SUPPORT MATERIAL

AISSCE, 2024-25

DAV

ODISHA
 TABLE OF CONTENTS

Sl Topic Page No.

No.
1. MCQs with Answer 2-12
(Chapter wise)
2. Assertion-Reason & Case 13-34
Study Questions with
Answer (Chapter wise)
3. Competency Based 35-80
Questions With Answer
4. Common Errors 81-87
5. 3-Unsolved Sample 88-107
Papers
6. CBSE Sample Paper- 108
2024-25

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 MULTIPLE CHOICE QUESTIONS
Relations and Functions
2𝑥
1. Consider the function f: R −{1}→R −{2} given by f(x) = , then:
𝑥−1

(A) f is one-one but not onto (B) f is onto but not one-one

(C) f is neither one-one nor onto (D) f is both one-one and onto

2. The function f: N −{1}→N defined by f(n)= the highest prime factor of n , is:

(A) both one-one and onto (B) one-one only (C) onto only (D) neither one-one nor onto

3. Let R be a relation on Z×Z defined by (a, b) R(c, d) if and only if ad−bc is divisible by 5,

Then R is

(A) Reflexive and symmetric but not transitive

(B) Reflexive but neither symmetric nor transitive

(C) Reflexive, symmetric and transitive

(D) Reflexive and transitive but not symmetric

4. f: A→B , A={1,2,3….,8}, B={1,2,3….,8} , find the number of one-one function from A to B

such that f (1) +f (3) =14

(A) 6! (B) 3×6! (C) 2×6! (D) None

5. A relation is (x1, y1) R(x2, y2) is defined as {(x, y) ∈N, x1≤ x2 ,y1≤y2} then the relation is:

(A) Reflexive and symmetric (B) Symmetric and transitive

(C) transitive and reflexive (D) none

6. The mapping f: N → N is given by f(n) = 1 + n2, n ∈ N when N is the set of natural numbers is:
(A) one-one and onto (B) onto but not one-one

(C) one-one but not onto (D) neither one-one nor onto
7. Let A={1,2,3,4} and B={1,2} Then, the number of onto functions from A to B is:

(A) 14 (B) 16 (C) 12 (D) 8

8. Let f: [2,∞)→R the function defined by f(x)=x2-4x+5 , then the range of f is:

(A) R (B) [1, ∞) (C) [4, ∞) (D) [5, ∞)

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9. If R is the smallest equivalence relation on the set{1,2,3,4} such that {(1,2),(1,3)}⊆ R, then the
number of elements in R is:

(A) 10 (B) 12 (C) 8 (D) 15

10. Let S= {1, 2, 3….20} be a given set. Relation R1 is defined as R1= {(x, y): 2x-3y=2, x, y∈ S} and

R2 = {(x, y): 4x=5y ,x, y∈S}. If m denotes number of elements required to make R1 symmetric and n
denote the number of elements to make R2 symmetric. Then the value of m+n is:

(A) 10 (B) 6 (C) 4 (D) 12

Answer Key:

1 2 3 4 5 6 7 8 9 10

D C A C C C A B A A

INVERSE TRIGONOMETRIC FUNCTIONS

1. Solution set of [sin−1 𝑥] > [cos −1 𝑥], where [ .] denotes the greatest integer function, is:
1
(A) [ , 1] (B) [cos1, sin1] (C) [sin1,1] (D)[cos1,1]
√2

√3
2. The value of sin {tan−1(−√3) + cos−1 (− )} is:
2

(A) 1 (B) − 1 (C) 0 (D) 2

3. In a triangle ABC, if A =tan−1 2 and B= tan−1 3 , then C is equal to:

𝜋 𝜋 𝜋 3𝜋
(A) 3 (B) 4 (C) 6 (D) 4

4. The value of sin−1 (𝑠𝑖𝑛 12) + cos−1 (𝑐𝑜𝑠 12)is equal to:

(A) 0 (B) 24 −2𝜋 (C) 4 𝜋 – 24 (D) 24

𝑛 𝜋
5. If cot −1 𝜋 > , n is a natural number, then the maximum value of n is:
6

(A) 1 (B) 5 (C) 9 (D) 7

1 4
6. Sin (2 cos−1 5 ) is equal to:
1 2 1 2
(A) (B) (C) (D)
√5 √5 √10 √10
−1 17𝜋
7. The value of 𝑠𝑖𝑛 {sin (− )} is:
8

𝜋 𝜋 𝜋 𝜋
(A) (B) − 4 (C) (D) − 8
4 8

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 Answer Key:

1 2 3 4 5 6 7

C A B A B C D

Matrices and Determinant

1. If A = 
2 - 1 2
 and A −4A −n = 0, where  is the unit matrix of order 2, then
- 1 2 
1 1
(A) n = − (B) n = (C) n = 3 (D) n = − 3
3 3

cos  sin  
2. If A =  and A.(adj(A)) =  
1 0
 then  is equal to:
- sin  cos 
  0 1  
(A) 1 (B) 2 (C) 3 (D) 1
2

3. If A is a square matrix such that A2 = A, then | A | equals:

(A) 0 or 1 (B) -2 or 2 (C) -3 or 3 (D) none of these

4. If A is a square matrix such that | A | = 2, then for any positive integer n, | An | is equal to:

(A) 0 (B) 2n (C) 2n (D) n2

2 0 0
5. If A = 0 2 0 then A5 is:
 
0 0 2
(A) 5A (B) 10A (C) 16A (D) 32A

6. If A and B are square matrices of order 3 such that | A | = −1, | B | = 3, then | 3AB | =

(A) −9 (B) −81 (C) −27 (D) 81

2 a 1 b 1 
7. If A = 
a b
 and A =   , then
b a  b1 a 1 
(A) a1 = a2 + b2, b1 = a2 - b2 (B) a1 = 2ab, b1 = a2 + b2

(C) a1 = a2 + b2, b1 = ab (D) a1 = a2 + b2, b1 = 2ab


8. If A = 
0
and B = 
1 0
  , then the value of  for which A2 = B is:
1 1 5 1

(A) 1 (B) − 1 (C) 4 (D) no real values

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 Answer Key:

1 2 3 4 5 6 7 8

C A A C A A D D

Continuity and Differentiability

 1 
tan x . tan -1  
 x -1
1. The set of points of discontinuity of the function f(x) = is:
x(x - 3)
   
(A) {0, 3, 5} (B)  
(2n + 1) , n  N  (C) {0, 3, 1} (2n + 1) n  N  (D) none of these
 2   2 

2. The function f(x) = x ( x − x + 1 ) is:

(A) Continuous but not differentiable at x = 0 (B) differentiable at x = 0

(C) Differentiable but not continuous at x = 0 (D) not differentiable at x = 0

 
tan  - x 
3. If f(x) = 4  , x   then the value which should be assigned to f at x =  so that it is
cot 2x 4 4
continuous everywhere is:
(A) 1 (B) 1 (C) 2 (D) none of these
2
 x
4. The set of all points where the function f(x) = 

, x 0 is differentiable is:
1 + e1 + x

 0 , x= 0
(A) (0, ) (B) R − {0} (C) (− , 0) (D) R
𝑑2 𝑥
5. If y = ex + sin x, then𝑑𝑦 2 is:
sin x - e x sin x - e x
(D) sin x + ex 3
x
(A) (−sin x + ex)-1 (B) (C)
(cos x + e x ) 2 (cos x + e x )3 (cos x + e )
6. If y = log | x |, then dy is:
dx
(A) 1 (B) − 1 (C) 1 (D) none of these
x x |x|
𝜋 𝑑𝑦
7. If sin-1 x + sin-1 y = 2 , then 𝑑𝑥 is:
(A) x (B) y (C) − x (D) − y
y x y x

8. If y = f 
2x - 1  2 dy
 and f(x) = sin x , then dx is:
 
2
x + 1

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 2 2
(A) cos x2. f (x) (B) −cos x2. f (x) (C) 2 (1 +2 x - 2x ) sin  2x2 - 1  (D) none of these
(x + 1) x +1  

9. If y2 = p(x), a polynomial of degree 3, then 2 d  y3 d y2  is equal to:

2

dx  dx 
(A) p(x) + p(x) (B) p(x) + p(x) (C) p(x) p(x) (D) a constant

10. The points on the curve y = 12x - x3 at which the gradient is zero, are:

(A) (0, 2), (2, 16) (B) (0, −2), (2, − 16) (C) (2, −16), (−2, 16) (D) (2, 16), (−2, −16)

Answer Key:

1 2 3 4 5 6 7 8 9 10

C B A B C A C C C D

Application of Derivatives

1. The value of a for which function f(x) = sin x − cos x − ax + b decreases for x  R is given by
(A) a  2 (B) a  1 (C) a < 2 (D) a < 1
2. The function f(x) = a sin x + 2 cos x is increasing for x  R when
sin x + cos x
(A) a < 1 (B) a > 1 (C) a < 2 (D) a > 2
3 2 2
3. If f(x) = x − ax + bx + 5 sin x is increasing on R, then a and b satisfy
(A) a2 −3b −15 > 0 (B) a2 − 3b + 15 < 0 (C) a2 −3b + 15 > 0 (D) a > 0, b > 0
4. If f(x) = (a + 2)x - 3ax + 9ax - 1 is decreasing on R, then ‘a’ lies in the interval
3 2

(A) (−,−2) (B) (−2, ) (C) (−3, 0) (D) (−,−3)

2
5. If x lies in [0, 1] then minimum value of x + x + 1 is:
(A) 3 b) 1 c) 3 d) none of these
4
6. On the interval [0, 1] the function f(x) = x25(1− x)75 takes its maximum value at the point
(A) 0 (B) 1 (C) 1 (D) 1
4 2 3
7. The minimum value of 2 + 27 is:
(x 2 - 3) 3

(A) 1 (B) 2 (C) 227 (D) none of these

8. The greatest value of f(x) = cos (xe[x] + 7x2 - 3x), x  [−1,) is:
(A) -1 (B) 1 (C) 0 (D) none of these

9. If f '(x) = (x − a)2n (x − b)2m + 1 where m, n  N, then

(A) x = b is a point of minimum (B) x = b is a point of maxima
(C) x = b is a point of inflexion (D) none of these
10. Maximum value of log x in [2,) is:
x
(A) log 2 (B) 0 (C) 1 (D) 1
2 e

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Answer Key:

1 2 3 4 5 6 7 8 9 10

A D B D B B A B A C

INTEGRALS
1−𝑥 2
1. ∫ 𝑒 𝑥 (1+𝑥 2 ) 𝑑𝑥 is equal to:
𝑒𝑥 −𝑒 𝑥 𝑒𝑥 −𝑒 𝑥
(A) 1+𝑥 2 + 𝑐 (B) 1+𝑥 2 + 𝑐 (C) (1+𝑥 2)2 + 𝑐 (D) (1+𝑥2 )2 + 𝑐
2
2. ∫−1|𝑥 − 1| 𝑑𝑥 is:
3 2
(A) 2 (B) 1 (C) 3 (D) None of these
(𝑥−4)
3. ∫ 𝑒 𝑥 (𝑥−2)3 𝑑𝑥 is equal to:

𝑒𝑥 𝑒𝑥 𝑒𝑥 𝑒𝑥
(A) +𝑐 (B)− (𝑥−2)2 + 𝑐 (C) (𝑥−2)3 + 𝑐 (D)(𝑥−2)2 + 𝑐
𝑥−2
𝜋
𝑡𝑎𝑛𝑥
4. The value of ∫02 𝑡𝑎𝑛𝑥+𝑐𝑜𝑡𝑥 dx is equal to:
𝜋 𝜋 7
(A) (B) (C) (D) none
2 4 8
1
5. ∫−1 𝑠𝑖𝑛5 𝑥𝑐𝑜𝑠 6 𝑥 𝑑𝑥 is equal to:

(A)0 (B) 1 (C) −1 (D) 𝜋

𝑏+𝑐
6. ∫𝑎+𝑐 𝑓(𝑥)𝑑𝑥 is equal to:
𝑏 𝑏 𝑏 𝑏−𝑐
(A) ∫𝑎 𝑓(𝑥 − 𝑐)𝑑𝑥 (B) ∫𝑎 𝑓(𝑥 + 𝑐)𝑑𝑥 (C) ∫𝑎 𝑓(𝑥)𝑑𝑥 (D) ∫𝑎−𝑐 𝑓(𝑥)𝑑𝑥

7. If 𝑓 and 𝑔 are continuous functions in [0, a] satisfying 𝑓(𝑥) = 𝑓(𝑎 – 𝑥)

𝑎
and 𝑔(𝑥) + 𝑔(𝑎 – 𝑥) = 𝑎 then ∫0 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 is equal to:
𝑎 𝑎 𝑎 𝑎 𝑎
(A) 2 (B) 2 ∫0 𝑓(𝑥)𝑑𝑥 (C) ∫0 𝑓(𝑥)𝑑𝑥 (D) 𝑎 ∫0 𝑓(𝑥)𝑑𝑥
2𝑎
8. If g(x) is a continuous function satisfying 𝑔(−𝑥) = − 𝑔(𝑥), then ∫0 𝑔(𝑥)𝑑𝑥 is equal to:
𝑎 𝑎 0
(A) 0 (B) 2 ∫0 𝑔(𝑥)𝑑𝑥 (C) ∫−𝑎 𝑔(𝑥)𝑑𝑥 (D) − ∫−2𝑎 𝑔(𝑥)𝑑𝑥
2
9. Integrate ∫0 cosec 7 x dx is equal to:

(A) 1 (B) 0 (C) 1 (D) 2

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 Answer Key:

1 2 3 4 5 6 7 8 9

C A D B A B C D B

APPLICATION OF INTEGRALS

1. The area bounded by the region {(𝑥, 𝑦): 𝑥 2 + 𝑦 2 ≤ 1 ≤ 𝑥 + 𝑦} is:

π 1 π 1 π 1 π 1
(A) 4 − 2 (B) 4 + 2 (C) 2 + 4 (D) 2 − 4

2. Area of the region in first quadrant enclosed by the x-axis, the line y=x and the circle
x 2 + y 2 = 32 is:
(A) 16π sq. units (B) 4π sq. units (C) 32π sq. units (D) 124π sq. units
2
3. The area of the region bounded by the curve y = 2x − x and the line 𝑦 = 𝑥 is:
1 1 1 7
(A) 6 sq. units (B) 2 sq. units (C) 3 sq. units (D) 6 sq. units
4. The area of the region bounded by the curve y = √32 − x 2 and x-axis in the first quadrant is
(A) 32π (B) 4π (C) 8π (D) 4√2π
5. The area (in sq. units) of the part of the circle x + y 2 = 36, which is outside the parabola
2

y 2 = 9x, is:
(A) 24π + 3√3 (B) 12π + 3√3 (C) 12π − 3√3 (D) 24π − 3√3
2
6. The area of the parabola x = y bounded by the line y = 1 is
1 2 4
(A) 3 (B) 3 (C) 3 (D)2
7. The area bounded by the parabola x = 4−y 2 and y- axis, in square units, is
2 32 33
(A) 32 (B) 3 (C) 2 (D) None of these
8. The area under the curve y = x 2 and the lines 𝑥 = −1, 𝑥 = 2 and x-axis is:
(A) 3 sq. units (B) 5 sq. units (C) 7 sq. units (D) 9 sq. units

Answer Key:

1 2 3 4 5 6 7 8

A B A C D C B A

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 DIFFRENETIAL EQUATION

1. The Sum of order and degree of the differential equation is:

1
𝑑𝑦 2 𝑑2 𝑦 3
√1 + ( ) = ( 2 ) is:
𝑑𝑥 𝑑𝑥
(A) 3 (B) 4 (C) 2 (D)1
2. The order of the differential equation whose general solution is given by
𝑦 = (𝑐1 + 𝑐2 ) 𝑐𝑜𝑠 (𝑥 + 𝑐3 )𝑥 − 𝑐4 𝑒 𝑥+𝑐5 then is:
(A) 3 (B) 5 (C) 2 (D) 1
𝑑𝑦 1+𝑦 2
3. The general solution of the differential equation 𝑑𝑥 = 1+𝑥 2 is:

(A)tan 𝑦 = tan 𝑥 + 𝐶 (B) tan 𝑦 = 𝐶 (C) tanx= C (D)𝑡𝑎𝑛−1 𝑦 = 𝑡𝑎𝑛−1 𝑥 + 𝐶

𝑑2 𝑦
4. If 𝑦 = 𝑎 sin 𝑚𝑥 + 𝑏 cos 𝑚𝑥, then is equal to:
𝑑𝑥 2
(A) – m y
2
(B) m y2
(C) −my (D) 𝑚𝑦
𝑑𝑦
5. The integrating factor of the differential equation 𝑑𝑥 (𝑥𝑙𝑜𝑔𝑥) + 𝑦 = 2𝑙𝑜𝑔𝑥 is:
(A) 𝑒 𝑥 (B) 𝑙𝑜𝑔𝑥 (C) log (𝑙𝑜𝑔𝑥) (D) 𝑥
𝑛 𝑑𝑦
6. The value of n, such that the differential equation 𝑥 𝑑𝑥 = 𝑦(𝑙𝑜𝑔𝑦 − 𝑙𝑜𝑔𝑥 + 1); (where
𝑥, 𝑦 ∈ 𝑅 + ) is homogeneous, is:
(A) 2 (B) 1 (C) 0 (D) 3
𝑑𝑦
7. The general solution of the differential equation 𝑒 = 𝑥 is: 𝑑𝑥

(A) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑐 (B) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 − 𝑥 + 𝑐 (C) 𝑦 = 𝑥𝑙𝑜𝑔𝑥 + 𝑥 + 𝑐 (D) 𝑦 = 𝑥 + 𝑐

𝑑𝑦
8. The general solution of the differential equation 2𝑥 𝑑𝑥 − 𝑦 = 3, represents a family of
(A) straight lines (B) parabolas (C) circles (D) ellipses
9. The general solution of the differential equation, 𝑠𝑖𝑛(𝑥 + 𝑦)𝑑𝑦 = 𝑑𝑥 is:
(A) 𝑠𝑒𝑐(𝑥 + 𝑦) − 𝑡𝑎𝑛(𝑥 + 𝑦) + 𝑦 = 𝐶
(B) 𝑠𝑒𝑐(𝑥 + 𝑦) + 𝑡𝑎𝑛(𝑥 + 𝑦) + 𝑦 = 𝐶
(C) 𝑠𝑒𝑐(𝑥 + 𝑦) − 𝑡𝑎𝑛(𝑥 + 𝑦) − 𝑦 = 𝐶
(D) 𝑠𝑒𝑐(𝑥 + 𝑦) + 𝑡𝑎𝑛(𝑥 + 𝑦) − 𝑦 = 𝐶
Answer Key:

1 2 3 4 5 6 7 8 9

B A D A B B B B A

VECTORS

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 8
1. For the vectors 𝑎⃗ = 2𝑖̂ + 6𝑗̂ + 3𝑘̂ and ⃗⃗
b = 𝜆𝑖̂ + 𝑗̂ − 4𝑘̂ . If the projection of ⃗⃗
b on 𝑎⃗ is , then
7
the value of 𝜆 is:
(A) −7 (B) 7 (C)1 (D) 8
2. Let 𝑎⃗ , 𝑏⃗⃗ and 𝑐⃗ be three mutually perpendicular vectors of same magnitude and equally inclined
at an angle 𝜃 with the vector𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗. Then 36 𝑐𝑜𝑠 2 2θ is equal to:
(A) 3 (B) 5 (C) 4 (D) 8
2
3. If a⃗⃗ is any vector, then |a⃗⃗ × î|2 + |a⃗⃗ × ĵ|2 + |a⃗⃗ × k̂| is:
(A) |a⃗⃗|2 (B) 2|a⃗⃗|2 (C) 3|a⃗⃗|2 (D) 4|a⃗⃗|2
4. If two vectors a⃗⃗ and ⃗⃗b are such that |a⃗⃗| = 2, |b ⃗⃗| = 3 and a⃗⃗. ⃗⃗
b = 4, then |a⃗⃗ − ⃗⃗⃗⃗⃗
2b| is equal to:

(A) √2 (B) 2√6 (C) 24 (D) 2√2

5. If |𝑎⃗| = 28, |𝑏⃗⃗| = 3 and |𝑎⃗ × 𝑏⃗⃗| = 12 then the value of 𝑎

⃗⃗⃗⃗. 𝑏⃗⃗ is :
(A) 6√3 (B) 4√3 (C) 12√3 (D) None of these

Answer Key:

1 2 3 4 5

B C B B D

THREE DIMENTIONAL GEOMETRY

𝑥−1 2𝑧+1
1. Direction ratios of a vector parallel to line = −𝑦 = are:
2 6

(A) 2, −1,6 (B)2,1,6 (C) 2,1,3 (D) 2, −1,3

𝑥 𝑦 𝑧
2. The angle which the line = −1 = 0 makes with the positive direction of Y − axis is:
1
5𝜋 3𝜋 5𝜋 7𝜋
(A) (B) (C) (D)
6 4 4 4
x−5 y−7 z+2 x+3 y−3 z−6
3. The point of intersection of the lines = = and = = is:
3 −1 1 −36 2 4
5 10
(A) (2, 10,4) (B) (21, 3 , 3 ) (C) (5,7,-2) (D) (−3,3,6)
4. The equation of a line passing through the point (−3,2, −4) and equally inclined to the axes are
(A) x − 3 = y + 2 = z − 4 (B) x + 3 = y − 2 = z + 4
x+3 y−2 z+4
(C) = = (D) None of these
1 2 3
5. If the direction ratios of a line are (𝜆 + 1,1 − 𝜆, 2) and the line makes an angle 60° with the y-
axis then a value of 𝜆 is:
(A) 1 + √3 (B) 2 − √3 (C) 3 + √5 (D) 2 − √5
6. If a line makes angles α, β, γ with the positive directions of the coordinate axes, then the value
of sin2𝛼 + sin2β + sin2𝛾 is
(A) 2 (B) 1 (C) −1 (D) −2
7. A line makes equal angles with coordinate axis. Direction cosines of this line are

Page | 10
 1 1 1 1 1 1 1 −1 −1
(A) ±(1,1,1) (B) ±( , , ) (C) ± ( , , ) (D) ± ( , , )
√3 √3 √3 3 3 3 √3 √3 √3

Answer Key:

1 2 3 4 5 6 7

D D B B D A B

LINEAR PROGRAMMING PROBLEM

1. The constraints of a linear programming problem are 𝑥 + 2𝑦 ≥ 3, 𝑥 ≥ 10, 𝑦 ≥ 0

Which of the following objective function has an optimal solution with respect to
above set of constraints?
(A) Minimize Z = 𝑥 + 𝑦
(B) Minimize Z = 0.5 𝑥 + 𝑦
(C) Maximize Z = 𝑥 + 𝑦
(D) Maximize Z = 2𝑥 + 𝑦
2. The point which does not lie in the half plane of 3𝑥 + 5𝑦 < 26
(A) (1, 2) (B) (1,5) (C) (1,4) (D) (3,2)
3. The corner points of the shaded unbounded feasible region of an LPP are

(0, 4).(0.6,1.6) and (3,0) as shown in figure.

The maximum value of the objective function Z = 4 x + 6 y occurs at

(A) (0.6,1.6) only
(B) (3, 0) only

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 (C) (0.6, 1.6) and (3, 0) only
(D) at every point of the – segment joining the points (0.6,1.6) and (3,0) .
4. The feasible region for an LPP is always a
(A) Regular polygon (B) Convex region (C) Unbounded region (D) Polyhedron
5. Corner points of the feasible region determined by the system of linear constraints are (0, 3),
(1, 1) and (3, 0). Let 𝑍 = 𝑝𝑥 + 𝑞𝑦, where 𝑝, 𝑞 > 0. Condition on 𝑝 and 𝑞, so that the
minimum of 𝑍 occurs at (1, 1) and (3,0) is:
(A) 𝑝 = 2𝑞 (B) 2𝑝 = 𝑞 (C) 𝑝 = 3𝑞 (D) 𝑝 = 𝑞
6. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is:
1 1 1
(A) 0 (B) 3 (C) 12 (D) 36
7. If the minimum value of an objective function 𝑍 = 𝑎𝑥 + 𝑏𝑦 occurs at two points (3, 4) and
(4, 3) then:
(A) 𝑎 + 𝑏 = 0 (B) 𝑎 = 𝑏 (C) 3𝑎 = 𝑏 (D) 𝑎 = 3𝑏

Answer Key:

1 2 3 4 5 6 7

A B D B B D B

PROBABILITY

1. Two events E and F are independent. If P(E)=0.3 and 𝑃(𝐸 ∪ 𝐹) = 0.5, then 𝑃(𝐸/𝐹) − 𝑃(𝐹/𝐸)
equals to
(A) 2/7 (B) 3/35 (C) 1/70 (D) 1/7

2. A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random
from the box without replacement, probability of drawing 2 green balls and one blue ball is
3 2 1 167
(A) 28 (B) 21 (C) 28 (D) 168
3 1 4
3. If P(B)=5, P(A/B)=2 and 𝑃(𝐴 ∪ 𝐵) = 5, then 𝑃(𝐴 ∪ 𝐵)′ + 𝑃(𝐴′ ∪ 𝐵) is equal to:
1 4 1
(A) 5 (B) 5 (C) 2 (D) 1
1 1 1
4. If A and B are two events such that P(A) =2, P(B)=3 and P(A/B)= 4 then 𝑃(𝐴′ ∩ 𝐵 ′ ) equals to:
1 3 1 3
(A) 12 (B) 4 (C) 4 (D) 16
2 5
5. If A and B are such that 𝑃(𝐴′ ∪ 𝐵 ′ ) = 3 and 𝑃(𝐴 ∪ 𝐵) = 9, then 𝑃(𝐴′ ) + 𝑃(𝐵 ′ ) is:
5 1 10 8
(A) 9 (B) 9 (C) (D) 9
9
6. Three persons A, B and C ,fire at a target in turn, starting with A. Their probability of hitting the
target are 0.4, 0.3 and 0.2, respectively. The probability two person hits the target is:
(A) 0.024 (B) 0.188 (C) 0.336 (D) 0.452

Page | 12
 7. If 𝑃(𝐴/𝐵) > 𝑃(𝐴), then which of the following is correct?
(A) P (B/A) < P (B) (B) P (A∩ 𝐵) < 𝑃(𝐴). 𝑃(𝐵) (C) P (B/A) > P (B) (D) P (B/A) = P (B)

Answer Key:

1 2 3 4 5 6 7

C A D C C B C

ASSERTION REASON AND CASE STUDY BASED QUESTIONS

Relation and function
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.
1. Assertion (A): The function 𝑓: 𝑅 + → 𝑅 𝑏𝑦 𝑓(𝑥) = 𝑥 2 + 1 is onto.
Reason (R): A function 𝑓 is onto if Range of the function is equal to Co-domain.
Answer: D
2. Assertion (A): The number of onto functions from a set P containing 5 elements to a set Q
containing 2 elements is 30.

