Items Machine time Craftsman time Profit (in `)
Pedestal lamp 1.5 hours 3 hours 30
Wooden shades 3 hours 1 hour 20

In a day, the factory has availability of not more than 42 hours of machine time
and 24 hours of craftsman time.
Assuming that all items manufactured are sold, how should the manufacture
schedule his daily production in order to maximize the profit? Formulate it as
an LPP and solve it graphically.
Ans: Let number of pedestal lamps = x y
number of wooden shades = y
24
1
Maximize Profit Z = 30x + 20y 20
2
16
Subject to constraints (4, 12)
12

8 1
1.5x + 3y ≤ 42 1
4
2
3x + y ≤ 24 x
O 4 8 12 14 16 20 24 28
x ≥ 0, y ≥ 0 correct graph 1

65/2/1 10

getting corners points & values of Z
(0, 0) 0
(8, 0) 240
(4, 12) 360
(0, 14) 280 1/2

Maximum profit = `360 where x = 4, y = 12 1/2

SECTION-D

Question numbers 33 to 36 carry 6 marks each.

⎡ 5 −1 4 ⎤
33. A = ⎢⎢ 2 3 5 ⎥⎥ , find A–1 and use it to solve the following system of equations:
⎢⎣ 5 −2 6 ⎥⎦

5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Ans: |A| = 51 1
A11 = 28 A12 = 13 A13 = −19
Cofactors: A 21 = −2 A 22 = 10 A 23 = 5 2
A 31 = −17 A 32 = −17 A 33 = 17

28 −2 −17 ⎤
1 ⎡⎢
A −1 = 13 10 −17 ⎥ 1
51 ⎢ −19 5 17 ⎥
⎣ ⎦

⎡x⎤ ⎡ 5⎤ 1
Given system is AX = B where X = y B = ⎢ 2 ⎥
⎢ ⎥
⎢z⎥ ⎢ −1⎥ 2
⎣ ⎦ ⎣ ⎦

28 −2 −17 ⎤ ⎡ 5 ⎤ ⎡ 3 ⎤
1 ⎡⎢
⇒X=A B= −1
13 10 −17 ⎥ ⎢ 2 ⎥ = ⎢ 2 ⎥ 1
51 ⎢ −19 5 17 ⎥ ⎢ −1⎥ ⎢ −2 ⎥
⎣ ⎦⎣ ⎦ ⎣ ⎦

1
x = 3, y = 2, z = –2
2

65/2/1 11

OR

x x 2 1 + x3
If x, y, z are different and y y 2 1 + y3 = 0 , then using properties of determinants
z z2 1 + z3

show that 1 + xyz = 0.

x x2 1 x x2 x3
Ans: Writing L.H.S as y y2 1 + y y2 y3 = 0 1
z z2 1 z z2 z3

x x2 1 1 x x2
⇒ y y 2 1 + xyz 1 y y2 = 0
1
z z2 1 1 z z2

getting

x x2 1
(1 + xyz) y y2 1 = 0
2
z z2 1

Applying R2 → R2 – R1, R3 → R3 – R1

x x2 1
(1 + xyz) y − x (y + x)(y − x) 0 = 0 1
z − x (z + x)(z − x) 0

Expanding and simplifying to get
⇒ (1 + xyz) (x – y) (y – z) (z – x) = 0 1/2
As x, y, z are different 1/2
∴ 1+ xyz = 0
34. Amongst all open (from the top) right circular cylindrical boxes of volume
125π cm3 , find the dimensions of the box which has the least surface area.
Ans: Let radius = r & height = h
125
πr 2 h = 125π or h=
r2
1

2
Surface Area, S = 2 π r h + π r
250 π
S= + π r2 1
r

65/2/1 12

dS −250 π
= + 2πr 1
dr r2
dS
=0 ⇒ r=5 1/2+1
dr

d 2S 500π
= + 2π > 0 so s is least 1
dr 2 r3

when r = 5 cm , gives h = 5 cm 1/2

35. Using integration, find the area lying above x-axis and included
between the circle x 2 + y 2 = 8x and inside the parabola y 2 = 4x .

Ans. Correct figure 1
x-coordinate of point of intersection is 4, 0 1

O 4 8
4 8

Required Area = ∫ 2 x dx + ∫ 16 − (x − 4) dx
2
2
0 4

4 8
4 32 x−4 x−4
= x + 16 − ( x − 4) 2 + 8sin −1 1
3 0 2 4 4

32
= + 4π 1
3
OR
Using the method of integration, find the area of the ΔABC, coordinates of
whose vertices are A(2, 0), B(4, 5) and C(6, 3).
B
Ans. Correct figure 5 1
4
5
Equation of AB : y = (x − 2) 3 1/2
2
Equation of BC : y = 9 – x 2 C 1/2
1
3 A
Equation of AC : y = (x − 2) 1/2
4 O 1 2 3 4 5 6

5 4 6 3 6
Required Area =
2 ∫2
(x − 2) dx + ∫ (9 − x) dx − ∫ (x − 2) dx
4 4 2
2

5 (x − 2) 2 4 (9 − x) 2 6 3 (x − 2) 2 6
= + − 1
2 2 2
−2 4 4 2 2

=7 1/2

65/2/1 13

36. Find the probability distribution of the random variable X, which denotes the
number of doublets in four throws of a pair of dice. Hence, find the mean of the
number of doublets (X).
1 5
Ans. Let X denote the number of doublets P(doublet) = , P(not a doublet) = 1
6 6
X 0 1 2 3 4 1
625 500 150 20 1 1
P(X) 2
1296 1296 1296 1296 1296 2

500 300 60 4
X·P(X) 0 1
1296 1296 1296 1296

864 2
mean = ∑ X·P(X) = or 1/2
1296 3

65/2/1 14

Strictly Confidential - (For Internal and Restricted Use Only)

