Math paper class 12 maths paper class 12

Class-XII/Sample Paper/2024-25/Mathematics/Page 1 of 9
SAMPLE QUESTION PAPER (2024 - 25)
CLASS- XII
SUBJECT: Mathematics (041)
Time: 3 Hours Maximum Marks: 80
General Instructions:
Read the following instructions very carefully and strictly follow them:
(i) This Question paper contains 38 questions. All questions are compulsory.
(ii) This Question paper is divided into five Sections - A, B, C, D and E.
(iii) In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs) and Questions no. 19 and
20 are Assertion-Reason based questions of 1 mark each.
(iv) In Section B, Questions no. 21 to 25 are Very Short Answer (VSA)-type questions, carrying 2 marks
each.
(v) In Section C, Questions no. 26 to 31 are Short Answer (SA)-type questions, carrying 3 marks each.
(vi) In Section D, Questions no. 32 to 35 are Long Answer (LA)-type questions, carrying 5 marks each.
(vii) In Section E, Questions no. 36 to 38 are Case study-based questions, carrying 4 marks each.
(viii) There is no overall choice. However, an internal choice has been provided in 2 questions in Section B,
3 questions in Section C, 2 questions in Section D and one subpart each in 2 questions of Section E.
(ix) Use of calculators is not allowed.
SECTION-A  
1 20 20
 =
(This section comprises of multiple choice questions (MCQs) of 1 mark each)
Select the correct option (Question 1 - Question 18):
Q.1. If for a square matrix ,
A 𝑨. (𝒂𝒅𝒋𝑨) = [
𝟐𝟎𝟐𝟓 𝟎 𝟎
𝟎 𝟐𝟎𝟐𝟓 𝟎
𝟎 𝟎 𝟐𝟎𝟐𝟓
], then the value of A adj A
+ is
equal to:
(A) 1 (B) 2025 1
+ (C) ( )
2
2025 45
+ (D) ( )
2
2025 2025
+
Q.2. Assume , , ,
X Y Z W and P are matrices of order 2 , 3 , 2 , 3
n k p n
    and ,
p k

respectively. Then the restriction on ,
n k and p so that PY WY
+ will be defined are:
(A) 3,
k p n
= = (B) is arbitrary, 2
k p =
(C) is arbitrary, 3
p k = (D) 2, 3
k p
= =
Q.3. The interval in which the function f defined by 𝑓(𝑥) = 𝑒𝑥
is strictly increasing, is
(A)[1, ∞) (B) ( )
,0
− (C) (−∞, ∞) (D) (0,∞)

Q.4. If and
A B are non-singular matrices of same order with 𝒅𝒆𝒕(𝑨) = 𝟓, then 𝒅𝒆𝒕(𝑩−𝟏
𝑨𝑩)
𝟐
is equal to
(A)5 (B)
2
5 (C)
4
5 (D) 5
5
Q.5. The value of ' '
n , such that the differential equation 𝒙𝒏 𝒅𝒚
𝒅𝒙
= 𝒚(𝒍𝒐𝒈𝒚 − 𝒍𝒐𝒈𝒙 + 𝟏);
(𝐰𝐡𝐞𝐫𝐞 𝒙, 𝒚 ∈ 𝑹+) is homogeneous, is
(A) 0 (B)1 (C) 2 (D) 3
Q.6. If the points (𝒙𝟏, 𝒚𝟏), (𝒙𝟐, 𝒚𝟐) and (𝒙𝟏 + 𝒙𝟐, 𝒚𝟏 + 𝒚𝟐)are collinear, then 𝒙𝟏𝒚𝟐 is equal to
(A) 2 1
x y (B) 1 1
x y (C) 2 2
x y (D) 1 2
x x
Q.7. If
0 1
1
2 3 0
c
A a b
 
 
= − −
 
 
 
is a skew-symmetric matrix then the value of a b c
+ + =
(A)1 (B) 2 (C) 3 (D) 4
Q.8. For any two events A and B , if ( ) 1
2
P A = , ( ) 2
3
P B = and ( )
1
,
4
P A B
 = then 𝑃 (𝐴̅
𝐵
̅
⁄ ) equals:
(A)
3
8
(B)
8
9
(C)
5
8
(D)
1
4
Q.9. The value of 𝛼 if the angle between 𝑝
⃗ = 2𝛼2
𝑖̂ − 3𝛼𝑗̂ + 𝑘
̂ and 𝑞
⃗ = 𝑖̂ + 𝑗̂ + 𝛼𝑘
̂ is obtuse, is
(A) 𝑅 − [0, 1] (B) (0, 1) (C) [0, ∞) (D) [1, ∞)
Q.10. If |𝑎
⃗| = 3, |𝑏
⃗⃗| = 4 and |𝑎
⃗ + 𝑏
⃗⃗| =5, then |𝑎
⃗ − 𝑏
⃗⃗| =
(A) 3 (B) 4 (C) 5 (D) 8
Q.11. For the linear programming problem (LPP), the objective function is 4 3
Z x y
= + and the feasible
region determined by a set of constraints is shown in the graph:

(Note: The figure is not to scale.)
Which of the following statements is true?
(A) Maximum value of Z is at ( )
40,0
R .
(B) Maximum value of Z is at ( )
30,20
Q .
(C) Value of Z at ( )
40,0
R is less than the value at ( )
0,40
P .
(D) The value of Z at ( )
30,20
Q is less than the value at ( )
40,0
R .
Q.12. ∫
𝒅𝒙
𝒙𝟑(𝟏+𝒙𝟒)
𝟏
𝟐
equals
(A) −
1
2𝑥2 √1 + 𝑥4 + 𝑐 (B)
1
2𝑥
√1 + 𝑥4 + 𝑐
(C) −
1
4𝑥
√1 + 𝑥4 + 𝑐 (D)
1
4𝑥2 √1 + 𝑥4 + 𝑐
Q.13. ∫ 𝒄𝒐𝒔𝒆𝒄𝟕
𝒙
𝟐𝝅
𝟎
𝒅𝒙 =
(A) 0 (B)1 (C) 4 (D) 2
Q.14. What is the general solution of the differential equation ey′
= x?
(A)𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑐 (B) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 − 𝑥 + 𝑐 (C) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑥 + 𝑐 (D) 𝑦 = 𝑥 + 𝑐
Q.15. The graph drawn below depicts
(A) y = 𝑠𝑖𝑛−1
𝑥 (B) y = 𝑐𝑜𝑠−1
𝑥 (C) y = 𝑐𝑜𝑠 𝑒 𝑐−1
𝑥 (D) y = 𝑐𝑜𝑡−1
𝑥
Q.16. A linear programming problem (LPP) along with the graph of its constraints is shown below.

