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SECTION A
This section comprises multiple choice questions (MCQs) of 1 mark each.
1. If x = at, y a , dy
is :
= dx
then
t
(A) t2 (B) – t2

(C) 1 1
(D) –
t2 t2
2. The solution of the differential dy 1 is
equation : log y
=
dx
(A) log y = x + c (B) y log y – y = x + c
(C) log y – y = x + c (D) y log y + y = x + c

3. The vector with terminal point A (2, – 3, 5) and initial point B (3, –
4, 7) is :
^ ^ ^ ^ ^
(A) i – j +2 (B) i + j +2k
^ ^ ^ ^
k (D) – i + j – 2 k
^ ^
(C) – i – j –
^
2k

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1. The distance of point P(a, b, c) from y-axis
is :
(A) b (B) b2
2
(C) a + (D) a2 + c2
2
c
2. The number of corner points of the feasible region
determined by constraints x ³ 0, y ³ 0, x + y ³ 4 is :
(A) 0 (B) 1
(C) 2 (D) 3

3. If matrices A and B are of order 1 ´ 3 and 3 ´ 1 respectively, then

the order of A'B' is :
(A) 1´1 (B) 3´1
(C) 1´3 (D) 3 ´ 3

4. A relation R defined on a set of human

beings as R = {(x, y) : x is 5 cm
shorter than y}
is :
(A) reflexive only
(B) reflexive and transitive
(C) symmetric and transitive
(D) neither transitive, nor symmetric, nor reflexive

5. If a matrix has 36 elements, the number of possible orders it can

have, is :
(A) 13 (B) 3
(C) 5 (D) 9

6. Which of the following statements is true for the function

f(x) ìï x 2 + 3, x s ?
= 0í
ïî 1 , x =
0
(A) f(x) is continuous and differentiable ∀ x ϵ ℝ
(B) f(x) is continuous ∀ x ϵ ℝ
(C) f(x) is continuous and differentiable ∀ x ϵ ℝ – {0}
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(D) f(x) is discontinuous at infinitely many points

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4. Let f(x) be a continuous function on [a, b] and differentiable on (a, b).
Then, this function f(x) is strictly increasing in (a, b) if
(A) f '(x) < 0, ∀ x ϵ (a, b)
(B) f '(x) > 0, ∀ x ϵ (a, b)
(C) f '(x) = 0, ∀ x ϵ (a, b)
(D) f(x) > 0, ∀ x ϵ (a, b)
「x + y 2 「6 2ù ( 24 24 ⎞
ù
5. If = , then the value of | + | is :
| xy]| 8] x y
L 5 ⎝ ⎠
| 5 |
L
(A) 7 (B) 6
(C) 8 (D) 18
π/ 2
6. If f(x) is an odd function,
then ∫ f (x) cos x
dx
3
equals :
– π/
2
π/ 2
(A) 3
f (x) cos x (B) 0
2
dx

∫ 0
π/ 2

π/
2
(C)
(D)
2
∫ f (x) 2
∫
0
cos3 x dx

dx
0

7. Let θ be the angle between two unit vectors a^ such that

and
^
b
sin θ 3
. Then, .
is equal to :
= ^a ^
5 b

(A) ± 3 3
(B) ±
5 4
(C) ±
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4 4
(D) ±
5 3
8. The integrating factor of the differential equation (1 dy
+ xy = ax,
– x 2) dx
– 1 < x < 1, is :

(A) 1 1
(B)
x2 – 1
x2 –
(C) 1 11
(D)
2 2
1–x 1–x

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3 3
If the direction cosines of a line 3 k, k, k, then the value of k
are is :
(E) ±1 (B) ± 3
1
(C) ±3 (D) ±
3

7. A linear programming problem deals with the optimization of a/an :

(A) logarithmic function (B) linear function
(C) quadratic function (D) exponential function

8. If P(A|B) = P(A'|B), then which of the following statements is true ?

(A) P(A) = P(A') (B) P(A) = 2 P(B)

(C) P(A 3 B) 1
= P(B) (D) P(A 3 B) = 2 P(B)
2

x +1 x–1
9. is equal to :
2 2
x +x +1 x – x +
1
(A) 2x3 (B) 2
(C) 0 (D) 2x3 –
2

Questions number 19 and 20 are Assertion and Reason based questions.

Two statements are given, one labelled Assertion (A) and the other labelled
Reason
(R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is
the correct explanation of the Assertion (A).

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not
the correct explanation of the Assertion (A).

(C) Assertion (A) is true, but Reason (R) is false.

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(D) Assertion (A) is false, but Reason (R) is true.

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「 1 cos θ 1 ù
9. Assertion (A) : For matrix A | |
|– cos θ 1 cos θ| , where θ ∊ [0, 2π],
=
|L – – cos 1 |]
|A| ∊ [2, 1 θ
4].
Reason (R) : cos θ ∊ [– 1, 1], ∀ θ ∊ [0, 2p].

10. Assertion (A) : A line in space cannot be drawn perpendicular to x,

y and z axes simultaneously.

Reason (R) : For any line making angles, α, β, y with the positive
directions of x, y and z axes
respectively, cos2 α + cos2 β + cos2 y = 1.