Senior School Certificate Examination-2020
Marking Scheme - MATHEMATICS
Subject Code: 041 Paper Code: 65/2/2
General instructions:-
1. You are aware that evaluation is the most important process in the actual and correct assessment of the candidates. A
small mistake in evaluation may lead to serious problems which may affect the future of the candidates, education
system and teaching profession. To avoid mistakes, it is requested that before starting evaluation, you must read and
understand the spot evaluation guidelines carefully. Evaluation is a 10-12 days mission for all of us. Hence, it is
necessary that you put in your best efforts in this process.
2. Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done according to one's
own interpretation or any other consideration. Marking Scheme should be strictly adhered to and religiously followed.
However, while evaluating, answers which are based on latest information or knowledge and/or are innovative,
they may be assessed for their correctness otherwise and marks be awarded to them.
3. The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first day, to
ensure that evaluation has been carried out as per the instructions given in the Marking Scheme. The remaining
answer books meant for evaluation shall be given only after ensuring that there is no significant variation in the
marking of individual evaluators.
4. Evaluators will mark( √ ) wherever answer is correct. For wrong answer 'X"be marked. Evaluators will not put right
kind of mark while evaluating which gives an impression that answer is correct and no marks are awarded. This is
most common mistake which evaluators are committing.
5. If a question has parts, please award marks on the right-hand side for each part. Marks awarded for different parts of
the question should then be totaled up and written in the left-hand margin and encircled. This may be followed
strictly.
6. If a question does not have any parts, marks must be awarded in the left-hand margin and encircled. This may also be
followed strictly.
7. If a student has attempted an extra question, answer of the question deserving more marks should be retained and the
other answer scored out.
8. No marks to be deducted for the cumulative effect of an error. It should be penalized only once.
9. A full scale of marks 0 - 80 has to be used. Please do not hesitate to award full marks if the answer deserves
it.
10. Every examiner has to necessarily do evaluation work for full working hours i.e. 8 hours every day and evaluate 20
answer books per day in main subjects and 25 answer books per day in other subjects (Details are given in Spot
Guidelines).
11. Ensure that you do not make the following common types of errors committed by the Examiner in the past:-
• Leaving answer or part thereof unassessed in an answer book.
• Giving more marks for an answer than assigned to it.
• Wrong totaling of marks awarded on a reply
• Wrong transfer of marks from the inside pages of the answer book to the title page.
• Wrong question wise totaling on the title page.
• Wrong totaling of marks of the two columns on the title page.
• Wrong grand total.
• Marks in words and figures not tallying.
• Wrong transfer of marks from the answer book to online award list.
• Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly and clearly
indicated. It should merely be a line. Same is with the X for incorrect answer.)
• Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
12. While evaluating the answer books if the answer is found to be totally incorrect, it should be marked as cross (X) and
awarded zero (0)Marks.
13. Any unassessed portion, non-carrying over of marks to the title page, or totaling error detected by the candidate shall
damage the prestige of all the personnel engaged in the evaluation work as also of the Board. Hence, in order to
uphold the prestige of all concerned, it is again reiterated that the instructions be followed meticulously and judiciously.
14. The Examiners should acquaint themselves with the guidelines given in the Guidelines for spot Evaluation before
starting the actual evaluation.
15. Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title page, correctly
totaled and written in figures and words.
16. The Board permits candidates to obtain photocopy of the Answer Book on request in an RTI application and also
separately as a part of the re-evaluation process on payment of the processing charges.

1 P.T.O.

QUESTION PAPER CODE 65/2/2
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. The area of a triangle formed by vertices O, A and B, where
uuur uuur
OA = ˆi + 2ˆj + 3kˆ ^j and OB = −3iˆ − 2ˆj + kˆ is

(A) 3 5 sq. units (B) 5 5 sq. units
(C) 6 5 sq. units (D) 4 sq. units

Ans: (A) 3 5 sq. units 1

⎛ −1 2 ⎞
2. If cos ⎜ sin + cos −1 x ⎟ = 0 , then x is equal to
⎝ 5 ⎠
1 2 2
(A) (B) − (C) (D) 1
5 5 5
2
Ans: (C) 1
5
3. The interval in which the function f given by f (x) = x 2 e − x is strictly
increasing, is
(A) (–∞, ∞) (B) (–∞, 0) (C) (2, ∞) (D) (0, 2)
Ans: (D) (0, 2) 1

x −1
The function f (x) = is discontinuous at
( )
4.
x x2 −1

(A) exactly one point (B) exactly two point
(C) exactly three point (D) no point
Ans: Solution not provided 1

(One mark to be given to all)

5. The function f : R → [–1, 1] defined by f(x) = cos x is
(A) both one-one and onto (B) not one-one, but onto
(C) one-one, but not onto (D) neither one-one, nor onto
Ans: (B) not one-one, but onto 1

65/2/2 2

6. The coordinates of the foot of the perpendicular drawn from the point
(2, –3, 4) on the y-axis is
(A) (2, 3, 4) (B) (–2, –3, –4)
(C) (0, –3, 0) (D) (2, 0, 4)
Ans: (C) (0, –3, 0) 1

7. The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1), (1, 1)} is
(A) symmetric and transitive, but not reflexive
(B) reflexive and symmetric, but not transitive
(C) symmetric, but neither reflexive nor transitive
(D) an equivalence relation
Ans: (C) symmetric, but neither reflexive nor transitive 1

8. The angle between the vectors ^i − ^j and ˆj − kˆ is

π π 2π
(A) − (B) 0 (C) (D)
3 3 3
2π
Ans: (D) 1
3

9. If A is a non-singular square matrix of order 3 such that A 2 = 3A, then
value of |A| is
(A) –3 (B) 3 (C) 9 (D) 27
Ans: (D) 27 1
r r
10. If a = 4 and −3 ≤ λ ≤ 2, then λa lies in

(A) [0, 12] (B) [2, 3] (C) [8, 12] (D) [–12, 8]
Ans: (A) [0, 12] 1

Fill in the blanks in questions numbers 11 to 15
11. If the radius of the circle is increasing at the rate of 0.5 cm/s, then the rate
of increase of its circumference is ________
Ans: π cm/s 1

2x − 9 −4 8
12. If −2 x = 1 − 2 , the value of x is ____________.
Ans: 3 or – 3 1

65/2/2 3

13. The corner points of the feasible region of an LPP are (0,0), (0,8), (2,7), (5,4)
and (6,0). The maximum profit P = 3x + 2y occurs at the point ________
Ans: (5, 4) 1

14. The range of the principal value branch of the function y = sec–1 x
is __________.
⎧π⎫
Ans: [ 0, π ] − ⎨ ⎬ 1
⎩2⎭
OR

−1 ⎛ 1 ⎞
The principal value of cos ⎜ − ⎟ is _____________.
⎝ 2⎠

2π
Ans: 1
3

15. The distance between parallel planes 2x + y − 2z − 6 = 0 and
4x + 2y − 4z = 0 ______ units.
Ans: 2 1

OR

If P(1,0,–3) is the foot of the perpendicular from the origin to the plane,
then the cartesian equation of the plane is ________
Ans: x – 3z = 10 1
Question numbers 16 to 20 are very short answer type questions