The corresponding objective function is: 18 10
Z x y
= + , which has to be minimized. The smallest value of
the objective function Z is 134 and is obtained at the corner point ( )
3,8 ,
(Note: The figure is not to scale.)
The optimal solution of the above linear programming problem __________.
(A) does not exist as the feasible region is unbounded.
(B) does not exist as the inequality 18 10 134
x y
+  does not have any point in common with the feasible
region.
(C) exists as the inequality 18 10 134
x y
+  has infinitely many points in common with the feasible region.
(D) exists as the inequality 18 10 134
x y
+  does not have any point in common with the feasible region.
Q.17. The function 𝑓: 𝑅 → 𝑍 defined by ( )  ;
f x x
= where  
. denotes the greatest integer function, is
(A)Continuous at 2.5
x = but not differentiable at 2.5
x =
(B) Not Continuous at 2.5
x = but differentiable at 2.5
x =
(C) Not Continuous at 2.5
x = and not differentiable at 2.5
x =
(D)Continuous as well as differentiable at 2.5
x =
Q.18. A student observes an open-air Honeybee nest on the branch of a tree, whose plane figure is parabolic
shape given by 2
4
x y
= . Then the area (in sq units) of the region bounded by parabola 2
4
x y
= and the line
4
y = is
(A)
32
3 (B)
64
3 (C)
128
3 (D)
256
3
ASSERTION-REASON BASED QUESTIONS
(Question numbers 19 and 20 are Assertion-Reason based questions carrying 1 mark each. Two
statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct
answer from the options (A), (B), (C) and (D) as given below.)

(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Q.19. Assertion (A): Consider the function defined as 𝑓(𝑥) = |𝑥| + |𝑥 − 1|, 𝑥 ∈ 𝑅. Then ( )
f x
is not differentiable at 0 and 1
x x
= = .
Reason (R): Suppose f be defined and continuous on ( )
,
a b and ( )
,
c a b
 , then ( )
f x is not
differentiable at 𝑥 = 𝑐 if lim
ℎ→0−
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
≠ lim
ℎ→0+
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
.
Q.20. Assertion (A): The function 𝑓: 𝑅 − {(2𝑛 + 1)
𝜋
2
: 𝑛 ∈ 𝑍 } → (−∞, −1] ∪ [1, ∞) defined by
( ) sec
f x x
= is not one-one function in its domain.
Reason (R): The line 2
y = meets the graph of the function at more than one point.
SECTION B  
2 5 10
 =
(This section comprises of 5 very short answer (VSA) type questions of 2 marks each.)
Q.21. If 𝑐𝑜𝑡−1(3𝑥 + 5) >
𝜋
4
, then find the range of the values of .
x
Q.22. The cost (in rupees) of producing x items in factory, each day is given by
𝐶(𝑥) = 0.00013𝑥3
+ 0.002 𝑥2
+ 5𝑥 + 2200
Find the marginal cost when 150 items are produced.
Q.23. (a) Find the derivative of 1
tan−
x with respect to log ;
x (where ( )
1,
x  ).
OR
Q.23. (b) Differentiate the following function with respect to x : (𝑐𝑜𝑠 𝑥)𝑥
; (where𝑥 ∈ (0,
𝜋
2
)).
Q.24. (a) If vectors 𝑎
⃗ = 2ı̂ + 2ȷ̂ + 3k
̂ , 𝑏
⃗⃗ = − ı̂ + 2ȷ̂ + k
̂ and 𝑐
⃗ = 3ı̂ + ȷ̂ are such that 𝑏
⃗⃗ + λ𝑐
⃗ is perpendicular
to 𝑎
⃗ , then find the value of λ.
OR
Q.24. (b) A person standing at O( )
0 0 0
, , is watching an aeroplane which is at the coordinate point
( )
A , ,
4 0 3 . At the same time he saw a bird at the coordinate point ( )
B , , .
0 0 1 Find the angles which
𝐵𝐴
⃗⃗⃗⃗⃗⃗ makes with the x,y and z axes.
Q.25. The two co-initial adjacent sides of a parallelogram are 2ı̂ − 4ȷ̂ − 5k
̂ and 2ı̂ + 2ȷ̂ + 3k
̂ . Find its
diagonals and use them to find the area of the parallelogram.

SECTION C  
3 6 18
 =
(This section comprises of 6 short answer (SA) type questions of 3 marks each.)
Q.26. A kite is flying at a height of 3 metres and 5 metres of string is out. If the kite is moving away
horizontally at the rate of 200 cm/s, find the rate at which the string is being released.
Q.27. According to a psychologist, the ability of a person to understand spatial concepts is given by
1
3
A t
= , where t is the age in years,  
5,18
t  . Show that the rate of increase of the ability to
understand spatial concepts decreases with age in between 5 and 18.
Q.28. (a) An ant is moving along the vector 𝑙1
⃗⃗⃗ = 𝑖̂ − 2𝑗̂ + 3𝑘
̂. Few sugar crystals are kept along the vector
𝑙2
⃗⃗⃗⃗ = 3𝑖̂ − 2𝑗̂ + 𝑘
̂ which is inclined at an angle  with the vector 𝑙1
⃗⃗⃗. Then find the angle  . Also find
the scalar projection of 𝑙1
⃗⃗⃗ 𝑜𝑛 𝑙2
⃗⃗⃗⃗.
OR
Q.28. (b) Find the vector and the cartesian equation of the line that passes through (−1, 2, 7) and is
perpendicular to the lines 𝑟
⃗ = 2ı̂ + ȷ̂ − 3k
̂ + λ(ı̂ + 2ȷ̂ + 5k
̂ ) and 𝑟
⃗ = 3ı̂ + 3ȷ̂ − 7k
̂ + μ(3ı̂ − 2ȷ̂ + 5k
̂ ).
Q.29. (a) Evaluate: ∫ {
𝟏
𝒍𝒐𝒈 𝒙
−
𝟏
(𝒍𝒐𝒈 𝒙)𝟐
} 𝒅𝒙; (where𝒙 > 𝟏).
OR
Q.29. (b) Evaluate : ∫ 𝒙(𝟏 − 𝒙)𝒏
𝒅𝒙; (𝐰𝐡𝐞𝐫𝐞 𝒏 ∈ 𝑵).
𝟏
𝟎
Q.30. Consider the following Linear Programming Problem:
Minimise 2
Z x y
= +
Subject to 2 3, 2 6, , 0.
x y x y x y
+  +  
Show graphically that the minimum of Z occurs at more than two points
Q.31. (a) The probability that it rains today is 0.4. If it rains today, the probability that it will rain tomorrow
is 0.8. If it does not rain today, the probability that it will rain tomorrow is 0.7. If
𝑃1: denotes the probability that it does not rain today.
𝑃2: denotes the probability that it will not rain tomorrow, if it rains today.
𝑃3: denotes the probability that it will rain tomorrow, if it does not rain today.
𝑃4: denotes the probability that it will not rain tomorrow, if it does not rain today.
(i) Find the value of 1 4 2 3 .
P P P P
 −   
2Marks
(ii) Calculate the probability of raining tomorrow.  
1Mark
OR
Q.31. (b) A random variable X can take all non – negative integral values and the probability that X takes