Page | 13
 Reason (R): Number of onto functions from a set containing m elements to a set containing n
elements is nm.
Answer: (C)
3. Assertion (A): Let A = {1, -1, 2}, then define a relation on A as R = {(1, 2)} is transitive.
Reason (R): A relation R defined on a non-empty set A is said to be transitive if (a, b), (b, c)∈
R ⇒ (a, c) ∈ 𝑅
Answer: (A)
4. Assertion (A): The relation 𝑅 = {(𝑥, 𝑦): 𝑥 + 𝑦 is a prime number and 𝑥, 𝑦 ∈ 𝑁} is not a
reflexive relation.
Reason (R): The number “2n” is composite for all natural numbers n.
Answer: (B)

Case based Questions:

5. Students of a school are taken to a railway museum to learn about railway heritage and its
history.

An exhibit in the museum depicted many rail lines on the track near the railway station. Let L
be the set of all rail lines on the railway track and R be the relation on L defined by
R = {(l1, l2): l1 is parallel to l2}.
Based on the above information, answer the following questions.
(i) Find whether the relation is reflexive or not. [1
(ii) Find whether the relation is transitive or not. [1
(iii) If one of the rail lines on the rail track is represented by the equation
y = 2x + 1, then find the set of rail lines in R related to it. [2
OR
Let S be the relation defined by S = {(l1, l2): l1 is perpendicular to l2}. Check whether the
relation S is symmetric and transitive.
Answer:
(i) It is reflexive as each line is parallel to itself
(ii) It is transitive as if a line a is parallel to b and line b is parallel to a line c, then a is parallel to c.
(iii) y = 2x + c; c ∈ R
OR
It is transitive as a line a perpendicular to a line b ⇒ line b is perpendicular to line a
It is not transitive as if a line a is perpendicular to a line b and b is perpendicular to a line c, then
line a is parallel to line c.

Page | 14
6. An organization conducted bike race under two different categories – Boys and Girls.
There were 28 participants in all. 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 B = {b1, b2 , b3} and G = {g1 , g2}, where B represents
the set of Boys selected and G the set of Girls selected for the final race.

Based on the above information, answer the following questions.

(i) How many relations are possible from B×B to G? [1
(ii) Among all the possible relations from G to B, how many one-one functions can be formed
from G to B? [1
(iii) Let R: B→B be defined by R = {(x, y): x and y are students of the same sex}.
Check if R is an equivalence relation. [2
OR
A function f: B→G be defined by f = {(b1, g1), (b2 , g2), (b3 , g2)}.
Check if f is bijective. Justify your answer.
Answer:
(i) No. of relations from B to G = 29×2 = 218
(ii) No. of one-one functions from G to B = 0
(iii) Let x ∈B. That means, x is a boy.
Clearly (x, x) ∈ R for all x ∈B. Therefore, R is a reflexive relation.
Let x, y ∈ B. That means, x and y are boys.
If (x, y) ∈ R, then (y, x) ∈ R. Therefore, R is symmetric relation.
Let x, y, z ∈ B. That means, x, y and z are boys.
If (x, y) ∈R and (y, z) ∈R, then (x, z) ∈R. Therefore, R is transitive relation.
Hence, R is an equivalence relation.
OR
It is not bijective as it is not 1-1 as two elements of B makes relation with one element of B.

Inverse trigonometric function

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).

Page | 15
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
1. Assertion (A): 𝑠𝑖𝑛−1 (−𝑥) = −𝑠𝑖𝑛−1 𝑥; 𝑥 ∈ [−1,1]
𝜋 𝜋
Reason (R): 𝑠𝑖𝑛−1 : [−1,1] → [− 2 , 2 ] is a bijective function.
Answer: B
3𝜋 𝜋 5𝜋
2. Assertion (A): The range of the function f(x) = 2𝑠𝑖𝑛−1 𝑥 + 2 ; 𝑥 ∈ [−1,1] is [ 2 , 2 ]
𝜋 𝜋
Reason (R): The range of the principal value branch of 𝑠𝑖𝑛−1 𝑥 is [− 2 , 2 ]
Answer: (A)
Case based Questions
3. The Government of Odisha is planning to fix a hoarding board at the face of a building on the
road of a busy market for awareness on SUBHADRA YOJANA. Ram, Robert and Rahim are
the three engineers who are working on this project. “A” is considered to be a person viewing
the hoarding board 20 meters away from the building, standing at the edge of a pathway nearby.
Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different
locations namely C, D and E. “C” is at the height of 10 meters from the ground level. For the
viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of
elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure
given and based on the above information answer the following:

(i) Find measure ∠𝐷𝐴𝐵 [2

(ii) Find measure ∠𝐸𝐴𝐵 [2
Answer:
4
(i) 𝑡𝑎𝑛−1 3
11
(ii) 𝑡𝑎𝑛−1 2
4. In the school project Shrishti was asked to construct a triangle and name it as ABC. Two angles
1 1
A and B were given to be equal to 𝑡𝑎𝑛−1 2 and 𝑡𝑎𝑛−1 3 respectively.

Page | 16
 Based on the above information answer the following:
(i) Find sinA [2
(ii) If B =𝑐𝑜𝑠 −1 𝑥, find x. [2
Answer:
1
(i)
√5
3
(ii)
√10
Matrices

ASSERTION AND REASONING QUESTIONS

Questions given below are Assertion and 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 codes (A), (B), (C) and (D) are given below.
(A)Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C) Assertion (A) is true but Reason(R) is false.
(D) Assertion (A) is false but Reason(R) is true.
1. Assertion (A): For any symmetric matrix 𝐴, 𝐵 / 𝐴𝐵 is a skew-symmetric matrix.
Reason (R): A square matrix A is skew-symmetric if 𝐴/ = −𝐴
Answer: (D)
2. Assertion (A): Every scalar matrix is a diagonal matrix.
Reason (R): In a diagonal matrix, all the diagonal elements are 0.
Answer: (C)
3. Assertion (A): A matrix A = [ 1 2 3 4] is a row matrix of order 1× 4.
Reason (R): A matrix having one row and any number of columns is a row matrix.
Answer: (C)
4. Assertion (A): A and B are symmetric matrices of same order then AB-BA is also a symmetric
matrix.
Reason (R): Any square matrix A is said to be symmetric if A = AT.
Answer: (A)
Case based Questions
5. Read the text carefully and answer the questions:
A manufacturer produces three stationery products Pencil, Eraser and Sharpener which he sells in
two markets. Annual sales are indicated below:
Market Products (in numbers)

Pencil Eraser Sharpener

I 10,000 2,000 18,000

II 6000 20,000 8,000

Page | 17
 The he unit Sale price of Pencil, Eraser and Sharpener are₹ 2.50, ₹ 1.50 and ₹ 1.00 respectively.
Based on the information given above, answer the following questions:
(i) What is the total revenue collected from Market - I? [1
(ii) What is the total revenue collected from Market - II? [1
(iii) What is the gross profit from both markets considering the unit costs of the three
Commodities as₹ 2.00, ₹ 1.00, and 50 paise respectively? [2
OR
If matrix A =[𝑎𝑖𝑗 ]2×2 , where a 𝑖𝑗 = 1, if i ≠ j and a 𝑖𝑗 = 0 if i = j, then what isthe value of A 2 ?
[2
Answer
(i) Let A be the 2 × 3 matrix representing the annual sales of products in two markets.
𝑥 𝑦 𝑧
∴ 𝐴 = [10000 2000 18000]
6000 20000 8000

Let B be the column matrix representing the sale price of each unit of products x, y, z.
2.5
∴ 𝐵 = [1.5]
1
(i) Now, revenue = sale price × number of items sold
Therefore, the revenue collected from Market I = ₹ 46000.
(ii) The revenue collected from Market II = ₹ 53000.
(iii) Let C be the column matrix representing cost price of each unit of products x, y, z.

2
Then, 𝐶 = [ 1 ]
0.5
Total cost in each market is given by
2
10000 2000 18000 31000
= 𝐴𝐶 = [ ][ 1 ] = [ ]
6000 20000 8000 36000
0.5
Now, Profit matrix = Revenue matrix - Cost matrix
46000 31000 15000
=[ ]−[ ]=[ ]
53000 36000 17000

Therefore, the gross profit from both the markets = ₹ 15000 + ₹ 17000 = ₹ 32000
OR
0 1 2 0 1 0 1 1 0
Here 𝐴 = [ ], 𝐴 = [ ][ ]=[ ]
1 0 1 0 1 0 0 1
6. Rahul wants to invest Rs 30,000/- in two different types of bonds. The first bond pays 5% p.a
interest while other will give him 7%. Rahul will get Rs 1800/- annual interest.

Page | 18
 Based on above information answer the following questions:
(i) Represent the given condition in matrix form. [1
(ii) If he invests Rs 15,000 for two years at 7% per annum simple interest then how much
interest will he get? [1
(iii) If he invests Rs 15,000 for two years at 5% per annum simple interest then how much
interest will he get? [2
OR

What amount the Rahul will invest at 5% per annum interest rate? [2
Answer:
5%
(i) Matrix form is (x 30000-x)(( ) =1800
7%
(ii) Rs 2100

(iii) Rs 1500
OR
Rs 15000

Determinant
Questions number given below are Assertion and 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 codes (A), (B), (C) and (D) are given below.
(A)Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B)Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C)Assertion (A) is true but Reason(R) is false.
(D)Assertion (A) is false but Reason(R) is true.
1. Assertion (A): If A is a square matrix of order 3× 3, then the value of |2𝐴| = 8|𝐴|.
Reason (R): If k is a scalar and A is a square matrix of n × n, then |𝑘𝐴| = 𝑘 𝑛−1 |𝐴|.
Answer: (C)
5−𝑥 𝑥+1
2. Assertion (A): If A =[ ], then matrix A is singular if x = 3.
2 4
Reason (R): A square matrix is singular if its determinant is zero.
Answer: (A)
3. Assertion(A): If A is any invertible square matrix, then AT is invertible.
Reason (R): Inverse of invertible symmetric matrix is symmetric matrix.
Answer: (B)
1 0 3
4. Assertion(A): If 𝐴 = [2 1 1], then |𝑎𝑑𝑗(𝑎𝑑𝑗𝐴)| = 16.
0 0 2
2
Reason (R):|𝑎𝑑𝑗(𝑎𝑑𝑗𝐴)| = |𝐴|(𝑛−1) .
Answer: (A)

Page | 19
 Case based Questions:
5. Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of Rs 160. From the same
shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of Rs 190. Also Ankur
buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of Rs 250.
Based on the above information, answer the following questions.
(i) Convert the given above situation into a matrix equation of the form AX = B.
(ii) Find |𝐴|
(iii) Find A-1
OR
2
Determine P = 𝐴 − 5𝐴
Answer:
(i) Let cost of one pen, one bag and one instrument box be x, y and z (in Rs) respectively.
So, 5x + 3y + z =160, 2x + y + 3z =190, x + 2y + 4z = 250
5 3 1 𝑥 160
𝑦
(2 1 3) ( ) = (190)
1 2 4 𝑧 250
(ii) |𝐴| = −22
2 10 −8
1
(iii) A-1 = 22 ( 5 −19 13 )
−3 7 1
OR
7 5 13
P = A2 – 5A = (5 8 2 )
8 3 3
6. A scholarship is a sum of money provided to a student to help him or her pay for
education. Some students are granted scholarships based on their academic achievements, while
others are rewarded based on their financial needs.

Every year a school offers scholarships to girl children and meritorious achievers based on certain
criteria. In the session 2022-23, the school offered monthly scholarships of Rs 3,000 each to some
girl students and Rs 4, 000 each meritorious achievers in academic as well as sports.
In all, 50 students were given the scholarships and monthly expenditure incurred by the school on
scholarships was Rs 1, 80, 000.
Based on the above information, answer the following questions.
(i) Express the given information algebraically using matrices
(ii) Check whether the system of matrix equations so obtained is consistent or not.
(iii) Find the number of scholarships of each kind given by the school, using matrices.
OR
Had the amount of scholarship given to each girl child and meritorious student been
interchanged, what would be the monthly expenditure incurred by the school?

Page | 20
 Answer:
(i) Let No. of girl child scholarships = x
No. of meritorious achievers = y
𝑥 + 𝑦 = 50,3000𝑥 + 4000𝑦 = 18000 𝑜𝑟 3𝑥 + 4𝑦 = 180
1 1 𝑥 50
[ ][ ] = [ ]
3 4 𝑦 180
1 1
(ii) | | = 1 ≠ 0 ⇒ system is consistent.
3 4
(iii) 𝑥 = 20, 𝑦 = 30
OR
Required expenditure = ₹[30(3000)+20(4000)] = ₹1,70,000

Continuity and differentiability

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.
3x if x < 0
1. Assertion (A): The function f(x) = 0 if 0 ≤ x ≤ 1 is continuous everywhere except at
2x if x > 1
x=1
Reason(R): Polynomial and constants functions are always continuous.
Answer: A
2. Assertion (A): Every differentiable function is continuous but converse is not true.
Reason(R): f(x) = |x| is continuous.
Answer: B
3. Assertion (A): The function f(x) = |sinx| is continuous function.
Reason(R): The function f(x) = sin|𝑥| is continuous function.
Answer: B
4. f(x) = [x-1] + |x − 2|, where [.] denotes the greatest integer function.
Assertion (A): f(x) is discontinuous at x = 2.
Reason(R): f(x) is not derivable at x = 2.
Answer: B
Case based Questions
5. A potter made a mud vessel, where the shape of the pot is based on f(x)=|x−3|+|x−2|, where f(x)
represents the height of the pot.
Based on the above information, answer the following questions
(i) When x > 4 what will be the height in terms of x. [1
𝑑𝑦
(ii) Find 𝑑𝑥 at x = 2.5 [1
𝑑𝑦
(iii) Find 𝑑𝑥 at x = 3 [2
OR
If the potter is trying to make a pot using the function f(x) = [x], will he get pot or not? why?
Answer

Page | 21
 (i) 2x – 5
(ii) 0
(iii) Not exists
OR
It is discontinuous

6. Let f (x) be a real valued function. Then its

𝑓(𝑎−ℎ)−𝑓(𝑎)
• Left Hand Derivative (L.H.D.): 𝐿𝑓 / (𝑎) = lim
ℎ→0 −ℎ
𝑓(𝑎+ℎ)−𝑓(𝑎)
• Right Hand Derivative (R.H.D.): 𝑅𝑓 / (𝑎) = lim
ℎ→0 ℎ
Also, a function f (x) is said to be differentiable at x = a if its L.H.D. and R.H.D. at x = a exist
and both are equal.
|𝑥 − 3| 𝑥≥1
For the function f(x) = 𝑥 2 3𝑥 13 answer the following questions
− + 𝑥 < 1
4 2 4
(i) What is R.H.D. of f(x) at x =1?
(ii) What is L.H.D. of f(x) at x = 1?
(iii)Check if the function f (x) is differentiable at x = 1
OR
/ (2) / (−1)
Find 𝑓 and 𝑓
Answer:
𝑥2 3𝑥 13
|𝑥 − 3| 𝑥≥1 − + 𝑥<1
4 2 4
(i) f(x) = 𝑥2 3𝑥 13 −(𝑥 − 2) = 1≤𝑥<2
− + 𝑥<1
4 2 4
𝑥−2 𝑥≥2
𝑓(1+ℎ)−𝑓(1) −(1+ℎ)+3−2
RHD = lim = lim = -1
ℎ→0 ℎ ℎ ℎ→0
(1−ℎ)2 3(1−ℎ) 13
𝑓(1−ℎ)−𝑓(1) − + −2
4 2 4
(ii) LHD = lim = lim = -1
ℎ→0 −ℎ ℎ→0 ℎ
(iii) As LHD and RHD are equal, so function f (x) is differentiable at x = 1
OR
/ (2) / (−1)
𝑓 = 1, 𝑓 =-2

Application of derivative
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.
𝑑
1. Let f (x) be a polynomial function of degree 6 such that 𝑑𝑥 (𝑓(𝑥)) = (𝑥 − 1)3 (𝑥 − 3)2 then
Assertion (A): f(x) has a minimum at x = 1
𝑑 𝑑
Reason (R): When 𝑑𝑥 (𝑓(𝑥)) < 0 for all x ∈ (𝑎 − ℎ, 𝑎) and 𝑑𝑥 (𝑓(𝑥)) > 0 for all x ∈ (𝑎, 𝑎 + ℎ)
where 'h' is an infinitesimally small positive quantity, then f(x) has a minimum at x = a. provided
f(x) is continuous at x = a.

Page | 22
 Answer: (A)
2. Assertion (A): 𝑓(𝑥) = −2 + |𝑥 − 1| has minimum value at x = 1.
𝑑 𝑑
Reason (R): When 𝑑𝑥 (𝑓(𝑥)) < 0 for all x ∈ (𝑎 − ℎ, 𝑎) and 𝑑𝑥 (𝑓(𝑥)) > 0 for all x ∈ (𝑎, 𝑎 + ℎ)
where 'h' is an infinitesimally small positive quantity, then f(x) has a minimum at x = a. provided
f(x) is continuous at x = a.
Answer: (A)
Case based Questions
3. An owner of a car rental company has determined that if they charge customers Rs. x per day to
rent a car, where 50 ≤x≤ 200, then number of cars (n), they rent per day can be shown by linear
function n(x) = 1000 – 5x. If they charge Rs. 50 per day or less they will rent all their cars. If they
charge Rs.200 or more per day they will not rent any car.

Based on the above information, answer the following question.

(i) Express total revenue R as a function of x. [1
(ii) At x = 230, find the revenue collected by the company. [1
(iii) If R(x) denote the revenue, then find x for maximum value of R(x). [2
OR
Find maximum revenue collected by company.
Answer
Let x be the price charge per car per day and n be the number of cars rented per day.
(i) R(x) = n.x = (1000 – 5x).x = –5x2 + 1000x
(ii) If company charge Rs.200 or more, they will not rent any car. Then revenue collected by
him will be zero.
(iii) We have, R(x) = 1000x – 5x2 ⇒R′(x)= 1000 –10x
For R(x) to be maximum or minimum, R′(x)=0 ⇒1000 –10x=0⇒x=100
Also, R″ (100) = –10 < 0, Thus, R(x) is maximum at x = 100
OR
At x = 100, R(x) is maximum.
Maximum revenue = R (100) = –5(100)2 + 1000(100) =Rs. 50,000
4. The temperature of a person during an intestinal illness is given by 𝑓(𝑥) = −0.1𝑥 2 + 𝑚𝑥 + 98.6;
0 ≤ 𝑥 ≤ 12 , m being a constant, where f(x) is the temperature in °F at x days.

Page | 23
 Based on the above information, answer the following questions
(i) Is the function differentiable in the interval (0, 12)? Justify your answer. [1
(ii) If 6 is the critical point of the function, then find the value of m. [1
(iii) Find the intervals in which the function is strictly increasing or strictly decreasing. [2
OR
Find the points of local maximum/local minimum, if any, in the interval (0, 12) as well as the
points of absolute maximum/absolute minimum in the interval [0, 12]. Also, find the corresponding
local maximum/local minimum and the absolute maximum/absolute minimum values of the
function.
Answer:
(i) Being a polynomial function, is differentiable everywhere, hence, differentiable in (0, 12)
(ii) 𝑓 / (𝑥) = −0.2𝑥 + 𝑚, 𝑓 / (6) = 0 ⇒ m = 1.2
(iii) 𝑓 / (𝑥) = −0.2𝑥 + 1.2 = - 0.2(x-6) > 0 ⇒ x < 6 ⇒ It is increasing in (0,6). It is decreasing
in(6,12).
OR
/ (𝑥)
𝑓 = - 0.2(x-6) = 0 ⇒ x = 6
𝑓 // (𝑥) = −0.2 ⇒ 𝑓 // (6) = −0.2 < 0
x = 6 is a point of local maximum. The local maximum value = (6) = 102.2
We have (0) = 98.6, (6) = 102.2, (12) = 98.6.
6 is the point of absolute maximum and the absolute maximum value of the
function = 102.2. 0 and 12 both are the points of absolute minimum and the absolute
minimum value of the function = 98.6.
Integrals
Questions number given below are Assertion and 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 codes (A), (B), (C) and (D) are given below.
(A)Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B)Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C)Assertion (A) is true but Reason(R) is false.
(D)Assertion (A) is false but Reason(R) is true.
𝑑𝑥 1 𝑥−3
1. Assertion (A): The value of∫ 𝑥 2 −6𝑥+13 = 2 tan−1 ( 2 ) + 𝑐
1 1 𝑥
Reason (R): 𝑥 2+𝑎2 = 𝑎 tan−1 (𝑎) + 𝑐
Answer-(A)
π/3 dx π
2. Assertion (A): The value of the integral ∫π/6 dx is 6
1+√ tanx
𝑏 𝑏
Reason (R): ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑎 + 𝑏 − 𝑥)𝑑𝑥.
Answer: (A)
𝜋
3. Assertion(A): The value of ∫ 2 𝜋 si𝑛7 x 𝑑𝑥 is 0
−
2
7
Reason(R): f(x) = sin x is an even function
Answer: (C)

Page | 24
 𝑎
4. Assertion (A): Let 𝑎 be a positive real number such that ∫0 𝑒 𝑥−[𝑥] 𝑑𝑥 = 10𝑒 − 9, [𝑥] is the
greatest integer less than or equal to 𝑥. Then value of 𝑎 is 10 + log 𝑒 2
Reason (R): 𝑥 − [𝑥] is a periodic function whose period is unity
Answer: (A)
Case based Questions
5. The definite integral are used to calculate the area bounded by the curve of the given function
𝑎
within the given limit. The definite integral is given by ∫𝑏 𝑓(𝑥)𝑑𝑥 =g(b)-g(a).
Based on the above information answer the following
2 𝑘
i) If ∫0 (𝑥 2 + 2𝑥)𝑑𝑥 = 3 then find the value of k. [1
𝑎
ii) If a=b then find the value of the definite integral ∫𝑏 𝑓(𝑥)𝑑𝑥. [1
1
iii) Find ∫−1 𝑥 3 𝑑𝑥 using the properties of definite integral. [2
𝜋
iv) Find ∫−𝜋|𝑐𝑜𝑠𝑥| + |𝑠𝑖𝑛𝑥|dx [2
Answer:
i) k=20 ii) 0 iii) 0 iv) 0 as odd function
6. If f(x) is a continuous function defined on[-a, a]
𝑎
𝑎 ∫−𝑎 𝑓(𝑥)𝑑𝑥 𝑖𝑓 𝑓(𝑥)𝑖𝑠 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
then ∫−𝑎 𝑓(𝑥)𝑑𝑥 =2 {
0 𝑖𝑓 𝑓(𝑥)𝑖𝑠 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
Based on the above information, answer the following questions:
𝜋
i) Evaluate ∫ 2𝜋|𝑐𝑜𝑠𝑥| + |𝑠𝑖𝑛𝑥|dx [2
−
2
𝜋
3 𝑠𝑒𝑐 2 𝑥
ii) Evaluate ∫ 𝜋 dx [2
− 𝑡𝑎𝑛2 𝑥+3
3
Answer
𝜋
i) 4 ii) 2√3
APPLICATION OF INTEGRALS
Questions number given below are Assertion and 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 codes (a), (b), (c) and (d) are given below.
(A)Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B)Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C)Assertion (A) is true but Reason(R) is false.
(D)Assertion (A) is false but Reason(R) is true.
1. Assertion (A): Area bounded by the curve 𝑦 = 𝑓(𝑥), the x-axis and between x = a and
𝑏
x = b is ∫𝑎 𝑓(𝑥)𝑑𝑥.
Reason (R): For area bounded by the curve and given ordinates we use y in terms of x.
Answer: (A)
2. Assertion (A): Area bounded by the curve 9x2 +25y2 = 225, is 15𝜋𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
Reason (R): Area of ellipse whose length major axis is 2a and length of minor axis is 2b is 𝜋𝑎𝑏
sq.units.
Answer: (A)

Page | 25
3. Assertion (A): The closest to the area under the parabola 𝑦 = 4𝑥 2 , bounded by the x-axis, and
the lines x = - 1 and x = - 2 is 9
Reason (R): The area bounded by the curves 𝒚 = |𝒙 − 𝟏| + |𝒙 − 𝟐| and 𝒚 = 𝟑 is 4
Answer: (B)
Case based Questions
4. An architect designs a building whose lift (elevator) is from outside of the building attached to
the walls. The floor (base) of the lift (elevator) is in semicircular shape. The floor of the elevator
(lift) whose circular edge is given by the equation 𝑥 2 + 𝑦 2 = 4 and the straight edge (line) is
given by the equation 𝑦 = 0.

Based on the above situation answer the following questions.

i) Find the point of intersection of the circular edge and straight-line edge. [1
ii) Find the length of each vertical strip of the region bounded by the given curves.[1
iii) Find the area of a vertical strip between given circular edge and straight edge. [2
OR
Find the area of the region of the floor of the lift of the building (in square units).
Ans: i) (2,0), (-2, 0) ii) √4 − 𝑥 2 iii) √4 − 𝑥 2 𝑑𝑥 iv) 2𝜋

5. In the common wealth games event 2010, a player won the gold medal in his game. The shape
of the gold medal is as shown in the figure given below that is elliptical in shape.
𝑥2 𝑦2
Assume that the gold medal is described by the ellipse 36 + 25 = 1, where it is kept on xy-
plane.

Based on the above situation answer the following questions.

Page | 26
 (i) Find the area of ABCOA using integration. [2
(ii) Find the area of elliptical gold medal using integration. [2

Answer:
(i) 15𝜋sq. unit (ii) 30𝜋 sq unit
Differential equation
Questions number given below are Assertion and 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 codes (a), (b), (c) and (d) are given below.
(A) Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B) Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C) Assertion (A) is true but Reason(R) is false.
(D) Assertion (A) is false but Reason(R) is true.
𝑑𝑦 𝑥 4 −𝑦 4
1. Assertion (A): The degree of the differential equation = (𝑥 2 +𝑦 2)𝑥𝑦 is 1.
𝑑𝑥
Reason (R): The degree of a differential equation is the degree of the highest order derivative
when differential coefficients are free from radicals and fractions.
Answer: (A)
2. Assertion (A): The order of the differential equation of all circles of given radius 𝑎 is 2
Reason (R): The equation of given family be (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑎2
Answer: (A)
𝑑𝑦
3. Assertion(A): The general solution of the differential equation 𝑑𝑥 = 1 + 2(𝑦/𝑥) is
𝑥 + 𝑦 = 𝐶𝑥²
𝑑𝑦 𝑦
Reason(R): The solution of the differential equation 𝑑𝑥 = 𝑓 (𝑥 ) is obtained by substituting 𝑦 =
𝑣𝑥.

Answer: (A)

Case based Questions

dy
4. If an equation is of the form dx + P(x)y = Q(x), where P, Q function of x, then such equation is
known as linear differential equation. Its solution is given by y. (I. F) = ∫ Q. (I. F)dx +
C , where I. F = e∫ Pdx .
dy
Now suppose the given equation is (1 + sinx) dx + y cosx + x = 0.