π
2

∫π x cos xdx .
16. Evaluate: 2

−
2

Ans: Let f(x) = x cos2 x f(–x) = – f(x) or f is odd 1/2
π/2

∫ x cos xdx = 0
2
∴ 1/2
− π/2

x −1 y + 4 z + 4
17. Find the coordinates of the point where the line = =
3 7 2
cuts the line xy-plane.
x −1 y + 4
Ans: Putting z = 0 in given equation gives = =2 1/2
3 7
Coordinates of required point are (7, 10, 0) 1/2

65/2/2 4

⎧⎪ kx 2 + 5, if x ≤1
18. Find the value of k, so that the function f (x) = ⎨
⎪⎩ 2 , if x >1

is continuous at x = 1.
Ans: L.H.L. is k + 5 1/2
getting k = –3 1/2

19. Find the integrating factor of the differential equation
dy
x = 2x 2 + y
dx

∫
−1 ⎧ writing given equation as
Ans: Integrating factor is e x
dx ⎪
or ⎨ dy y 1/2
⎪⎩ dx − x = 2x

1
= 1/2
x

20. Differentiate sec 2 (x 2 ) with respect to x2.
Ans: Let x2 = u, differentiating sec2 u w.r.t u, putting = x2 1/2
2 2 2
2sec x tan x 1/2
OR

2 dy
If y = f (x ) and f ′(x) = e x , then find .
dx

dy
Ans: = f ' ( x 2 ) 2x 1/2
dx
= 2xe x 1/2

SECTION-B

Question numbers 21 to 26 carry 2 marks.
r
21. Find a vector r equally inclined to the three axes and whose magnitude
is 3 3 units.
r
Ans: Let the vector r = aiˆ + ajˆ + akˆ 1/2

∴ 3a 2 = 3 3 1

required vector is 3iˆ + 3jˆ + 3kˆ or − 3iˆ − 3jˆ − 3kˆ 1/2

OR
r r r r
Find the angle between unit vectors a and b so that 3 a – b is also a
unit vector.

65/2/2 5

r r r r2
Ans: Using 3 a – b = 1 i.e. 3 a – b = 1 1/2

r r 3
& getting a · b = 1
2

π
getting angle or 30° 1/2
6

⎡ −3 2 ⎤ ⎡1 0 ⎤
22. If A = ⎢ ⎥ and I = ⎢ ⎥ , find scalar k so that A 2 + I = kA .
⎣ 1 −1⎦ ⎣0 1 ⎦

⎡ 11 −8⎤
Ans: A2 = ⎢ −4 3 ⎥ 1
⎣ ⎦

⎡12 −8⎤ ⎡ −3k 2k ⎤
A2 + I = kA ⇒ ⎢ −4 4 ⎥ = ⎢ k − k ⎥⎦ 1/2
⎣ ⎦ ⎣
k = –4 1/2

sec x − 1 ⎛π⎞
23. If f (x) = , find f ′ ⎜ ⎟ .
sec x + 1 ⎝3⎠

1
−1
cos x x
Ans: f(x) = = tan 1
1 2
+1
cos x

1 x
f ’(x) = sec 2 1/2
2 2
⎛π⎞ 2
f′⎜ ⎟ = 1/2
⎝3⎠ 3
OR

Find f ′ ( x ) if f ( x ) = (tan x) tan x .
Ans: Taking log on both sides. log f(x) = tan x log tan x 1/2

f '(x)
differentiating to get = sec2 x + sec2 x log tan x 1
f (x)
Thus, f ’(x) = (tan x)tan x · sec2 x(1 + log tan x) 1/2

tan 3 x
24. Find ∫ dx .
cos3 x

sin 3 x
Ans: Given Integral is I = ∫ dx 1/2
cos6 x
Put cos x = t
sin x dx = – d t 1/2

65/2/2 6

⎡ −1 1 ⎤
= ∫ ⎢ 6 + 4 ⎥ dt
⎣t t ⎦

t −5 t −3
= − +c 1/2
5 3
1 1 sec5 x sec3 x
= 5(cos x)5 − 3(cos x)3 + c or − +c 1/2
5 3

x −5
25. Show that the plane x – 5y – 2z = 1 contains the line = y = 2−z
3
Ans: Given point on line 5, 0, 2
Putting in equation of plane 5 – 0 – 4 = 1
1 = 1 ∴ point lies on plane 1
dr’s of line 3, 1, –1
dr’s of normal to the plane 1, –5, –2
As 3 · 1 + 1 · –5 + –1 · –2 = 0 1
∴ given plane contains given line.

26. A fair dice is thrown two times. find the probability distribution of the
number of sixes. Also determine the mean of the number of sixes.
1 5
Ans: X denote number of sixes P(6) = , P ( 6 ) = 1/2
6 6
X 0 1 2
25 10 1
P(X) … 1
36 36 36
10 2
XP(X) 0
36 36
1
mean = ∑ XP(X) = 1/2
3
SECTION-C

Question numbers 27 to 32 carry 4 marks.

27. Solve the following differential equation:

(1 + e ) dy + e
y/x y/x ⎛ y⎞
⎜ 1 − ⎟ dx = 0 (x ≠ 0)
⎝ x⎠

dy e y/ x (y / x − 1)
Ans: Writing given differential equation as = 1
dx 1 + e y/ x
dy dv 1
Putting y = vx & =v+x
dx dx 2

65/2/2 7

dv − ( e +v )
v

to get x = 1
dx 1 + ev

ev + 1 dx
⇒ ∫ ev + v dv = − ∫ x
log e v + v = – log | x | + log c 1

y c 1
e y/x + = or x e y/x + y = c
x x 2
28. A cotton industry manufactures pedestal lamps and wooden shades. Both the
products require machine time as well as craftsman time in the making. The
number of hour(s) required for producing 1 unit of each and the corresponding
profit is given in the following table

Items Machine time Craftsman time Profit (in `)
Pedestal lamp 1.5 hours 3 hours 30
Wooden shades 3 hours 1 hour 20

In a day, the factory has availability of not more than 42 hours of machine time
and 24 hours of craftsman time.
Assuming that all items manufactured are sold, how should the manufacture
schedule his daily production in order to maximize the profit? Formulate it as
an LPP and solve it graphically.
Ans: Let number of pedestal lamps = x y
number of wooden shades = y
24
1
Maximize Profit Z = 30x + 20y 20
2
16
Subject to constraints (4, 12)
12