SECTION B

This section comprises very short answer (VSA) type questions of 2 marks each.

→
11. In the given figure, ABCD is a parallelogram. If = 2 ^i – 4 ^j and
AB
^
+5k
→ ^ ^ ^
DB = 3 i – 6 j + 2 k , then AD and hence find the area of
→
find

parallelogram ABCD.

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10. (a) Check the differentiability of function f(x) = [x] at x = – 3,

where [.] denotes greatest integer function.

OR
(b)
find If x1/3 + y1/3 = 1, dy
at the (1 1 ⎞
point , .
| |
dx ⎝ 8 8⎠

11. Find local maximum value and local minimum value (whichever exists)
for the function f(x) = 1
(x s 0).
2
4x + x

12. (a) Find :

1+
∫ x 2x dx

OR
(b) Evaluate
:
π2 sin x
dx
∫
4
0 x
→ → →
13. If → and are two non-zero vectors such that →
+ ) z
( and
a b a b a
(2 + ) z → , then prove that |→ | =
→ → | → |.
a b b b 2 a

SECTION C

This section comprises short answer (SA) type questions of 3 marks each.

14. Solve the following linear programming problem graphically :

Minimise z = 5x – 2y
subject to the
constraints
x + 2y £ 120
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x + y ³
60 x –
2y ³ 0 x,
y?0

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12. –
E and F are two independent events such that P( E ) = 0·6 and
– –
E 4F ).
P(E 4 F) = 0·6. Find P(F) and P(

13. (a) A relation R on set A = {1, 2, 3, 4, 5} is

defined as R = {(x, y) : |x2 – y2| < 8}. Check whether the

relation R is reflexive, symmetric and transitive.

OR

(b) A function f is defined from R → R as f(x) = ax + b, such

that f(1) = 1 and f(2) = 3. Find function f(x). Hence, check

whether function f(x) is one-one and onto or not.

14. (a) If 2
1– 1–y
+ 1– = a (x – y), prove dy
x
2
y
2
that dx = 1 – x2 .

OR

(b) If y = (tan x)x, then dy

find dx .

15. (a) Find :

x2
∫ 2 2
(x + 4) (x +
dx

9)
OR

(b) Evaluate :
3

∫ 1
(|x – 1|+|x – 2|+|x – 3|) dx

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16. Solve the following differential equation :

(tan–1 y – x) dy = (1 + y2) dx

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SECTION D

This section comprises long answer type questions (LA) of 5 marks each.

15. Find the equation of a line l2 which is the mirror image of the line l1 with
z–2
respect to line l : y– = , given that line l passes through the
x =
1 3
1
2
1
point P(1, 6, 3) and parallel to line l.

「1 – 0ù
16. (a) If 2 |
– 1 , find A–1 and use it to solve the following
| |
A= 2 –1
| 1 |]
|L 0 – 2
system of equations :
x – 2y = 10, 2x – y – z = 8, – 2y + z = 7

OR

「– 1 a 2ù 「 1 –1 1ù
(b) If | x| | |
A = | 1 2 | and A–1 = |– 8 7 – 5| ,
|L 3 1 |L b y 3 |]

1|]

find the value of (a + x) – (b + y).

17. (a) Find :

(3 cos x – 2) x
∫ sin 2
d
5 – sin x – 4 cosxx

OR
(b) Evaluate
: 3
2 x +|x|+ 1
∫ 2
x + 4|x|+ 4
dx

–2
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666
2
2 y
18. Using integration, find the area of the x = 1, included
4
+
ellipse between the lines x = – 2 and x = 2. 16

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SECTION E

This section comprises 3 case study based questions of 4 marks each.

Case Study – 1

17. If a function f : X → Y defined as f(x) = y is one-one and onto, then

we can define a unique function g : Y → X such that g(y) = x,
where x ϵ X and y = f(x), y ϵ Y. Function g is called the inverse of
function f.

The domain of sine function is R and function sine : R → R is

neither one-one nor onto. The following graph shows the sine
function.

Let sine function be defined from set A to [– 1, 1] such that inverse

of sine function exists, i.e., sin–1 x is defined from [– 1, 1] to A.

On the basis of the above information, answer the following questions :

(i) If A is the interval other than principal value branch, give an

example of one such interval. 1
(ii) If sin–1 (x) is defined from [– 1, 1] to its principal value branch, find
1⎞
the value of sin–1 – sin–1 (1). 1
( | – 2|
⎝ ⎠
(iii) (a) Draw the graph of sin–1 x from [– 1, 1] to its principal value
branch.

2
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666
OR
(iii) (b) Find the domain and range of f(x) = 2 sin–1 (1 – x). 2

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Case Study –
2
19. The traffic police has installed Over Speed Violation Detection
(OSVD) system at various locations in a city. These cameras can
capture a speeding vehicle from a distance of 300 m and even
function in the dark.