the value r is proportional to 5−𝑟
. Find ( )
3 .
P X 
SECTION D
 
5 4 20
 =
(This section comprises of 4 long answer (LA) type questions of 5 marks each)
Q.32. Draw the rough sketch of the curve 𝑦 = 20 𝑐𝑜𝑠 2 𝑥; (where
𝜋
6
≤ 𝑥 ≤
𝜋
3
).
Using integration, find the area of the region bounded by the curve y = 20 cos2x from the ordinates
𝑥 =
𝜋
6
to 𝑥 =
𝜋
3
and the x − axis.
Q.33. The equation of the path traversed by the ball headed by the footballer is
𝑦 = 𝑎𝑥2
+ 𝑏𝑥 + 𝑐; (𝑤ℎ𝑒𝑟𝑒 0 ≤ 𝑥 ≤ 14 𝑎𝑛𝑑 𝑎, 𝑏, 𝑐 ∈ 𝑅 𝑎𝑛𝑑 𝑎 ≠ 0) with respect to a XY-coordinate
system in the vertical plane. The ball passes through the points ( ) ( ) ( )
2,15 , 4,25 and 14,15 . Determine
the values of a, b and c by solving the system of linear equations in a, b and c, using matrix method.
Also find the equation of the path traversed by the ball.
Q.34. (a) If 𝑓: 𝑅 → 𝑅 is defined by 𝑓(𝑥) = |𝑥|3
, show that 𝑓"(𝑥) exists for all real x and find it.
OR
Q.34. (b) If (𝑥 − 𝑎)2
+ (𝑦 − 𝑏)2
= 𝑐2
, for some c 0,
 prove that
[1+(
𝑑𝑦
𝑑𝑥
)
2
]
3
2
𝑑2𝑦
𝑑𝑥2
is a constant independent
of a and b .
Q.35. (a) Find the shortest distance between the lines 1
l and 2
l whose vector equations are
𝑟
⃗ = (−𝑖̂ − 𝑗̂ − 𝑘
̂) + 𝜆(7𝑖̂ − 6𝑗̂ + 𝑘
̂) and 𝑟
⃗ = (3𝑖̂ + 5𝑗̂ + 7𝑘
̂) + 𝜇(𝑖̂ − 2𝑗̂ + 𝑘
̂)
where  and  are parameters.
OR
Q.35. (b) Find the image of the point ( )
1,2, 1 with respect to the line
𝑥−3
1
=
𝑦+1
2
=
𝑧−1
3
. Also find the
equation of the line joining the given point and its image.
SECTION- E  
4 3 12
 =
(This section comprises of 3 case-study/passage-based questions of 4 marks each with subparts. The first
two case study questions have three subparts (i), (ii), (iii) of marks 1, 1, 2 respectively. The third case study
question has two subparts of 2 marks each)
Case Study-1
Q.36. Ramesh, the owner of a sweet selling shop, purchased some rectangular card board sheets of
dimension 25 by 40
cm cm to make container packets without top. Let xcm be the length of the side of the
square to be cut out from each corner to give that sheet the shape of the container by folding up the flaps.
Based on the above information answer the following questions.

(i) Express the volume (V) of each container as function of x only.  
1Mark
(ii) Find
𝑑𝑉
𝑑𝑥
 
1Mark
(iii) (a) For what value of ,
x the volume of each container is maximum?  
2Marks
OR
(iii) (b) Check whether V has a point of inflection at x =
65
6
or not?  
2Marks
Case Study-2
Q.37. An organization conducted bike race under 2 different categories-boys and girls. In all, there were 250
participants. Among all of them finally three from Category 1 and two from Category 2 were selected for the
final race. Ravi forms two sets B and G with these participants for his college project.
Let    
1 2 3 1 2
, , , ,
B b b b G g g
= = where B represents the set of boys selected and G the set of girls who
were selected for the final race.
Ravi decides to explore these sets for various types of relations and functions.
On the basis of the above information, answer the following questions:
(i) Ravi wishes to form all the relations possible from B to G . How many such relations are possible?
 
1Mark
(ii) Among these relations, how many are functions from B to G ?  
1Mark
(iii) (a) Find the total number of one-one and onto functions which can be defined from B to G .
 
2Marks
OR
(iii) (b) If the track of the final race (for the biker 1
b ) follows the curve
𝑥2
= 4𝑦; (where0 ≤ 𝑥 ≤ 20√2&0 ≤ 𝑦 ≤ 200), then state whether the track represents a
one-one and onto function or not. (Justify).  
2Marks
Case Study- 3
Q.38. Arka bought two cages of birds: Cage-I contains 5 parrots and 1 owl and Cage –II contains 6
parrots. One day Arka forgot to lock both cages and two birds flew from Cage-I to Cage-II (simultaneously).
Then two birds flew back from cage-II to cage-I(simultaneously).
Assume that all the birds have equal chances of flying.
On the basis of the above information, answer the following questions:-
(i) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I then
find the probability that the owl is still in Cage-I.  
2Marks

(ii) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I, the
owl is still seen in Cage-I, what is the probability that one parrot and the owl flew from Cage-I to
Cage-II?  
2Marks
*******************************************************************************

of 15
MARKING SCHEME
CLASS XII
MATHEMATICS (CODE-041)
SECTION: A (Solution of MCQs of 1 Mark each)
Q no. ANS HINTS/SOLUTION
1. (D) For a square matrix A of order n n
 , we have ( )
. ,
n
A adj A A I
= where n
I is the identity matrix of
order .
n n

So, ( ) 3
2025 0 0
. 0 2025 0 2025
0 0 2025
A adj A I
 
 
= =
 
 
 
2025
A
 = & ( )
2
3 1
2025
adj A A
−
= =
( )
2
2025 2025 .
A adj A
 + = +
2. (A)
3. (C)
𝑦 = 𝑒𝑥
= >
𝑑𝑦
𝑑𝑥
= 𝑒𝑥
In the domain (R) of the function,
𝑑𝑦
𝑑𝑥
> 0 , hence the function is strictly increasing in (−∞, ∞)
4. (B)
5,
A = ( )
2
2 2
1 1 2
5 .
B AB B A B A
− −
= = =
5. (B)
A differential equation of the form ( )
,
dy
f x y
dx
= is said to be homogeneous, if ( )
,
f x y is a
homogeneous function of degree 0.
Now, log log
n
e e
dy y
x y e
dx x
 
= +
 
 
( ) ( )
log . , ;
e
n
dy y y
e f x y Let
dx x x
 
 
 = =
 
 
 