Based on the above information, answer the following question:

(i) Find the value of P and Q. [1

(ii) Evaluate I.F. [1
dy 𝑦
(iii) Solve the LDE +𝑥 =𝑥 [2
dx

Answer:

Page | 27
 𝑥3
𝑐𝑜𝑠𝑥 +𝑐
3
i) 1+𝑠𝑖𝑛𝑥 ii) 1+sinx iii) 𝑥
5. A bacteria sample of certain number of bacteria is observed to grow exponentially in a given
amount of time. Using exponential growth model, the rate of growth of this sample of bacteria
𝑑𝑝
is calculated. The differential equation representing the growth of bacteria is given as : =
𝑑𝑡
𝑘𝑝, where p is the population of bacteria at any time t.

Based on the above information, answer the following questions:

(i) Obtain the general solution of the given differential equation and express it as an
exponential function of t. [2
(ii) If the population of bacteria is 1000 at t = 0, and 2000 at t = 1, find the value of k. [2
Answer:
𝑑𝑃
(i) Given that: i) = 𝑘𝑃
𝑑𝑡
𝑑𝑃 𝑑𝑃
⇒ 𝑃 = 𝑘𝑑𝑡 ⇒ ∫ = ∫ 𝑘𝑑𝑡 ⇒log|𝑃| = 𝑘𝑡 + 𝑐 ⇒ 𝑝 = 𝑒 𝑘𝑡+𝑐
𝑃
(ii) When t = 0, P = 1000
⇒ log1000 = c
1
∴ log|𝑃| = 𝑡 + 𝑙𝑜𝑔1000
20
When t = 1, P = 2000
log 2000 = 𝑘 + 𝑙𝑜𝑔1000 ⇒ k = log2
Vector
Questions number given below are Assertion and 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 codes (a), (b), (c) and (d) are given below.
(A)Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B)Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C)Assertion (A) is true but Reason(R) is false.
(D)Assertion (A) is false but Reason(R) is true.
2
1. Assertion (A): If 𝑎⃗ is any vector, then the value of |𝑎⃗ × 𝑖̂|2 + |𝑎⃗ × 𝑗̂|2 + |𝑎⃗ × 𝑘̂| is 2𝑎2
Reason (R): Cross product of 𝑎⃗ × 𝑏⃗⃗ is ab cos𝜃.
Answer: (C)
2 2
2. Assertion (A): If |𝑎⃗ × 𝑏⃗⃗| + |𝑎⃗. 𝑏⃗⃗| =144 and |𝑎⃗| = 4 then |𝑏⃗⃗| is equal to 9
Reason (R): If 𝑎⃗ and 𝑏⃗⃗ are any two vectors then is any vector then

Page | 28
 2 2 2
|𝑎⃗ × 𝑏⃗⃗| = |𝑎⃗|2 |𝑏⃗⃗| − |𝑎⃗. 𝑏⃗⃗|
Answer: (C)
3. Assertion (A): For any two non-zero, unit vector 𝑎⃗ and 𝑏⃗⃗, 𝑎⃗. 𝑏⃗⃗ = 𝑏⃗⃗ . 𝑎⃗
Reason (R): For any two non-zero, unit vector 𝑎⃗ and 𝑏⃗⃗, 𝑎⃗ × 𝑏⃗⃗ = 𝑏⃗⃗ × 𝑎⃗
Answer: (D)

Case based Questions

4. Read the following passage and answer the questions given below.

Employee in a office are following social distance and during lunch they are sitting at places
( ) (
marked by points A, B and C. Each one is representing position as A ˆi − 2ˆj + 4kˆ ,B 5iˆ + 2kˆ )
(
and C 3iˆ + 2ˆj + 4kˆ )
(i) Find the distance between B and C. [2
(ii) Find the area of a triangle with vertices A, B and C. [2
Answer:
(i) 2√3 (ii) 4√14
5. A barge is pulled into harbour by two tug boats as shown in the figure

Based on the above information, answer the following questions

(i) Find the position vector of A. [1
(ii) Find the position vector of B. [2
⃗⃗⃗⃗⃗⃗
(iii) Find the vector 𝐴𝐵 [2
OR
Find vector ⃗⃗⃗⃗⃗⃗
𝐴𝐶 .
Answer:
i) Price = Rs (300- 3x), Quantity = 80 + x quintals
ii) Revenue = Quantity × Price = (80 + x)(300-3x) = 24000 + 60x - 3𝑥 2

Page | 29
 iii) R/(x) = 60 – 6x = 0 ⇒ x = 10, R//(10) < 0, So x = 10
OR
Maximum revenue = R(10) = 24300

Three-dimensional geometry
Questions number given below are Assertion and 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 codes (a), (b), (c) and (d) are given below.
(A)Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B)Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C)Assertion (A) is true but Reason(R) is false.
(D)Assertion (A) is false but Reason(R) is true.
1. Assertion (A): If a line makes an angle 𝛼, 𝛽 𝑎𝑛𝑑 𝛾 with X-axis, Y-axis and Z-axis respectively,
then cos2𝛼 + 𝑐𝑜𝑠2𝛽 + 𝑐𝑜𝑠2𝛾 = -1
Reason(R): If l, m , n are direction cosines of a line, then 𝑙 2 + 𝑚2 + 𝑛2 =1.
Answer: (A)
2. Assertion (A): The direction ratios of the line 3𝑥 + 1 = 6𝑦 − 2 = 1 − 𝑧 are 2, 1, 6
Reason(R): If equation of a line passing through (1,2,3) with drs (a,b,c) then equation of a line is
𝒙−𝟏 𝒚−𝟐 𝒛−𝟑
= =
𝒂 𝒃 𝒄
Answer: (D)
3. Assertion (A): The value of λ for which the angle between the lines
𝜋
𝑟⃗ = 𝑖̂ + 𝑗̂ + 𝑘̂ + p(2𝑖̂ + 𝑗̂ + 2𝑘̂) and 𝑟⃗ = (1 + 𝑞)𝑖̂ + (1 + 𝑞𝜆)𝑗̂ + (1 + 𝑞)𝑘̂ is 2 is – 4.
Reason R: Let 𝜃 be the angle between the lines
𝑟⃗=𝑥1 𝑖̂ + 𝑦1 𝑗̂ + 𝑧1 𝑘̂ + λ(𝑎1 𝑖̂ + 𝑏1 𝑗̂ + 𝑐1 𝑘̂) and
|𝑎1 𝑎2 +𝑏1 𝑏2 +𝑐1 𝑐2 |
𝑟⃗ = 𝑥2 𝑖̂ + 𝑦2 𝑗̂ + 𝑧2 𝑘̂ + λ(𝑎2 𝑖̂ + 𝑏2 𝑗̂ + 𝑐2 𝑘̂) is co𝑠 𝜃 =
√𝑎1 2 +𝑏1 2 +𝑐1 2 √𝑎2 2 +𝑏2 2 +𝑐2 2

Answer: (A)
4. Assertion (A): The vector equation of the line passing through the points (6,-4,5) and (3,4,1)
is 𝑟⃗ = 6𝑖̂ − 4𝑗̂ + 5𝑘̂ + λ (-3𝑖̂ + 8𝑗̂ + 4𝑘̂).
Reason (R): The vector equation of the line passing through the points 𝑎⃗ and 𝑏⃗⃗ is
𝑟⃗ = 𝑎⃗ + 𝜆𝑏⃗⃗.
Answer: (D)
Case based Questions
5. Read the following passage and answer the questions given below.

Two motorcycles A and B are running on the road along two lines 𝑟⃗ = 𝜆(𝑖̂ + 2𝑗̂ − 𝑘̂) and

𝑟⃗ = 3𝑖̂ + 3𝑗̂ + 𝜇(2𝑖̂ + 𝑗̂ + 𝑘̂) respectively.

Page | 30
 (i) Find the shortest distance between the given lines. [2
(ii) Find the point at which motorcycles will meet with an accident. [2
Answer:
i) 𝑎⃗2 − 𝑎⃗1 = (3𝑖̂ + 3𝑗̂) − (0𝑖̂ + 0𝑗̂ + 0𝑘̂) = 3𝑖̂ + 3𝑗̂
𝑖̂ 𝑗̂ 𝑘̂
𝑏2 × 𝑏1 = |1 2 −1| = 3𝑖̂ − 3𝑗̂ − 3𝑘̂
⃗⃗ ⃗⃗
2 1 1
(𝑎⃗⃗2 −𝑎⃗⃗1 ).(𝑏⃗⃗1 ×𝑏
⃗⃗2 )
Shortest distance = ⃗⃗ ⃗
|𝑏1 ×𝑏2 |⃗
=0
(ii) 𝜆(𝑖̂ + 2𝑗̂ − 𝑘̂) = (3 + 2𝜇)𝑖̂ + (3 + 𝜇)𝑗̂ + 𝜇𝑘̂

𝜆 = 3 + 2𝜇; 2𝜆 = 3 + 𝜇; −𝜆 = 𝜇

On solving above equation, we get point of intersection (1, 2, –1).

6. Read the following passage and answer the questions given below
An enemy country fired a missile along the straight line 𝑟⃗ = (−2𝑖̂ + 3𝑗̂ + 5𝑘̂) + λ(9𝑖̂ − 3𝑗̂ −
6𝑘̂). Another iron dome missile moving along the straight line 𝑟⃗ = (−3𝑖̂ − 2𝑗̂ − 5𝑘̂) +
µ(6𝑖̂ + 6𝑗̂ + 12𝑘̂ )

(i) Find the directions of both missile and iron drone. [2

(ii) Find the point where iron drone will destroy the missile. [2
Answer:
9 −3 −6 6 6 12
(i) Direction of missile are , , and Direction of drone are , ,
√126 √126 √126 √216 √216 √216
(ii) At the point of intersection
(−2𝑖̂ + 3𝑗̂ + 5𝑘̂) + λ(9𝑖̂ − 3𝑗̂ − 6𝑘̂) = (−3𝑖̂ − 2𝑗̂ − 5𝑘̂) + µ(6𝑖̂ + 6𝑗̂ + 12𝑘̂)
⇒ -2 + 9 λ = -3 + 6µ, 3 -3 λ = - 2 + 6 µ, 5 - 6 λ = - 5 + 12µ
1
⇒ λ=3
putting it in equation of first line, we get point of intersection as (1, 2, 3).

Page | 31
 Linear Programming Problem
Questions number given below are Assertion and 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 codes (a), (b), (c) and (d) are given below.
(A)Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B)Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C)Assertion (A) is true but Reason(R) is false.
(D)Assertion (A) is false but Reason(R) is true.
1. Assertion (A): Based on the given shaded region as the feasible region in the graph, at point D
the objective function Z = 3x + 9y is maximum.

Reason (R): The optimal solution of a LPP having bounded feasible region must occur at corner
points
Answer: (D)
2. Assertion (A): The shaded region in the figure represents the solution of the system of
inequalities 𝑥 + 𝑦 ≤ 7, 2𝑥 − 3𝑦 + 6 ≥ 0, 𝑥 ≥ 0, 𝑦 ≥ 0

Reason (R): LPP is a process of finding optimum value of objective functions which lies in
feasible region.
Answer: (A)
Case based Questions
3. Read the following passage and answer the questions given below
The feasible region for an LPP is shown shaded in the figure. Let F = 3x-4y be the objective
function.

Page | 32
 (i) Find maximum value of F
(ii) Find the value of maximum F + minimum F.
Answer: (i) 0 (ii) 0 + (-46) = -46
4. Read the following passage and answer the questions given below
The feasible region of a linear programming problem is shown in the figure below:

(i) Find all the constraints. [2

(ii) If the corner points of the bounded feasible region determined by a system of linear
constraints are (0, 3), (1, 1) and (3, 0). Let 𝑍 = 𝑝𝑥 + 𝑞𝑦; p, q > 0. Find condition on p and
q so that minimum of Z occurs at (3, 0) and (1, 1). [2
Answer:
(i) 𝑥 + 2𝑦 ≥ 4, 𝑥 + 𝑦 ≥ 3, 𝑥 ≥ 0, 𝑥 ≥ 0
(ii) 2𝑝 = 𝑞

Probability
Questions number given below are Assertion and 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 codes (a), (b), (c) and (d) are given below.
(A)Both Assertion (A) and Reason(R) are true and Reason(R) is the correct explanation of
Assertion (A).
(B)Both Assertion (A) and Reason(R) are true but Reason(R) is not the correct explanation of A.
(C)Assertion (A) is true but Reason(R) is false.
(D)Assertion (A) is false but Reason(R) is true.
1. Assertion (A): If A and B are independent event then 𝑃(𝐴 ∩ 𝐵)′ = P(𝐴′ )P(𝐵 ′ )
Reason (R): 𝑃(𝐴 ∪ 𝐵)′ = 1+P(A∪B)

Answer: (D)
2. Assertion (A): Two dice are rolled that and it is given that the sum of the number on both the
5
dice is greater than 8. The probability of getting a doublet is
16
𝑃(𝐸∩𝐹)
Reason (R): Probability of an event E when F is given is P(E/F) = 𝑃(𝐹)
Answer: (D)
Case based Questions
3. Read the following passage and answer the questions given below.

Page | 33
 Ragini was doing a project on a school survey, on the
average number of hours spent on study by students in
hostel selected at random. At the end of survey, Ragini
prepared the following report related to the data.

Let X denotes the average number of hours spent on study

by students. The probability that X can take the values x,
has the following form, where k is some unknown
constant.
0.2, 𝑖𝑓 𝑥 = 0
𝑘𝑥, 𝑖𝑓 𝑥 = 1 𝑜𝑟 2
𝑃(𝑋 = 𝑥) = {
𝑘(6 − 𝑥), 𝑖𝑓 𝑥 = 3 𝑜𝑟 4
0. 𝑂𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

(i) Find the value of k.

(ii) What is the probability that the average study time of students is not more than 1 hour?
(iii) What is the probability that the average study time of students is at least 3 hours?
OR
What is the probability that the average study time of students is exactly 2 hours?
Answer:

(i) We know that ∑ 𝑃𝑖 = 1

Then, 0.2 + k + 2k + 3k + 2k + 0 = 1 ⇒ 8k = 1 – 0.2 = 0.8 ⇒ k = 0.1

(ii) P (Average study time is not more than 1 hour)

= P (X ≤ 1) = P (X = 0) + P (X = 1) = 0.2 + 0.1 = 0.3
(iii) P (Average study time is at least 3 hours)
= P (X ≥ 3) = P (X = 3) + P (X = 4) = 0.3 + 0.2 = 0.5
OR
P (Average study time is exactly 2 hours) = P (X = 2) = 0.2
4. Read the following passage and answer the questions given below.

A family has two children. Assume that each born child is equally likely to be a boy or a girl.
(i) Find the probability that youngest child is a girl.
(ii) Find the probability that both children are girls.
(iii) What is the conditional probability that both are girls? Given that the youngest is a girl.

Page | 34
 OR

What is the conditional probability that both are girls? Given that atleast one is girl.

Answer:

(i) All possible cases are 𝑆 = {𝐵𝑏, 𝐵𝑔, 𝐺𝑔, 𝐺𝑏}. ∴ 𝑛(𝑠) = 4
2 1
The event that youngest child is a girl: E1 = {Bg, Gg}, P(E1 ) = 4 = 2

1
(ii) The event that both children are girls: 𝐹 = {𝐺𝑔} = 𝑃(𝐹) = 4

(iii) 𝐹 ∩ 𝐸1 = {𝐺𝑔}

𝑃(𝐴∩𝐸1 ) 1/4 1
∴ 𝑃(𝐹/𝐸1 ) = = 1/2 = 2
𝑃(𝐸1 )

OR

The event that at least one is girl 𝐸2 = {𝐵𝑔, 𝐺𝑔, 𝐺𝑏} , 𝐹 ∩ 𝐸2 = { 𝐺𝑔}

1 3
𝑃(𝐹 ∩ 𝐸2 ) = 4 ; 𝑃(𝐸2 ) = 4

𝑃(𝐹∩𝐸2 ) 1/4 1
𝑃(𝐹/𝐸2 ) = = 3/4 = 3
𝑃(𝐸2 )

Page | 35
 COMPETENCY BASED QUESTIONS
RELATIONS AND FUNCTIONS
VSA – Type (2 marks each)
1. Verify whether the relation R in the set R of real numbers, defined as
R = {(𝑎, 𝑏) ∶ 𝑎 ≤ 𝑏 2 } is reflexive, symmetric or transitive.
n+1
, if n is odd
2. Let f: N → N be defined by f(n) = { n2 for all n ∈ N, state whether
, if n is even
2
the function f is bijective. Justify your answer.
3. If f: R→R be a function defined as f (𝑥) = 𝑥 2 +3. Then find whether f is bijective or
not.
4. Let the mapping f: N → N, where N is the set of natural numbers is defined by
𝑥2, 𝑥 is odd
f(𝑥) = { for 𝑥 ∈ N. Then state whether f is injective, surjective or both.
2𝑥 + 1, 𝑥 is even

SA – Type (3 marks each)

1. The relation R in the set Z of integers given by R = {(a, b): 2 divides a – b} is an

equivalence relation. Also, find the distinct number of equivalence classes of 𝑍.
2. Let 𝐴 = { 𝑥 ∈ 𝑍 ∶ 0 ≤ 𝑥 ≤ 12 }. Show that 𝑅 = {(𝑎, 𝑏): 𝑎, 𝑏 ∈ 𝐴 , |𝑎 − 𝑏| is divisible by 4} is an
equivalence relation. Also find [1].
3. Let A = R − {3} 𝑎𝑛𝑑 𝐵 = 𝑅 − {1} . Consider the function 𝑓: 𝐴 → 𝐵 defined by
𝑥−2
f(𝑥) = 𝑥−3 is one-one and onto.
𝑛 + 1, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
4. Let f: N→N be defined by f(n) = {
𝑛 − 1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
Show that f is bijective.
𝑥
5. Show that the function f: R → R defined by f(x) = 1+𝑥 2 , ∀ x 𝜖 R is neither
one-one nor onto.
LA – Type (5 marks each)
1
1. Let 𝑓: 𝑅+ → [−5, ∞) be a function defined as f(𝑥) = 9𝑥 2 + 6𝑥 − 5. Show that f is bijective.
2. Prove that the relation R on the set Z – {0}, defined as
𝑎
R = {(𝑎, 𝑏) ∶𝑏 is equal to an integral power of 5} is an equivalence relation.
3. Prove that the relation R on the set A×A, where 𝐴 = {1, 2, 3, 4, 5, 6, 7, 8, 9}, defined by
(𝑎, 𝑏) 𝑅 (𝑐, 𝑑) iff 𝑎 + 𝑑 = 𝑏 + 𝑐 for all (a, b), (c, d) ϵ A × A is an equivalence relation. Also, find
the equivalence class of (3, 7), i. e [(3, 7)].
4. Let N be the set of all natural numbers and R be a relation on N × N defined by
(a, b)R (c, d) ⇔ ad = bc for all (a, b), (c, d) ∈ N × N.Show that R is an equivalence relation on
N × N.
5. Prove that the relation R on the set N×N defined by (𝑎, 𝑏) 𝑅 (𝑐, 𝑑) iff ad(b + c) = bc(a + d)
for all (a, b), (c, d) ϵ N × N is an equivalence relation.

Page | 36
 I.T.F
VSA - Type(2 marks each)
43𝜋
1. Find the principal value of 𝑠𝑖𝑛−1 (𝑐𝑜𝑠 ( )).
5
1
2. Find the principal value of 𝑐𝑜𝑡 −1 (− ).
√3
√3 π
3. Evaluate: cos [ cos−1 (− ) + 6 ].
2
1 2
4. Prove that: tan (2 cos−1 ) = √5 − 2
√5
−1 −1 1
5. Evaluate: tan {2𝑐𝑜𝑠 (2 sin )}.
2
𝜋 1
6. Find the value of sin ( 3 − sin−1 (− 2))
7. Which is greater, tan 1 or tan−1 1 ? Justify.
8. What is the domain of the function sin−1 √𝑥 − 1 ?
SA - Type(3 marks each)
𝜋
(1) Solve: s𝑖𝑛−1 (1 − 𝑥) − 2𝑠𝑖𝑛−1 𝑥 = 2

𝑐𝑜𝑠𝑥 𝜋 𝑥 𝜋 𝜋
(2) Prove that: ta𝑛−1 (1+𝑠𝑖𝑛𝑥) = 4 − 2 , 𝑥 ∈ (− 2 , 2 )

√1+𝑥−√1−𝑥 𝜋 𝑐𝑜𝑠 −1 𝑥 1
(3) Prove that: ta𝑛−1 ( )=4− ,− ≤𝑥≤1
√1+𝑥+√1−𝑥 2 √2

MATRICES AND DETERMINANTS

VSA - Type(2 marks each)
0 𝑎 3
(1) If the matrix A=[2 𝑏 −1] is a skew symmetric matrix, then find the value of (a+b-c).
𝑐 1 0
(2) If A and B are symmetric matrices, show that (AB+BA) is symmetric and (AB−BA) is skew
symmetric.
𝑥 −3 1
(3) If the matrix A=[2 𝑦 1] and 𝑥𝑦𝑧 = 7, 𝑥 + 𝑦 − 6𝑧 = 11, then find A. adj A.
1 1 𝑧
(4) If the area of a triangle with vertices (−3, 0), (3, 0) and (0, 𝑘) is 9 sq. unit. Then find the value of
𝑘.
SA - Type(3 marks each)
𝛼
0 −𝑡𝑎𝑛 2
(1) If A= [ 𝛼 ] and I is the identity matrix of order 2, show that
𝑡𝑎𝑛 2 0
cosα −sinα
I + A = (I– A ) [ ].
sinα cosα
𝑐𝑜𝑠 ∝ −𝑠𝑖𝑛 ∝ 0
(2) If f(∝) = [ 𝑠𝑖𝑛 ∝ 𝑐𝑜𝑠 ∝ 0] , Prove that f(∝). f(−β) = f(∝ −𝛽)
0 0 1
2 −3
(3) If A= [ ] and I is the identity matrix of order 2, compute A−1 and show that
−4 7
2A−1 = 9𝐼 − 𝐴

Page | 37
 LA- Type(5 marks each)
5 0 4 1 3 3
(1) Given A = [2 3 2] , B−1 = [1 4 3], compute (AB)−1
1 2 1 1 3 4
1 2 0
(2) If A= [−2 −1 −2] find A−1 and hence solve the system of linear equations.
0 −1 1
𝑥 − 2𝑦 = 10, 2𝑥 − 𝑦 − 𝑧 = 8, −2𝑦 + 𝑧 = 7
2 2 − 4 1 − 1 0 
 
(3) Given A = − 4 2 − 4 and B = 2 3 4 . Find BA and use this to solve the system of
 2 − 1 5  0 1 2
equations: 𝑥 − 𝑦 = 3, 2𝑥 + 3𝑦 + 4𝑧 = 17. 𝑦 + 2𝑧 = 7
(4) The sum of three numbers is 6. Twice the third number when added to the first number gives
7.On adding the sum of the second and third numbers to thrice the first number, we get 12. Find
the numbers, using matrix method.
(5) Solve the following system of equations by matrix method when x≠ 0, 𝑦 ≠ 0, 𝑧 ≠ 0.
2 3 3
− + = 10
𝑥 𝑦 𝑧
1 1 1
+ + = 10
𝑥 𝑦 𝑧
3 1 2
− + = 13
𝑥 𝑦 𝑧

CONTINUITY AND DIFFERENTIABILITY

VSA - Type(2 marks each)
√2𝑐𝑜𝑠𝑥−1 𝜋 𝜋
(1) Given 𝑓(𝑥) = is continuous at x= 4 , find f ( 4 ).
𝑐𝑜𝑡𝑥−1
(2) If 𝑓(𝑥) = 𝑒 𝑥 𝑔(𝑥), 𝑔(0) = 2 and 𝑔′ (0) = 1. Find 𝑓 ′ (0).
𝑑2 𝑦 𝑑𝑦
(3) If y = 3 𝑒 2𝑥 + 2 𝑒 3𝑥 , 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 − 5 𝑑𝑥 + 6𝑦 = 0.
𝑑𝑥 2
−1 −1 dy y
(4) If x = a sin t
, y = a cos t
, then show that =−
dx x
SA - Type(3 marks each)
𝑑𝑦 1
(1) If 𝑥√1 + 𝑦 + y√1 + 𝑥 = 0, prove that 𝑑𝑥 = − (1+𝑥)2
𝑑𝑦 𝑙𝑜𝑔𝑥
(2) If 𝑥 𝑦 = 𝑒 𝑥−𝑦 , Prove that =
𝑑𝑥 (1+𝑙𝑜𝑔𝑥)2
𝑑𝑦 𝜋
(3)If sin2 𝑦 + cos 𝑥𝑦 = 𝐾. Find at 𝑥 = 1 and 𝑦 = 4 .
𝑑𝑥
LA - Type(5 marks each)
𝑥−4
|𝑥−4|
+ 𝑎, if 𝑥 < 4
(1) f(𝑥) = 𝑎 + 𝑏, if 𝑥 = 4 is continuous at 𝑥 = 4.
𝑥−4
+ 𝑏, if 𝑥 > 4
{|𝑥−4|

Page | 38
 1−𝑐𝑜𝑠4𝑥
, when 𝑥 < 0
𝑥2
(2) If f(𝑥) = 𝑎 , when 𝑥 = 0 is continuous at 𝑥 = 0 ,then find the value of ‘a’
√𝑥
, when 𝑥 > 0
{√16+√𝑥−4
𝑥 2 + 3𝑥 + 𝑎, 𝑥 ≤ 1
(3) Find the values of a and b so that the function 𝑓(𝑥) = {
𝑏𝑥 + 2, 𝑥 > 1
is differentiable at 𝑥 ∈ R.

APPLICATION OF DERIVATIVE
VSA - Type(2 marks each)
1 𝑥
(1) Find the maximum value of the function y = (𝑥)
(2) If the function f(𝑥) = 𝑥 2 + 𝑎𝑥 + 1, then find the least value of a such that f is
strictly increasing on (1,2).
(3) The volume of a cube is increasing at the rate of 9cm3/sec. How fast is its surface
area increasing when the length of an edge is 10 cm?
(4) A balloon which always remains spherical on inflation is being inflated by pumping
in 900 cubic centimeters of gas per second. Find the rate at which the radius of the
balloon increases, when the radius is 15cm.