8
1
1
1.5x + 3y ≤ 42 4
2
3x + y ≤ 24 x
O 4 8 12 14 16 20 24 28
x ≥ 0, y ≥ 0 correct graph 1
getting corners points & values of Z
(0, 0) 0
(8, 0) 240
(4, 12) 360
(0, 14) 280 1/2

Maximum profit = `360 where x = 4, y = 12 1/2

65/2/2 8

π
2
29. Evaluate : ∫ sin 2x tan −1 (sin x)dx
0

π
2
−1
Ans: Writing I = ∫ 2sin x cos x tan (sin x)dx 1/2
0

putting sinx = t
cosxdx = dt 1/2
1
−1
⇒ I = ∫ 2t tan t dt 1/2
0

1
−1 t2 ⎤
2
= t tan t−∫ dt ⎥ 1
1 + t 2 ⎥⎦ 0

−1 −1 1
= t tan t − t + tan t ⎤⎦
2
1
0

π
= −1 1/2
2
30. Check whether the relation R in the set N of natural numbers given by
R = {(a, b) : a is divisor of b}
is reflexive, symmetric or transitive. Also determine whether R is an
equivalence relation.
Ans: For reflexive
Let a ∈ N clearly a divides a ∴ (a, a) ∈ R
∴ R is reflexive 1
For symmetric
(1, 2) ∈ R but (2, 1) ∉ R 1
∴ R is not symmetric
For transitive
Let (a, b) , (b, c) ∈ R
∴ a divides b and b divides c
⇒ a divides c ∴ (a, c) ∈ R 1
R is transitive
As R is not symmetric ∴ It is not an equivalence relation 1

65/2/2 9

OR

−1 1 2 1 −1 ⎛ 4 ⎞
Prove that tan + tan −1 = sin ⎜ ⎟ .
4 9 2 ⎝5⎠

⎛ 1 2 ⎞
−1
⎜ 4+9 ⎟ −1 1
Ans: LHS = tan ⎜ 1 2 ⎟ = tan 2
⎜ 1− · ⎟ 2
⎝ 4 9⎠

1 1
= · 2 tan −1 1
2 2

⎛ 1⎞
⎜ 2× ⎟
1
sin −1 ⎜ 2 = 1 sin −1 ⎛ 4 ⎞
= 2 1⎟ ⎜ ⎟
⎜ 1+ ⎟ 2 ⎝5⎠ 1
⎝ 4⎠
= RHS

31. Find the equation of the plane passing through the points (1, 0, –2), (3, –1, 0) and
perpendicular to the plane 2x – y + z = 8. Also find the distance of the plane thus
obtained from the origin.
Ans: Let P(1, 0, –2) and Q(3, –1, 0)
dr’s of line through PQ 2, –1, 2 1/2

Let dr’s of Normal of required plane be A, B, C
∴ 2A − B + 2C = 0
& 2A − B + C = 0 } 1

Solving we get A = 1, B = 2, C = 0 1
Equation of plane is 1(x – 1) + 2 (y – 0) + 0(z + 2) = 0
i.e. x + 2y = 1 1/2

0 + 0 + 0 −1
Distance of above plane from origin is
1+ 4

1 5
= or 1
5 5

−1 ⎛ y ⎞ 2 2 dy x + y
32. If tan ⎜ ⎟ = log x + y , prove that = .
⎝x⎠ dx x − y

Ans: Differentiating both sides w.r.t. x to get
dy dy
x −y 2x + 2y
1 dx 1 dx 1 1
· = · 1 +1
y2 x2 x + y 2 x + y2
2 2 2 2 2
1+ 2
x

65/2/2 10

dy dy 1
Simplyfying we get x −y=x+y
dx dx 2
dy x+y 1
getting =
dx x − y 2
OR
d2 y dy 2
a cos −1 x
, −1 < x < 1 , then show that (1 − x ) −x −a y = 0
2
If y = e 2
dx dx
-l
dy −ae a cos x
Ans: = 1
dx 1− x2
dy -l 1
1− x2 = −ae a cos x
dx 2
Differentiating again & getting
-l
d2 y x dy a 2 e a cos x
1− x −2
· = 2
dx 2 1 − x 2 dx 1− x2

d2 y dy 2
(1 − x 2 )
1
2
−x −a y = 0
dx dx 2

SECTION-D

Question numbers 33 to 36 carry 6 marks.

33. Amongst all open (from the top) right circular cylindrical boxes of volume
125π cm3 , find the dimensions of the box which has the least surface area.
Ans: Let radius = r & height = h
125
πr 2 h = 125π or h=
r2
1

2
Surface Area, S = 2 π r h + π r
250 π
S= + π r2 1
r
dS −250 π
= + 2πr 1
dr r2
dS
=0 ⇒ r=5 1/2+1
dr

d 2S 500π
= 3 + 2π > 0 so s is least 1
dr 2 r

when r = 5 cm , gives h = 5 cm 1/2

65/2/2 11

34. Using integration, find the area lying above x-axis and included
between the circle x 2 + y 2 = 8x and inside the parabola y 2 = 4x .

Ans. Correct figure 1
x-coordinate of point of intersection is 4, 0 1

O 4 8
4 8

Required Area = ∫ 2 x dx + ∫ 16 − (x − 4) dx
2
2
0 4

4 8
4 32 x−4 −1 x − 4
= 3 x + 2 16 − ( x − 4) + 8sin
2

4 4 1
0

32
= + 4π 1
3
OR
Using the method of integration, find the area of the ΔABC, coordinates of
whose vertices are A(2, 0), B(4, 5) and C(6, 3).
B
5
Ans. Correct figure 1
4
5
Equation of AB : y = (x − 2) 3 1/2
2
2 C
Equation of BC : y = 9 – x 1/2
1
3 A
Equation of AC : y = (x − 2) O 1 2 3 4 5 6 1/2
4
5 4 6 3 6
Required Area = ∫
2 2
(x − 2) dx + ∫ (9 − x) dx − ∫ (x − 2) dx
4 4 2
2

5 (x − 2) 2 4 (9 − x) 2 6 3 (x − 2) 2 6
= + − 1
2 2 2
−2 4 4 2 2

=7 1/2

⎡ 5 −1 4 ⎤
35. A = ⎢⎢ 2 3 5 ⎥⎥ , find A–1 and use it to solve the following system of equations:
⎢⎣ 5 −2 6 ⎥⎦