RADAR SPEED DETECTION

AVERAGE SPEED DETECTION
Distance RADAR RADAR measures the change
Speed
= Time B – Time in the frequency of returned
POINT A POINT B
A radio waves to precisely
measure the speed of vehicles
TIME TIME (the Doppler effect)
A B
Radio waves
emitted by the
RADAR bounce back
to confirm an object
was detected

A camera is installed on a pole at the height of 5 m. It detects a car

travelling away from the pole at the speed of 20 m/s. At any point, x
m away from the base of the pole, the angle of elevation of the
speed camera from the car C is θ.

On the basis of the above information, answer the following questions :

(i) Express θ in terms of height of the camera installed on the pole

and x. 1

(ii) Find
dθ
. 1
dx

(iii) (a) Find the rate of change of angle of elevation with respect
to time at an instant when the car is 50 m away from the
pole. 2

OR

(iii) (b) If the rate of change of angle of elevation with respect to

time of another car at a distance of 50 m from the base

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of the pole
3
is rad/s, then find the speed of the car. 2
101

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Case Study –
2

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Case Study –
3

18. According to recent research, air turbulence has increased in

various regions around the world due to climate change. Turbulence
makes flights bumpy and often delays the flights.

Assume that, an airplane observes severe turbulence, moderate

turbulence or light turbulence with equal probabilities. Further, the
chance of an airplane reaching late to the destination are 55%, 37%
and 17% due to severe, moderate and light turbulence respectively.

On the basis of the above information, answer the following questions :

(i) Find the probability that an airplane reached its destination late. 2

(ii) If the airplane reached its destination late, find the probability
that it was due to moderate turbulence. 2

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.
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.
2 “Evaluation policy is a confidential policy as it is related to the confidentiality of the
examinations conducted, Evaluation done and several other aspects. Its’ leakage to
public in any manner could lead to derailment of the examination system and affect
the life and future of millions of candidates. Sharing this policy/document to anyone,
publishing in any magazine and printing in News Paper/Website etc may invite
action under various rules of the Board and IPC.”
3 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 due marks be
awarded to them.
4 The Marking scheme carries only suggested value points for the answers
These are in the nature of Guidelines only and do not constitute the complete answer. The
students can have their own expression and if the expression is correct, the due marks
should be awarded accordingly.
5 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. If there is any variation, the same should be
zero after deliberation and discussion. 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.
6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer CROSS ‘X” be
marked. Evaluators will not put right (✓)while evaluating which gives an impression that
answer is correct and no marks are awarded. This is most common mistake which
evaluators are committing.

7 If a question has parts, please award marks on the right-hand side for each part. Marks

65 /1/3 24 P.T.O.
 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.
8 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.

9 In Q1-Q20, if a candidate attempts the question more than once (without canceling the
previous attempt), marks shall be awarded for the first attempt only and the other answer
scored out `with a note “Extra Question”.

10 In Q21-Q38, if a student has attempted an extra question, answer of the question deserving
more marks should be retained and the other answer scored out with a note “Extra
Question”.

11 No marks to be deducted for the cumulative effect of an error. It should be penalized only once.

12 A full scale of marks (example 0 to 80/70/60/50/40/30 marks as given in

Question Paper) has to be used. Please do not hesitate to award full marks if the
answer deserves it.
13 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).This is in view of the reduced
syllabus and number of questions in question paper.
14 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 an answer.
● 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/not same.
● 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.
15 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.
16 Any un assessed 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.

65 /1/3 2 P.T.O.
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Case Study –
3

65 /1/3 3 P.T.O.
17 The Examiners should acquaint themselves with the guidelines given in the “Guidelines
for spot Evaluation” before starting the actual evaluation.
18 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.
19 The candidates are entitled to obtain photocopy of the Answer Book on request on payment
of the prescribed processing fee. All Examiners/Additional Head Examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out
strictly as per value points for each answer as given in the Marking Scheme.

65 /1/3 4 P.T.O.
 MARKING SCHEME
MATHEMATICS (Subject Code–041)
(PAPER CODE: 65/1/3)

Q.No. EXPECTED ANSWER / VALUE POINTS Marks

SECTION-A
(Question nos. 1 to 18 are Multiple choice Questions carrying 1 mark each)

1.

Ans (D) −1
𝑡2
1

2.

Ans (B) y log y − y = x + c 1

3.

Ans (D) −𝜾̂ + 𝗝̂ − 𝟐𝒌̂ 1

4.

Ans (C) √𝒂𝟐 + 𝒄𝟐 1

5.

Ans (C) 2 1

65 /1/3 4 P.T.O.
6.

Ans (D) 3 x 3 1

7.

Ans (D) Neither transitive, nor symmetric, nor reflexive 1

8.

Ans (D) 9 1
9.

Ans (C) f(x) is continuous and differentiable ∀ 𝒙 ∈ 𝑹 − {𝟎} 1

10.

Ans (B) f ′(x) > 0, ∀ x ∈ (a, b) 1

11.

Ans (D) 18 1

65 /1/3 5 P.T.O.
12.

Ans (B) 0 1

13.

𝟒
(C) ±
𝟓
Ans 1

14.

𝟏
√𝟏− 𝒙𝟐
Ans (D) 1

15.

𝟏
(D) ±
𝟑
Ans 1

16.

65 /1/3 6 P.T.O.
Ans (B) linear function 1

17.