SA - Type(3 marks each)

(1) Find both maximum value and the minimum value of the function
𝑓(𝑥) = 3𝑥 4 – 8𝑥 3 + 12𝑥 2 − 48𝑥 + 25 in the interval [0, 3]
(2) An Apache helicopter of enemy is flying along the path 𝑦 = 𝑥 2 + 7. A soldier is placed at
A(3, 7) wants to shoot down the helicopter when it is nearest to him. Find the nearest point
and minimum distance of the helicopter from the point A.
(3) The median of an equilateral triangle is increasing at the rate of 2√3 cm/s .Find the rate at
which its area is increasing when the length of its side is 3cm.
LA - Type(5 marks each)
(1) A rectangular cake dish is made by cutting the congruent squares from each corner of a 40 cm
by 25 cm rectangular tin plate and then folding the metal to form a container. Find the
maximum volume of the container and area of tin plate used.
(2) Prove that the volume of the largest cone that can be inscribed in a sphere of radius
R is 8/27 times the volume of the sphere.
4sin𝑥−2𝑥−𝑥cos𝑥
(3) Find the intervals in which the function f given by f(𝑥) = , where 0 ≤ 𝑥 < 2π is
2+cos𝑥
(i)increasing (ii) decreasing.
Chapter- Integrals
2 Marks
𝑥2
1. Evaluate: ∫ 𝑑𝑥
√𝑥 6 +𝑏6
√cos 𝑥
2. Evaluate:∫ 𝑑𝑥
sin 𝑥

Page | 39
 𝜋
3. Evaluate:∫−𝜋(cos 𝑎𝑥 − sin 𝑏𝑥)2 𝑑𝑥
𝑥2
4. Evaluate:∫ 𝑑𝑥
1−𝑥 4

3 Marks
𝜋 𝜋 𝜋
1. If ∫ 𝑒 −𝑥 |𝑐𝑜𝑠2𝑥|𝑑𝑥 = 𝛼 (𝑒 −4 + 𝑒 4 ) then find the value of 𝛼.
4
𝜋
−
4
𝑥
2. Find :∫ √ 𝑑𝑥; 𝑥 ∈ (0,1).
1−𝑥 3
𝜋
𝑥 𝑑𝑥
3. Evaluate:∫04
1+𝑐𝑜𝑠2𝑥+𝑠𝑖𝑛2𝑥
√1+𝑠𝑖𝑛2𝑥
4. Find : ∫ 𝑒 𝑥 𝑑𝑥
1+𝑐𝑜𝑠2𝑥
5 Marks
𝑥2
1. Evaluate:∫ 4 2 𝑑𝑥
𝑥 −𝑥 −12
𝜋
4
2. Evaluate:∫ log(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥) 𝑑𝑥
𝜋
−
4
2𝜋 𝑥 𝑠𝑖𝑛2𝑛 𝑥
3. Evaluate :∫0 𝑑𝑥
𝑠𝑖𝑛2𝑛 𝑥+ 𝑐𝑜𝑠 2𝑛 𝑥
Chapter: Application of Integrals
2 Marks
1
1. Find area bounded by the curve y= 𝑎𝑛𝑑 2 ≤ 𝑥 ≤ 4.
𝑥
2. Find area bounded by the curve y= 𝑥|𝑥|, 𝑤ℎ𝑒𝑛 − 1 ≤ 𝑥 ≤ 2.
3. Find area bounded by the lines x=0 , y=0 and x-y=2.

3 Marks
1. Find area of the region enclosed between the parabola 𝑦 2 = 4𝑎𝑥 and the line y=mx.
2. Find area of the region bounded by the curve 𝑦 = 8𝑥 3, and the lines y=27 and x=0.
3. Sketch the region {(𝑥, 𝑦): 𝑦 = √4 − 𝑥 2 }. Find area of the region using integration.

5 Marks
1. Using method of integration , find area bounded by the curve |𝑥| + |𝑦| = 1.
2. Draw a rough sketch of the curves 𝑦 = 1 + |𝑥 + 1|, 𝑥 = −3, 𝑥 = 3, 𝑦 = 0 and find
area of the region bounded by them using integration.
3. Using integration, find area of the region bounded by the curves 𝑦 = |log 𝑥|,
1
𝑥 = , x=2 and y=0.
2

Page | 40
 DIFFERENTIAL EQUATION

2 Marks
dy
1. Solve: tan ( ) = x.
dx
dy
2. Solve: (x + 3y 2 ) = y (y > 0).
dx
3 2
d2 y 2 d3 y 2 dy 3
3. If x and y are order and degree of the equation ( + 3xy) = ( +x ),
dx2 dx3 dx
find x+y.
3 Marks
π
4. Find the equation of a Curve passing through (1, ) if the slope of the Tangent to
4
y y
the Curve at any point P (x, y) is ( ) − cos 2 ( ).
x x
dy y
5. Solve: x 2 − xy = 1 + cos .
dx x
5
6. Solve y x. dx + y. dx − x. dy = 0.
5 Marks

7. Solve the Initial Value Problem (x + y + 1)2 dy = dx, y(-1) = 0.

dx
8. Solve = x 2 y 3 + xy.
dy
9. In a bank, principal increases at the rate of 5% per year. In how many years ₹1000
double itself.
CHAPTER-VECTORS

2 marks
𝜋
⃗⃗&𝑐⃗be three vectors of equal magnitude, the angle between each pair be
1. Let 𝑎⃗, 𝑏
3
And|𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = √12. Then find |𝑎⃗|.
⃗⃗are position vectors of A & B respectively, find the position vector of a
2. If 𝑎⃗&𝑏
point C in BA produced such that BC = 1.5 BA.
3. If 𝑎⃗&𝑏⃗⃗ are unit vectors, then find the the greatest value of |𝑎⃗⃗ + 𝑏⃗⃗| + |𝑎⃗ − 𝑏⃗⃗|

3 marks

1. For three vectors𝑎⃗, 𝑏⃗⃗ & 𝑐⃗, if𝑎⃗ × 𝑏⃗⃗ = 𝑐⃗ &⃗⃗⃗⃗

𝑎 × 𝑐⃗ = 𝑏⃗⃗, prove that 𝑎⃗, 𝑏⃗⃗&𝑐⃗are
mutually perpendicular vectors, |𝑏⃗⃗| = |𝑐⃗|and|𝑎⃗| = 1.

2. Let 𝑎⃗ = 𝑥 2 𝑖̑ + 2𝑗̑ − 2𝑘̑, 𝑏⃗⃗ = 𝑖̑ − 𝑗̑ + 2𝑘̑&𝑐⃗ = 𝑥 2 𝑖̑ + 5𝑗̑ − 4𝑘̑be three vectors.

Page | 41
 Find the value of x for which the angle between 𝑎⃗&𝑏⃗⃗is acute and the angle
between 𝑏⃗⃗&𝑐⃗is obtuse.

3. Given that vectors 𝑎⃗, 𝑏⃗⃗&𝑐⃗ form a triangle such that 𝑎⃗ = 𝑏⃗⃗ + 𝑐⃗. Find p, q, r ,s such
that area of triangle is 5√6 sq units where
  
a = piˆ + qˆj + rkˆ, b = s î + 3ĵ + 4k̂ and c = 3î + ĵ - 2k̂.

5 marks

1. Let A, B, C and D be four points in space. Prove that

→ → → → → →
|𝐴𝐵 × 𝐶𝐷 + 𝐵𝐶 × 𝐴𝐷 + 𝐶𝐴 × 𝐵𝐷 | = 4(Area of 𝛥ABC)
3. Find a unit vector 𝑐⃗ if −𝑖̑ + 𝑗̑ − 𝑘̑ bisects the angle between the vectors
𝑐⃗and3𝑖̑ + 4𝑗̑.
3. A barge is pulled into harbor by two tug boats as shown

Based on the above information, answer the following questions:

(a) Find position vector of A.
(b) Find the unit vector along position vector of A.
(c) Find position vector of B.
→
(d) Find the vector𝐴𝐶 .
→ →
(e) If 𝐴 = 4𝑖̑ + 3𝑗̑and𝐵 = 3𝑖̑ + 4𝑗̑, then find |𝐴⃗| + |𝐵
⃗⃗|&|𝐴⃗ + 𝐵
⃗⃗|
3-DIMENSIONAL GEOMETRY
2 marks
Q1. Let A(2,3,-1) and B(1,-1,2) be two points. Find a point C on Y-axis such that AB is
perpendicular to BC.
Q2. A line makes the same angle 𝜃 with each of X & Z axes. If the angle 𝛽 which it
makes with Y-axis is such that sin2 𝛽 = 3sin2 𝜃, find the value of cos 2 𝜃.
Q3. Find the direction ratios of the line 𝑥 − 𝑦 + 𝑧 − 5 = 0 = 𝑥 − 3𝑦 − 6.

Page | 42
 3 marks

Q1. Find the points on the line through the points A (1, 2, 3) and B (3, 5, 9) at a distance
of 14 units from the midpoint of line segment AB.
Q2. A line with direction ratios 2, 1, 2 meets each of the lines given by the equations
𝑥 = 𝑦 + 1 = 𝑧 and 𝑥 + 1 = 2𝑦 = 2𝑧. Find the co-ordinates of each of these points
of intersection.
Q3. Find the equation of the line which bisects the join of the points (-1,2,3) and (3,0,-1)
𝑦−4 2−𝑧
and parallel to the line 𝑥 = −3, =
3 1

5 marks
𝑥−1 𝑦−2 𝑧−3
Q1. Find the equations of the line which intersects the lines = = and
2 3 4
𝑥+2 𝑦−3 𝑧+1
= = and passes through (1, 1, 1).
1 2 4
Q2. The points A(4, 5, 10) B(2, 3, 4) and C(1, 2, –1) are three vertices of a parallelogram
ABCD. Find the vector equation of the sides AB and CD. Find the distance between
these two lines.
Q3. Find the equation of the lines passing through the origin and intersects the line
𝑥−3 𝑦−3 𝑧 2𝜋
= = at the angle .
2 1 1 3

LINEAR PROGRAMMING PROBLEM

2 marks:
1. The corner points of the feasible region determined by the set of constraints
(linear inequalities) are P(0,5), Q(3,5), R(5,0) &S(4, 1) and the objective function
Z = ax+2by where a, b > 0. Find the condition on a and b such that the maximum
Z occurs at Q and S.
2. Solve the following LPP graphically:
Maximize Z = 5x+8y subject to the constraints: x + y = 5, x  4, y  2, x, y  0
3. Find the number of solutions of the system of inequalities
x + 2 y  3, 3x + 4 y  12 , y  1, x  0
3 marks:
1. Solve the following LPP graphically:
Minimize and Maximize Z = x+2y subject to the constraints:
x + 2 y  100,2x − y  0,2x + y  200, x, y  0
2. Solve the following LPP graphically:
Maximize Z = 8x+9y subject to the constraints: 2x + 3 y  6, 3x − 2 y  6, y  1, x, y  0
3. Solve the following LPP graphically:
Minimize Z = 3x+5y subject to the constraints:
x + 3 y  3, x + y  2, x, y  0

Page | 43
 5 marks:
1. In the graph, O is the origin, A is the point (2, 1), B is the point (2, 7) and C is the
point (6, 3). The shaded region R is defined by three inequalities one of which is x +
y = 9.

Based on the above information, answer the following questions:

(i) Find the other two inequalities.
(ii) Find the maximum value of Z = 3x + y if the point is in the region R.
(iii) Find the minimum value of P = 5x - 2y if the point is in the region R.
(iv) Find the area of the region R.
2. A company has formulated a model for LPP with feasible solution as follows:

(i) Find the constraint related to the points A and B.

(ii) Find the constraint related to the points C and D.
(iii) Find the relation between p and q if the value attained by the objective
function Z = px + qy, p > 0 and q > 0 at B(4,3)is thrice the value at
C(3,4)
(iv) Find the value of Z at points D and A if Z =4x –y.
3. Solve the following LPP graphically:
Maximize and minimize Z = 3x-5y
Subject to the constraints: 3x − 2 y + 6  0, x − 2 y + 4  0, 3x + 2 y − 6  0, x, y  0
4. Solve the following LPP graphically:
Maximize and minimize Z = 3x-4y
Subject to the constraints: x − 2 y  0, - 3x + y  4, x − y  6, x, y  0

Page | 44
 PROBABILITY

2 marks:
1. The probability of finding a green signal on a busy crossing X is 30%. What is the
probability of finding a green signal on X on two consecutive days out of three?
2. A dice is thrown three times. If the first thrown dice is 4, find the chance of
getting 15 as the sum.
3. The odds against A solving a given problem are 4:3 and the odds in favour of B
solving the same problem are 7:5. Find the probability that the problem will be
solved.
3 marks:
1
1. A and B are two students. Their chances of solving a problem correctly are and
3
1 1
respectively. If the probability of their making a common error is and they
4 20
obtain the same answer, then find the probability of their answer to be correct.
2. Two cards are drawn without replacement from a well shuffled deck of 52 cards.
Find the probability distribution of the no. of face cards.
3. A committee of 4 students is selected at random from a group consisting of 7 boys
and 4 girls. Given that there is at least one girl on the committee, calculate the
probability that there are exactly 2 boys on the committee.

5 marks:
1. A bag contains 4 balls. Two balls are drawn at random, and are found to be white.
Find the probability that all balls are white.
2. A random variable X has the following probability distribution:
X: 0 1 2 3 4 5 6 7
2 2
P(X): 0 k 2k 2k 3k k 2k 7 k2+k

Find (i) k (ii) P(X < 3) (iii) P(X > 6) (iv) P (0 < X <3)
3. Find the probability distribution of the random variable X, which denotes the no. of
doublets in four throws of a pair of dice. Also find the mean of the no. of doublets.
ANSWERS
RELATIONS AND FUNCTIONS
VSA – Type (2 marks each)

1 1 2 1 1
1. As 2 ≰ (2) , so (2 , 2) ∉ 𝑅. So 𝑅 is not reflexive.
(1, 2) ∈ 𝑅 as 1 < 22 = 4, but (2, 1) ∉ 𝑅 as 2 ≮ 12 = 1. So 𝑅 is not symmetric.
3 3 2 9
(3, 2) ∈ 𝑅 𝑎𝑠 3 < 22 = 4 𝑎𝑛𝑑 (2, ) ∈ 𝑅 𝑎𝑠 2 < ( ) = = 2.25,
2 2 4
3 3 2 9
But (3, 2) ∉ 𝑅 𝑎𝑠 3 ≮ (2) = 4 = 2.25. So, 𝑅 is not transitive.
So, 𝑅 is neither reflexive, nor symmetric nor transitive.

Page | 45
 𝑛+1
, if n is odd
2
2. Given f(𝑛) = { 𝑛
, if n is even
2
Let 𝑛1 = 1 𝑎𝑛𝑑 𝑛2 = 2 be two elements of 𝑁
1+1 2
∴ f(𝑛1 ) = 𝑓(1) = = 1 and f(𝑛2 ) = 𝑓(2) = 2 = 1
2
f(𝑛1 ) = f(𝑛2 ) for 𝑛1 ≠ 𝑛2
⇒ f: N → N is not one-one.
As f is not one-one.
∴ f is not a bijective function.
3. Let 𝑥1 , 𝑥2 ∈ dom f(R)
Consider, f (𝑥1 ) = f (𝑥2 )
𝑥1 2 +3 =𝑥2 2 +3
⟹ 𝑥1 = ±𝑥2 , so, it is not injective (one-one)
Let 𝑦 ∈ codom f(R)
y = f (𝑥) = 𝑥 2 +3.
𝑥 = √𝑦 − 3 ∉ 𝑅 for 𝑦 < 3.
Here, range 𝑓 = [3, ∞)
∴ codom f ≠ Range f.
∴ f is not onto.
Hence f is not bijective.
4. For injective
f(3) = 32 = 9
f (4) = 2× 4 + 1 = 9
f(3) = f (4) = 9(many-one)
f is not injective.
For surjective
For 𝑥 is odd, f(𝑥) is odd
For 𝑥 is even, f(𝑥) is also odd
So range = set of odd natural numbers ≠ co-domain (N)
f is not surjective.
f is neither injective nor surjective hence not bijective.
SA – Type (3 marks each)
1. Given R = {(a, b): 2 divides (𝑎 − 𝑏)}
Reflexive:
As a– a=0, which is divisible by
2,∀a∈A
So (a, a) ∈A,∀a∈A, ∴ R is reflexive.
Symmetric:
Let a, b∈A and (a, b) ∈ R, ⇒ 2 divides (a–b), say a –b=2m
⇒b –a = –2m, i. e 2 divides (b–a) ⇒ (b, a) ∈R, ∴ Ris symmetric.
Transitive:
Let a, b, 𝑐 ∈A and (a, b), (b, c) ∈R
⇒2 divides (a– b), let a–b=2m

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 ⇒2divides (b–c), let a–b = 2n
⇒ a– b + b – c = 2m + 2n
⇒ a– c =2(m+n)
⇒2 divides (a– c)
⇒(a, c)∈R ∴R is Transitive.
As R is reflexive, symmetric and transitive,
∴ R is an equivalence relation.
2. Since |𝑎 − 𝑎| = 0, which is divisible by 4.
So, 𝑅 is reflexive.
As |𝑎 − 𝑏| is divisible by 4 , so |𝑏 − 𝑎| is divisible by 4.
Thus, 𝑅 is symmetric.
Let |𝑎 − 𝑏| and |𝑏 − 𝑐| are divisible by 4.
⇒ |𝑎 − 𝑏| = 4𝑘 and |𝑏 − 𝑐| = 4𝑙
⇒ 𝑎 − 𝑏 = ±4𝑘 and 𝑏 − 𝑐 = ±4𝑙
⇒ 𝑎 − 𝑐 = ±4(𝑘 + 𝑙) = ±4𝑚
⇒ |𝑎 − 𝑐| = 4𝑚, which is divisible by 4.
So, 𝑅 is transitive.
Therefore, 𝑅 is an equivalence relation.
Also, [1] = {1, 5, 9}
𝑥−2
3. f: A → B is given by, 𝑓(𝑥) = 𝑥−3
Let 𝑥1 , 𝑥1 ∈ A
Consider f(𝑥1 ) = f(𝑥2 )
𝑥 −2 𝑥 −2
⇒ 𝑥1 −3 = 𝑥2−3
1 2
⟹ 𝑥1 𝑥2 − 3𝑥1 − 2𝑥2 + 6 = 𝑥1 𝑥2 − 3𝑥2 − 2𝑥1 + 6
⟹ 𝑥1 = 𝑥2
⇒ 𝑓 is one-to-one
Let 𝑦 ∈ B and y = 𝑓(𝑥) = 𝑥−2
𝑥−3
⇒ 𝑥𝑦 − 3𝑦 = 𝑥 − 2 ⇒ 𝑥 = 3𝑦−2
𝑦−1
∈A
Therefore, for any 𝑦 ∈ 𝐵, there exists 3𝑦−2
𝑦−1
=𝑥 ∈ 𝐴. Also, here range 𝑓 = 𝑅 − {1} = 𝐵 = 𝑐𝑜𝑑𝑜𝑚 𝑓.

⇒ 𝑓 is onto
∴ 𝑓 is one-one and onto.
4. One-One: -
Case - 1
If x and y both are even
𝑓(𝑥) = 𝑓(𝑦)
⟹ 𝑥−1=𝑦−1⇒𝑥 =𝑦
Case-2
If x and y both are even
𝑓(𝑥) = 𝑓(𝑦)
⟹ 𝑥+1=𝑦+1⇒𝑥 =𝑦
Case-3
If one is even and one is odd, 𝑥 ≠ 𝑦 ⇒ 𝑓(𝑥) ≠ 𝑓(𝑦)
∴ f is one-one.

Page | 47
 Onto: -
Also, any odd number 2𝑟 + 1 in the codomain 𝑁 is the image of of 2𝑟 + 2 in the domain 𝑁 and any
even number 2𝑟 in the codomain 𝑁 is the image of of 2𝑟 − 1 in the domain 𝑁. So 𝑓 is onto.
Therefore, 𝑓 is bijective.
5. One-one: -
𝑥
The given function f: R → R defined by 𝑓(𝑥) = 1+𝑥 2 , ∀ x 𝜖 R
1 2
Here, as 𝑓(2) = 𝑓 (2) = 5, so 𝑓 is not one-one.
Onto: -
𝑥 −(−1)±√(−1)2 −4𝑦2 −(−1)±√1−4𝑦 2
Let 𝑦 = 1+𝑥 2 ⇒ 𝑦(1 + 𝑥 2 ) = 𝑥 ⇒ 𝑦𝑥 2 − 𝑥 + 𝑦 = 0 ⇒ 𝑥 = = .
2𝑦 2𝑦
1 1 1 1
1 − 4𝑦 2 ≥ 0 ⇒ 4𝑦 2 ≤ 1 ⇒ − 2 ≤ 𝑦 ≤ 2. So, range 𝑓 = [− 2 , 2] ≠ 𝑐𝑜𝑑𝑜𝑚 𝑓.
So, 𝑓 is not onto.
Hence, 𝑓 is not a bijective function.

LA – Type (5 marks each)

1. Given 𝑓: 𝑅+ → [−5, ∞) be a function defined as f(𝑥) = 9𝑥 2 + 6𝑥 − 5.
Injective:
Let 𝑥, 𝑦 ∈ domf (𝑅+ )
𝑓(𝑥) = 𝑓(𝑦) ⇒ 9𝑥 2 + 6𝑥 − 5 = 9𝑦 2 + 6𝑦 − 5
⇒ 9(𝑥 2 − 𝑦 2 ) + 6(𝑥 − 𝑦) = 0
⇒ (𝑥 − 𝑦)(9𝑥 + 9𝑦 + 6) = 0
⇒ 𝑥 = 𝑦. [ as (9𝑥 + 9𝑦 + 6) ≠ 0
So f is one one.
Surjective:
𝑦 ∈ rngf
Let f(𝑥) = 𝑦 ⇒ 9𝑥 2 + 6𝑥 − 5 = (3𝑥 + 1)2 − 6 = 𝑦.
√𝑦+6 −1
⇒𝑥= 3
Here ∈ 𝑅+ , when 𝑦 ∈ [−5, ∞). Here rng f = [−5, ∞) = codomain. So f is onto.
∴ f is bijective.
2. Reflexive
Let a ∈ Z – {0}
𝑎
=1= 50 , ∴ (a, a) ∈ R and R is reflexive.
𝑎
Symmetric
Let a, b ∈ Z – {0} &
(a, b) ∈ R
𝑎 𝑏
= 5𝑘 ⇒𝑎=5−𝑘 , ∴(b, a) ∈ R and R is symmetric.
𝑏
Transitive
Let a, b , c ∈ Z – {0}
𝒂 𝒃 𝒂
(a, b) ∈ R and (b, c) ∈ R ⇒ 𝒃 =𝟓𝒌 , 𝒄 = 𝟓𝒌𝟏 ⇒ 𝒄 =𝟓𝒌+𝒌𝟏
(a, c) ∈ R ∴ R is transitive.
As R is reflexive, Symmetric and transitive ∴ R is equivalence relation.

Page | 48
3. Given A = {1, 2, 3, ………9} and R is a relation in A × A defined by
(a, b) R (c, d) iff a + d = b + c.
For all (a, b) ∈ A × A, a + b = b + a
⇒ (a, b) R (a, b)
Hence R is reflexive.
Consider (a, b) R (c, d) for all (a, b), (c, d) ∈ A × A
i. e 𝑎 + 𝑑 = 𝑏 + 𝑐 ⇒ 𝑐 + 𝑏 = 𝑑 + 𝑎
⇒ (c, d) R (a, b).
Hence R is Symmetric relation.
Let (a, b) R (c, d) and (c, d) R (e, f), for all (a, b), (c, d), (e, f) ∈ A X A
On adding a + d + c + f = b + c + d + e
⇒ a+f=b+e
⇒ (a, b) R (e, f)
So, R is transitive
Hence, R is an equivalence relation as it is reflexive, symmetric & transitive.
[(3,7)] = {(1,5), (2, 6), (3,7), (4,8), (5,9)}.
𝑎 𝑐
4. Given, (𝑎, 𝑏)R(𝑐, 𝑑) iff ad= bc ⇒ 𝑏 = 𝑑, for (𝑎, 𝑏), (𝑐, 𝑑) ∈ 𝑁 × 𝑁
Reflexive: -
𝑎 𝑎
As 𝑏 = 𝑏, so (𝑎, 𝑏)𝑅(𝑎, 𝑏). So, 𝑅 is reflexive.
𝑎 𝑐 𝑐 𝑎
Symmetric: - Let (𝑎, 𝑏)R(𝑐, 𝑑) ⇒ 𝑏 = 𝑑 ⇒ 𝑑 = 𝑏 ⇒ (𝑐, 𝑑)𝑅(𝑎, 𝑏). So 𝑅 is symmetric.
𝑎 𝑐 𝑐 𝑒 𝑎 𝑒
Transitive: - Let (𝑎, 𝑏)R(𝑐, 𝑑) and (𝑐, 𝑑)𝑅(𝑒, 𝑓) ⇒ 𝑏 = 𝑑 and = 𝑓 ⇒ 𝑏 = 𝑓.
𝑑
So, 𝑅 be an equivalence relation.
5. Given, (𝑎, 𝑏)R(𝑐, 𝑑) iff ad(𝑏 + 𝑐) = 𝑏𝑐(𝑎 + 𝑑)
Reflexive: As ab(b + a)=𝑏𝑎(𝑎 + 𝑏)
⇒ (𝑎, 𝑏)𝑅(𝑎, 𝑏)
Symmetry:(𝑎, 𝑏)𝑅(𝑐, 𝑑) ⇒ 𝑎 𝑑(𝑏 + 𝑐) = 𝑏𝑐(𝑎 + 𝑑)
⇒ 𝑐𝑏(𝑑 + 𝑎) = 𝑑𝑎(𝑐 + 𝑏)
⇒ (𝑐, 𝑑)𝑅(𝑎, 𝑏)
1 1 1 1
Transitivity: (𝑎, 𝑏)𝑅(𝑐, 𝑑) ⇒ 𝑐 + 𝑏 = 𝑑 + 𝑎

1 1 1 1
(𝑐, 𝑑)𝑅(𝑒, 𝑓) ⇒ + = +
𝑒 𝑑 𝑓 𝑐

1 1 1 1
On adding above equations, we get: 𝑒 + 𝑏 = 𝑓 + 𝑎

⇒ (𝑎, 𝑏)𝑅(𝑒, 𝑓)

I.T.F
VSA - Type(2 marks each)
43𝜋 3𝜋 3𝜋 2𝜋
1. 𝑠𝑖𝑛−1 (𝑐𝑜𝑠 ( )) = 𝑠𝑖𝑛−1 (𝑐𝑜𝑠 (8𝜋 + )) = 𝑠𝑖𝑛−1 (𝑐𝑜𝑠 ( 5 )) = 𝑠𝑖𝑛−1 (𝑐𝑜𝑠 (𝜋 − )) =
5 5 5
2𝜋 2𝜋 𝜋 2𝜋 𝜋
𝑠𝑖𝑛−1 (−𝑐𝑜𝑠 ( 5 )) = −𝑠𝑖𝑛−1 (𝑐𝑜𝑠 ( 5 )) = −𝑠𝑖𝑛−1 (𝑠𝑖𝑛 (2 − )) = − 10.
5
1 1 𝜋 2𝜋
2. 𝑐𝑜𝑡 −1 (− ) = 𝜋 − 𝑐𝑜𝑡 −1 ( ) = 𝜋 − 3 = 3
.
√3 √3

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 √3 π
3. cos[ cos−1 (− ) + 6]
2
√3 π
= cos[ π − cos −1 ( 2 ) + 6 ]
π π
= cos[ π − cos−1 cos 6 + 6 ]
π π
= cos[ π − 6 + 6 ]
= cos π = −1
1 2
4. tan (2 𝑐𝑜𝑠 −1 )
√5
1 −1 2
Let 2 𝑐𝑜𝑠 =𝛼
√5