 
. ( )
,
f x y will be a
homogeneous function of degree 0, if 1.
n =
6. (A) Method 1: ( Short cut)
When the points ( ) ( )
1 1 2 2
, , ,
x y x y and ( )
1 2 1 2
,
x x y y
+ + are collinear in the Cartesian plane then
( ) ( )
( )
1 2 1 2 1 2 1 2
1 2 2 2 2 1 2 2
1 1 2 1 1 2 2 2
0 0
x x y y x x y y
x y x y x y x y
x x x y y y x y
− − − −
=  = − + + − =
− + − + − −
2 1 1 2 .
x y x y
 =

of 15
Method 2:
When the points ( ) ( )
1 1 2 2
, , ,
x y x y and ( )
1 2 1 2
,
x x y y
+ + are collinear in the Cartesian plane then
1 1
2 2
1 2 1 2
1
1 0
1
x y
x y
x x y y
=
+ +
( ) ( ) ( )
2 1 2 2 1 2 2 2 1 1 1 2 1 1 2 1 1 2 2 1
1. 1 0
x y x y x y x y x y x y x y x y x y x y
 + − − − + − − + − =
2 1 1 2 .
x y x y
 =
7. (A) 0 1
1
2 3 0
c
A a b
 
 
= − −
 
 
 
When the matrix A is skew symmetric then ;
T
ij ji
A A a a
= −  = −
2; 0 and 3
c a b
 = − = =
So , 0 3 2 1.
a b c
+ + = + − =
8. (C)
( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
1 2 1
; ;
2 3 4
1 1
;
2 3
1 1 1 7
Wehave,
2 3 4 12
P A P B P A B
P A P B
P A B P A P B P A B
= =  =
 = =
 = + −  = + − =
( )
( )
( )
( )
( )
( )
7
1
1 5
12 .
2 8
3
P A B P A B P A B
A
P
B P B P B P B
−

   − 
= = = = =
 
 
9. (B) For obtuse angle, cos 𝜃 < 0 => 𝑝
⃗. 𝑞
⃗ < 0
𝟐𝜶𝟐
− 𝟑𝜶 + 𝜶 < 𝟎 => 𝟐𝜶𝟐
− 𝟐𝜶 < 𝟎 => 𝜶 ∈ (𝟎, 𝟏)
10. (C) 3, 4, 5
a b a b
= = + =
r r r r
We have , ( ) ( )
2 2 2 2
2 2 9 16 50 5.
a b a b a b a b
+ + − = + = + =  − =
r r r r r r r r
11. (B) Corner point Value of the objective function 4 3
Z x y
= +
1. ( )
0,0
O 0
z =
2. ( )
40,0
R 160
z =
3. ( )
30,20
Q 120 60 180
z = + =
4. ( )
0,40
P 120
z =
Since , the feasible region is bounded so the maximum value of the objective function 180
z = is at
( )
30,20 .
Q

of 15
12. (A)
( )
1
3 4 2
1
dx
x x
+
 1
2
5
4
1
1
dx
x
x
=
 
+
 
 

( Let 4
4
1
1 1
x t
x
−
+ = + = , 5
5 5
4 1
4
4
dx
dt x dx dx dt
x x
−
= − = −  = − )
1
2
1 1
2
4 4
dt
t c
t
= − = −   +
 , where ' '
c denotes any arbitrary constant of integration.
4
1 1
1
2
c
x
= − + +
13. (A) We know, ( ) ( ) ( )
2
0
0, if 2
a
f x dx f a x f x
= − = −

Let ( ) 7
cos
f x ec x
= .
Now, ( ) ( ) ( )
7 7
2 cos 2 cos
f x ec x ec x f x
 
− = − = − = −
2
7
0
cos 0;
ec x dx

 =
 Using the property ( ) ( ) ( )
2
0
0, if 2 .
a
f x dx f a x f x
= − = −

14. (B) The given differential equation 𝑒𝑦′
= 𝑥 =>
𝑑𝑦
𝑑𝑥
= log 𝑥
𝑑𝑦 = log 𝑥 𝑑𝑥 => ∫ 𝑑𝑦 = ∫ log 𝑥 𝑑𝑥
𝑦 = 𝑥 log 𝑥 − 𝑥 + 𝑐
hence the correct option is (B).
15. (B) The graph represents 1
cos
y x
−
= whose domain is  
1,1
− and range is  
0, .

16. (D) Since the inequality 18 10 134
Z x y
= +  has no point in common with the feasible region hence
the minimum value of the objective function 18 10
Z x y
= + is 134 at ( )
3,8
P .
17. (D) The graph of the function 𝑓: 𝑅 → 𝑅 defined by ( )  ;
f x x
=  
( )
where . denotes . .
G I F is a straight
line ( )
2.5 ,2.5
x h h
  − + , ' '
h is an infinitesimally small positive quantity. Hence, the function is
continuous and differentiable at 2.5 .
x =
18. (B) The required region is symmetric about the y − axis.
So, required area (in sq units ) is
4
3
4 2
0
0
64
2 2 4 .
3 3
2
y
ydy
 
 
= = =
 
 
 

19. (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
20. (A) Both (A) and (R) are true and (R) is the correct explanation of (A).

of 15
Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]
21
cot−1(3𝑥 + 5) >
𝜋
4
= cot−1
1
=>3x + 5 < 1 ( as cot−1
𝑥 is strictly decreasing function in its domain)
=> 3x < – 4
=> 𝑥 < −
4
3
⸫ 𝑥 ∈ (−∞, −
4
3
)
1
2
1
2
1
22. The marginal cost function is ( ) 2
' 0.00039 0.004 5
C x x x
= + + .
( )
' 150
C = ₹ 14.375 .
1
1
23.(a) 1
tan
y x
−
= and loge
z x
=
.
Then 2
1
1
dy
dx x
=
+
and
1
dz
dx x
=
So,
2
2
1
1 .
1 1
dy
dy dx
dz
dz
dx
x
x
x
x
=
+
= =
+
1
2
1
2
1
2
1
2
OR
23.(b)
Let x
y x
(cos ) .
= Then, log cosx
x e
y e
=
On differentiating both sides with respect to ,
x we get
log cos
( log cos )
x x
e
e
dy d
e x x
dx dx
=
(cos ) log cos ( ) (log cos )
x
e e
dy d d
x x x x x
dx dx dx
 
 = +
 
 
1
(cos ) log cos . ( sin )
cos
x
e
dy
x x x x
dx x
 
 = + −
 
 
(cos ) (log cos tan ) .
x
e
dy
x x x x
dx
 = −
1
2
1
2
1
24.(a) We have b
⃗⃗ + λc
⃗ = (−1 + 3λ)î + (2 + λ )ĵ + k
̂
(b
⃗⃗ + λc
⃗) . a
⃗⃗ = 0 => 2(−1 + 3λ ) + 2 (2 + λ ) + 3 = 0
1
2
1