𝟏
(C) P(A ∩ 𝑩) = 𝑷 (𝑩)
𝟐
Ans 1

18.

Ans (B) 2 1

(Question Nos. 19 & 20 are Assertion-Reason based questions of 1 mark each)

19.

Ans (A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of 1
Assertion (A).

20.

Ans (A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of 1
Assertion (A).
SECTION-B
(Question nos. 21 to 25 are very short Answer type questions carrying 2 marks each)

65 /1/3 7 P.T.O.
 21.

Ans 𝑨⃗⃗⃗⃗𝑫⃗⃗→ + 𝑫⃗⃗⃗⃗𝑩⃗⃗→ = 𝑨⃗⃗⃗⃗𝑩⃗⃗→

𝑨⃗⃗⃗⃗𝑫⃗⃗→ = (2 𝜾̂ − 𝟒 𝗝̂ + 𝟓 𝒌̂) − (3 𝜾̂ − 𝟔 𝗝̂ + 𝟐 𝒌̂)
𝟏
= − 𝜾̂ + 𝟐𝗝̂ + 𝟑 𝒌̂
𝜾̂ 𝗝̂ 𝒌̂ 𝟐
𝑨⃗⃗⃗𝑫⃗⃗⃗⃗→ × 𝑨⃗⃗⃗⃗𝑩⃗⃗→ = |−𝟏 𝟐 𝟑| = 𝟐𝟐 𝜾̂ + 𝟏𝟏𝗝̂ 𝟏
𝟐 −𝟒 𝟓
Area = |𝑨⃗⃗⃗⃗𝑫⃗⃗→ × 𝑨⃗⃗⃗⃗𝑩⃗⃗→| = |𝟐𝟐 𝜾̂ + 𝟏𝟏 𝗝̂ | 𝟏
= √𝟔𝟎𝟓 𝒐𝒓 𝟏𝟏 √𝟓 𝟐

22(a).

Ans f(x) = [x] at x = −3

RHD = lim
𝑓(−3+ℎ)
−𝑓(−3) ℎ→0 ℎ
𝟏
𝟐
= ℎ→0
lim =0
−3−(−3)
ℎ

lim
𝑓(−3−ℎ)−𝑓(−3)
LHD = ℎ→0 −ℎ

= lim ( )
= lim
−4−(−3) −1
ℎ→0 ℎ ℎ→0 ℎ 𝟏
𝟐

𝟏
= not defined

∵ LHD ≠ RHD 𝟐

𝟏
So f is not differentiable at x = −3
𝟐

OR

22(b).

65 /1/3 8 P.T.O.
65 /1/3 9 P.T.O.
 −𝟐 −𝟐
𝒙𝟑 + 𝒚 =𝟎
𝟏 𝟏 𝒅𝒚
𝟑
Ans
𝟑 𝟑 𝒅𝒙

−𝟐
𝟏
𝒅𝒚 −𝒙𝟑 𝟏
𝒅𝒙 −𝟐 𝟐
𝒚𝟑
=

𝒅𝒚 −𝟒
( 𝒙) = = −𝟏 𝟏
𝒅 (𝟏 ,𝟏 ) 𝟒
𝟖 𝟖 𝟐

23.

f(x) = 4 𝑥2 + ( x ≠ 0)
1

𝑥
Ans

𝑓′ (x) = 8x −
1
=0 1
𝑥2
2
⇒ 𝑥 = ⇒𝑥 =
3 1 1
1
8 2
2
𝑓 ( 𝑥) = 8 + > 0 at x =
′′ 2 1
1
𝑥3 2
2
∴ Local minimum value = f ( ) = 3 1
1
2
2

24(a).

1 + 2x = 𝒕𝟐 𝟏
2 𝒅𝒙 = 2t 𝒅𝒕 𝟐
Sol.

∫(𝒕𝟒 − 𝒕𝟐)𝒅𝒕 = [ − ]+𝑪

𝟏 𝟏 𝒕𝟓 𝒕𝟑
𝟐 𝟐 𝟓 𝟑 𝟏
𝟏
𝟓 𝟑

− +𝑪
( 𝟏+𝟐𝒙)𝟐 ( 𝟏+𝟐𝒙)𝟐
= 𝟏𝟎 𝟔 𝟐
OR
24(b).

𝝅𝟐𝒔𝒊𝒏√𝒙
∫𝟎𝟒 𝒅𝒙 Put √𝒙 = 𝒕 ⇒ 𝒅𝒙 = 𝟐𝒕 𝒅𝒕
Sol.
𝟏
√𝒙
𝟐
2 ∫ 𝐬𝐢𝐧 𝒕 𝒅𝒕 = 𝟐 [− 𝐜𝐨𝐬 𝒕]𝟐
𝝅 𝝅
𝟐
𝟎 𝟎
𝟏
65 /1/3 1 P.T.O.
0
65 /1/3 1 P.T.O.
1
 𝟏
=2 𝟐

25.