2 2
𝑐𝑜𝑠 −1 = 2𝛼 , cos2𝛼 =
√5 √5

2 √5−2 √5−2
1−2𝑠𝑖𝑛2 𝛼 = ⟹ 𝑠𝑖𝑛2 𝛼 = ⟹ 𝑠𝑖𝑛𝛼 = √ 2√5
√5 2√5

√5+2 √5−2
cos𝛼 = √ 2√5 ,tan𝛼 = √ = √5 − 2
√5+2

𝜋 𝜋
5. tan−1 {2𝑐𝑜𝑠 (2 × 6 )}= tan−1 {2 × 1/2}= 4
𝜋 1 𝜋 1 𝜋 𝜋 𝜋
6. sin (3 − sin−1 (− 2)) = sin ( 3 + sin−1 (2)) = sin ( 3 + 6 ) = 𝑠𝑖𝑛 2 = 1

7. Tan1 is greater.
𝜋 𝜋
Reason: 4 < 1, tan 4 < tan1,1<tan1
𝜋
tan1>1> 4 , tan1 > tan−11
8. 𝑥 − 1 ≥ 0
⇒ 𝑥 ≥ 1 ---------------------(i)
⇒ −1 ≤ √𝑥 − 1 ≤ 1
⇒ 0≤𝑥−1≤1
⇒ 1 ≤ 𝑥 ≤ 2---------------(ii)
From (i) and (ii)
Domain is [1, 2]
SA - Type(3 marks each)
−1 (1 𝜋 −1
(1) Given, s𝑖𝑛 − 𝑥) − 2𝑠𝑖𝑛 𝑥 = 2
𝜋
⇒ s𝑖𝑛−1 (1 − 𝑥) = 2 + 2𝑠𝑖𝑛−1 𝑥
𝜋
⇒ (1 − 𝑥) = sin ( + 2𝑠𝑖𝑛−1 𝑥)
2
⇒ (1 − 𝑥) = 𝑐𝑜𝑠( 2𝑠𝑖𝑛−1 𝑥)
⇒(1 − 𝑥) = 1 − 2𝑠𝑖𝑛2 (𝑠𝑖𝑛−1 𝑥 ⇒ (1 − 𝑥) = 1 − 2𝑥 2 ⇒ 2𝑥 2 − 𝑥 = 0
1
𝑥 = 0 or 𝑥 = 2(Impossible as it doesn’t satisfy the given equation)
∴𝑥=0
𝜋 𝜋 𝑥 𝜋 𝑥
𝑐𝑜𝑠𝑥 sin ( −𝑥) 2sin ( − )𝑐𝑜𝑠( − )
−1 −1 2 −1 4 2 4 2
(2) L. H. S = ta𝑛 (1+𝑠𝑖𝑛𝑥) = ta𝑛 [ 𝜋 ] = ta𝑛 [ 𝜋 𝑥 ]
1+𝑐𝑜𝑠( −𝑥) 2𝑐𝑜𝑠2 ( − )
2 4 2

Page | 50
 𝜋 𝑥
sin ( − ) 𝜋 𝑥 𝜋 𝑥
−1
= ta𝑛 [ 4 2
𝜋 𝑥 ] = 𝑡𝑎𝑛−1 𝑡𝑎𝑛 ( 4 − 2) = ( 4 − 2) = R. H. S.
𝑐𝑜𝑠( − )
4 2
√1+𝑥−√1−𝑥 𝑐𝑜𝑠−1 𝑥
(3) L. H. S = ta𝑛−1 ( ) put 𝑥 = 𝑐𝑜𝑠2𝜃, 𝜃 =
√1+𝑥+√1−𝑥 2

√1+𝑐𝑜𝑠2𝜃−√1−𝑐𝑜𝑠2𝜃 √2𝑐𝑜𝑠2 𝜃−√2𝑠𝑖𝑛2 𝜃 √2𝑐𝑜𝑠𝜃−√2𝑠𝑖𝑛𝜃

= ta𝑛−1 ( ) = ta𝑛−1 (√2𝑐𝑜𝑠2 ) = ta𝑛−1 ( )=
√1+𝑐𝑜𝑠2𝜃+√1−𝑐𝑜𝑠2𝜃 𝜃+√2𝑠𝑖𝑛2 𝜃 √2𝑐𝑜𝑠𝜃+√2𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃−𝑠𝑖𝑛𝜃 1−𝑡𝑎𝑛𝜃 𝜋 𝜋 𝜋 𝑐𝑜𝑠−1 𝑥
ta𝑛−1 (𝑐𝑜𝑠𝜃+𝑠𝑖𝑛𝜃) = ta𝑛−1 (1+𝑡𝑎𝑛𝜃) = ta𝑛−1 [tan ( 4 − 𝜃)] = ( 4 − 𝜃) = − = R.H.S
4 2

MATRICES AND DETERMINANTS

VSA - Type(2 marks each)
0 𝑎 3
(1) Given, A=[2 𝑏 −1] is a skew symmetric matrix
𝑐 1 0
AT = −A
0 2 𝑐 0 𝑎 3 0 −𝑎 −3
[𝑎 𝑏 1] = − [2 𝑏 −1] = [−2 −𝑏 1 ]
3 −1 0 𝑐 1 0 −𝑐 −1 0
On comparing,
𝑎 = −2 , b = 0 and c = −3
a + b −𝑐 = 1
(2) Given, AT = A and BT = B
Now (AB + BA)T = (AB)T + (BA)T = BT AT + AT B T = (BA + AB)
∴ (AB+BA) is symmetric
Again (AB − BA)T = (AB)T − (BA)T = BT AT − AT BT = (BA − AB) = −(AB − BA)
∴ (AB−BA) is skew symmetric.
𝑥 −3 1
(3) Given, the matrix A=[2 𝑦 1] and 𝑥𝑦𝑧 = 7, 𝑥 + 𝑦 − 6𝑧 = 11
1 1 𝑧
𝑥 −3 1
Now,|𝐴| = |2 𝑦 1| = 𝑥(𝑦𝑧 − 1) + 3(2𝑧 − 1) + 1(2 − 𝑦)=𝑥𝑦𝑧 − 𝑥 + 6𝑧 − 3 + 2 − 𝑦
1 1 𝑧
= 𝑥𝑦𝑧 − (𝑥 + 𝑦 − 6𝑧) − 1 = 7 − 11 − 1 = −5 (by using given condition)
A.(adj A)= |𝐴|I = 5I.
(4) Given the area of a triangle with vertices (−3, 0), (3, 0) and (0, 𝑘) is 9 sq. unit.

−3 3 0
1
So, 2 [ 0 0 𝑘] = ±9 ⇒ 6𝑘 = ±18 ⇒ 𝑘 = ±3.
1 1 1
SA - Type(3 marks each)

𝛼
0 −𝑡𝑎𝑛 2 𝛼
(1) A= [ 𝛼 ] Let 𝑡𝑎𝑛 2 = 𝑡
𝑡𝑎𝑛 2 0
0 −𝑡
A=[ ],
𝑡 0

Page | 51
 𝛼
1 −𝑡𝑎𝑛 1 −𝑡
2
L.H.S = I +A = [ 𝛼 ]=[ ]
𝑡𝑎𝑛 2 1 𝑡 1
cosα −sinα
RHS = ( I −A) [ ]
sinα cosα
2 1−𝑡 −2𝑡
1 𝑡 1+𝑡 2 1+𝑡 2
=[ ][ 1−𝑡 2
]
−𝑡 1 2𝑡
1+𝑡 2 1+𝑡 2

1−𝑡 2 +2𝑡 2 −2𝑡+𝑡−𝑡 3

1+𝑡 2 1+𝑡 2
= [−𝑡+𝑡 3+2𝑡 2𝑡 2 +1−𝑡 2
]
1+𝑡 2 1+𝑡 2

1+𝑡 2 −𝑡−𝑡 3
2 1+𝑡 2
=[1+𝑡
𝑡+𝑡 3 1+𝑡 2
]
1+𝑡 2 1+𝑡 2
1 −𝑡
=[ ] = L.H.S
𝑡 1
L.H.S = R.H.S
𝑐𝑜𝑠 ∝ −𝑠𝑖𝑛 ∝ 0
(2) 𝐺𝑖𝑣𝑒𝑛, f(∝) = [ 𝑠𝑖𝑛 ∝ 𝑐𝑜𝑠 ∝ 0] , Prove that f(∝). f(−β) = f(∝ −𝛽)
0 0 1
𝑐𝑜𝑠 ∝ −𝑠𝑖𝑛 ∝ 0 cos (−β) −sin (−β) 0
L.H.S = f(∝). f(−β) = [ 𝑠𝑖𝑛 ∝ 𝑐𝑜𝑠 ∝ 0] [ sin (−β) cos (−β) 0]
0 0 1 0 0 1
𝑐𝑜𝑠 ∝ −𝑠𝑖𝑛 ∝ 0 cos β sin β 0
= [ 𝑠𝑖𝑛 ∝ 𝑐𝑜𝑠 ∝ 0] [−sin β cos β 0]
0 0 1 0 0 1
𝑐𝑜𝑠 ∝. cos β − 𝑠𝑖𝑛 ∝ sin β 𝑐𝑜𝑠 ∝ sin β − 𝑠𝑖𝑛 ∝ cos β 0
=[ 𝑠𝑖𝑛 ∝. cos β − 𝑐𝑜𝑠 ∝ sin β 𝑠𝑖𝑛 ∝ sin β + 𝑐𝑜𝑠 ∝ cos β 0]
0 0 1
cos (∝ −β) −sin (∝ −β) 0
= [ sin (∝ −β) cos (∝ −β) 0] = f(∝ −𝛽) = R. H. S
0 0 1
2 −3
(3) Given, A= [ ]
−4 7
|A| = 14 − 12 = 2 ≠ 0 , A−1 exists
7 3
Adj A= [ ]
4 2
𝑎𝑑𝑗 𝐴 1 7 3
A−1 = |A| = 2 [ ]
4 2
1 0 2 −3 7 3
Now, R.H.S= 9I − 𝐴= 9[ ]−[ ]=[ ] = 2A−1 =L.H.S
0 1 −4 7 4 2
LA – Type (5 marks each)

(1) |A| = 5(−1) + 4(1) = −1

−1 8 −12
Adj(A) = [ 0 1 −2 ]
1 −10 15

Page | 52
 1 1 −8 12
A−1 = Adj(A) = [ 0 −1 2 ]
|A|
−1 10 −15
1 3 3 1 −8 12 −2 19 −27
(AB)−1 = B −1 A−1 = [1 4 3] [ 0 −1 2 ] = [ −2 18 −25]
1 3 4 −1 10 −15 −3 29 −42
(2) |A| = 1≠ 0 ,

−3 −2 −4
Adj(A) = [ 2 1 2]
2 1 3
−3 −2 −4
−1 1
A = |A|
Adj(A) =[ 2 1 2]
2 1 3
Writing the equations in matrix form
1 −2 0 𝑥 10
CX = D, where 𝐶 = [2 −1 −1] , 𝑋 = [𝑦] , 𝐷 = ⌈ 8 ⌉
0 −2 1 𝑧 7
−3 2 2
C = AT and C-1 = (𝐴𝑇 )−1 = (𝐴−1 )𝑇 = [−2 1 1]
−4 2 3
−3 2 2 10 0
X = C-1D =[−2 1 1] ⌈ 8 ⌉ = ⌈−5⌉
−4 2 3 7 −3
∴ 𝑥 = 0, 𝑦 = −5 and 𝑧 = −3
2 2 − 4 1 − 1 0 
 
(3) Given A = − 4 2 − 4 and B = 2 3 4
 2 − 1 5  0 1 2
1 −1 0 2 2 −4 6 0 0
BA= [2 3 4] [−4 2 −4] = [0 6 0]= 6I
0 1 2 2 −1 5 0 0 6
2 2 −4
1 1
𝐵 −1 = 6 𝐴 = 6 [−4 2 −4]
2 −1 5
Given, the system of equations is 𝑥 − 𝑦 = 3, 2𝑥 + 3𝑦 + 4𝑧 = 17, 𝑦 + 2𝑧 = 7
𝑥 3
BX = C where X= [𝑦] and C = [17]
𝑧 7
2 2 −4 3 6 + 34 − 28 12 2
1 1 1
X = B-1C = 6 [−4 2 −4] [17] = 6 [−12 + 34 − 28] = 6 [−6] = [−1]
2 −1 5 7 6 − 17 + 35 24 4
𝑥 2
[𝑦] = [−1] ∴ 𝑥 = 2, 𝑦 = −1, 𝑧 = 4
𝑧 4
(4) Let the first, second and third numbers be x, y, z respectively.
Then,

𝑥+𝑦+𝑧 = 6

Page | 53
 𝑥 + 2𝑧 = 7

3𝑥 + 𝑦 + 𝑧 = 12
The given linear equations can be written as
AX=B
1 1 1 𝑥 6
Where A=[1 0 2] , X=[𝑦] and B=[ 7 ]
3 1 1 𝑧 12
1 1 1
A=[1 0 2]
3 1 1
|𝐴|= 4≠0
𝐴−1 exists with unique solution.
After Finding the cofactors of each element
−2 0 2
Adj A= [ 5 −2 −1]
1 2 −1
−2 0 2
𝑎𝑑𝑗 𝐴 1
𝐴−1 = |𝐴| = 4 [ 5 −2 −1]
1 2 −1
−2 0 2 6 −12 + 0 + 24
−1 1 1
X= 𝐴 𝐵 = 4 [ 5 −2 −1] [ 7 ]= 4 [ 30 − 14 − 12 ]
1 2 −1 12 6 + 14 − 12
𝑥 12 3
1
[𝑦] =4 [ 4 ]= [1]
𝑧 8 2
𝑥 = 3, 𝑦 = 1, 𝑧 = 2

The first number is 3, second number is 1 and third number is 2.

(5) Given system of equations can be written in matrix form as

1

2 −3 3 𝑥 10
1
AX=B, Where A=[1 1 1] , X= 𝑦
and B=[10]
3 −1 2 1 13
[𝑧 ]
|𝐴| = −9 ≠ 0
𝐴−1 exists with unique solution

Cofactors are

𝐶11 = 3, 𝐶12 = 1, 𝐶13 = -4

𝐶21 = 3, 𝐶22 = -5, 𝐶23 = -7

𝐶31 = -6, 𝐶32 = 1, 𝐶33 = 5

Page | 54
 3 3 −6
Adj A= [ 1 −5 1 ]
−4 −7 5
3 3 −6
𝑎𝑑𝑗 𝐴 1
𝐴−1 = |𝐴|
= − 9 [ 1 −5 1 ]
−4 −7 5
3 3 −6 10 30 + 30 − 78
1 1
X= 𝐴−1 𝐵 = − 9 [ 1 −5 1 ] [10]= − 9 [ 10 − 50 + 13 ]
−4 −7 5 13 −40 − 70 + 65
1
𝑥 2
1 1 1 1
𝑦
= [3] 𝑥 = 2 , 𝑦 = ,𝑧 = 5
3
1 5
[𝑧 ]

CONTINUITY AND DIFFERENTIABILITY

VSA - Type(2 marks each)
 
(1) f (x) will be continuous at x = if lim f ( x ) = f  
4 x → 4 4

 2 cos x − 1
 f   = lim
 4 x→ 4 cot x − 1

= lim
( 2 cos x − 1) sinx
x→ 4 cos x − sin x

= lim
( 2 cos x − 1)( 2 cos x + 1) (cos x + sin x )
x→ 4 ( 2 cos x + 1)(cos x − sin x ) . (cos x + sin x ) . sin x
 2 cos 2 x − 1  (cos x + sin x ) sin x
= lim  
x →  4  cos 2 x − sin 2 x 
  2 cos x + 1 ( )
sin x (cos x + sin x )
= lim , since cos 2 x + sin 2 x = 1
x→ 4 2 cos x + 1

1  1 1 
 + 
 
2  2 2 1
= =
1 2
2 +1
2
(2)f 1 (𝑥) = 𝑒 𝑥 𝑔1 (𝑥)+g(x) 𝑒 𝑥

f 1 (0) = 𝑒 0 𝑔1 (0)+g(0) 𝑒 0 =1.1+2.1=3

𝑑𝑦
(3) = 6𝑒 2𝑥 + 6𝑒 3𝑥
𝑑𝑥

𝑑2 𝑦
= 12𝑒 2𝑥 + 18𝑒 3𝑥
𝑑𝑥 2

Page | 55
 𝑑2 𝑦 𝑑𝑦
L.H.S.= −5 +6y = 0 = R.H.S
𝑑𝑥 2 𝑑𝑥

𝑑𝑥 1 −1 𝑡 1
(4) 𝑑𝑡 = −1
. 𝑎 𝑠𝑖𝑛 . log 𝑎 . √1−𝑡 2
2√𝑎𝑠𝑖𝑛 𝑡

𝑑𝑦 1 −1 𝑡 −1
= . 𝑎𝑐𝑜𝑠 . log 𝑎 . √1−𝑡 2
𝑑𝑡 2√𝑎 𝑐𝑜𝑠−1 𝑡

1 −1 𝑡 −1
.𝑎𝑐𝑜𝑠 .log 𝑎 .
−1 √1−𝑡2
𝑑𝑦 𝑑𝑦/𝑑𝑡 2√𝑎𝑐𝑜𝑠 𝑡 𝑦
∴ 𝑑𝑥 = − 𝑑𝑥/𝑑𝑡 = 1 −1 1 = −𝑥
.𝑎𝑠𝑖𝑛 𝑡 .log 𝑎 .
−1 √1−𝑡2
2√𝑎𝑠𝑖𝑛 𝑡

SA – Type (3 marks each)

(1) Given, 𝑥√1 + 𝑦 + y√1 + 𝑥 = 0
𝑥√1 + 𝑦 = −𝑦√1 + 𝑥

Squaring both sides,

𝑥 2(1+y) = 𝑦 2 (1 + 𝑥)

𝑥 2+ 𝑥 2y = 𝑦 2 + 𝑦 2 𝑥

𝑥 2 −𝑦 2 = 𝑦 2 𝑥 − 𝑥 2 y

(𝑥 − 𝑦)(𝑥 + 𝑦) = −𝑥𝑦(𝑥 − 𝑦)

𝑥 + 𝑦 + 𝑥𝑦=0, (𝑥 ≠ y)

𝑦(1 + 𝑥) = −𝑥
𝑥
y =− 1+𝑥
𝑑𝑦 1
= − (1+𝑥)2
𝑑𝑥
(2) Given,𝑥 𝑦 = 𝑒 𝑥−𝑦
𝑦 𝑙𝑜𝑔𝑥 = 𝑥 − 𝑦
𝑦 + 𝑦𝑙𝑜𝑔𝑥 = 𝑥
𝑦(1 + 𝑙𝑜𝑔𝑥) = 𝑥
𝑥
𝑦 = 1+𝑙𝑜𝑔𝑥
𝑑𝑦 𝑙𝑜𝑔𝑥
= (1+𝑙𝑜𝑔𝑥)2
𝑑𝑥
(3) sin2 𝑦 + cos 𝑥𝑦 = 𝐾. Differentiating w.r.t. 𝑥 we get
𝑑𝑦 𝑑𝑦
⇒ 2 sin 𝑦 cos 𝑦 + (− sin 𝑥𝑦) (𝑥 + 𝑦) = 0
𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑑𝑦
⇒ sin 2𝑦 𝑑𝑥 − 𝑥 sin 𝑥𝑦 ⋅ 𝑑𝑥 − 𝑦 sin 𝑥𝑦 = 0
𝜋 1
𝑑𝑦 𝑦 sin 𝑥𝑦 𝑑𝑦 𝑦 sin 𝑥𝑦 ⋅ 𝜋
4 √2
⇒ 𝑑𝑥 = (sin 2𝑦−𝑥 sin 𝑥𝑦) ⇒ [𝑑𝑥 ] 𝜋 = (sin 2𝑦−𝑥 sin 𝑥𝑦) = 1 = 4(√2−1)
𝑥=1,𝑦= 1−
4 √2
LA – Type (5 marks each)
(1) L.H. L=a−1
R.H.L.= b+1

Page | 56
 f(4)= a + b
As f(𝑥) is a continuous function L. V= F. V = f (4)
i, e. a−1= b+1=a+b
a=1, b= −1
1−𝑐𝑜𝑠4𝑥
, when 𝑥 < 0
𝑥2
(2) If f(𝑥) = 𝑎 , when 𝑥 = 0 is continuous at 𝑥 = 0 ,then find the
√𝑥
, when 𝑥 > 0
{√16+√𝑥−4
value of ‘a’
1−𝑐𝑜𝑠4𝑥 1−𝑐𝑜𝑠4𝑥 1
L.H.L= lim− f(𝑥) = lim ( ) = 16 × lim ( ) = 16 × 2 = 8
𝑥→0 𝑥→0 𝑥2 𝑥→0 16𝑥 2
√𝑥 √
√𝑥( 16+√𝑥+4)
R.H.L.= lim+ f(𝑥) = lim ( ) = lim ( )=8
𝑥→0 𝑥→0 √16+√𝑥−4 𝑥→0 16+√𝑥−16
f(0)= a
As f(𝑥) is a continuous function L.V= F.V = f (0) ∴a=8
2
𝑥 + 3𝑥 + 𝑎, 𝑥 ≤ 1
(3) Clearly the function 𝑓(𝑥) = { is differentiable at 𝑥 = 1
𝑏𝑥 + 2, 𝑥 > 1
The function is continuous as well as differentiable at 𝑥 = 1.
L.H.L= lim− f(𝑥) = lim f(1 − ℎ) =
𝑥→1 ℎ→0
lim (1 − ℎ)2 + 3(1 − ℎ) + 𝑎 = 4+a
ℎ→0
R.H.L= lim+ f(𝑥) = lim b(1 + ℎ) + 2 = 𝑏 + 2
𝑥→1 ℎ→0
As the function is continuous,
∴ L.H.L= R.H.L
lim− f(𝑥) = lim− f(𝑥)
𝑥→1 𝑥→1
4+a=b+2----------------------(1)
𝑓(1−ℎ)−𝑓(1) (1−ℎ)2 +3(1−ℎ)+𝑎−(4+𝑎) (1−2ℎ+ℎ2 +3−3ℎ+𝑎−4−𝑎)
Here, L𝑓 ′ (1) = lim = lim = lim
ℎ→0 −ℎ ℎ→0 −ℎ ℎ→0 −ℎ
(−5ℎ+ℎ2
= lim = lim (5 − ℎ) = 5
ℎ→0 −ℎ ℎ→0

𝑓(1+ℎ)−𝑓(1) 𝑏(1+ℎ)+2−(4+𝑎) 𝑏(1+ℎ)+2−(4+𝑎) 𝑏+𝑏ℎ+2−(4+𝑎)

𝑅𝑓 ′ (1) = lim = lim = lim = lim
ℎ→0 ℎ ℎ→0 ℎ ℎ→0 ℎ ℎ→0 ℎ

Using b=2+a from equation-(1)

𝑏ℎ
= lim ℎ = b
ℎ→0
As the function is differentiable at 𝑥 = 1.
∴ L𝑓 ′ (1) = 𝑅𝑓 ′ (1)
b = 5 and 2+a=5, ∴ 𝑎 = 3

APPLICATION OF DERIVATIVE
VSA - Type(2 marks each)
1 𝑥
(1) Let y = (𝑥)
1
Log y = 𝑥log 𝑥
1 𝑑𝑦 −1 1
= 𝑥 2 . 𝑥 2 + 𝑙𝑜𝑔 𝑥
𝑦 𝑑𝑥

Page | 57
𝑑𝑦/𝑑𝑥 = 𝑦(𝑙𝑜𝑔 1/𝑥 − 1)
For maximum y,
𝑑𝑦 1
= 𝑦 (𝑙𝑜𝑔 − 1) = 0
𝑑𝑥 𝑥
1
log 𝑥 − 1 = 0
1
log 𝑥 = 1 = loge
1 1
=e⟹𝑥=
𝑥 𝑒
1
The maximum value is (𝑒)𝑒 .
(2) f / (x ) = 2 x + a
f is increasing in (1,2) if f (x )  0 in (1,2)
 2x + a  0
or, a> −2
(3) Let v and s be the volume and surface of a cube of side x cm
Given dv/dt=9cm2/sec
Now v= x3 ⟹ dv/dt=3x2 d x /dt

⟹9/3 x 2 =d x /dt
⟹3/ x 2 =d x /dt
Again s=6 x 2, so ds/dt]x=10cm = 12 x × dx/dt = 12 × 10 ×3/100 =3.6cm2 /sec
(4) Let the radius of the spherical balloon at any time be ‘r’
4
Volume v= 3 𝜋𝑟 3
𝑑𝑣 𝑑𝑟
= 4𝜋𝑟 2 𝑑𝑡
𝑑𝑡
𝑑𝑣
For r = 15, 𝑑𝑡 = 900
𝑑𝑟
 900 = 4𝜋(15)2
𝑑𝑡
𝑑𝑟 1
 = cm/sec
𝑑𝑡 𝜋
SA – Type (3 marks each)

(1) Given, f(x) = 3x4 − 8x3 + 12x2 − 48x + 25

f′(x) = 12x3 − 24x2 + 24x − 48
=12(x3 −2x2 +2x −4)
=12[x2(x − 2) + 2(x − 2)]
=12(x2 + 2) (x − 2)
For maxima and minima, f′(x) = 0
⟹ 12(x2 + 2) (x − 2) = 0
⟹ x = 2, x2 = −2
Since x2 =− 2 is not possible
So, x = 2 ∈ [0, 3]
Now we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3]
f(0) = 25
f(2) = 3 × 24 – 8 × 23 + 12 × 22 – 48× 2 + 25

Page | 58
 = 48 – 64 + 48 – 96 + 25 = −39
f(3) = 3 × 34 – 8 × 33 + 12 × 32 – 48 × 3 + 25
= 243 – 216 + 108 −144 + 2 = 16
Hence, at x = 0, Maximum value = 25
At x = 2, Minimum value = -39
(2) Distance 𝑆(𝑥) = √(𝑥 − 3)2 + 𝑥 4
𝑑𝑆 2(𝑥−3)+4𝑥 3
= = 0 ⇒ 2(𝑥 − 1)(2𝑥 2 + 2𝑥 + 3) = 0
𝑑𝑥 2√(𝑥−3)2 +𝑥 4
⇒ 𝑥 = 1 is the only critical point
𝑑𝑆 𝑑𝑆
< 0 for 𝑥 < 1 and 𝑑𝑥 > 0 for 𝑥 > 1
𝑑𝑥
Hence 𝑥 = 1 is a point of local minimum. The position nearest to A is 𝑃(1,8) and the minimum
distance is 𝑆(1) = √(1 − 3)2 + 14 = √5
(3) Let the side of the equilateral triangle = 𝑥
√3
Median (M) = 𝑥
2
𝑑𝑀 √3 𝑑𝑥
= = 2√3 cm/s
𝑑𝑡 2 𝑑𝑡
𝑑𝑥
= 4 𝑐𝑚/𝑠
𝑑𝑡
√3 2
Area(A) = 𝑥
4
𝑑𝐴 √3 𝑑𝑥 √3
= × 2𝑥 𝑑𝑡 = × 2 × 3 × 4 = 6√3 𝑐𝑚2 /𝑠𝑒𝑐.
𝑑𝑡 4 4
LA – Type (5 marks each)
(1) Let the side of the square piece to be cut off be 𝑥 cm.
Volume of the dish V=(40 − 2𝑥)(25 − 2𝑥)𝑥 = (1000𝑥 − 130𝑥 2 + 4𝑥 3 ) cm3
𝑑𝑣
= 12𝑥 2 − 260𝑥 + 1000
𝑑𝑥
𝑑𝑣
= 0,
𝑑𝑥
2
12𝑥 − 260𝑥 + 1000 = 0
∴ 𝑥=5
Length of the square cut off = 5cm
𝑑2 𝑦
= 24𝑥 − 260
𝑑𝑥 2