of 15
λ = −
5
8
1
2
OR
24.(b) 𝐵𝐴
⃗⃗⃗⃗⃗⃗ = 𝑂𝐴
⃗⃗⃗⃗⃗⃗ − 𝑂𝐵
⃗⃗⃗⃗⃗⃗ = (4𝑖̂ + 3𝑘
̂) − 𝑘
̂ = 4𝑖̂ + 2𝑘
̂
𝐵𝐴
̂ =
4
2√5
𝑖̂ +
2
2√5
𝑘
̂ =
2
√5
𝑖̂ +
1
√5
𝑘
̂
So, the angles made by the vector 𝐵𝐴
⃗⃗⃗⃗⃗⃗ with the ,
x y and the z axes are respectively
𝑐𝑜𝑠−1
(
2
√5
) ,
𝜋
2
, 𝑐𝑜𝑠−1
(
1
√5
).
1
2
1
2
1
25. 𝑑1
⃗⃗⃗⃗⃗ = 𝑎
⃗ + 𝑏
⃗⃗ = 4𝑖̂ − 2𝑗̂ − 2𝑘
̂ , 𝑑2
⃗⃗⃗⃗⃗ = 𝑎
⃗ − 𝑏
⃗⃗ = −6𝑗̂ − 8𝑘
̂
Area of the parallelogram =
1
2
|𝑑1
⃗⃗⃗⃗⃗ × 𝑑2
⃗⃗⃗⃗⃗| =
1
2
||
𝑖̂ 𝑗̂ 𝑘
̂
4 −2 −2
0 −6 −8
|| = 2|𝑖̂ + 8𝑗̂ − 6𝑘
̂|
Area of the parallelogram = 2√101 sq. units.
1
2
1
1
2
Section –C
[This section comprises of solution short answer type questions (SA) of 3 marks each]
26.
𝑥2
+ 32
= 𝑦2
𝑊ℎ𝑒𝑛 𝑦 = 5 𝑡ℎ𝑒𝑛 𝑥 = 4, 𝑛𝑜𝑤 2𝑥
𝑑𝑥
𝑑𝑡
= 2𝑦
𝑑𝑦
𝑑𝑡
4 (200) = 5
𝑑𝑦
𝑑𝑡
=>
𝑑𝑦
𝑑𝑡
= 160 cm/s
1
2
1
2
1
1
27. 𝐴 =
1
3
√𝑡 ∴
𝑑𝐴
𝑑𝑡
=
1
6
𝑡−
1
2 =
1
6√𝑡
; ∀𝑡 ∈ (5,18)
𝑑𝐴
𝑑𝑡
=
1
6√𝑡
∴
𝑑2𝐴
𝑑𝑡2
= −
1
12𝑡√𝑡
So,
𝑑2𝐴
𝑑𝑡2
< 0, ∀𝑡 ∈ (5,18)
This means that the rate of change of the ability to understand spatial concepts decreases
(slows down) with age.
1
1
1
2
1
2
28(a)
(i) 𝜽 = 𝐜𝐨𝐬−𝟏
(
𝒍𝟏
⃗⃗⃗⃗.𝒍𝟐
⃗⃗⃗⃗
|𝒍𝟏
⃗⃗⃗⃗|.|𝒍𝟐
⃗⃗⃗⃗|
) = 𝐜𝐨𝐬−𝟏
(
(ı̂−2ȷ̂+3k
̂).(3ı̂−2ȷ̂ + 𝑘
̂ )
|(ı̂−2ȷ̂+3k
̂)|| (3ı̂−2ȷ̂ + 𝑘
̂ )|
)
= 𝒄𝒐𝒔−𝟏
(
𝟑+𝟒+𝟑
√𝟏+𝟒+𝟗√𝟗+𝟒+𝟏
) = 𝒄𝒐𝒔−𝟏
(
𝟏𝟎
𝟏𝟒
) = 𝒄𝒐𝒔−𝟏
(
𝟓
𝟕
).
1
1
2
3
3
y
x

of 15
(ii) Scalar projection of 𝒍𝟏
⃗⃗⃗⃗ on 𝒍𝟐
⃗⃗⃗⃗ =
𝒍𝟏
⃗⃗⃗⃗.𝒍𝟐
⃗⃗⃗⃗
|𝒍𝟐
⃗⃗⃗⃗|
=
(ı̂−2ȷ̂+3k
̂).(3ı̂−2ȷ̂ + 𝑘
̂ )
| (3ı̂−2ȷ̂ + 𝑘
̂ )|
=
3+4+3
√9+4+1
=
10
√14
.
1
1
2
28(b) Line perpendicular to the lines
𝑟
⃗ = 2ı̂ + ȷ̂ − 3k
̂ + λ(ı̂ + 2ȷ̂ + 5k
̂ ) and 𝑟
⃗ = 3ı̂ + 3ȷ̂ − 7k
̂ + μ(3ı̂ − 2ȷ̂ + 5k
̂ ).
has a vector parallel it is given by 𝑏
⃗⃗ = 𝑏1
⃗⃗⃗⃗ × 𝑏2
⃗⃗⃗⃗⃗ = |
𝑖̂ 𝑗̂ 𝑘
̂
1 2 5
3 −2 5
| = 20î + 10ĵ − 8k̂
⸫ equation of line in vector form is 𝑟
⃗ = − ı̂ + 2 ȷ̂ + 7k
̂ + a(10ı̂ + 5ȷ̂ − 4k
̂ )
And equation of line in cartesian form is
𝑥+1
10
=
𝑦−2
5
=
𝑧−7
−4
1
1
1
29.(a)
∫ {
1
𝑙𝑜𝑔𝑒 𝑥
−
1
(𝑙𝑜𝑔𝑒 𝑥)2}𝑑𝑥
= ∫
𝑑𝑥
− ∫
1
(𝑙𝑜𝑔𝑒 𝑥)2
𝑑𝑥 =
1
∫ 𝑑𝑥 − ∫ {
𝑑
𝑑𝑥
(
1
) ∫ 𝑑𝑥} 𝑑𝑥 − ∫
1
𝑑𝑥
=
𝑥
+ ∫
1
1
𝑥
. 𝑥. 𝑑𝑥 − ∫
1
𝑑𝑥
=
𝑥
+ ∫
1
𝑑𝑥 − ∫
𝑑𝑥
=
𝑥
+ 𝑐;
where′𝑐′is any arbitary constant of integration.
1
1
1
OR
29.(b)
( )
1
0
1
n
x x dx
−

= ∫ (1 − 𝑥){1 − (1 − 𝑥)}𝑛
1
0
𝑑𝑥, (𝑎𝑠, ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥
𝑎
0
𝑎
0
)
= ∫ 𝑥𝑛
1
0
(1 − 𝑥)𝑑𝑥
= ∫ 𝑥𝑛
𝑑𝑥
1
0
− ∫ 𝑥𝑛+1
𝑑𝑥
1
0
=
1
𝑛 + 1
[𝑥𝑛+1]1
0 −
1
𝑛 + 2
[𝑥𝑛+2]0
1
=
1
𝑛+1
−
1
𝑛+2
=
1
(𝑛+1)(𝑛+2)
.
1
1
2
1
2
1
30. The feasible region determined by the constraints, 2 3, 2 6, 0, 0
x y x y x y
+  +    is as shown.