(𝒂⃗→ + 𝒃⃗→). 𝒂⃗→ = 𝟎 ⇒ |𝒂⃗→|𝟐 + 𝒃⃗→. 𝒂⃗→ = 𝟎---(1)

𝟏
𝟐
Sol.
𝟐
(𝟐𝒂⃗→ + 𝒃⃗→). 𝒃⃗→ = 𝟎 ⇒ 𝟐𝒂⃗→. 𝒃⃗→ + |𝒃⃗→|------= 𝟎 𝟏
𝟐
(2)

2 (-|𝒂⃗→|𝟐) + |⃗𝒃→| = 𝟎 {Using (1) and (2)}

𝟐
𝟏
𝟐
|⃗𝒃→| = 2|𝒂⃗→|𝟐 ⇒ |⃗𝒃→| = √𝟐|𝒂⃗→|
𝟐
𝟏
𝟐
SECTION-C
(Question nos. 26 to 31 are short Answer type questions carrying 3 marks each)

26.

Ans Min z = 5 x -2 y

Correct
graph-
𝟏𝟐
𝟏

Corner Points Z = 5x −2y Correct

𝟏
A(60, 0) 300 table-
B(40, 20) 160
C( 60, 30) 240
D(120, 0) 600
𝟏
Min Z = 160 at x = 40, y = 20 𝟐

65 /1/3 10 P.T.O.
 27.

P (𝑬̅) = 0.6 ⇒ 𝑷(𝑬) = 0.4 𝟏

𝟐
Sol.

P (E ∪ F) = P(E) + P(F) – P(E∩F) 𝟏

⇒ 0.6 = 0.4 +P(F) -0.4 P(F) ⇒ P(F) = 𝟏 𝟐
𝟑 𝟏
𝟏
P (𝑬̅ ∪ 𝑭̅) = 𝟏 − 𝑷 (𝑬 ∩ 𝑭)
𝟐
𝟏
𝟏 𝟏𝟑
= 1 – 0.4 × =
𝟑 𝟏𝟓 𝟐

28(a).

∵ |𝒙𝟐 − 𝒙𝟐| < 𝟖 ∀ 𝒙 ∈ 𝑨 ⇒ (𝒙, 𝒙) ∈ 𝑹 ∴ 𝑹 is reflexive . 𝟏

Ans (a) Reflexive:

𝟐
Let (x,y) ∈ 𝑹 for some x,y ∈ A
(b)Symmetric:

∴ |𝒙𝟐 − 𝒚𝟐| < 𝟖 ⇒ |𝒚𝟐 − 𝒙𝟐| < 𝟖 ⇒ (𝒚, 𝒙) ∈ 𝑹

Hence R is symmetric. 𝟏

(1,2) , (2,3) ∈ 𝑹 as |𝟏𝟐 − 𝟐𝟐| < 𝟖 , |𝟐𝟐 − 𝟑𝟐| < 𝟖

(c)Transitive:

respectively But |𝟏𝟐 − 𝟑𝟐| 𝕍 𝟖 ⇒ (𝟏, 𝟑) ∉ 𝑹 𝟏

𝟏
Hence 𝑹 is not transitive. 𝟐

OR

28(b).

𝟏
Ans f(x) = ax + b

= −1 f(x) = 2 x – 1
Solving a+ b = 1 and 2a + b = 3 to get a=2 , b

Let f (𝒙𝟏) = f (𝒙𝟐) for some 𝒙𝟏 , 𝒙𝟐 ∈ 𝑹

2 𝒙𝟏 − 𝟏 = 2 𝒙𝟐 − 𝟏 ⇒ 𝒙𝟏 = 𝒙𝟐
𝟏
Let y = 2x -1, y ∈ 𝑹 (Codomain)
Hence f is one – one.

⇒ x = 𝒚+𝟏 ∈ 𝑹 (domain)
𝟐
Also, f(x) = f (𝒚+𝟏) = y
𝟐
∴ f is onto. 𝟏

29(a).

65 /1/3 11 P.T.O.
65 /1/3 12 P.T.O.
Ans √𝟏 − 𝒙𝟐 + √𝟏 − 𝒚𝟐 = 𝒂(𝒙 − 𝒚)
𝟏
Put x = sin 𝜽 , y = sin 𝝓
𝟐
⇒ cos 𝜽 + cos 𝝓 = a (sin 𝜽 – sin 𝝓 )
⇒ 2 cos (𝜽+ 𝝓) cos (𝜽− 𝝓) = 2 a sin (𝜽− 𝝓) cos (𝜽+ 𝝓) 𝟏
𝟐 𝟐 𝟐 𝟐
𝟐
⇒ cot (𝜽− 𝝓) = a
𝟐
⇒ 𝜽 − 𝝓 = 2 𝐜𝐨𝐭−𝟏 𝒂
⇒ 𝐬𝐢𝐧−𝟏 𝒙 − 𝐬𝐢𝐧−𝟏 𝒚 = 𝟐 𝐜𝐨𝐭−𝟏 𝒂 𝟏
𝟐
⇒ √𝟏− 𝒙𝟐− √𝟏− 𝒚𝟐 𝒅𝒙=
𝟏 𝟏 𝒅𝒚

𝟏
0

√ 𝒙𝟐
𝟏
⇒ 𝒅𝒙 = 𝟏−
𝒅𝒚 𝟏− 𝒚𝟐
𝟐
OR

y = (𝐭𝐚𝐧 𝒙)𝒙
29(b).