5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1

65/2/2 12

Ans: |A| = 51 1
A11 = 28 A12 = 13 A13 = −19
Cofactors: A 21 = −2 A 22 = 10 A 23 = 5 2
A 31 = −17 A 32 = −17 A 33 = 17

28 −2 −17 ⎤
1 ⎡⎢
−1
A = 13 10 −17 ⎥ 1
51 ⎢ −19 5 17 ⎥
⎣ ⎦

⎡x⎤ ⎡ 5⎤ 1
Given system is AX = B where X = ⎢ y ⎥ B = ⎢ 2 ⎥
⎢z⎥ ⎢ −1⎥ 2
⎣ ⎦ ⎣ ⎦

28 −2 −17 ⎤ ⎡ 5 ⎤ ⎡ 3 ⎤
1 ⎡⎢
⇒ X = A −1B = 13 10 −17 ⎥ ⎢ 2 ⎥ = ⎢ 2 ⎥ 1
51 ⎢ −19 5 17 ⎥ ⎢ −1⎥ ⎢ −2 ⎥
⎣ ⎦⎣ ⎦ ⎣ ⎦

1
x = 3, y = 2, z = –2
2
OR

x x 2 1 + x3
If x, y, z are different and y y 2 1 + y3 = 0 , then using properties of determinants
z z2 1 + z3

show that 1 + xyz = 0.

x x2 1 x x2 x3
Ans: Writing L.H.S as y y2 1 + y y2 y3 = 0 1
z z2 1 z z2 z3

x x2 1 1 x x2
⇒ y y 2 1 + xyz 1 y y2 = 0
1
z z2 1 1 z z2

getting

x x2 1
(1 + xyz) y y2 1 = 0
2
2
z z 1

65/2/2 13

Applying R2 → R2 – R1, R3 → R3 – R1

x x2 1
(1 + xyz) y − x (y + x)(y − x) 0 = 0 1
z − x (z + x)(z − x) 0

Expanding and simplifying to get
⇒ (1 + xyz) (x – y) (y – z) (z – x) = 0 1/2

As x, y, z are different 1/2

∴ 1+ xyz = 0

36. A card from a pack of 52 cards is lost. From the remaining cards of the
pack, two cards are drawn randomly one-by-one without replacement and
are found to be both kings. Find the probability of the lost card being a king.
Ans: Let E1 : Lost card is king

Let E2 : Lost card is not a king 1/2

A : Two cards drawn are kings
1 12
P ( E1 ) = , P ( E2 ) = 1
13 13
3 4
C2 C
P ( A/E1 ) = 51 , P ( A/E 2 ) = 51 2 1+1
C2 C2

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

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

3 1
= or 1
75 25

65/2/2 14

Strictly Confidential - (For Internal and Restricted Use Only)

Senior School Certificate Examination-2020
Marking Scheme - MATHEMATICS
Subject Code: 041 Paper Code: 65/2/3
General instructions:-
1. You are aware that evaluation is the most important process in the actual and correct assessment of the candidates. A
small mistake in evaluation may lead to serious problems which may affect the future of the candidates, education
system and teaching profession. To avoid mistakes, it is requested that before starting evaluation, you must read and
understand the spot evaluation guidelines carefully. Evaluation is a 10-12 days mission for all of us. Hence, it is
necessary that you put in your best efforts in this process.
2. Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done according to one's
own interpretation or any other consideration. Marking Scheme should be strictly adhered to and religiously followed.
However, while evaluating, answers which are based on latest information or knowledge and/or are innovative,
they may be assessed for their correctness otherwise and marks be awarded to them.
3. The Head-Examiner must go through the first five answer books evaluated by each evaluator on the first day, to
ensure that evaluation has been carried out as per the instructions given in the Marking Scheme. The remaining
answer books meant for evaluation shall be given only after ensuring that there is no significant variation in the
marking of individual evaluators.
4. Evaluators will mark( √ ) wherever answer is correct. For wrong answer 'X"be marked. Evaluators will not put right
kind of mark while evaluating which gives an impression that answer is correct and no marks are awarded. This is
most common mistake which evaluators are committing.
5. If a question has parts, please award marks on the right-hand side for each part. Marks awarded for different parts of
the question should then be totaled up and written in the left-hand margin and encircled. This may be followed
strictly.
6. If a question does not have any parts, marks must be awarded in the left-hand margin and encircled. This may also be
followed strictly.
7. If a student has attempted an extra question, answer of the question deserving more marks should be retained and the
other answer scored out.
8. No marks to be deducted for the cumulative effect of an error. It should be penalized only once.
9. A full scale of marks 0 - 80 has to be used. Please do not hesitate to award full marks if the answer deserves
it.
10. Every examiner has to necessarily do evaluation work for full working hours i.e. 8 hours every day and evaluate 20
answer books per day in main subjects and 25 answer books per day in other subjects (Details are given in Spot
Guidelines).
11. Ensure that you do not make the following common types of errors committed by the Examiner in the past:-
• Leaving answer or part thereof unassessed in an answer book.
• Giving more marks for an answer than assigned to it.
• Wrong totaling of marks awarded on a reply
• Wrong transfer of marks from the inside pages of the answer book to the title page.
• Wrong question wise totaling on the title page.
• Wrong totaling of marks of the two columns on the title page.
• Wrong grand total.
• Marks in words and figures not tallying.
• Wrong transfer of marks from the answer book to online award list.
• Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly and clearly
indicated. It should merely be a line. Same is with the X for incorrect answer.)
• Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
12. While evaluating the answer books if the answer is found to be totally incorrect, it should be marked as cross (X) and
awarded zero (0)Marks.
13. Any unassessed portion, non-carrying over of marks to the title page, or totaling error detected by the candidate shall
damage the prestige of all the personnel engaged in the evaluation work as also of the Board. Hence, in order to
uphold the prestige of all concerned, it is again reiterated that the instructions be followed meticulously and judiciously.
14. The Examiners should acquaint themselves with the guidelines given in the Guidelines for spot Evaluation before
starting the actual evaluation.
15. Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title page, correctly
totaled and written in figures and words.
16. The Board permits candidates to obtain photocopy of the Answer Book on request in an RTI application and also
separately as a part of the re-evaluation process on payment of the processing charges.