of 15
The corner points of the unbounded feasible region are ( )
6,0
A and ( )
0,3
B .
The values of Z at these corner points are as follows:
Corner point
Value of the objective function
2
Z x y
= +
( )
6,0
A 6
( )
0,3
B 6
We observe the region 2 6
x y
+  have no points in common with the unbounded feasible region. Hence
the minimum value of 6
z = .
It can be seen that the value of Z at points A and B is same. If we take any other point on the line
2 6
x y
+ = such as (2,2) on line 2 6,
x y
+ = then 6
Z = .
Thus, the minimum value of Z occurs for more than 2 points, and is equal to 6.
1
1
1

of 15
31.(a) Since the event of raining today and not raining today are complementary events so if the probability
that it rains today is 0.4 then the probability that it does not rain today is 1 0.4 0.6
− = 1 0.6
P
 =
If it rains today, the probability that it will rain tomorrow is 0.8 then the probability that it will not rain
tomorrow is 1 0.8 0.2
− = .
If it does not rain today, the probability that it will rain tomorrow is 0.7 then the probability that it will
not rain tomorrow is 1 0.7 0.3
− =
(i) 1 4 2 3 0.6 0.3 0.2 0.7 0.04.
P P P P
 −  =  −  =
(ii) Let 1
E and 2
E be the events that it will rain today and it will not rain today respectively.
( ) ( )
1 2
0.4 & 0.6
P E P E
= =
A be the event that it will rain tomorrow. 𝑃 (
𝐴
𝐸1
) = 0.8 & 𝑃 (
𝐴
𝐸2
) = 0.7
We have, 𝑃(𝐴) = 𝑃(𝐸1)𝑃 (
𝐴
𝐸1
) + 𝑃(𝐸2)𝑃 (
𝐴
𝐸2
) = 0.4 × 0.8 + 0.6 × 0.7 = 0.74.
The probability of rain tomorrow is 0.74.
1
1
1
2
1
2
OR
31.(b)
Given 𝑃(𝑋 = 𝑟)𝛼
1
5𝑟
𝑃(𝑋 = 𝑟) = 𝑘
1
5𝑟,
( where k is a non-zero constant )
𝑃(𝑟 = 0) = 𝑘.
1
50
𝑃(𝑟 = 1) = 𝑘.
1
51
𝑃(𝑟 = 2) = 𝑘.
1
52
𝑃(𝑟 = 3) = 𝑘.
1
53
1
2
1
2

of 15
Section –D
[This section comprises of solution of long answer type questions (LA) of 5 marks each]
………………………….
………………………….
We have, 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)+. . . . . . . . . . . . . . . = 1
⇒ 𝑘 (1 +
1
5
+
1
52
+
1
53
+. . . . . . . . . . . . . ) = 1
⇒ 𝑘 (
1
1 −
1
5
) = 1 ⇒ 𝑘 =
4
5
So, 𝑃(𝑋 < 3) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)
=
4
5
(1 +
1
5
+
1
52
) =
4
5
(
25 + 5 + 1
25
) =
124
125
.
1
2
1
2
1
32.
Required area = 20 ∫ 𝑐𝑜𝑠2𝑥 𝑑𝑥 + |20 ∫ 𝑐𝑜𝑠2𝑥 𝑑𝑥
𝜋
3
𝜋
4
|
𝜋
4
𝜋
6
= 20 [
sin 2𝑥
2
]𝜋
6
𝜋
4
+ |20 [
sin 2𝑥
2
]𝜋
4
𝜋
3
|
= 10 (1 −
√3
2
) + 10 (1 −
√3
2
) = 20 (1 −
√3
2
) sq. units.
1
1+1
1
1
33. 2
y ax bx c
= + +
15 4 2
a b c
= + +
25 16 4
a b c
= + +
15 196 14
a b c
= + +
The set of equations can be represented in the matrix form as AX B
= ,
where 𝐴 = [
4 2 1
16 4 1
196 14 1
]’ 𝑋 = [
𝑎
𝑏
𝑐
] and 𝐵 = [
15
25
15
] ⇒ [
4 2 1
16 4 1
196 14 1
] [
𝑎
𝑏
𝑐
] = [
15
25
15
].
1
1
2

of 15
|𝐴| = 4(4 − 14) − 2(16 − 196) + (224 − 784) = −40 + 360 − 560 = −240 ≠ 0. Hence 1
A−
exists.
Now,𝑎𝑑𝑗(𝐴) = [
−10 180 −560
12 −192 336
−2 12 −16
]
𝑇
= [
−10 12 −2
180 −192 12
−560 336 −16
]
[
𝑎
𝑏
𝑐
] = −
1
240
[
−10 12 −2
180 −192 12
−560 336 −16
][
15
25
15
] = −
5
240
[
−10 12 −2
180 −192 12
−560 336 −16
][
3
5
3
] = −
5
240
[
24
−384
−48
]
1
, 8, 1
2
a b c
 = − = =
So, the equation becomes 2
1
8 1
2
y x x
= − + +
1
2
1
1
1
2
1
2
34.(a)
We have, 𝑓(𝑥) = |𝑥|3
, {
𝑥3
,if 𝑥 ≥ 0
(−𝑥)3
= −𝑥3
,if𝑥 < 0
Now, (𝐿𝐻𝐷𝑎𝑡𝑥 = 0) = 𝑙𝑖𝑚
𝑥→0−
𝑓(𝑥)−𝑓(0)
𝑥−0
= 𝑙𝑖𝑚
𝑥→0−
(
−𝑥3−0
𝑥
) = 𝑙𝑖𝑚
𝑥→0−
(−𝑥2) = 0
(𝑅𝐻𝐷𝑎𝑡𝑥 = 0) 𝑙𝑖𝑚
𝑥→0+
𝑓(𝑥)−𝑓(0)
𝑥−0
= 𝑙𝑖𝑚
𝑥→0+
(
𝑥3−0
𝑥
) = 𝑙𝑖𝑚
𝑥→0
(−𝑥2) = 0
∴ (𝐿𝐻𝐷𝑜𝑓𝑓(𝑥)𝑎𝑡𝑥 = 0) = (𝑅𝐻𝐷𝑜𝑓𝑓(𝑥)𝑎𝑡𝑥 = 0)
So, ( )
f x is differentiable at 0
x = and the derivative of ( )
f x is given by
𝑓′(𝑥) = {
3𝑥2
,if𝑥 ≥ 0
−3𝑥2
,if𝑥 < 0
Now, (𝐿𝐻𝐷𝑜𝑓𝑓′(𝑥)𝑎𝑡𝑥 = 0) = 𝑙𝑖𝑚
𝑥→0−
𝑓′(𝑥)−𝑓′(0)
𝑥−0
= 𝑙𝑖𝑚
𝑥→0−
(
−3𝑥2−0
𝑥
) = 𝑙𝑖𝑚
𝑥→0−
(−3𝑥) = 0
(𝑅𝐻𝐷𝑜𝑓𝑓′(𝑥)𝑎𝑡𝑥 = 0) = 𝑙𝑖𝑚
𝑥→0+
𝑓′(𝑥)−𝑓′(0)
𝑥−0
= 𝑙𝑖𝑚
𝑥→0+
(
3𝑥2−0
𝑥−0
) = 𝑙𝑖𝑚
𝑥→0+
(3𝑥) = 0
∴ (𝐿𝐻𝐷𝑜𝑓𝑓′(𝑥)𝑎𝑡𝑥 = 0) = (𝑅𝐻𝐷𝑜𝑓𝑓′(𝑥)𝑎𝑡𝑥 = 0)
So, 𝑓′(𝑥)is differentiable at 0.
x =
Hence, 𝑓′′(𝑥) = {
6𝑥,if𝑥 ≥ 0
−6𝑥,if𝑥 < 0.
1
2
1
2
1
2
1
1
2
1
2
1
2
1
2
1
2
OR
34 .(b)
Given relation is (𝑥 − 𝑎)2
+ (𝑦 − 𝑏)2
= 𝑐2
, 𝑐 > 0.
Let cos
x a c
− =  and 𝑦 − 𝑏 = 𝑐 𝑠𝑖𝑛 𝜃.
Therefore,
𝑑𝑥
𝑑𝜃
= −𝑐 𝑠𝑖𝑛 𝜃 And
𝑑𝑦
𝑑𝜃
= 𝑐 𝑐𝑜𝑠𝜃
∴
𝑑𝑦
𝑑𝑥
= − 𝑐𝑜𝑡 𝜃
Differentiate both sides with respect to ,
 we get
𝑑
𝑑𝜃
(
𝑑𝑦
𝑑𝑥
) =
𝑑
𝑑𝜃
(− 𝑐𝑜𝑡 𝜃)
1
2
1
2
1
1
2