𝟏
Ans

𝟐
log y = x log (tan x)
𝟐

=( x 𝐭𝐚𝐧
) + 𝐥𝐨𝐠(𝐭𝐚𝐧 𝒙)
𝟏 𝒅𝒚 𝒔𝒆𝒄 𝒙
𝒚 𝒅𝒙 𝒙 𝟐
𝒙 𝒔𝒆𝒄𝟐𝒙
= (𝐭𝐚𝐧 𝒙) [ + 𝐥𝐨𝐠(𝐭𝐚𝐧 𝒙)]
( )
𝒅𝒚 𝒙

𝒅𝒙 𝐭𝐚𝐧 𝒙 𝟏
𝟐

30(a).

𝒅𝒙
𝒙𝟐
Sol. Let I = ∫
(𝒙𝟐+𝟒)(𝒙𝟐+𝟗) 𝟏
𝟐
Put 𝒙𝟐 = 𝒕

= + ⇒A=
𝒕 𝑨 𝑩 −𝟒
𝟗
(𝒕+𝟒)(𝒕+𝟗) 𝒕+ 𝟒 𝒕+𝟗
, B
𝟓 𝟓
=
𝟏
𝟏
I = −𝟒 ∫ 𝒅𝒙 + 𝒅𝒙
𝟏 𝟗 𝟏
∫ 𝟐
𝟓 𝟐𝟐+ 𝒙𝟐 𝟓 𝟑𝟐+ 𝒙𝟐
−𝟐 𝒙 𝟑
𝐭𝐚𝐧−𝟏 ( ) + 𝐭𝐚𝐧−𝟏 ( ) + 𝑪
𝒙
𝟏
𝟓 𝟐 𝟓 𝟑
=
OR

30(b).

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

65 /1/3 12 P.T.O.
65 /1/3 13 P.T.O.
 = ∫ 𝟐 𝒅𝒙 + ∫ (𝟐 − 𝒙) 𝒅𝒙 + ∫ (𝒙 − 𝟐)𝒅𝒙
𝟑 𝟐 𝟑

𝟏 𝟏 𝟐
𝟐 𝟐 𝟐 𝟑
= [𝟐𝒙] + [ ] + [ ]
𝟑 (𝟐−𝒙) (𝒙−𝟐)

𝟏 −𝟐 𝟐
𝟏 𝟐 𝟏
+
𝟏 𝟏 𝟏
𝟐
𝟐 𝟐
=4+ = 5

31.

Sol. (tan−1 𝑦 − 𝑥 )𝑑𝑦 = (1 + 𝑦2) 𝑑𝑥

𝟏
+1+ 𝑦2 𝑥 = 1+ 𝑦2
𝑑𝑥 1 tan−1 𝑦
𝑑𝑦

1
1
∫ 𝑑𝑦
I. F = 𝑒 = 𝑒tan 𝑦
−1
1+ 𝑦2
2

𝟏
−1

𝑥 × 𝑒 ∫ 𝑦2 𝑒tan 𝑑𝑦
tan−1 𝑦 tan 𝑦 −1
𝑦
= 1+
1
⇒ 𝑥𝑒tan 𝑦
= (tan−1 𝑦) 𝑒tan 𝑦
− 𝑒tan 𝑦
+𝐶
−1 −1 −1
2

OR

⇒ 𝑥 = tan−1 𝑦 − 1 + 𝐶 𝑒− 𝑦
−1
tan

SECTION-D
(Question nos. 32 to 35 are Long Answer type questions carrying 5 marks each)

32.

𝑦−1 𝑧−2
D ratios of the line 𝑙 i.e. 𝑥 = = 1
1 2 3 2
Sol. are 1, 2, 3
1
Let coordinates of foot of perpendicular M on line 𝑙 be (𝜆, 2𝜆 + 1, 3𝜆 + 2) 2

D.ratios of PM are 𝜆 − 1, 2𝜆 − 5, 3𝜆 − 1 1
2

1 (𝜆 − 1) + 2 (2𝜆 − 5) + 3(3𝜆 − 1) = 0 (∵ PM ⊥ 𝑙) 1
2

⇒ 𝜆=1 1
2
Coordinates of M are (1, 3, 5) 1

Since M is midpoint of PQ ∴ Coordinates of Q are (1, 0, 7) 1

65 /1/3 14 P.T.O.
 𝑧−7
Equation of line 2𝑙 is 𝑥−1 = 𝑦 =
1 2 3 1
2

33(a).

|A| = 1 ≠ 𝟎 hence 𝑨−𝟏 exists. 𝟏

−𝟑 𝟐 𝟐
Ans

𝑨𝒅𝒋 𝑨 = [−𝟐 𝟏 𝟐
𝟏]
−𝟒 𝟐 𝟑 𝟏
−𝟑 𝟐 𝟐
𝑨−𝟏 = [−𝟐 𝟏 𝟐
𝟏]
−𝟒 𝟐 𝟑
𝟏 −𝟐 𝟎 𝒙 𝟏𝟎
AX = B ⇒ [𝟐 −𝟏 −𝟏] [𝒚] = [ 𝟏
𝟏
𝟖] 𝟐
𝟎 −𝟐 𝟏 𝒛 𝟕
𝒙 −𝟑 𝟐 𝟐 𝟏𝟎 𝟎
𝒚
X = 𝑨−𝟏𝑩 ⇒ [ ] = [−𝟐 𝟏 𝟏] [
𝟖 ] = [−𝟓]
𝒛 −𝟒 𝟐 𝟑 𝟕 −𝟑

⇒ 𝒙 = 𝟎, 𝒚 = −𝟓, 𝒛 = −𝟑
OR

33(b).