1 P.T.O.

QUESTION PAPER CODE 65/2/3
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
1. The area of a triangle formed by vertices O, A and B, where
uuur uuur
OA = ˆi + 2ˆj + 3kˆ ^j and OB = −3iˆ − 2ˆj + kˆ is

(A) 3 5 sq. units (B) 5 5 sq. units
(C) 6 5 sq. units (D) 4 sq. units

Ans: (A) 3 5 sq. units 1

2. The domain of the function f(x) = sin–1(2x) is
(A) [0, 1] (b) [–1, 1]
⎡ 1 1⎤
(C) ⎢ − , ⎥ (d) [–2, 2]
⎣ 2 2⎦
⎡ 1 1⎤
Ans: (C) ⎢ − , ⎥ 1
⎣ 2 2⎦

3. The interval in which the function f given by f (x) = x 2 e − x is strictly
increasing, is

(A) (–∞, ∞) (B) (–∞, 0) (C) (2, ∞) (D) (0, 2)
Ans: (D) (0, 2) 1
4. The value of k so that f defined by

⎧ 2 ⎛1⎞
⎪ x sin ⎜ ⎟ if x ≠ 0
f (x) = ⎨ ⎝x⎠
⎪⎩ k if x ≠ 0
is continuous at x = 0 is
1
(A) 0 (B) (C) 1 (D) 2
2
Ans: (A) 0 1
1
5. The derivative of log x with respect to is
x
1 1 1
(A) − 3 (B) − (C) –x (D)
x x x
Ans: (C) –x 1

65/2/3 2

6. The coordinates of the foot of the perpendicular drawn from the point
(2, –3, 4) on the y-axis is
(A) (2, 3, 4) (B) (–2, –3, –4)
(C) (0, –3, 0) (D) (2, 0, 4)
Ans: (C) (0, –3, 0) 1

7. The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1), (1, 1)} is
(A) symmetric and transitive, but not reflexive
(B) reflexive and symmetric, but not transitive
(C) symmetric, but neither reflexive nor transitive
(D) an equivalence relation
Ans: (C) symmetric, but neither reflexive nor transitive 1

r r r r r r
8. If | a | = 3, | b | = 4 and | a × b | = 6, the value of a . b is
(A) 12 (B) 6

(C) 3 3 (D) 6 3

Ans: (D) 6 3 1

9. If A is a non-singular square matrix of order 3 such that A 2 = 3A, then
value of |A| is

(A) –3 (B) 3 (C) 9 (D) 27
Ans: (D) 27 1
r r
10. If a = 4 and −3 ≤ λ ≤ 2, then λa lies in

(A) [0, 12] (B) [2, 3] (C) [8, 12] (D) [–12, 8]
Ans: (A) [0, 12] 1

Fill in the blanks in questions numbers 11 to 15
11. If the radius of the circle is increasing at the rate of 0.5 cm/s, then the rate
of increase of its circumference is ________
Ans: π cm/s 1

⎡ 2⎤
⎢ ⎥
12. If [3 − 2 0] ⎢ k ⎥ = O, where O is the null matrix, then the value of k is __________.
⎢⎣ −5 ⎥⎦

Ans: 3 1

65/2/3 3

13. The corner points of the feasible region of an LPP are (0,0), (0,8), (2,7), (5,4)
and (6,0). The maximum profit P = 3x + 2y occurs at the point ________

Ans: (5, 4) 1

14. The range of the principal value branch of the function y = sec–1 x
is __________.
⎧π⎫
Ans: [ 0, π ] − ⎨ ⎬ 1
⎩2⎭
OR

−1 ⎛ 1 ⎞
The principal value of cos ⎜ − ⎟ is _____________.
⎝ 2⎠

2π
Ans: 1
3

15. The distance between parallel planes 2x + y − 2z − 6 = 0 and
4x + 2y − 4z = 0 ______ units.

Ans: 2 1

OR

If P(1,0,–3) is the foot of the perpendicular from the origin to the plane,
then the cartesian equation of the plane is ________
Ans: x – 3z = 10 1
Question numbers 16 to 20 are very short answer type questions

π
2

∫π x cos xdx .
16. Evaluate: 2

−
2

Ans: Let f(x) = x cos2 x f(–x) = – f(x) or f is odd 1/2
π/2

∫ x cos xdx = 0
2
∴ 1/2
− π/2

x −1 y + 4 z + 4
17. Find the coordinates of the point where the line = =
3 7 2
cuts the line xy-plane.
x −1 y + 4
Ans: Putting z = 0 in given equation gives = =2 1/2
3 7
Coordinates of required point are (7, 10, 0) 1/2

65/2/3 4

⎧⎪ kx 2 + 5, if x ≤1
18. Find the value of k, so that the function f (x) = ⎨
⎪⎩ 2 , if x >1

is continuous at x = 1.
Ans: L.H.L. is k + 5 1/2
getting k = –3 1/2

19. Form the differential equation representing the family of curves y = mx, where m is an
arbitrary constant.
dy
Ans: Differentiating to get =m 1/2
dx
dy y dy
required differential equation is = or x −y=0 1/2
dx x dx

20. Differentiate sec 2 (x 2 ) with respect to x2.
Ans: Let x2 = u, differentiating sec2 u w.r.t u, putting = x2 1/2
2 2 2
2sec x tan x 1/2
OR

2 dy
If y = f (x ) and f ′(x) = e x , then find .
dx

dy
Ans: = f ' ( x 2 ) 2x 1/2
dx
= 2xe x 1/2

SECTION-B

Question numbers 21 to 26 carry 2 marks.
r
21. Find a vector r equally inclined to the three axes and whose magnitude
is 3 3 units.
r
Ans: Let the vector r = aiˆ + ajˆ + akˆ 1/2

∴ 3a 2 = 3 3 1

required vector is 3iˆ + 3jˆ + 3kˆ or − 3iˆ − 3jˆ − 3kˆ 1/2

65/2/3 5

OR
r r r r
Find the angle between unit vectors a and b so that 3 a – b is also a
unit vector.
r r r r2
Ans: Using 3 a – b = 1 i.e. 3 a – b = 1 1/2

r r 3
& getting a · b = 1
2

π
getting angle or 30° 1/2
6

⎡ −3 2 ⎤ ⎡1 0 ⎤
22. If A = ⎢ ⎥ and I = ⎢ ⎥ , find scalar k so that A 2 + I = kA .
⎣ 1 −1⎦ ⎣0 1 ⎦