of 15
Or,
𝑑
𝑑𝑥
(
𝑑𝑦
𝑑𝑥
)
𝑑𝑥
𝑑𝜃
= 𝑐𝑜𝑠 𝑒 𝑐2
𝜃
Or,
𝑑2𝑦
𝑑𝑥2
(−𝑐 𝑠𝑖𝑛 𝜃) = cosec2
𝜃
𝑑2𝑦
𝑑𝑥2
= −
𝑐𝑜𝑠𝑒𝑐3𝜃
𝑐
∴
[1+(
𝑑𝑦
𝑑𝑥
)
2
]
3
2
𝑑2𝑦
𝑑𝑥2
=
𝑐[1+𝑐𝑜𝑡2 𝜃]
3
2
− 𝑐𝑜𝑠 𝑒𝑐3𝜃
=
− 𝑐(𝑐𝑜𝑠 𝑒𝑐2𝜃)
3
2
cosec3𝜃
= −𝑐,
Which is constant and is independent of a and b .
1
2
1
2
1
1
2
35.(a)
Given that equation of lines are
𝑟
⃗ = (−𝑖̂ − 𝑗̂ − 𝑘
̂) + 𝜆(7𝑖̂ − 6𝑗̂ + 𝑘
̂). . . . . . . . . . . . . . . . (𝑖) and
𝑟
⃗ = (3𝑖̂ + 5𝑗̂ + 7𝑘
̂) + 𝜇(𝑖̂ − 2𝑗̂ + 𝑘
̂). . . . . . . . . . . . . . . . . . (𝑖𝑖)
The given lines are non-parallel lines as vectors 7𝑖̂ − 6𝑗̂ + 𝑘
̂and 𝑖̂ − 2𝑗̂ + 𝑘
̂are not parallel. There is a
unique line segment PQ ( P lying on line ( )
i and Q on the other line( )
ii ), which is at right angles
to both the lines PQ is the shortest distance between the lines.
Hence, the shortest possible distance between the lines PQ
= .
Let the position vector of the point P lying on the line𝑟
⃗ = (−𝑖̂ − 𝑗̂ − 𝑘
̂) + 𝜆(7𝑖̂ − 6𝑗̂ + 𝑘
̂) where' '

is a scalar, is (7𝜆 − 1)𝑖̂ − (6𝜆 + 1)𝑗̂ + (𝜆 − 1)𝑘
̂, for some and the position vector of the point Q
lying on the line 𝑟
⃗ = (3𝑖̂ + 5𝑗̂ + 7𝑘
̂) + 𝜇(𝑖̂ − 2𝑗̂ + 𝑘
̂)where ' '
 is a scalar, is
(𝜇 + 3)𝑖̂ + (−2𝜇 + 5)𝑗̂ + (𝜇 + 7)𝑘
̂, for some .
 Now, the vector
𝑃𝑄
⃗⃗⃗⃗⃗⃗ = 𝑂𝑄
⃗⃗⃗⃗⃗⃗⃗ − 𝑂𝑃
⃗⃗⃗⃗⃗⃗ = (𝜇 + 3 − 7𝜆 + 1)𝑖̂ + (−2𝜇 + 5 + 6𝜆 + 1)𝑗̂ + (𝜇 + 7 − 𝜆 + 1)𝑘
̂
𝑖. 𝑒. , 𝑃𝑄
⃗⃗⃗⃗⃗⃗ = (𝜇 − 7𝜆 + 4)𝑖̂ + (−2𝜇 + 6𝜆 + 6)𝑗̂ + (𝜇 − 𝜆 + 8)𝑘
̂ ; (where ' '
O is the origin), is
perpendicular to both the lines, so the vector 𝑃𝑄
⃗⃗⃗⃗⃗⃗ is perpendicular to both the vectors 7𝑖̂ − 6𝑗̂ + 𝑘
̂ and
𝑖̂ − 2𝑗̂ + 𝑘
̂.
 (𝜇 − 7𝜆 + 4). 7 + (−2𝜇 + 6𝜆 + 6). (−6) + (𝜇 − 𝜆 + 8). 1 = 0
1
2
1
2
1

of 15
&(𝜇 − 7𝜆 + 4). 1 + (−2𝜇 + 6𝜆 + 6). (−2) + (𝜇 − 𝜆 + 8). 1 = 0
 𝟐𝟎𝝁 − 𝟖𝟔𝝀 = 𝟎 => 𝟏𝟎𝝁 − 𝟒𝟑𝝀 = 𝟎&6𝜇 − 20𝜆 = 0 ⇒ 3𝜇 − 10𝜆 = 0
On solving the above equations, we get 0
 
= =
So, the position vector of the points P and Q are −𝑖̂ − 𝑗̂ − 𝑘
̂and 3𝑖̂ + 5𝑗̂ + 7𝑘
̂respectively.
𝑃𝑄
⃗⃗⃗⃗⃗⃗ = 4𝑖̂ + 6𝑗̂ + 8𝑘
̂ and
|𝑃𝑄
⃗⃗⃗⃗⃗⃗| = √42 + 62 + 82 = √116 = 2√29𝑢𝑛𝑖𝑡𝑠.
1
1
2
1
2
1
OR
35.(b)
Let ( )
1, 2 , 1
P be the given point and Lbe the foot of the perpendicular from P to the given line AB
(𝑎𝑠 𝑠ℎ𝑜𝑤𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒 𝑎𝑏𝑜𝑣𝑒).
Let’s put
3 1 1
.
1 2 3
x y z

− + −
= = = Then, 3, 2 1, 3 1
x y z
  
= + = − = +
Let the coordinates of the point Lbe( )
3,2 1,3 1 .
  