𝑑2 𝑦
For, 𝑥 = 5, 𝑑𝑥 2 = 24 × 5 − 260 = 120 − 260 = −140 < 0

∴ volume is maximum for 𝑥 = 5

Maximum volume of the dish V= (40 − 2 × 5)(25 − 2 × 5)5 = 30× 15 × 5 = 2250cm3
The area of the tin sheet used =Total area of tin sheet – Area of unused sheet=
(40 × 25) − 4 × 52 = 900cm2
(2) Let r be the radius of cone. R be the radius of sphere.
Height of cone is (R + x)

1
Volume of cone 𝑉 = 3 𝜋(𝑅 2 − 𝑥 2 )(𝑅 + 𝑥)

Page | 59
 1
= 𝜋[𝑅 3 + 𝑅 2 𝑥 − 𝑅𝑥 2 − 𝑥 3 ]
3
𝑑𝑉 1
= 𝜋[𝑅 2 − 2𝑅𝑥 − 3𝑥 2 ]
𝑑𝑥 3
𝑑𝑉
For 𝑑𝑥 = 0 ⇒ 3𝑥 2 + 2𝑅𝑥 − 𝑅 2 = 0

⇒ (𝑥 + 𝑅)(3𝑥 − 𝑅) = 0
𝑅
⇒𝑥 = 𝑜𝑟 𝑥 = −𝑅(not possible)
3

𝑑2𝑆 1
= 𝜋[−2𝑅 − 6𝑥] < 0
𝑑𝑥 2 3
𝑅
Hence the volume is maximum when 𝑥 = .
3

1 𝑅2 𝑅 1 8𝑅 2 4𝑅
Volume of cone = 3 𝜋 (𝑅 2 − ) (𝑅 + 3 ) = 3 𝜋 . .
9 9 3

8 4𝜋𝑅 3 8
( ) = 27(Volume of sphere)
27 3

4sin𝑥−2𝑥−𝑥cos𝑥
(3) f(𝑥) = 2+cos𝑥

4sin𝑥−𝑥(2+cos𝑥) 4sin𝑥
= = −𝑥
2+cos𝑥 2+cos𝑥

(2 + cos𝑥)4cos𝑥 − 4sin𝑥(−𝑠𝑖𝑛𝑥)
𝑓 ′ (𝑥) = −1
(2 + cos𝑥)2
(2+cos𝑥)4cos𝑥−4sin𝑥(−𝑠𝑖𝑛𝑥) 8cos𝑥+4−(4+cos2 𝑥+4𝑐𝑜𝑠𝑥)
= − 1=
(2+cos𝑥)2 (2+cos𝑥)2

4cos𝑥−cos2 𝑥 cos𝑥(4−cos𝑥)
= =
(2+cos𝑥)2 (2+cos𝑥)2

π 3π
For increasing, 𝑓 ′ (𝑥) ≥ 0 , cos𝑥 ≥ 0⇒𝑥 ∈ [0, 2 ) ∪ ( 2 , 2π]
π 3π
For decreasing, 𝑓 ′ ≤ 0 , cos𝑥 ≤ 0⇒𝑥 ∈ ( 2 , 2
)
INTEGRALS
Answers
2 Marks
𝑥2
1. Let I=∫ 𝑑𝑥
√𝑥 6 +𝑏6
𝑥2
=∫ 𝑑𝑥
√(𝑥 3 )2 +𝑏6
Put 𝑥 3 = 𝑡 ⟹ 3𝑥 2 𝑑𝑥 = 𝑑𝑡
1 𝑑𝑡
⟹ I= ∫
3 √𝑡 2 +(𝑏3 )2

Page | 60
 1
= [𝑙𝑜𝑔|𝑡 + √𝑡 2 + 𝑏 6 |] + 𝐶
3
1
= [𝑙𝑜𝑔|𝑥 3 + √𝑥 6 + 𝑏 6 |] + 𝐶
3
√cos 𝑥 . 𝑠𝑖𝑛𝑥
2. ∫ 𝑑𝑥
1−𝑐𝑜𝑠 2 𝑥
2𝑡 2
Put cos x=𝑡 2 = ∫ 𝑑𝑡
𝑡 4 −1
(𝑡 2 + 1) + (𝑡 2 − 1)
=∫ 2 𝑑𝑡
(𝑡 + 1) + (𝑡 2 − 1)
1 1
=∫ 2 𝑑𝑡 + ∫ 2 𝑑𝑡
𝑡 −1 𝑡 +1
1 √𝑐𝑜𝑠𝑥 − 1
= log | | + 𝑡𝑎𝑛−1 √𝑐𝑜𝑠𝑥 + 𝐶
2 √𝑐𝑜𝑠𝑥 + 1
𝜋 2 2
3. 𝐼 = ∫−𝜋(𝑐𝑜𝑠 𝑎𝑥 + 𝑠𝑖𝑛 𝑏𝑥 − 2 cos 𝑎𝑥 sin 𝑏𝑥)𝑑𝑥
𝜋 𝜋
𝐼 = 2 ∫0 𝑐𝑜𝑠 2 𝑎𝑥 𝑑𝑥 + 2 ∫0 𝑠𝑖𝑛2 𝑏𝑥 𝑑𝑥 − 0 [First two integrands are even function
while third is odd function.]
𝑠𝑖𝑛2 𝑎𝜋 sin 2𝑏𝜋
𝐼 = 2𝜋 + −
2𝑎 2𝑏
1 𝑥2 1 𝑥2
𝑥2 + − +
2 2 2 2
4. Let 𝐼 = ∫ 𝑑𝑥 = ∫ 𝑑𝑥
1−𝑥 4 (1−𝑥 2 )(1+𝑥 2 )
1 1 1 1 1 1 1+𝑥 1
= ∫ 𝑑𝑥 − ∫ 𝑑𝑥 = . 𝑙𝑜𝑔 | | + 𝐶1 − 𝑡𝑎𝑛−1 𝑥 + 𝐶2
2 1−𝑥 2 2 1+𝑥 2 2 2 1−𝑥 2
1 1+𝑥 1
= 𝑙𝑜𝑔 | | − 𝑡𝑎𝑛−1 𝑥 + 𝐶
4 1−𝑥 2

3 Marks

1. Let I=∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥 𝑑𝑥

1 1
I= 𝑒 −𝑥 𝑠𝑖𝑛2𝑥 − ∫(−1)𝑒 −𝑥 × 𝑠𝑖𝑛2𝑥 𝑑𝑥
2 2
1 −𝑥 1 1 −𝑥 (−𝑐𝑜𝑠2𝑥) 1
I= 𝑒 𝑠𝑖𝑛2𝑥 + { 𝑒 − ∫(−𝑒 −𝑥 ) × (−𝑐𝑜𝑠2𝑥)𝑑𝑥}
2 2 2 2
1 1 −𝑥 1
I= 𝑒 −𝑥 𝑠𝑖𝑛2𝑥 − 𝑒 𝑐𝑜𝑠2𝑥 − ∫ 𝑒 −𝑥 𝑐𝑜𝑠2𝑥 𝑑𝑥
2 4 4
1 1
I+ I= 𝑒 −𝑥 (2𝑠𝑖𝑛2𝑥 − 𝑐𝑜𝑠2𝑥)
4 4
5 1
I= 𝑒 −𝑥 (2𝑠𝑖𝑛2𝑥 − 𝑐𝑜𝑠2𝑥)
4 4
𝑒 −𝑥
I= (2𝑠𝑖𝑛2𝑥 − 𝑐𝑜𝑠2𝑥) + 𝐶
5
𝜋
1 −𝑥 (2𝑠𝑖𝑛2𝑥 4
∴ 𝐼 =[ 𝑒 − 𝑐𝑜𝑠2𝑥)] 𝜋
5 −
4
𝜋 𝜋
1 𝜋 𝜋 𝜋 𝜋
= [𝑒 −4 (2𝑠𝑖𝑛 − 𝑐𝑜𝑠 ) − 𝑒 4 (2𝑠𝑖𝑛 (− ) − 𝑐𝑜𝑠 ( ))]
5 2 2 2 2

Page | 61
 𝜋 𝜋
1
= [𝑒 −4 (2 − 0) − 𝑒 4 (−2 − 0)]
5
𝜋 𝜋
1
= [2𝑒 −4 + 2𝑒 4 ]
5
𝜋 𝜋
2
= [𝑒 −4 + 𝑒 4 ]
5
2
Hence, value of 𝛼 = .
5
3
2. Let 𝑥 =t 2

3 1
⟹ 𝑑𝑡 = 𝑥 2 𝑑𝑥
2
𝑥 2 𝑑𝑡
∫√ 𝑑𝑥 = ∫
1 − 𝑥3 3 √1 − 𝑡 2
2
= 𝑠𝑖𝑛−1 (𝑡) + 𝑐
3
3
2
= 𝑠𝑖𝑛−1 (𝑥 2 ) + 𝑐
3
Where ‘c’ is an arbitrary constant of integration.

𝜋
𝑥 𝑑𝑥
3. Let I= ∫04 … … … . (𝑖)
1+𝑐𝑜𝑠2𝑥+𝑠𝑖𝑛2𝑥
𝜋
Replace x by [ − 𝑥]
4
𝜋 𝜋
( −𝑥)
4
I=∫0 4
𝜋 𝜋 𝑑𝑥
1+𝑐𝑜𝑠2[ −𝑥]+𝑠𝑖𝑛2[ −𝑥]
4 4
𝜋 𝜋
( −𝑥)
4
=∫0 4 𝑑𝑥 … … … … … (𝑖𝑖)
1+𝑐𝑜𝑠2𝑥+𝑠𝑖𝑛2𝑥
Adding (i) and (ii)
𝜋
𝜋 1
2I= ∫04 𝑑𝑥
4 1+𝑐𝑜𝑠2𝑥+𝑠𝑖𝑛2𝑥
𝜋
𝜋 1
I= ∫04 1−𝑡𝑎𝑛2 𝑥 2 𝑡𝑎𝑛𝑥
𝑑𝑥 [𝐵𝑦 ℎ𝑎𝑙𝑓 𝑎𝑛𝑔𝑙𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑒]
8 1+ +
1+𝑡𝑎𝑛2 𝑥 1+𝑡𝑎𝑛2 𝑥
𝜋
𝜋 (1+𝑡𝑎𝑛2 𝑥)
I= ∫0 4 𝑑𝑥
8 1+𝑡𝑎𝑛2 𝑥+1−𝑡𝑎𝑛2 𝑥+2𝑡𝑎𝑛𝑥
𝜋 𝜋
𝜋 1+𝑡𝑎𝑛2 𝑥 𝜋 4 𝑠𝑒𝑐 2 𝑥
I= ∫0 4 𝑑𝑥 = ∫0 𝑑𝑥
8 2+2𝑡𝑎𝑛𝑥 8 2[1+𝑡𝑎𝑛𝑥]
2
Put , 1+tanx=t , 𝑠𝑒𝑐 𝑥 𝑑𝑥=dt
𝜋 2 𝑑𝑡 𝜋
I= ∫0 = [𝑙𝑜𝑔]20
16 𝑡 16
𝜋
= 𝑙𝑜𝑔2
16
√(𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥)2 sin 𝑥+cos 𝑥
4. 𝐼 = ∫ 𝑒 𝑥 𝑑𝑥 = ∫ 𝑒 𝑥
1+𝑐𝑜𝑠2𝑥 2 𝑐𝑜𝑠 2 𝑥
1
= ∫ 𝑒 𝑥 (𝑠𝑒𝑐𝑥 + sec 𝑥. 𝑡𝑎𝑛𝑥)𝑑𝑥
2

Page | 62
 1
= 𝑒 𝑥 . 𝑠𝑒𝑐𝑥 + 𝐶
2

5 Marks
𝑥2
1. Let I=∫ 𝑑𝑥
𝑥 4 −𝑥 2 −12
𝑥2
=∫ 𝑑𝑥
𝑥 4 −4𝑥 2 +3𝑥 2 −12
𝑥2
=∫ 𝑑𝑥
𝑥 2 (𝑥 2 −4)+3(𝑥 2 −4)
𝑥2
=∫ (𝑥 2 𝑑𝑥
+3)(𝑥 2 −4)
Let 𝑥 2 = 𝑡
𝑥2 𝑡 𝐴 𝐵
Now, (𝑥 2 =(𝑡−4)(𝑡+3) = +
+3)(𝑥 2 −4) 𝑡−4 𝑡+3
⇒ 𝑡 = 𝐴(𝑡 + 3) + 𝐵(𝑡 − 4)
On comparing the coefficient of t and constant terms on both sides, we get
⇒𝐴+𝐵 =1 … . . (𝑖)
And 3𝐴 − 4𝐵 = 0 … … (𝑖𝑖)
On solving (i) and (ii), we get
4 3
A= and B=
7 7
𝑥2 4 3
∴ 2 = +
(𝑥 − 4)(𝑥 2 + 3) 7(𝑥 2 − 4) 7(𝑥 2 + 3)
4 𝑑𝑥 3 1
∴I= ∫ 2 2 + ∫ 2 2 𝑑𝑥
7 𝑥 −2 7 𝑥 +(√3)
=
4 1 𝑥−2 3 1 𝑥
× 𝑙𝑜𝑔 | | × 𝑡𝑎𝑛−1 ( ) + 𝐶
7 4 𝑥 + 2 7 √3 √3

1 𝑥−2 √3 𝑥
= 𝑙𝑜𝑔 | |+ 𝑡𝑎𝑛−1 ( ) +𝐶
7 𝑥+2 7 √3

𝜋
4
2. Let ,I= ∫ log(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥) 𝑑𝑥
𝜋 … …. (𝑖)
−
4
𝜋 𝜋
Replace x by ( − − 𝑥)
4 4
𝜋
4
I=∫ log (cos 𝑥 − 𝑠𝑖𝑛𝑥)𝑑𝑥
𝜋 … .. (𝑖𝑖)
−
4
Adding equations (i) and (ii), we get
𝜋
4
⟹2I=∫ [log (cos 𝑥 + 𝑠𝑖𝑛𝑥) + log(𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥)]𝑑𝑥
𝜋
−
4

Page | 63
 𝜋

⟹2I=∫ log (cos2 𝑥 − 𝑠𝑖𝑛2 𝑥)𝑑𝑥

4
𝜋
−
4
𝜋
4
⟹2I=∫ log(𝑐𝑜𝑠2𝑥) 𝑑𝑥𝜋
−
4
𝜋
⟹2I=2 ∫04 log(𝑐𝑜𝑠2𝑥) 𝑑𝑥 … … . . (𝑖𝑖𝑖)
𝑎 𝑎

[∵ ∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥), 𝑖𝑓 𝑓(−𝑥) = 𝑓(𝑥)]

−𝑎 0
Put 2x=t
𝑑𝑡
⟹ 𝑑𝑥 =
2
𝑤ℎ𝑒𝑛 𝑥 = 0 , 𝑡ℎ𝑒𝑛 𝑡 = 0
𝜋 𝜋
And when x= 𝑡ℎ𝑒𝑛 , 𝑡 =
4 2
𝜋
∴ 2I=∫0 log(cos 𝑡)𝑑𝑡
4 … … . (𝑖𝑣)
𝜋
Replace t by ( − 𝑡)
2
𝜋
⟹ 2I= ∫0 log(sin 𝑡)𝑑𝑡 4 … … … . (𝑣)
On adding equations (iv) and (v), we get
𝜋
4I=∫0 (log 𝑠𝑖𝑛𝑡 + log 𝑐𝑜𝑠𝑡 𝑑𝑡
4

𝜋
2𝑠𝑖𝑛𝑡 𝑐𝑜𝑠𝑡
=∫04 log ( ) 𝑑𝑡
2
𝜋
=∫0 (log sin 2𝑡 − 𝑙𝑜𝑔2) 𝑑𝑡
4

Put 2t=u
1
⟹ 2𝑑𝑡 = 𝑑𝑢 𝑜𝑟 𝑑𝑡 = 𝑑𝑢
2
When t=0, u=0
𝜋
When t= ,u=𝜋
2

1 𝜋 𝜋
∴ 4I= ∫ log(𝑠𝑖𝑛𝑢) 𝑑𝑢 − 2 𝑙𝑜𝑔2
2 0
𝜋
1 𝜋
⟹ 4I= × 2 ∫0 log(𝑠𝑖𝑛𝑢) 𝑑𝑢 − 𝑙𝑜𝑔2
2
2 2
𝜋
𝜋
⟹ 4I= ∫0 log(𝑠𝑖𝑛𝑡) 𝑑𝑡 − 𝑙𝑜𝑔2
2
2

[∵ ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑡)𝑑𝑡]

Page | 64
 𝜋
⟹ 4I=2I− 𝑙𝑜𝑔2 [Using (v)]
2
𝜋
⟹I= − 𝑙𝑜𝑔2
4

3. Replace x by (2𝜋 − 𝑥) and add

2𝜋 2𝜋 .𝑠𝑖𝑛2𝑛 𝑥
2𝐼 = ∫0 𝑑𝑥
𝑠𝑖𝑛2𝑛 𝑥+ 𝑐𝑜𝑠 2𝑛 𝑥
2𝜋
𝑠𝑖𝑛2𝑛 𝑥
∴ 𝐼 = 𝜋∫ 𝑑𝑥
𝑠𝑖𝑛2𝑛 𝑥 + 𝑐𝑜𝑠 2𝑛 𝑥
0
𝜋
𝑠𝑖𝑛2𝑛 𝑥
= 2𝜋 ∫ 𝑑𝑥
𝑠𝑖𝑛2𝑛 𝑥 + 𝑐𝑜𝑠 2𝑛 𝑥
0
𝜋
2
𝑠𝑖𝑛2𝑛 𝑥
∴ 𝐼 = 4𝜋 ∫ 𝑑𝑥
𝑠𝑖𝑛2𝑛 𝑥 + 𝑐𝑜𝑠 2𝑛 𝑥
0
𝜋
Replace x by ( − 𝑥)
2
𝜋
2
𝑐𝑜𝑠 2𝑛 𝑥
∴ 𝐼 = 4𝜋 ∫ 𝑑𝑥
𝑐𝑜𝑠 2𝑛 𝑥 + 𝑠𝑖𝑛2𝑛 𝑥
0
𝜋
On adding above two : 2𝐼 = 4𝜋 ∫0 𝑑𝑥 = 2𝜋 2 2

⟹ 𝐼 = 𝜋2
Chapter: APPLICATION OF INTEGRALS
Answers
2 Marks
41
1. Area= ∫2 = 𝑙𝑜𝑔4 − 𝑙𝑜𝑔2
𝑥
=log 2
0 2
2. Area=|∫−1 −𝑥 2 𝑑𝑥 | + ∫0 𝑥 2 𝑑𝑥
1 8
= +
3 3
=3
2
3. Area=|∫0 (𝑥 − 2)𝑑𝑥 | = 2 𝑠𝑞. 𝑢𝑛𝑖𝑡

Page | 65
 3 Marks
1.

4𝑎 4𝑎
Point of intersection P( , )
𝑚2 𝑚

4𝑎⁄ 4𝑎⁄
Area=∫0 𝑚2 √4𝑎𝑥 − ∫0 𝑚2 𝑚𝑥 𝑑𝑥

32 𝑎2 8𝑎2
= −
𝑚3 𝑚3

24𝑎2
= 𝑠𝑞. 𝑢𝑛𝑖𝑡
𝑚3

2.

1 27 1⁄ 243
Area= ∫0 𝑦 3 𝑑𝑦 =
2 8

Page | 66
 3.

2
Area=∫−2 √4 − 𝑥 2 𝑑𝑥
2
𝑥 22 −1 𝑥
=[ √22 − 𝑥2 + 𝑠𝑖𝑛 ]
2 2 2 −2
𝜋 𝜋
=2 + 2 . = 2𝜋 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠
2 2
5 Marks
1.

1
Area= 4.∫0 (1 − 𝑥)𝑑𝑥
= 2 sq. unit

2.

−1 3
Area=∫−3 −𝑥 𝑑𝑥 + ∫−1(𝑥 + 2)𝑑𝑥
= 16 sq. Unit

Page | 67
3.

1 2
Area=∫1 − log 𝑥 𝑑𝑥 + ∫1 log 𝑥 𝑑𝑥
2
1 1
= log + 2 log 2
2 2

DIFFERENTIAL EQUATION

SOLUTION

1. Solution:
∫ dy = ∫ tan−1 x. dx

d
⇒ y = tan−1 x ∫ dx − ∫ tan−1 x. ∫ dx + C
dx

1 2x
⇒ y = xtan−1 x − ∫ 2 . dx + C
2 1+x

1
⇒ y = xtan−1 x − log|1 + x 2 | + C
2

2. Solution:
dy
y = x + 3y 2
dx
dx x
⇒ − = 3y
dy y

−1
P(xy)= , Q(y)= 3y
y

−1
.dy 1
I.F = ∫ e y =
y

x. I.F = ∫ Q(y). dy
x 1
⇒ = ∫ 3y. . dy + C
y y

Page | 68
 x
⇒ = 3y + C
y

⇒ x = 3y 2 + Cy

3. Solution:
x = 3 and y = 4

x+y = 7

4. Solution:
dy y y
= ( ) − cos 2 ( )--- (1)
dx x x

Put y = vx.
dy dv
⇒ = v+x --- (2)
dx dx

Putting (2) in (1).

dv
v+x = v − cos 2 v
dx
dv
⇒x = cos 2 v
dx

dx
⇒ sec 2 v. dv = −
x

Upon Integrating we get

tan v = − log x + C
y
⇒ tan = − log x + C
x
π
The Curve passes through (1, ).
4

π
tan = − log 1 + C
4
⇒C=1
y
Therefore, the Curve is tan = − log x + 1.
x

Page | 69
 5. Solution:
dy y
x2 − xy = 2cos 2 ( )
dx 2x
y
sec2 ( ) dy
⇒
2
2x
(x 2 dx − xy) = 1

Dividing both sides by x 3 , we get

y dy
sec2 ( ) (x −y) 1
2x dx
⇒ 2
=
2 x x3

d y 1
⇒ (tan (2x)) = x3
dx

y 1
⇒ tan ( ) = − 2 + C.
2x 2x

6. Solution:

4
x 3 y. dx − x. dy
x . dx + 3 ( )=0
y y2
x 3 x
⇒ x 4 . dx + ( ) d ( ) = 0
y y

On Integrating it
x 4
x5 ( )
y
⇒ + =C
5 4

x5 x4
⇒ + =C
5 4y4

7. Solution:
dy 1
=
dx (x + y + 1)2

Putting x+y+1 = v
dy dv
= −1
dx dx
dv 1
−1= 2
dx v

Page | 70
 dv 1+v2
⇒ =
dx v2

v2
⇒∫ dv = ∫ dx
1+v2

v2 +1−1
⇒∫ dv = ∫ dx
1+v2

⇒ v − tan−1 v = x + C

⇒ (x + y + 1) − tan−1 (x + y + 1) = x + C

⇒ y + 1 − tan−1 (x + y + 1) = C

Putting x = -1, y = 0

1 − tan−1 0 = C

⇒C=1

Hence, the required solution is

y = tan−1 (x + y + 1) or x + y + 1 = tan y.

8. Solution:
dx
= x 2 y 3 + xy
dy
1 dx 𝑦
⇒ = y3 +
𝑥 2 dy 𝑥

1 dx 1
⇒ − y = y3
x2 dy x

1 1 dx du
Put − = u and = .
x x2 dy dy

du
⇒ + uy = y 3 (Linear D.E.)
dy

y2
I.F = e 2

y2 y2
u. e =∫ e . y 3 . dy + C
2 2

y2 y2
2
=− (y . e − 2e ) + C
2 2

Page | 71
 y2 y2 y2
1 2
Or . e +y . e − 2e
2 2 2 =C
x

9. Solution:
Let P be the principal at any time t. Then,
dP 5P
=
dt 100
dP P
⇒ =
dt 20

1 1
⇒ dP = dt
p 20

Integrating both sides, we get

1 1
∫ p dP = ∫ 20 dt

1
⇒ log P = t + log C
20

P 1
⇒ log = t
C 20

t
⇒ P = Ce ⁄20 ---(i)

It is given that P = 1000 when t = 0.

Substituting these values in (i), we get

1000 = C

Substituting C = 1000 in (i), we get

t
P = 1000 e ⁄20

Let t1 years be the time required to double the principal i.e. at t = t1, P = 2000.

Substituting these values in (i), we get

t1⁄ t1⁄ t1
2000 = 1000 e 20 ⇒e 20 =2⇒ = log e 2 ⇒ t1 = 20 log e 2
20

Hence, the principal doubles in 20 log e 2 years.