Ans 𝑨𝑨−𝟏 = 𝑰 𝟏
−𝟏 𝒂 𝟐 𝟏 −𝟏 𝟏 𝟏 𝟎 𝟎
[ 𝟏 𝟐 𝒙] [−𝟖 𝟕 −𝟓] =
[𝟎 𝟏 𝟎]
𝟑 𝟏 𝟏 𝒃 𝒚 𝟑 𝟎 𝟎 𝟏 𝟏
−𝟏 − 𝟖𝒂 + 𝟐𝒃 𝟏 + 𝟕𝒂 + 𝟐𝒚 𝟓 − 𝟓𝒂 𝟏
𝟐
𝟏 𝟎𝟎 [ −𝟏𝟓 + 𝒃𝒙 𝟏𝟑 +
𝒙𝒚 𝟑𝒙 − 𝟗] = [𝟎 𝟏 𝟎] 𝟏
−𝟓 + 𝒃 𝟒+𝒚 𝟏 𝟎 𝟎 𝟏
𝟏
-5 + b = 0 ⇒ 𝒃 = 𝟓 , 𝟓 − 𝟓𝒂 = 𝟎 ⇒ 𝒂 = 𝟏
4 + y = 0 ⇒ 𝒚 = −𝟒 , 𝟑𝒙 − 𝟗 = 𝟎 ⇒ 𝒙 = 𝟑
𝟏
𝟐

65 /1/3 14 P.T.O.
∴ (a + x) – (b + y) = ( 1 + 3) – (5 - 4) = 3

65 /1/3 15 P.T.O.
34(a).

Put cos x = t so that, −sin x dx = dt

(3 cos 𝑥−2) sin 𝑥
∫ 𝑑𝑥 , 𝟏
𝟐
Ans
5− 𝑠𝑖𝑛2𝑥−4 cos 𝑥

2−3𝑡
= ∫5− (1−𝑡2)−4𝑡 𝑑𝑡

2−3𝑡
𝟏
= ∫ (𝑡−2)2 𝑑𝑡

𝑑𝑡 = −3 ∫ 𝑑𝑡 − 4 ∫ dt 𝟏
2−3𝑡 1 1
∫
(𝑡−2)2 𝑡−2 (𝑡−2)2

= −3 log|𝑡 − 2| − 4 ( ) + 𝐶
−1 𝟐
𝑡−2
𝟏
= −3 log|cos 𝑥 − 2| +
4
+𝐶
cos 𝑥−2 𝟐

OR

34(b).

𝑑𝑥
𝑥 +|𝑥|+1
∫−2 𝑥2+4|𝑥|+4
Ans 2
I=

65 /1/3 16 P.T.O.
 𝟏
3

𝑥2+4|𝑥|+4 𝑑𝑥 𝑥2+4|𝑥|+4 𝑑𝑥
|𝑥|+1
∫−2 ∫
2 𝑥 2
= + −2

= 𝐼1 + 𝐼2 (𝑠𝑎𝑦)-----(1)

𝐼1 = 0 (∵ 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛)
𝑥3
𝟏
𝑥2+4|𝑥|+4

2 𝑥+1 |𝑥|+1
𝐼2 = 2 ∫0 𝑑𝑥 (∵ 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛. ) 𝟏
𝑥2+4𝑥+4 𝑥2+4|𝑥|+4
2
𝑑𝑥
𝑥+1
=2∫
0 (𝑥+2)2

Put x + 2 = t , so that dx = dt
4 𝑡−1 𝟏
=2∫ 𝑑𝑡 𝟐
2 𝑡2
4
= 2 [∫ (1 − 1 ) 𝑑𝑡]
2 𝑡 𝑡2

= 2 [log|𝑡| + ] 𝟏
14

𝑡 2

= 2 [log 4 + − log 2 ]−
1 1

4 2

𝟏
= 2 log 2 −
1

2 𝟐

35.

Sol.
𝐶𝑜𝑟𝑟𝑒𝑐𝑡
𝑔𝑟𝑎𝑝ℎ
−1

𝟐
Area = 4 ∫ 𝒚 𝒅𝒙
𝟏
𝟎
𝟐
= 4 [ ∫ √𝟒𝟐 − 𝒙𝟐 𝒅𝒙]
𝟏

𝟐 𝟎
𝟐 𝒙
= 2 [ √𝟒𝟐 − 𝒙𝟐 + 𝟖 𝐬𝐢𝐧−𝟏( )] 𝟐
𝒙

𝟐 𝟒 𝟎

= 2 [√𝟏𝟐 + ] = 𝟒 √𝟑 +
𝟖𝝅 𝟖𝝅

𝟔 𝟑
𝟏

65 /1/3 16 P.T.O.
 SECTION-E
(Question nos. 36 to 38 are source based/case based/passage based/integrated units of assessment questions
carrying 4 marks each)

36.