⎡ 11 −8⎤
Ans: A2 = ⎢ −4 3 ⎥ 1
⎣ ⎦

⎡12 −8⎤ ⎡ −3k 2k ⎤
A2 + I = kA ⇒ ⎢ −4 4 ⎥ = ⎢ k − k ⎥⎦ 1/2
⎣ ⎦ ⎣
k = –4 1/2

sec x − 1 ⎛π⎞
23. If f (x) = , find f ′ ⎜ ⎟ .
sec x + 1 ⎝3⎠

1
−1
cos x x
Ans: f(x) = = tan 1
1 2
+1
cos x

1 x
f ’(x) = sec 2 1/2
2 2
⎛π⎞ 2
f′⎜ ⎟ = 1/2
⎝3⎠ 3
OR

Find f ′ ( x ) if f ( x ) = (tan x) tan x .
Ans: Taking log on both sides. log f(x) = tan x log tan x 1/2

f '(x)
differentiating to get = sec2 x + sec2 x log tan x 1
f (x)
Thus, f ’(x) = (tan x)tan x · sec2 x(1 + log tan x) 1/2

65/2/3 6

tan 3 x
24. Find ∫ dx .
cos3 x

sin 3 x
Ans: Given Integral is I = ∫ dx 1/2
cos6 x
Put cos x = t
sin x dx = – d t 1/2
⎡ −1 1 ⎤
= ∫ ⎢ 6 + 4 ⎥ dt
⎣t t ⎦

t −5 t −3
= − +c 1/2
5 3
1 1 sec5 x sec3 x
−
= 5(cos x)5 3(cos x)3 + c or − +c 1/2
5 3

25.
r
() r
( )
Find the angle between the line r = 2iˆ − 3jˆ + λ kˆ and the plane r . ˆj − kˆ = 7
Ans: Let θ be angle between line & plane

(
ˆ ˆj − kˆ
k. )
∴ sin θ = ˆ | ˆj − kˆ |
|k| 1

−1
= 1/2
2

π
θ = 1/2
4
26. A and B throw a pair of dice alternately till one of them gets the sum of
the numbers as multiples of 6 and wins the game. If A starts first, find the
probability of B winning the game.
Ans: Let S denote Sum of numbers as multiples of six
6 1 5
P(S) = or , P (S) = 1
36 6 6
P(B winning) = P(A) P(B) + P(A)P(B) P(A).P(B) + …
5 1 5 5 5 1
= ⋅ + ⋅ ⋅ ⋅ +… 1/2
6 6 6 6 6 6
5
= 1/2
11

65/2/3 7

SECTION-C

Question numbers 27 to 32 carry 4 marks.

27. Solve the following differential equation:

(1 + e ) dy + e
y/x y/x ⎛ y⎞
⎜ 1 − ⎟ dx = 0 (x ≠ 0)
⎝ x⎠

dy e y/ x (y / x − 1)
Ans: Writing given differential equation as = 1
dx 1 + e y/ x

dy dv 1
Putting y = vx & =v+x
dx dx 2

dv − ( e +v )
v

to get x = 1
dx 1 + ev

ev + 1 dx
⇒ ∫ ev + v dv = − ∫ x
log e v + v = – log | x | + log c 1

y c 1
e y/x + = or x e y/x + y = c
x x 2

28. A cotton industry manufactures pedestal lamps and wooden shades. Both the
products require machine time as well as craftsman time in the making. The
number of hour(s) required for producing 1 unit of each and the corresponding
profit is given in the following table

Items Machine time Craftsman time Profit (in `)
Pedestal lamp 1.5 hours 3 hours 30
Wooden shades 3 hours 1 hour 20

In a day, the factory has availability of not more than 42 hours of machine time
and 24 hours of craftsman time.
Assuming that all items manufactured are sold, how should the manufacture
schedule his daily production in order to maximize the profit? Formulate it as
an LPP and solve it graphically.

65/2/3 8

Ans: Let number of pedestal lamps = x y
number of wooden shades = y 24

20 1
Maximize Profit Z = 30x + 20y
16 2
Subject to constraints 12
(4, 12)

8 1
1.5x + 3y ≤ 42 1
4 2
3x + y ≤ 24 x
O 4 8 12 14 16 20 24 28
x ≥ 0, y ≥ 0 correct graph 1

getting corners points & values of Z
(0, 0) 0
(8, 0) 240
(4, 12) 360
(0, 14) 280 1/2

Maximum profit = `360 where x = 4, y = 12 1/2

⎛ 1 ⎞
29. Find: ∫ ⎜⎜ cot x + ⎟ dx
⎝ cot x ⎟⎠

sin x + cos x
Ans: Writing given Integral as I = ∫ dx 1
sin x · cos x

Let sin x – cos x = t 1/2

1− t2
Squaring & getting sin x cos x = 1/2
2

dt
∴ I = 2∫ 1/2
1− t2
–1
= 2 sin t + C 1
–1
= 2 sin (sin x – cos x) + C 1/2

30. Check whether the relation R in the set N of natural numbers given by
R = {(a, b) : a is divisor of b}
is reflexive, symmetric or transitive. Also determine whether R is an
equivalence relation.
Ans: For reflexive
Let a ∈ N clearly a divides a ∴ (a, a) ∈ R
∴ R is reflexive 1

65/2/3 9

For symmetric
(1, 2) ∈ R but (2, 1) ∉ R 1
∴ R is not symmetric
For transitive
Let (a, b) , (b, c) ∈ R
∴ a divides b and b divides c
⇒ a divides c ∴ (a, c) ∈ R 1
R is transitive
As R is not symmetric ∴ It is not an equivalence relation 1

OR

−1 1 2 1 −1 ⎛ 4 ⎞
Prove that tan + tan −1 = sin ⎜ ⎟ .
4 9 2 ⎝5⎠

⎛ 1 2 ⎞
⎜ 4+9 ⎟ −1 −1 1
Ans: LHS = tan ⎜ ⎟ = tan 2
⎜ 1− 1 ·2 ⎟ 2
⎝ 4 9⎠

1 1
= · 2 tan −1 1
2 2

⎛ 1⎞
⎜ 2× ⎟
1
sin −1 ⎜ 2 = 1 sin −1 ⎛ 4 ⎞
= 2 1⎟ 2 ⎜ ⎟
⎝5⎠ 1
⎜ 1+ ⎟
⎝ 4⎠
= RHS

31. Find the cartesian equation of the plane passing through the intersection of the planes
x + 2y – 3 = 0 and 2x – y + z = 1 and the origin i.e. (0, 0, 0). Also write the equation of
the plane so obtained in vector form.
Ans: Required equation of plane is given by
x + 2y – 3 + λ(2x – y + z – 1) = 0 1
(1 + 2λ)x + (2 – λ)y + λz = 3 + λ … (i) 1
Putting (0, 0, 0) in (i) we get λ = – 3 1
Thus required equation of plane is 5x – 5y + 3z = 0 1/2
r
(
Vector equation of plane is r · 5 ˆi − 5 ˆj + 3kˆ = 0 ) 1/2