Page | 72
 VECTORS-SOLUTIONS

2 marks
2 3
⃗⃗ + 𝑐⃗| = √12 ⇒ |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = 12 ⇒ 3|𝑎⃗|2 + 2 ( ) |𝑎⃗|2 = 12
1. Ans.√2|𝑎⃗ + 𝑏
2
⇒ 6|𝑎⃗|2 = 12 ⇒ |𝑎⃗| = √2
⃗⃗
3𝑎⃗⃗−𝑏
2. Ans. If 𝑐⃗is the position vector of C, then 𝑐⃗ =
2

3 3
𝐵𝐶 = 𝐵𝐴 ⇒ 𝑐⃗ − 𝑏⃗⃗ = (𝑎⃗ − 𝑏⃗⃗) ⇒ 2(𝑐⃗ − 𝑏⃗⃗) = 3(𝑎⃗ − 𝑏⃗⃗)
2 2
𝜃 𝜃
3. Greatest value of |𝑎⃗⃗ + 𝑏⃗⃗| + |𝑎⃗ − 𝑏⃗⃗| is 2 (𝑐𝑜𝑠 + 𝑠𝑖𝑛 ) = 2√2
2 2

3 marks

⃗⃗ = 𝑐⃗ ⇒ 𝑐⃗ ⊥ 𝑎⃗&𝑐⃗ ⊥ 𝑏⃗⃗
1. 𝑎⃗ × 𝑏
𝑎⃗ × 𝑐⃗ = 𝑏⃗⃗ ⇒ 𝑏⃗⃗ ⊥ 𝑎⃗&𝑏⃗⃗ ⊥ 𝑐⃗𝑖. 𝑒. 𝑎⃗, 𝑏⃗⃗&𝑐⃗are mutually perpendicular vectors.
Also, |𝑎⃗||𝑏⃗⃗||𝑎⃗||𝑐⃗| = |𝑏̄||𝑐⃗| ⇒ |𝑎⃗| = 1
∴ |𝑏⃗⃗| = |𝑐⃗|
⃗⃗ > 0 ⇒ 𝑥 2 > 4 ⇒ 𝑥 ∈ (−∞, −2) ∪ (2, ∞)
2. 𝑎⃗. 𝑏
𝑏⃗⃗. 𝑐⃗ < 0 ⇒ 𝑥 2 < 9 ⇒ 𝑥 ∈ (−3,3)So 𝑥 ∈ (−3, −2) ∪ (2,3)
3. 𝑎⃗ = 𝑏 ⃗⃗ + 𝑐⃗
 piˆ + qˆj + rkˆ = (sî + 3ĵ + 4k̂) + (3î + ĵ - 2k̂)
 s + 3 = p, q = 4, r = 2.
1  
Area of triangle = b  c
2

iˆ ˆj k
1 1
5 6 = s 3 4 5 6 = 325 + 5s 2 + 30 s  s = −11,5
2 2
3 1 −2
p=-8, q=4, r=2.

5 marks

1 .Let the position vectors of the four points A, B, C and D be 𝑎⃗, 𝑏⃗⃗, 𝑐⃗and
            →
𝑑⃗respectively AB = b − a, BC = c − b , AD = d − a, BD = d − b , CA = a − c , CD = d − c |𝐴𝐵 ×

Page | 73
 → → → → →
𝐶𝐷 + 𝐵𝐶 × 𝐴𝐷 + 𝐶𝐴 × 𝐵𝐷 | = 2(𝑎⃗ × 𝑏⃗⃗ + 𝑏⃗⃗ × 𝑐⃗ + 𝑐⃗ × 𝑎⃗) = 2 × 2(Area of the
𝛥𝐴𝐵𝐶)

2. Let 𝑐⃗ = 𝑥𝑖̑ + 𝑦𝑗̑ + 𝑧𝑘̑ As −𝑖̑ + 𝑗̑ − 𝑘̑ bisects the angle between 𝑐⃗&3𝑖̑ + 4𝑗̑,
3𝑖̑ + 4𝑗̑
−𝑖̑ + 𝑗̑ − 𝑘̑ = 𝜆 (𝑥𝑖̑ + 𝑦𝑗̑ + 𝑧𝑘̑ + )
5
3 4
⇒ −1 = 𝜆(𝑥 + ); 1 = 𝜆(𝑦 + ); −1 = 𝜆𝑧
5 5
⇒ 𝑥 = (−5 − 3𝜆)/5𝜆, 𝑦 = (5 − 4𝜆)/5𝜆, 𝑧 = −1/𝜆 But

𝑥2 + 𝑦2 + 𝑧2 = 1
15
⇒𝜆=
2
1
∴ 𝑐̑ = − (11𝑖̑ + 10𝑗̑ + 2𝑘̑)
15
→
4. (a) 𝑂𝐴 = 4𝑖̑ + 10𝑗̑
→ 1
(b)Unit vector along 𝑂𝐴 = (4𝑖̑ + 10𝑗̑)
√116

→
(c) 𝑂𝐵 = 9𝑖̑ + 7𝑗̑
→
(d) 𝐴𝐶 = −8𝑗̑

(e) |𝐴⃗| + |𝐵
⃗⃗| = √5 + √5 = 2√5&|𝐴⃗ + 𝐵
⃗⃗| = √72 + 72 = √98

3-DIMENSIONAL GEOMETRY

ANSWERS

Q1. Let the point C (0,y,0)

Drs of AB -1, -4, 3

Drs of BC -1, y+1, -2

As AB⊥BC

⇒ 1 – 4y – 4 – 6 = 0
9
⇒y=−
4

Page | 74
 9
Required point (0, − , 0)
4

Q2. cos 2 𝜃 + cos 2 𝛽 + cos 2 𝜃 = 1

⇒ cos 2 𝜃 + 1 − sin2 𝛽 + cos 2 𝜃 = 1

⇒ 2cos 2 𝜃 + 1 − 3sin2 𝜃 = 1
3
⇒ 2cos 2 𝜃 = 3(1 − cos 2 𝜃) ⇒ 5cos 2 𝜃 = 3 ⇒ cos 2 𝜃 =
5

Q3. 𝑥 − 3𝑦 − 6 = 0 ⇒ 𝑥 = 3𝑦 + 6
𝑥−6
⇒𝑦=
3

𝑥 − 𝑦 + 𝑧 − 5 = 0 ⇒ 3𝑦 + 6 − 𝑦 + 𝑧 − 5 = 0
−𝑧−1
⇒ 2𝑦 + 𝑧 + 1 = 0 ⇒ 𝑦 =
2

𝑥−6 𝑧+1
So =𝑦= drs 〈3, 1, −2〉
3 −2

3 MARKS
𝑥−1 𝑦−2 𝑧−3
Q1. Equation of line AB is = =
2 3 6

Any point on 𝐴𝐵(1 + 2𝜆, 2 + 3𝜆, 3 + 6𝜆)

7
Mid-point on 𝐴𝐵(2, , 6)
2

Distance = 14
3 2
⇒ (2𝜆 − 1)2 + (3𝜆 − ) + (6𝜆 − 3)2 = 196
2

⇒ 4𝜆2 − 4𝜆 − 15 = 0
5 3
⇒ 𝜆 = ,−
2 2

19 5
Required points (6, , 18), (−2, − , −6)
2 2

Q2. Co-ordinates of any point on L1𝑃(𝜆, 𝜆 − 1, 𝜆)

𝜇 𝜇
Co-ordinates of any point on L2𝑄(𝜇 − 1, , )
2 2

Page | 75
 𝜇 𝜇
Drs of 𝑃𝑄 〈𝜆 − 𝜇 + 1, 𝜆 − 1 − , 𝜆 − 〉
2 2

Given drs of PQ 2, 1, 2
𝜆−𝜇+1 𝜆−1−𝜇/2 𝜆−𝜇/2
= =
2 1 2

Solving 𝜆 = 3, 𝜇 = 2

P(3, 2, 3) Q(1, 1, 1)

Q3. Midpoint of line segment joining points (-1, 2, 3) and (3,0,-1) is (1,1,1).
𝑦−4 2−𝑧
𝑥 = −3, =
3 1
𝑥+3 𝑦−4 𝑧−2
= =
0 3 −1

Direction ratios of the line are 0, 3,-1. Hence equation of the line is
𝑥−1 𝑦−1 𝑧−2
= =
0 3 −1
5 MARKS

Q1. Equations of the line passing through (1, 1, 1) is

𝑥−1 𝑦−1 𝑧−1
= =
𝑎 𝑏 𝑐
The line intersects the line l1

(𝑎⃗2 − 𝑎⃗1 ). (𝑏⃗⃗1 𝑋𝑏⃗⃗2 ) = 0

1−1 2−1 3−1

⇒| 𝑎 𝑏 𝑐 |=0
2 3 4
⇒ 𝑎 − 2𝑏 + 𝑐 = 0 ________(1)

Also, it intersects the l2

−2 − 1 3 − 1 −1 − 1
⇒| 𝑎 𝑏 𝑐 |=0
1 2 4
⇒ 6𝑎 + 5𝑏 − 4𝑐 = 0 ________(2)

Page | 76
 𝑎 𝑏 𝑐
Solving = =
8−5 6+4 5+12

𝑎 𝑏 𝑐
⇒ = =
3 10 17

𝑥−1 𝑦−1 𝑧−1

So required equation is = =
3 10 17

Q2. Since ABCD is a parallelogram, AC and BD bisects each other.

Co-ordinates of D(3, 4, 5)

Equation of AB is 𝑟⃗ = 4𝑖̂ + 5𝑗̂ + 10𝑘̂ + 𝜆(−2𝑖̂ − 2𝑗̂ − 6𝑘̂)

Equation of CD is 𝑟⃗ = 𝑖̂ + 2𝑗̂ − 𝑘̂ + 𝜇(2𝑖̂ + 2𝑗̂ + 6𝑘̂)

These two lines are are parallel lines.

⃗⃗|
|(𝑎⃗⃗2 −𝑎⃗⃗1 )×𝑏
Distance between these two lines = ⃗⃗ |
|𝑏

|(−3𝑖̂ − 3𝑗̂ − 11𝑘̂) × (2𝑖̂ + 2𝑗̂ + 6𝑘̂)|

=
√4 + 4 + 36
|4𝑖̂ − 4𝑗̂| 4√2 4√2 2√2
= = = =
√44 √44 2√11 √11
Q3. Let 𝑃(2𝜆 + 3, 𝜆 + 3, 𝜆) be any point on the given line.
2𝜋
As OP makes an angle with the line with drs 2, 1, 1
3

and Drs of OP 2𝜆 + 3, 𝜆 + 3, 𝜆
2𝜋 (2𝜆 + 3)2 + (𝜆 + 3)1 + 𝜆
cos =
3 √(2𝜆 + 3)2 + (𝜆 + 3)2 + 𝜆2 √22 + 12 + 12
−1 6𝜆 + 9 −1 2𝜆 + 3
⇒ = ⇒ =
2 6√𝜆2 + 3𝜆 + 3 2 2√𝜆2 + 3𝜆 + 3
⇒ 𝜆2 + 3𝜆 + 3 = 4𝜆2 + 12𝜆 + 9 ⇒ 𝜆2 + 3𝜆 + 2 = 0

𝜆 = −1, −2

𝑃(1, 2, −1) 𝑜𝑟 𝑃(−1, 1, −2)

Page | 77
 𝑥 𝑦 𝑧 𝑥 𝑦 𝑧
Equation = = 𝑎𝑛𝑑 = =
1 2 −1 −1 1 −2

ANSWERS LINEAR PROGRAMMING PROBLEM

2 marks
1. As maximum of Z occurs at Q(3,5) and S(4,1), 3a +10b = 4a + 2b
i.e. a = 8b
2. Correct graph.
As x + y = 5 passes through the feasible region, maximum or minimum value of
objective function exists on end points of line i.e. (3,2) or (0,5).
Max. Z =40 at (0, 5).
3. As no feasible region, no. of solutions is 0.

3 marks
1.

The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100) and D(0,
200)values of Z at these points are as follows:

The maximum value of Z is 400 at D(0, 200) and the minimum value of Z is 100 at all
the points on the line segment joining A(0, 50) and B(20, 40).

Page | 78
2. Correct Graph (bounded)
30 6 3
Corner points are A(2,0), B( , ) , C( ,1) , D(0,1) and O(0,0) and maximum
13 13 2
value occurs at B. Max Z = 294/13=22.62 .

3.

As the feasible region is unbounded therefore, 7 may or may not be the minimum
value of Z.

For this, we draw the graph of the inequality, 3x+5y<7 and check whether the resulting
half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 5y < 7.
Therefore, the minimum value of Z is 7 at (3/2, 1/2).
5 marks
1. (i) x  2, 2y  x
(ii) Max Z = 21 at C (6, 3)
(iii) Min P = -4 at B (2, 7)
(iv) Area = 12 sq. units
2. (i) 3x − 2 y  6
(ii) x + 3y  15

Page | 79
 (iii) 5p + 9q =0
(iv) Z at A (2, 0) is 8 and at D (0, 5) is -5.
3. Correct bounded graph
7
Corner points are P (3, ) , Q(3,3) and R(2, 3) and max Z = -6 at Q(3, 3)
2
Min Z = -9 at R (2, 3)
4. Correct unbounded graph
Corner points are O (0, 0), A (0, 4) and B (12, 6) and Min Z = -16, Max Z = 12
PROBABILITY
2 marks
3 3 7 7 3 3 126
1. P( A) =   +   =
10 10 10 10 10 10 1000
2. P( E / A) = P(E  A) = 2 = 1 {465,456}
P( A) 1  6  6 18
P( A) = 3 / 7 P(B) = 7/12
3.
P(A  B) = 1 - P(A  B) = 1 − 4 / 7  5 / 12 = 16 / 21

3 marks
1. Let E1= A and B both solve.
E2 = A and B got incorrect solution
E= both got same answer

1 1 1 2 3 1
P( E1 ) =  = P(E 2 ) =  =
3 4 12 3 4 2
P( E / E1 ) = 1 P(E/E 2 ) = 1 / 20
10
P( E1 / E ) =
13
2.
X P(X)
0 C(40,2)/C(52,2) = 130/221
1 C(12,1).C(40,1)/ C(52,2)=80/221
2 C(12,2)/C(52,2) = 11/221

3. A = exactly 2 girls B = at least one girl,

P( A  B ) 21 / 55 126
P( A / B) = = =
P (B ) 59 / 66 295
C (7,4) 59 C(4,2)  C(7,2) 21
P( B) = 1 − P(no girl ) = 1 − = & P(A  B) = =
C (11,4) 66 C(11,4) 55
5 marks
1. E1 = 2 white balls E2 = 3 white balls E3 = 4 white balls
A = getting 2 white balls

Page | 80
 1
P( E1 ) = = P(E 2 ) = P( E3 )
3
C (2,2) C (3,2)
P( A / E1 ) = = 1/ 6 P(A/E 2 ) = = 1/ 2
C (4,2) C 4,2)
C (4,2) 3
P( A / E3 ) = =1  P(E 3 / A) =
C (4,2) 5
1
2. (i)  P( X ) = 1  10k 2 + 9k − 1 = 0  k =
10
3 17 3
(ii) P(X<3) = (iii) P(X>6) = (iv) P(0<X<3) =
10 100 10

3.
X 0 1 2 3 4
P(X) 625/1296 500/1296 150/1296 20/1296 1/1296
2
Also, Mean =  X P( X ) = 3

Page | 81
 COMMON ERRORS
CHAPTER-1
Relations and Functions
• Let A = {1, 2, 3}. Check whether R= {(1, 2), (2, 1), (1, 1), (1, 3)} is symmetric or not.
Common Error: Here (1, 2) ∈ R, (2, 1) ∈ R. So, it is symmetric.
Correction: The students thinks that only an ordered pair is to check for symmetric. So, he
forgets to check for (1, 3).
• If A = {1, 2, 3} check whether R = {(1, 1), (1, 2), (2, 1)} is transitive or not.
Common Error: (1, 2) ∈ R, (2, 1) ∈ R ⟹ that (1, 1) ∈ R. So it is transitive.
Correction: Here the students forget to see for (2, 1) ∈ R, (1, 2) ∈ R implies (2, 2) ∈ R or not

• Show that relation R in the set A = {1, 2, 3, 4, 5 } given by R= {(a, b):|a - b| is even}, is an
equivalence relation.
Common Error: Student forget to write while proving transitive
Correction: Transitive: Let (a, b) ∈ R and (b, c) ∈ R where a, b, c ∈ A

⟹|a - b| is even and |b - c| is even

⟹a – b = ±2k1 ±2k1 =±2 (k1 + k2)

⟹|a - b|=2 (k1 + 2k2)

⟹|a - b| is even

⟹ (a ,c) ∈ R, so R is transitive.

Chapter-2
INVERSE TRIGONOMETRIC FUNCTION
1 𝜋
• Error: cos-1 (− 2) = − 3
1 𝜋 2𝜋
Remedy : cos-1 (− ) = 𝜋 − =
2 3 3
Chapter-4
Determinants
• Common Error:
2 −1 4
|0 1 6| = 2 (2 + 18) −(0 − 6) + (0 − 1) = 42
1 −3 2

It is wrong answer

Correction : Correct Solution is = 2 (2 + 18) +1(0 − 6) + 4(0 − 1) = 30

• Find the value of k, such that area of triangle with vertices (3, k), (2, 2), (-4, 1) is 3 square units.
Common Error

Page | 82
 3 𝑘 1
1
Area of Triangle = 2 | 2 2 1| = 3
−4 1 1
3 𝑘 1
⟹| 2 2 1|=6
−4 1 1

−6𝑘 = 6 − 13

(2 − 1) − 𝑘 (2 + 4) + 1(2 + 8) = 6

3 − 6𝑘 + 10 = 6

−6𝑘 = −7

K=7/6 (Wrong Value)

Correction: Here area of triangle = 3 square unit 3(2 − 1) − 𝑘 (2 + 4) + 1(2 + 8) = ±6

So the values of determinant can be 3 or −3 3 − 6𝑘 + 10 = ±6

So we must take both the cases Either −6𝑘 = 6 − 13 𝑜𝑟 − 6𝑘 = −6 − 13

3 𝑘 1
1
Area of triangle = | 2 2 1| =±3 Either −6𝑘 = −7 𝑜𝑟 − 6𝑘 = −19
2
−4 1 1

Either k=7/6 or k=19/6

7 19
So k={6 , 6 } (Correct Values)

3 1
• Given matrix A=[ ]
−1 2
Prove that 𝐴2 − 5𝐴 + 7𝐼 = 0, 𝑎𝑛𝑑 ℎ𝑒𝑛𝑐𝑒 𝑓𝑖𝑛𝑑 𝐴−1

Common Error: For finding the value of 𝐴−1, some students apply the formula

1
𝐴−1 = × 𝑎𝑑𝑗𝑜𝑖𝑛𝑡 𝐴
|𝐴|

• Find the product of matrices A and B and hence solve the following system of equations
𝑥 − 𝑦 = 3, 2𝑥 + 3𝑦 + 4𝑧 = 17, 𝑦 + 2𝑧 = 7
Common Error: In this problem some of the students opts the following procedure
Step-1: Find AB
Step-2: Find 𝐴−1 by applying the following formula
1
𝐴−1 = |𝐴| × 𝐴𝑑𝑗𝑜𝑖𝑛𝑡 𝐴
Step-3: Find the value of x, y and z
Correction:

1 −1 0 2 2 −4 6 0 0
Step-1: Find the product 𝐴𝐵 = [2 3 4] [−4 2 −4] = [0 6 0] =6𝐼
0 1 2 2 −1 −5 0 0 6

Page | 83
 Step-2: Find 𝐴−1 by using AB=6I
Multiply on both side by 𝐴−1 we get
(𝐴−1 𝐴)𝐵 = 6(𝐴−1 𝐼)
IB= 6 𝐴−1 or B=6𝐴−1
2 2 −4
1 1
𝐴−1 = 6 𝐵= 6 [−4 2 −4]
2 −1 −5

Now given equations can be written in matrix form as AX=C where

1 −1 0 𝑥 3
A= [2 3 4] , X=[𝑦] , 𝐶 = [17]
0 1 2 𝑧 7

Step-3: Find the value of X by using the formula

X=𝐴−1 𝐶

2 2 −4 3
1
X= [−4 2 −4] [17]
6
2 −1 −5 7
𝑥 12 2
1
|𝑦| = 6 [−6]= [−1]
𝑧 24 4

⟹ x=2, y=-1, z=4

Continuity and Differentiability

Common Error: Finding second derivative when parametric function is given

𝑑2 𝑦
• If x= 𝑎(cos 𝑡 + 𝑡 sin 𝑡) 𝑎𝑛𝑑 𝑦 = 𝑎(sin 𝑡 − 𝑡 cos 𝑡) then find 𝑑𝑥 2

x=𝑎 (cos 𝑡 + sin 𝑡)

Differentiate w.r.t x we get

dx/dt= at cos t …………………….(1)

Y=𝑎 (sin 𝑡 − 𝑡 cos 𝑡)

Differentiate w.r.t x we get

dy/dt = at sin t
𝑑𝑦 𝑑𝑦/𝑑𝑡 𝑎𝑡 sin 𝑡 sin 𝑡
𝑑𝑥
= 𝑑𝑥/𝑑𝑡 = 𝑎𝑡 cos 𝑡 = cos 𝑡 = tan 𝑡

Again differentiating w.r.t x we get

𝑑2 𝑦 𝑑
𝑑𝑥 2
= 𝑑𝑥 (tan 𝑡) = 𝑠𝑒𝑐 2 𝑡 (Error)

Page | 84
 Correction:

𝑑2 𝑦 𝑑 𝑑𝑡
= (tan 𝑡) = 𝑠𝑒𝑐 2 𝑡 ×
𝑑𝑥 2 𝑑𝑥 𝑑𝑥

From Eqn. (1) dx/dt = at cos t Putting this value above we get

𝑑𝑡 1 sec 𝑡 𝑑2 𝑦 sec 𝑡 𝑠𝑒𝑐 3 𝑡

⟹ 𝑑𝑥 = 𝑎𝑡 cos 𝑡 = 𝑎𝑡 𝑑𝑥 2
= 𝑠𝑒𝑐 2 𝑡 × 𝑎𝑡
= 𝑎𝑡

• Common Error: improper use of log during logarithm differentiation. Students generally use wrong log
properties.

For Example: Log (a+b) = Log a + Log b

(This is not a property and is a wrong perception made by the students)

Common Error: For Example: Correction:

y = sin 𝑥 𝑥 + 𝑥 𝑥 Students should know the correct formula for logarithmic

Taking log on both side we get differentiation Correct formula is Log (ab) =Log a + Log b

Logy = Log(sin 𝑥 𝑥 + 𝑥 𝑥 ) y = sin 𝑥 𝑥 + 𝑥 𝑥

Log y = Log( (sin 𝑥 𝑥 ) + Log(𝑥 𝑥 ) (Error) Here teacher should insist the students to take y = u + v

Where u = sin 𝑥 𝑥 and v = 𝑥 𝑥

Chapter 6

Application of Derivatives
• Common Error: Incorrect identifications of intervals after obtaining the critical points.

Remedy: After obtaining the critical points it is better to draw the table for identifying the sign of f' (x)

in different intervals or students may use Wavy Curve Method.

• Common Error: Incorrect sign of f' * (x) to identify the increasing / decreasing functions.

Remedy: Correctly observing the sign of f' (x) in the interval before deciding the nature of the function.

• Common Error: Incorrect factorization of an algebraic function to obtain the critical point.

Remedy: Students should Recheck the answer after solving.

Chapter-7

Integrals

• Common Error: Solving Definite Integral Problems by using Substitution Method.

While making the substitution students forget to change the limit and committed mistake.

Page | 85
 1
For Example: ∫0 2𝑥(𝑥 2 + 3)1/2 𝑑𝑥

Putting 𝑥 2 + 3 = 𝑡 𝑆𝑜 2𝑥𝑑𝑥 = 𝑑𝑡
1 2 1 2
∫0 (𝑡)1/2 𝑑𝑥 = [3 (𝑡)3/2 ] = 3 [Error]
0

Correction:

While making the substitution students should remember to change the limit also
1
1/2
𝐼 = ∫ 2𝑥(𝑥2 + 3) 𝑑𝑥
0

Putting 𝑥 2 + 3 = 𝑡 𝑆𝑜 2𝑥𝑑𝑥 = 𝑑𝑡

If x=0, then t=3 and if x=1, then t=4

1
2 4
∫ (𝑡)1/2 𝑑𝑥 = [ (𝑡)3/2 ]
0 3 3

2
𝐼 = 3 (8 − 3√3) [Correct Ans.]

• Common Error: In indefinite integral students forget to write constant.

Remedy:- Students must learn to write constant in indefinite integral.
1
• Common Error: Dropping the absolute value when integrating 𝐼 = ∫ 𝑥 𝑑𝑥

Remedy: Recall that the following formula

1
𝐼 = ∫ 𝑥 𝑑𝑥 = 𝑙𝑜𝑔|𝑥| + 𝑐

Here value bars on the argument are required. It is certainly true that on occasion they can be
dropped after the integration is done. But in most of the cases they are required.

Let us take the following two examples

2𝑥
1) 𝐼 = ∫ 𝑥 2 +10 𝑑𝑥 = 𝑙𝑜𝑔|𝑥 2 + 10| + 𝑐

2𝑥
2) 𝐼 = ∫ 𝑥 2 −10 𝑑𝑥 = 𝑙𝑜𝑔|𝑥 2 − 10| + 𝑐

In first case 𝑥 2 + 10 𝑖𝑠 𝑎𝑙𝑤𝑎𝑦𝑠 > 𝑧𝑒𝑟𝑜. So here absolute bars are not required

But in second case 𝑥 2 − 10 may be + ve, -ve or 0. Its value depends upon the value of x. As the
domain of log function is (0, ∞). So 𝑥 2 − 10 should always be positive. So it is compulsory to apply
here absolute bars.

Page | 86
 Chapter 9

Differential Equations

• Common Error: In finding the degree of a differential equation when it is not defined.
Remedy: Students should learn," when a differential equation is not a polynomial in
derivatives its degree is not defined".
Students should have more practice and understanding of questions in which degree is not
defined.
• Common Error: Difference between General solution and Particular solution.
Remedy: Here students should learn General solution contains an arbitrary constant where as
Particular solution contains particular value in place of arbitrary constant.
Chapter 10

Vector Algebra

• Collinear Vectors
Show that the points 𝐴(−2𝑖̂ + 2𝑗̂ + 5𝑘̂), 𝐵(𝑖̂ + 2𝑗̂ + 3𝑘̂) and 𝐶(7𝑖̂ − 𝑘̂)are collinear.
• Common Error: Here students commonly use the following concept
Three points are collinear if ⃗⃗⃗⃗⃗⃗ 𝐴𝐵 + 𝐵𝐶 ⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗
𝐴𝐶
Remedy: Correct form is:
Three points are collinear if | ⃗⃗⃗⃗⃗⃗ 𝐴𝐵 |+|𝐵𝐶 ⃗⃗⃗⃗⃗⃗ |=|𝐴𝐶
⃗⃗⃗⃗⃗⃗ |
• Triangle inequality: For any two vectors 𝑎⃗ and 𝑏⃗⃗
Common Mistake: |𝑎 ⃗⃗⃗⃗ + 𝑏⃗⃗| = |𝑎 ⃗⃗⃗⃗| + |𝑏⃗⃗|
Correct Form: |𝑎 ⃗⃗⃗⃗ + 𝑏⃗⃗| ≤ |𝑎⃗⃗⃗⃗| + |𝑏⃗⃗|
⃗⃗⃗⃗ + 𝑏⃗⃗| = |𝑎
|𝑎 ⃗⃗⃗⃗| + |𝑏⃗⃗| is true only when points are collinear.
• Equality of two vectors: Find the value of x, y and z so that 𝑎⃗ and⃗⃗⃗⃗⃗ 𝑏 are equal vectors. Where

𝑎⃗ = 𝑥𝑖̂ + 2𝑗̂ + 𝑧𝑘̂ 𝑎𝑛𝑑 𝑏⃗⃗ = 2𝑖̂ + 𝑦𝑗̂ + 𝑘̂

• Common Error: Here students may use the concept |𝑎 ⃗⃗⃗⃗| = |𝑏⃗⃗|
Correct Form: If two vectors are equal then their components are equal
𝑥𝑖̂ + 2𝑗̂ + 𝑧𝑘̂ = 2𝑖̂ + 𝑦𝑗̂ + 𝑘̂
⟹ x=2, y=2, z=1
Chapter 11
Three Dimensional Geometry
• Common Mistake: Students start using the Cartesian form of equation of line without
converting them into standard form.
2𝑥−1 4−𝑦 𝑧+1
= =
2 7 2

• Direction line are 2, 7, 2 (wrong)

Correction: -
2𝑥−1/2 4−𝑦 𝑧−(−1)
= =
2 −7 2

Direction ratios of the line are 1, -7, 2 (correct)

Page | 87
 So students should learn and understand the standard form of equation of line
𝑥−𝑥1 4−𝑦 𝑧+𝑧
= 𝑏1= 𝑐1
𝑎
LPP
• In graphical solution of Linear Programming problems, proper shading of the
feasible region is very important. If the region is unbounded, don’t forget to draw
dotted line.
• When a line passing through origin, to identify the half plane, point other than
origin should be taken.
• In case of getting more than one optimal solution, students are making error .
PROBABILITY
• Error in conversion of language form to Symbolic form.
• Error in choosing partition events(E1,E2,….) and arbitrary event (E)in Bayes
theorem and Theorem of Total Probability.