(i) [ , ] or any other interval corresponding to the domain [-1,1] 𝟏

𝝅 𝟑𝝅
Ans
𝟐 𝟐
(ii) 𝐬𝐢𝐧−𝟏( ) − 𝐬𝐢𝐧−𝟏(𝟏)
−𝟏

−𝟐
−𝝅 𝝅
= 𝟔
−𝟒𝝅 −𝟐𝝅
= 𝟔 or 𝟑 𝟏

65 /1/3 17 P.T.O.
65 /1/3 18 P.T.O.
 𝐶𝑜𝑟𝑟𝑒𝑐
𝑡
𝑔𝑟𝑎𝑝ℎ
(iii) (a)

-𝟐

(b) f(x) = 2 𝒔𝒊𝒏−𝟏(𝟏 − 𝒙)

OR

-1≤ 𝟏 − 𝐱 ≤ 𝟏
⇒ −𝟐 ≤ −𝐱 ≤ 𝟎 𝟏

⇒𝟎 ≤ 𝐱 ≤𝟐 𝟐

Domain = [𝟎, 𝟐] 𝟏
𝟐

≤ 𝐬𝐢𝐧−𝟏(𝟏 − 𝐱) ≤
−𝛑 𝛑

𝟐 𝟐
-𝛑 ≤ 𝟐 𝐬𝐢𝐧 (𝟏 − 𝐱) ≤ 𝛑
−𝟏
𝟏
𝟐

So range = [ −𝝅 , 𝝅]
𝟏
𝟐

65 /1/3 18 P.T.O.
65 /1/3 19 P.T.O.
37.

(i) tan 𝜽 = ⇒ 𝜽 = 𝐭𝐚𝐧−𝟏

𝟓 𝟓
Ans ( ) 𝟏
𝒙 𝒙

(ii) 𝒅𝜽 =
−𝟓
𝟏
𝒅𝒙 𝟓𝟐+ 𝒙𝟐

(𝒂) = × = × 𝟐𝟎]
𝒅𝜽 𝒅𝜽 𝒅𝒙 −𝟓
𝟏
𝟏
(iii)
𝒅𝒕 𝒅𝒙 𝒅𝒕 𝟓𝟐+ 𝒙𝟐 𝒙=𝟓𝟎 𝟐
= 𝒓𝒂𝒅/𝒔
−𝟏𝟎𝟎 −𝟒

𝟐𝟓𝟐𝟓 𝟏𝟎𝟏 𝟏
or
𝟐
OR

(b) 𝒅𝒕 =𝒅𝒙 ×𝒅𝒕 ⇒ = ] ×

𝒅𝜽 𝒅𝜽 𝒅𝒙 𝟑 −𝟓 𝒅𝒙
𝟏
𝟏𝟎𝟏 𝟓𝟐+ 𝒙𝟐 𝒙=𝟓𝟎 𝒅𝒕 𝟏
𝟐

65 /1/3 20 P.T.O.
 ⇒ = × ⇒ = −𝟏𝟓 𝒎/𝒔
𝟑 −𝟓 𝒅𝒙 𝒅𝒙 𝟏
𝟏𝟎𝟏 𝟐𝟓𝟐𝟓 𝒅𝒕 𝒅𝒕 𝟐

Hence the speed is 15 m/s

38.

Sol. (i) Let A denote the event of airplane reaching its destination late
𝑬𝟏 = severe turbulence 𝟏
𝟐
𝑬𝟐 = moderate turbulence
}
𝑬𝟑 = light turbulence
P(A) = P (𝑬𝟏) 𝑷( 𝑨|𝑬𝟏) + 𝑷(𝑬𝟐)𝑷(𝑨|𝑬𝟐) + 𝑷(𝑬𝟑)𝑷(𝑨|𝑬𝟑)

𝟏
=𝟑 × 𝟏𝟎𝟎 +𝟑 ×𝟏𝟎𝟎 + ×
𝟏 𝟓𝟓 𝟏 𝟑𝟕 𝟏 𝟏𝟕
𝟑 𝟏𝟎𝟎
𝟏
=𝟑 (𝟏𝟎𝟎 ) =𝟑𝟎𝟎
𝟏 𝟏𝟎𝟗 𝟏𝟎𝟗
𝟐

(ii) P (𝑬 |𝑨) =
𝑷(𝑬𝟐)𝑷(𝑨|𝑬𝟐)
𝟐 𝑷(𝑨)

𝟏 𝟑𝟕
× 𝟏𝟎𝟎 𝟏
𝟑
𝟏𝟎𝟗 𝟏
𝟐
=
𝟑𝟎𝟎
𝟑𝟕 𝟏
𝟏𝟎𝟗 𝟐
=

65 /1/3 20 P.T.O.