65/2/3 10

⎛y⎞ dy x + y
32. If tan −1 ⎜ ⎟ = log x 2 + y 2 , prove that = .
⎝x⎠ dx x − y
Ans: Differentiating both sides w.r.t. x to get
dy dy
x −y 2x + 2y
1 dx 1 dx 1 1
· = · 1 +1
y2 x2 x + y 2 x + y2
2 2 2 2 2
1+ 2
x

dy dy 1
Simplyfying we get x −y=x+y
dx dx 2
dy x+y 1
getting =
dx x − y 2
OR
d2 y dy 2
, −1 < x < 1 , then show that (1 − x ) 2 − x −a y = 0
a cos −1 x 2
If y = e
dx dx
-l
dy −ae a cos x
Ans: = 1
dx 1− x2
dy -l 1
1− x2 = −ae a cos x
dx 2
Differentiating again & getting
-l
d2 y x dy a 2 e a cos x
1− x −2
· = 2
dx 2 1 − x 2 dx 1− x2

d2 y dy 2
(1 − x 2 )
1
2
−x −a y = 0
dx dx 2

SECTION-D

Question numbers 33 to 36 carry 6 marks.

33. Amongst all open (from the top) right circular cylindrical boxes of volume
125π cm3 , find the dimensions of the box which has the least surface area.
Ans: Let radius = r & height = h
125
πr 2 h = 125π or h=
r2
1

2
Surface Area, S = 2 π r h + π r
250 π
S= + π r2 1
r

65/2/3 11

dS −250 π
= + 2πr 1
dr r2
dS
=0 ⇒ r=5 1/2+1
dr

d 2S 500π
= + 2π > 0 so s is least 1
dr 2 r3

when r = 5 cm , gives h = 5 cm 1/2

34. Using integration, find the area lying above x-axis and included
between the circle x 2 + y 2 = 8x and inside the parabola y 2 = 4x .

Ans. Correct figure 1
x-coordinate of point of intersection is 4, 0 1

O 4 8

4 8

Required Area = ∫ 2 x dx + ∫ 16 − (x − 4) dx
2
2
0 4

4 8
4 32 x−4 −1 x − 4
= 3 x + 2 16 − ( x − 4) + 8sin
2

4 4 1
0

32
= + 4π 1
3
OR
Using the method of integration, find the area of the ΔABC, coordinates of
whose vertices are A(2, 0), B(4, 5) and C(6, 3).
B
Ans. Correct figure 5 1
4
5
Equation of AB : y = (x − 2) 3 1/2
2
2 C
Equation of BC : y = 9 – x 1/2
1
3 A
Equation of AC : y = (x − 2) 1/2
4 O 1 2 3 4 5 6

5 4 6 3 6
Required Area = ∫
2 2
(x − 2) dx + ∫ (9 − x) dx − ∫ (x − 2) dx
4 4 2
2

5 (x − 2) 2 4 (9 − x) 2 6 3 (x − 2) 2 6
= + − 1
2 2 2
−2 4 4 2 2

=7 1/2

65/2/3 12

⎡ 5 −1 4 ⎤
35. A = ⎢⎢ 2 3 5 ⎥⎥ , find A–1 and use it to solve the following system of equations:
⎢⎣ 5 −2 6 ⎥⎦

5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Ans: |A| = 51 1
A11 = 28 A12 = 13 A13 = −19
Cofactors: A 21 = −2 A 22 = 10 A 23 = 5 2
A 31 = −17 A 32 = −17 A 33 = 17

28 −2 −17 ⎤
1 ⎡⎢
A −1 = 13 10 −17 ⎥ 1
51 ⎢ −19 5 17 ⎥
⎣ ⎦

⎡x⎤ ⎡ 5⎤ 1
Given system is AX = B where X = y B = ⎢ 2 ⎥
⎢ ⎥
⎢z⎥ ⎢ −1⎥ 2
⎣ ⎦ ⎣ ⎦

28 −2 −17 ⎤ ⎡ 5 ⎤ ⎡ 3 ⎤
1 ⎡⎢
⇒X=A B= 13 10 −17 ⎥ ⎢ 2 ⎥ = ⎢ 2 ⎥
−1
1
51 ⎢ −19 5 17 ⎥ ⎢ −1⎥ ⎢ −2 ⎥
⎣ ⎦⎣ ⎦ ⎣ ⎦

1
x = 3, y = 2, z = –2
2
OR

x x 2 1 + x3
If x, y, z are different and y y 2 1 + y3 = 0 , then using properties of determinants
z z2 1 + z3

show that 1 + xyz = 0.

x x2 1 x x2 x3
Ans: Writing L.H.S as y y2 1 + y y2 y3 = 0 1
z z2 1 z z2 z3

x x2 1 1 x x2
⇒ y y 2 1 + xyz 1 y y2 = 0
1
2 2
z z 1 1 z z

65/2/3 13

getting

x x2 1
(1 + xyz) y y2 1 = 0
2
z z2 1

Applying R2 → R2 – R1, R3 → R3 – R1

x x2 1
(1 + xyz) y − x (y + x)(y − x) 0 = 0 1
z − x (z + x)(z − x) 0

Expanding and simplifying to get
⇒ (1 + xyz) (x – y) (y – z) (z – x) = 0 1/2

As x, y, z are different 1/2

∴ 1+ xyz = 0
36. A discrete random variable X has the following probability distribution:

x 0 1 2 3 4 5

P(X) 4C2 3C2 2C2 C2 C 2C

(a) Find the value of C.

(b) Find the mean of the distribution.

2
(c) Given Σ x i pi = 14 , find the variance of the distribution.

1
Ans. (a) Σ pi = 1 ⇒ 10C2 + 3C – 1 = 0 1
2

1 1
gives C = , − (rejected) 1
5 2
(b) For mean
Σ x i pi = 0 + 3C2 + 4C2 + 3C2 + 4C + 10C 1
= 10C2 + 14C
1 16
Putting C = we get mean = 1
5 5

(c) Variance = ∑ x i2 pi − (mean)2
256
= 14 − 1
25
94
= 1/2
25

65/2/3 14