+ − +
So, direction ratios of PL are(𝜆 + 3 − 1,2𝜆 − 1 − 2,3𝜆 + 1 − 1)𝑖. 𝑒. , (𝜆 + 2,2𝜆 − 3,3𝜆)
Direction ratios of the given line are 1, 2 and 3,which is perpendicular to PL. Therefore, we have,
(𝜆 + 2). 1 + (2𝜆 − 3). 2 + 3𝜆. 3 = 0 ⇒ 14𝜆 = 4 ⇒ 𝜆 =
2
7
Then, 𝜆 + 3 =
2
7
+ 3 =
23
7
; 2𝜆 − 1 = 2 (
2
7
) − 1 = −
3
7
; 3𝜆 + 1 = 3 (
2
7
) + 1 =
13
7
Therefore, coordinates of the point L are (
23
7
, −
3
7
,
13
7
).
Let 𝑄(𝑥1, 𝑦1, 𝑧1)be the image of ( )
1, 2 , 1
P with respect to the given line. Then, L is the mid-point
of .
PQ
Therefore,
1+𝑥1
2
=
23
7
,
2+𝑦1
2
= −
3
7
,
1+𝑧1
2
=
13
7
⇒ 𝑥1 =
39
7
, 𝑦1 = −
20
7
, 𝑧1 =
19
7
Hence, the image of the point ( )
1,2,1
P with respect to the given line 𝑄 (
39
7
, −
20
7
,
19
7
).
The equation of the line joining ( )
1,2,1
P and 𝑄 (
39
7
, −
20
7
,
19
7
)is
1
2
1
2
1
2
1
2
1
1

of 15
Section –E
[This section comprises solution of 3 case- study/passage based questions of 4 marks each with two sub
parts. Solution of the first two case study questions have three sub parts (i),(ii),(iii) of marks 1,1,2
respectively. Solution of the third case study question has two sub parts of 2 marks each.)
36. (i) 𝑉 = (40 − 2𝑥)(25 − 2𝑥)𝑥𝑐𝑚3
(ii)
𝑑𝑉
𝑑𝑥
= 4(3𝑥 − 50)(𝑥 − 5)
(iii)For extreme values
𝑑𝑉
𝑑𝑥
= 4(3𝑥 − 50)(𝑥 − 5) = 0
⇒ 𝑥 =
50
3
or 𝑥 = 5
𝑑2𝑉
𝑑𝑥2
= 24𝑥 − 260
∴
𝑑2𝑉
𝑑𝑥2
at 𝑥 = 5 is − 140 < 0
∴ 𝑉 is max 𝑤ℎ𝑒𝑛 𝑥 = 5
(iii) OR
For extreme values
𝑑𝑉
𝑑𝑥
= 4(3𝑥2
− 65𝑥 + 250)
𝑑2𝑉
𝑑𝑥2
= 4(6𝑥 − 65)
𝑑𝑉
𝑑𝑥
𝑎𝑡 𝑥 =
65
6
exists and
𝑑2𝑉
𝑑𝑥2
𝑎𝑡 𝑥 =
65
6
𝑖𝑠 0.
𝑑2
𝑉
𝑑𝑥2
𝑎𝑡 𝑥 = (
65
6
)
−
is negative and
𝑑2
𝑉
𝑑𝑥2
𝑎𝑡 𝑥 = (
65
6
)
+
is positive
⸫ 𝑥 =
65
6
is a point of inflection.
1
1
𝟏
𝟐
⁄
𝟏
𝟐
⁄
𝟏
𝟐
⁄
𝟏
𝟐
⁄
𝟏
𝟐
⁄
𝟏
𝟐
⁄
𝟏
𝟐
⁄
𝟏
𝟐
⁄
37. (i) Number of relations is equal to the number of subsets of the set B G
 = ( )
2
n B G

( ) ( ) 3 2 6
2 2 2
n B n G
 
= = =
( )
Where denotes the number of the elements in the finit
( e set
n A )
A
(ii) Number of functions =
( ) ( )
( )
( )
number of elements in the domain 3
numberof elementsintheco-domain 2
n B
n G
= =
(iii) (a) Number of one-one functions = 0, as number of elements in the co-domain ( )
G
is less than the number of elements in the domain ( )
B .
1
1
𝑥 − 1
32/7
=
𝑦 − 2
−34/7
=
𝑧 − 1
12/7
⇒
𝑥 − 1
16
=
𝑦 − 2
−17
=
𝑧 − 1
6
.
1

of 15
number of onto functions 3
2 2 6
= − =
Total number of functions =
( ) ( )
( )
( )
number of elements in the domain 3
numberof elementsintheco-domain 2
n B
n G
= =
Out of which there are two possibilities in which all the elements in the domain set
( )
B ( )
is mapped to the one element the co domain set G
−
So, number of onto functions 3
2 2 6
= − =
OR (iii) (b) One-one and onto function
( )
( ) ( )
( )( )
( )
 
2
2
2 2
1 1
1 2 1 2
2 2
1 2 1 2 1 2 1 2 1 2
4 . let
4
Let , 0,20 2 suchthat
4 4
0 as , 0,20 2
isone-onefunction
Now,0 200 hence the value of is nonnegative
and 2
for any arbitrary 0,200 , the pr
x
x y y f x
x x
x x f x f x
x x x x x x x x x x
f
y y
f y y
y
= = =
 
 =  =
 
 
 =  − + =  =   

 
=
  e image of existsin 0,20 2
hence is ontofunction.
y
f
 
 
1
1
1
1
38. Let 1
E be the event that the parrot and the owl fly from cage –I
𝐸2 be the event that two parrots fly from cage-I
𝐸3be the event that the owl is still in cage-I
𝐸3
𝑐
be the event that the owl is not in cage-I
𝑛(𝐸1 ∩ 𝐸3) = (5𝐶1
× 1𝐶1
)(7𝐶1
× 1𝐶1
) = 35
𝑛(𝐸1 ∩ 𝐸3
𝑐) = (5𝐶1
× 1𝐶1
)(7𝐶2
) = 105

of 15
𝑛(𝐸2 ∩ 𝐸3) = (5𝐶2
)(8𝐶2
) = 280
𝑛(𝐸2 ∩ 𝐸3
𝑐) = 0
n(S) = 35 + 105 + 280 + 0 = 420 (Total sample points)
(i) Probability that the owl is still in Cage–I = P(𝐸3) = P(𝐸1 ∩ 𝐸3) +P(𝐸2 ∩ 𝐸3)
=
35 + 280
420
=
315
420
=
3
4
(ii) The probability that one parrot and the owl flew from Cage-I to Cage-II given
that the owl is still in cage-I is 𝑃 (
𝐸1
𝐸3
)
𝑃 (
𝐸1
𝐸3
) =
P(𝐸1∩𝐸3)
P(𝐸3)
=
P(𝐸1∩𝐸3)
P(𝐸1∩𝐸3)+P(𝐸2∩𝐸3)
(by Baye’s Theorem)
=
35
420
315
420
=
1
9
1
1
1
1
2
1
2

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