Page | 88
 SUB: MATHEMATICS (041)
SAMPLE QUESTION PAPER-1
Time Allowed: 3 Hours Maximum Marks : 80
General Instructions:
1. This question paper contains – five sections A, B, C, D, and E. Each section is
compulsory. However, there are internal choices in some questions.
2. Section – A has 18 MCQ’s and Assertion-Reason based questions of 1 mark each.
3. Section – B has 5 very short Answer (VSA) – type questions of 2 marks each.
4. Section – C has 6 short Answer (SA)- type questions of 3 marks each.
5. Section – D has 4 long Answer (LA) – type questions of 5mark each.
6. Section –E has 3 source based/case based/passage based/integrated units of
assessment (4 mark each) with sub parts.
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) :
1. If A and B are square matrices of order 3 such that AB = 6I and |A| = 12, then |B| is :
A) 1/2 B) 2 C) 18 D) 54
2. The number of all possible matrices of order 3×3 with each entry 0 or 1 or – 1 is :
A) 27 B) 36 C) 81 D) 39
3. The total revenue in Rupees received from the sale of 𝑥 units of a product is given by 𝑅(𝑥) =
3𝑥 2 + 36𝑥 + 5. The marginal revenue, when 𝑥 = 15 is
(a) 116 (b) 96 (c) 90 (d) 126
4. If AB = A and BA = B, then :
A) B = I B) A = I C) A2 = A D) B2 = I
5. If m and n respectively are the order and the degree of the differential equation
𝑑 𝑑𝑦 4
[( )] = 0 𝑡ℎ𝑒𝑛 𝑚 + 𝑛 is :
𝑑𝑥 𝑑𝑥

A) 1 B) 2 C) 3 D) 4
1 1 1
6. If the direction cosines of a line are (𝑎 , 𝑎 , 𝑎), then

A) 0 < 𝑎 < 1 B) 𝑎 > 2 C) 𝑎 > 0 D) 𝑎 = ±√3

Page | 89
7. If A is a matrix of order m × n and B is a matrix such that ABT and BTA both are defined then
order of matrix B is :
A) m × m B) n × n C) m × n D) n × m
8. If P(A/B) > P(A), then which of the following is correct ?
A) P(B/A) < P(B) B) P(A∩ 𝐵) < 𝑃(𝐴). 𝑃(𝐵) C) P(B/A) > P(B) D) P(B/A) = P(B)

9. If 𝑎⃗ and 𝑏⃗⃗ are two vectors such that 𝑎⃗. 𝑏⃗⃗ > 0 and |𝑎⃗ . 𝑏⃗⃗ | = |𝑎⃗ × 𝑏⃗⃗|,then the angle between

𝑎⃗ and 𝑏⃗⃗ is
𝜋 𝜋 𝜋 3𝜋
(A) (B) (C) 2 3 (D)
4 3 4

10. If the projection of 𝑎⃗ = 𝑖̂ − 2𝑗̂ + 3𝑘̂ 𝑜𝑛 𝑏⃗⃗ = 2𝑖̂ + 𝜆𝑗̂ is zero, then the value of 𝛌 is :
A) 0 B) 1 C) - 2/3 D) -3/2
11. If the minimum value of an objective function Z = ax + by occurs at two points (3, 4) and
(4, 3) then :
A) a + b = 0 B) a = b C) 3a = b D) a = 3b
𝑒 𝑥 (1+𝑥)
12. ∫ 𝑐𝑜𝑠2 (𝑥𝑒 𝑥 ) 𝑑𝑥is dequal to :

A) 𝑡𝑎𝑛(𝑥𝑒 𝑥 ) + 𝑐, B) 𝑐𝑜𝑡(𝑥𝑒 𝑥 ) + 𝑐 C) 𝑐𝑜𝑡(𝑒 𝑥 ) + 𝑐 D) 𝑡𝑎𝑛[𝑒 𝑥 (1 + 𝑥)] +𝑐

13. If f and g are continues functions in [0, a] satisfying f(x) = f(a – x) and g(x) + g(a – x) = a then
𝑎
∫0 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜:
𝑎 𝑎 𝑎 𝑎 𝑎
A) B) 2 ∫0 𝑓(𝑥)𝑑𝑥 C) ∫0 𝑓(𝑥)𝑑𝑥 D) 𝑎 ∫0 𝑓(𝑥)𝑑𝑥
2
𝑑𝑦
14. The integrating factor of the differential equation (𝑥 + 3𝑦 2 ) 𝑑𝑥 = 𝑦 is:

A) y B) – y C) 1/y D) - 1/y

𝜋 1
15. 𝑠𝑖𝑛 ( 3 − 𝑠𝑖𝑛−1 (− 2)) is equal to
A) 1/2 B)1/3 C) 1/4 D) 1
16. The feasible region corresponding to the linear constraints of a LPP is given below:
Which of the following is not a

constraints to the given LPP ?

Page | 90
 A) 𝑥 + 𝑦 ≥ 2 B) 𝑥 + 2𝑦 ≤ 10 C) 𝑥 − 𝑦 ≥ 1 D) 𝑥 − 𝑦 ≤ 1
𝑑𝑦
17. If 𝑥 𝑦 . 𝑦 𝑥 = 16, then the value of at (2, 2) is :
𝑑𝑥

A) – 1 B) 0 C) 1 D) None of these
18. The area bounded by the curve 2𝑥 2 + 𝑦 2 = 2 is :
𝜋
A) 𝜋 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠 B) √2𝜋 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠 C) 2 𝑠𝑞. 𝑖𝑢𝑛𝑖𝑡𝑠 D) 2𝜋 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠

ASSERTION-REASON BASED QUESTIONS

In the following questions, a statement of Assertion(A) is followed by a statement of

Reason(R). Choose the correct answer out of the following choices:

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.

19. Assertion(A) : 𝑓(𝑥) = 𝑐𝑜𝑠|𝑥| is everywhere continuous.
Reason(R) : Composition of continuous function is continuous.
𝑛+1
, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
20. Assertion(A) : The function 𝑓: 𝑁 → 𝑁 be defined by 𝑓(𝑛) = { 𝑛2 is
, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
2

bijective.
Reason(R) : A function 𝑔: 𝐴 → 𝐵 is said to be bijective if it is one-one and onto.

SECTION-B (2×5 = 10)

[ This section comprises of 5 very short answer type questions of 2 marks each]

1+𝑥 2
21. Prove that : 𝑐𝑜𝑠[𝑡𝑎𝑛−1 {𝑠𝑖𝑛(𝑐𝑜𝑡 −1 𝑥)}] = √2+𝑥 2 .

22. Sand is pouring from a pipe at the rate of 12 cm3/sec. The falling sand forms a cone on the
ground in such a way that the height of the cone is always one-sixth of the radius of the
base. How fast is the height of the sand cone increased when the height is 4 cm?
OR
2𝑥
Show that 𝑦 = 𝑙𝑜𝑔(1 + 𝑥) − 2+𝑥 , 𝑥 > −1 is an increasing function of 𝑥 throughout its

domain.

Page | 91
23. For any three non-zero vectors 𝑎⃗, 𝑏⃗⃗ 𝑎𝑛𝑑 𝑐⃗ 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑎⃗. 𝑏⃗⃗ = 𝑎⃗. 𝑐⃗ 𝑎𝑛𝑑 𝑎⃗ × 𝑏⃗⃗ = 𝑎⃗ × 𝑐⃗, then

show that 𝑏⃗⃗ = 𝑐⃗ .

OR

If the vectors 𝑎⃗, 𝑏⃗⃗ 𝑎𝑛𝑑 𝑐⃗ represent the three sides of a triangle, then show that

𝑎⃗ × 𝑏⃗⃗ = 𝑏⃗⃗ × 𝑐⃗ = 𝑐⃗ × 𝑎⃗
24. Find the area of the parallelogram whose one side and a diagonal are represented by co
initial vectors 𝑖̂ − 𝑗̂ + 𝑘̂ 𝑎𝑛𝑑 4𝑖̂ + 5𝑘̂ respectively.
𝑑𝑦 𝑠𝑖𝑛2 (𝑎+𝑦)
25. If 𝑠𝑖𝑛𝑦 = 𝑥𝑠𝑖𝑛(𝑎 + 𝑦), prove that 𝑑𝑥 = .
𝑠𝑖𝑛𝑎

SECTION – C (3×6 = 18)

[This section comprises of 6 short answer type questions of 3 marks each.]
𝜋
26. Find whether the function 𝑓(𝑥) = 𝑐𝑜𝑠 (2𝑥 + 4 ), is increasing or decreasing in the interval
3𝜋 5𝜋
<𝑥< .
8 8

27. If 𝑓(𝑥) = 𝑎𝑙𝑜𝑔|𝑥| + 𝑏𝑥 2 + 𝑥 has extreme values at 𝑥 = −1 and at 𝑥 = 2, find a and b.

28. Find the shortest distance between the lines whose vector equations are :
𝑟⃗ = (1 − 𝑡)𝑖̂ + (𝑡 − 2)𝑗̂ + (3 − 2𝑡)𝑘̂
𝑟⃗ = (𝑠 + 1)𝑖̂ + (2𝑠 − 1)𝑗̂ − (2𝑠 + 1)𝑘̂
OR
Find the vector and Cartesian equations of the line passing through the point (1, 2, -4) and
𝑥−8 𝑦+19 𝑧−10 𝑥−15 𝑦−29 𝑧−5
perpendicular to the two lines = = 𝑎𝑛𝑑 = = .
3 −16 7 3 8 −5
1
29. Find ∫ [𝑙𝑜𝑔(𝑙𝑜𝑔𝑥) + (𝑙𝑜𝑔𝑥)2] 𝑑𝑥 .

OR
𝜋 𝑥𝑡𝑎𝑛𝑥
Evaluate : ∫0 𝑑𝑥 .
𝑠𝑒𝑐𝑥+𝑡𝑎𝑛𝑥

30. Solve the following LPP graphically

Minimize : 𝑍 = 𝑥 + 2𝑦

Subject to : 2𝑥 + 𝑦 ≥ 3, 𝑥 + 2𝑦 ≥ 6, 𝑥, 𝑦 ≥ 0
31. Two cards are drawn successively with replacement from a well shuffled deck of 52 cards.
Find the probability distribution of the number of aces and find its mean.
OR

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 A and B throw a die alternatively till one of them gets a 6 and win the game. Find the
probability of A win, if A starts first.

SECTION – D (4×5 = 20)

[This section comprises of 6 short answer type questions of 3 marks each.]
32. Find the area of the region by using integration {(𝑥, 𝑦): 0 ≤ 𝑦 ≤ 𝑥 2 + 1, 0 ≤ 𝑦 ≤ 𝑥 + 1, 0 ≤
𝑥 ≤ 2}.
1 −1 2
33. Find the inverse of the matrix 𝐴 = [0 2 −3]. Using the A-1 , solve the system of linear
3 −2 4
equations : 𝑥 − 𝑦 + 2𝑧 = 1, 2𝑦 − 3𝑧 = 1, 3𝑥 − 2𝑦 + 4𝑧 = 3.
OR
−3 −2 −4 1 2 0
If 𝐴 = [ 2 1 2 ] , 𝐵 = [−2 −1 −2], then find AB and use it to solve the
2 1 3 0 −1 1
following system of equations : 𝑥 − 2𝑦 = 3, 2𝑥 − 𝑦 − 𝑧 = 2, −2𝑦 + 𝑧 = 3 .

𝑑𝑦 1−𝑦 2
34. If √1 − 𝑥 2 + √1 − 𝑦 2 = 𝑎(𝑥 − 𝑦),Prove that 𝑑𝑥 = √1−𝑥 2 .

OR
−1 𝑥 𝑑2 𝑦 𝑑𝑦
If 𝑦 = 𝑒 𝑎 𝑐𝑜𝑠 , −1 ≤ 𝑥 ≤ 1, show that (1 − 𝑥 2 ) 𝑑𝑥 2 − 𝑥 𝑑𝑥 − 𝑎2 𝑦 = 0

35. Find the co-ordinates of the foot of perpendicular and the length of the perpendicular
drawn from the point P(5, 4, 2) to the line 𝑟⃗ = −𝑖̂ + 3𝑗̂ + 𝑘̂ + 𝜆(2𝑖̂ + 3𝑗̂ − 𝑘̂). Also find the
image of P in this line.
OR
𝑥+2 𝑦−1 𝑧
The vertices B and C of a ΔABC lies on the line = = 4 such that BC = 5 units. If the co-
3 0
ordinate of point A is (1, -1, 2) then find the area of the triangle ABC.
SECTION – E (4×3 = 12)
Case study-1
36.Romi and Rasmi are playing Ludo at home. While rolling the dice, Romi’s sister Raji observed
and noted the possible outcomes of the throw every time belongs to set
{1, 2, 3, 4, 5, 6}. Let A be the set of players while B be the set of all possible outcomes.
A = { S, D }, B = {1, 2, 3, 4, 5, 6 }

Page | 93
 i) Let R : B → B be defined by R = {(x, y) : y is divisible by x }. Verify that whether R is
reflexive, symmetric and transitive.
ii) Raji wants to know the number of relations from A to B. Find the number of all possible
relations.
iii) Let R be a relation on B defined by R = { (1, 2), (2, 2), (1, 3), (3, 4), (3, 1), (4, 3), (5, 5) .
Then R is which types of relation?
OR
Raji wants to know the number of relations possible from A to B. Find the number of
possible relations.

Case study - 2
37. An Apache helicopter of enemy is flying along the curve given by 𝑦 = 𝑥 2 + 7. A soldier placed
at point (3, 7) wants to shoot down the helicopter when it is nearest to him.

Based on above information answer the following:

i) Find the distance function from the soldier and helicopter in terms of x . [1
ii) Find the critical points of the function. [1
iii) Find the co-ordinate of the point of the helicopter, when it is nearest to the soldier. [2
OR
Using second derivative test, find the shortest distance between the helicopter and soldier.

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 Case study - 3
38. Read the following passage and answer the questions given below.
A doctor is to visit a patient. From the past experience, it is known that the probabilities that
he will come by bus, metro, bike or by other means of transport are respectively 0.3, 0.2, 0.1,
and 0.4. the probabilities that he will be late are 0.25, 0.3, 0.35 and 0.1when come by bus,
metro, bike, or by other means of transport respectively.

Based on above information, answer the following :

i) When the doctor arrives late, what is the probability that he come by metro? [2
ii) What is the probability that the doctor is late? [2

***

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 SAMPLE QUESTION PAPER-2
SECTION A
(Multiple Choice Questions)
Each question carries 1 mark
1. 2 3 1 3 2 1 4 6 8
If𝐴 = [ ], 𝐵 = [ ] , 𝐶 = [ ] 𝑎𝑛𝑑 𝐷 = [ ], then which of the following
1 2 4 3 1 2 5 7 9
is defined?

(A)𝐴 + 𝐵 (B) 𝐵 + 𝐶 (C) 𝐶 + 𝐷 (D) 𝐵 + 𝐷

2. 6 𝑥
The value of 𝑥 for which the matrix 𝐴 = [ ] is singular
12 4

(A) 2 (B) 1 (C) −1 (D) −2

3. 2 −1
The minor and cofactors of all the elements of the determinant | | are
3 5

(A)5,3, −1,2 ; 5, −3,1,2 (B) 5, −3,1,2 ; 5,3, −1,2

(C)5, −1,3,2 ; 5, −3,1,2 (D) 5,3, −1,2; 5,1, −3,2

4. The linear programming problem

Minimize 𝑍 = 3𝑥 + 2𝑦 subject to constraints

𝑥 + 𝑦 ≥ 8, 3𝑥 + 5𝑦 ≤ 15, 𝑥 ≥ 0 and 𝑦 ≥ 0 has

(a) one solution (b) no feasible solution

(c) tow solutions (d) infinitely many solutions

5. ⃗⃗⃗⃗ 𝑎𝑛𝑑 𝑏⃗⃗ 𝑎𝑟𝑒 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟𝑠, 𝑡ℎ𝑒𝑛 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑎
If 𝑎 ⃗⃗⃗⃗ 𝑎𝑛𝑑 𝑏⃗⃗ for √3𝑎⃗ − 𝑏⃗⃗ to be
a unit vector?

(A)30° (B) 45° (C) 60° (D) 90°

6. For every square matrix A, there exist an identity matrix of same order such that

(A) 𝐼𝐴 = 𝐴 only (B) 𝐴𝐼 = 𝐴 only

Page | 96
 (C) 𝐼𝐴 = 𝐴𝐼 = 𝐴 (D) None of these

7. 𝑥𝑑𝑦
Integrating factor of − 𝑦 = 𝑥 4 − 3𝑥 𝑖𝑠 ∶
𝑑𝑥

1
(A) 𝑥 (B) 𝑙𝑜𝑔𝑥 (C) 𝑥 (D) −𝑥

8. 4 x+2
If 𝐴 = [ ] is symmetric, then 𝑥 is equal to
2𝑥 − 3 x+1

(A) 2 (B) 3 (C) −1 (D) 5

9. If 𝑓(𝑥) = |𝑠𝑖𝑛𝑥|, then

(A) 𝑓is everywhere differentiable

(B) 𝑓 is everywhere continuous but not differentiable 𝑎𝑡 𝑥 = 𝑛𝜋, 𝑛 ∈ 𝑍

𝜋
(C) 𝑓 is everywhere continuous but not differentiable at 𝑥 = (2𝑛 + 1) 2 , 𝑛 ∈ 𝑍

(D) none of these.

𝑎 1 𝜋
10. 𝐼𝑓 ∫0 𝑑𝑥 = , 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑎 is
1+4𝑥 2 8

1 1 1 1
(A) (B) 4 (C) 2 (D)
3 5

11. The two vectors 𝑗̂ + 𝑘̂ 𝑎𝑛𝑑 3𝑖̂ − 𝑗̂ + 4𝑘̂ represent the two sides AB and AC, respectively of a
triangle ABC . The length of the median through A is

√34 √48
(A) (B) (C) √18 (D) None of these
2 2

12. The derivative of 𝑠𝑖𝑛2 𝑥 with respect to 𝑒 𝑐𝑜𝑠𝑥 is

2 cos 𝑥 2 cos 𝑥 2
(A) (B) – (C) 𝑒 𝑐𝑜𝑠𝑥 (D) none of these
𝑒 𝑐𝑜𝑠𝑥 𝑒 𝑐𝑜𝑠𝑥

13. The rate of change of the area of a circle with respect to its radius 𝑟 at 𝑟 = 6 𝑐𝑚 is

(A) 10𝜋 (B) 12𝜋 (C) 6𝜋 (D) 11𝜋

14. The corner points of the shaded unbounded feasible region of an LPP are (0, 4),

(0.6, 1.6) and (3, 0) as shown in the figure. The minimum value of the objective

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 function Z = 4x + 6y occurs at

(A) (0.6, 1.6) 𝑜𝑛𝑙𝑦 (B) (3, 0) only (C) (0.6, 1.6) and (3, 0) only

(D) at every point of the line-segment joining the points (0.6, 1.6) and (3, 0)
3
15. 𝑑𝑦 2 2 𝑑2 𝑦
The degree of the differential equation [1 + (𝑑𝑥 ) ] = 𝑑𝑥 2 is

(A) 4 (B) 2 (C) 3 (D) not defined

16. The function 𝑓(𝑥) = 𝑥 4 defined from R to R is:

(a) only one-one (b) only onto (c) one-one and onto (d) many-one and into

17. The area of the region bounded by the circle 𝑥 2 + 𝑦 2 = 4 and the lines 𝑥 = 0 and

𝑥 = 2 (in sq. unit) is

𝜋 𝜋 𝜋
(𝐴)𝜋 (b) (c) (d)
2 3 4

18. Let A and B be two events. If 𝑃(𝐴) = 0.2, 𝑃(𝐵) = 0.4, 𝑃(𝐴 ∪ 𝐵) = 0.6,then
𝑃(𝐴|𝐵)𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜

(𝐴)0.8 (𝐵) 0.5 (𝐶) 0.3 (𝐷)0

ASSERTION-REASON BASED QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of

Reason (R). Choose the correct answer out of the following choices.

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

Page | 98
(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.

19. 𝑑𝑥 1 3+𝑥
Assertion (A): The value of ∫ = 𝑙𝑜𝑔 | |+𝐶
√9−𝑥 2 6 3−𝑥

𝑑𝑥 1 𝑎+𝑥
Reason (R): Value of the Indefinite integral ∫ 2 2 𝑖𝑠 log | |+𝐶
𝑎 −𝑥 2𝑎 𝑎−𝑥

20. 1 𝜋
Assertion (A): The principal value of 𝑠𝑖𝑛−1 ( − 2) is − 6

Reason (R): The principal value branch of the function 𝑠𝑖𝑛−1 𝑥 𝑖𝑠 [0, 𝜋]

SECTION B

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

21. Find the principal value of 𝑐𝑜𝑠 [𝑐𝑜𝑠 −1 (

−√3
) + 6]
𝜋
2

OR

Show that the function f: N→N given by f (𝑥) = 𝑥 2 is one-one but not onto.

22. If r = xiˆ + yˆj + zkˆ, find (r  iˆ).(r  ˆj ) + xy.

23. A man 1.6 m tall walks at the rate of 0.3 m/sec away from a street light that is 4 m above the
ground. At what rate is the tip of his shadow moving? At what rate is his shadow lengthening?

24. d2y dy
If y = 3 cos(log x) + 4 sin(log x), then show that x 2 2
+x +y=0
dx dx
OR
d2y dy
If y = Aemx + Benx then prove that 2
− (m + n) + mny = 0
dx dx
25. The scalar product of the vector 𝑖̂ + 𝑗̂ + 𝑘̂ with a unit vector along the sum of the

vectors 2𝑖̂ + 4𝑗̂-5𝑘̂and  𝑖̂ + 2𝑗̂ + 3𝑘̂ is equal to one. Find the value of 

Page | 99
 SECTION C

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

26. 1+𝑠𝑖𝑛𝑥
Find the value of ∫ (1+𝑐𝑜𝑠𝑥) 𝑒 𝑥 𝑑𝑥

OR

𝑥2
Evaluate ∫ 𝑥 4 +𝑥 2−2 𝑑𝑥

27. Find the shortest distance between lines

x −3 y −5 z −7
= =
1 −2 1
x +1 y + 1 z +1
and = =
7 −6 1

28. 𝑑𝑦
Find the particular solution of the differential equation log(𝑑𝑥 ) = 3x + 4y. given that y = 0

when x = 0 .
OR
𝑥 𝑥
Solve the differential equation 𝑦𝑒 𝑦 𝑑𝑥 = (𝑥𝑒 𝑦 + 𝑦 2 ) 𝑑𝑦

29. Find graphically, the maximum value of Z=2x+5y, subject to constraints given below:
2𝑥 + 4𝑦 ≤ 8; 3𝑥 + 𝑦 ≤ 6, 𝑥 + 𝑦 ≤ 4; 𝑥 ≥ 0, 𝑦 ≥ 0
OR
Solve the following equation graphically
Minimum Z = 5x + 10y
Subject to x + 2y ≤ 120
x + y ≥ 60
x -2y ≥ 0
x, y ≥ 0
30. 𝑥3
Find the intervals in which the function 𝑓(𝑥) = 𝑥 4 − is increasing or decreasing.
3

31. Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards.
Find the probability distribution of the number of aces.

SECTION D

Page | 100
(This section comprises of long answer-type questions (LA) of 5 marks each)

32. Using matrix method, solve the following system of equations

2 3 3 1 1 1 3 1 2
− 𝑦 + 𝑧 = 10 , + 𝑦 + 𝑧 = 10 , 𝑥 − 𝑦 + 𝑧 = 13 ; x, y, z ≠ 0
𝑥 𝑥

OR

−4 4 4 1 −1 1
Determine the product of [−7 1 3 ]and [1 −2 −2]then use to solve the system of
5 −3 −1 2 1 3
equations 𝑥 − 𝑦 + 𝑧 = 4, 𝑥 – 2𝑦 − 2𝑧 = 9 𝑎𝑛𝑑 2𝑥 + 𝑦 + 3𝑧 = 1.

33. 𝑑2𝑦 𝑎 2
If(𝑎 + 𝑏𝑥)𝑒 𝑦/𝑥 = 𝑥 , then prove that: 𝑥 =( )
𝑑𝑥 2 𝑎+𝑏𝑥

34. Find area of the region {(𝑥, 𝑦) ∶ 𝑥 2 + 𝑦 2 ≤ 4 , 𝑥 + 𝑦 ≥ 2}.

OR

If the area between 𝑥 = 𝑦 2 and 𝑥 = 4 is divided into two equal parts by the line 𝑥 = a, find the
value of a.

35. A bird is flying along the line 𝑟⃗ = (𝑖̂ + 2𝑗̂ + 3𝑘̂) + 𝜆(𝑖̂ − 3𝑗̂ + 2𝑘̂)and another bird is flying
along the line 𝑟⃗ = (4𝑖̂ + 5𝑗̂ + 6𝑘̂) + 𝜇(2𝑖̂ + 3𝑗̂ + 𝑘̂).At what points on the lines they reach so
that the distance between them is the least?Find the shortest possible distance between them.

SECTION E

(This section comprises of with two sub-parts. First two case study questions have three sub -parts of
marks 1, 1, 2 respectively. The third case study question has two sub-parts of 2 marks each.)

36. Students of a school are taken to a railway museum to learn about railway
heritage and its history?

An exhibit in the museum depicted many rail lines on the track near the railway

Page | 101
 station. Let L be the set of all rail lines on the railway track and R be the relation
on L defined by
𝑅 = {(𝑙1 , 𝑙2 ): 𝑙1 𝑖𝑠 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑙2 }
On the basis of the above information, answer the following questions:
(i) Find whether the relation R is symmetric or not.
(ii) Find whether the relation R is transitive or not.
(iii) (a)If one of the rail lines on the railway track is represented by the
equation 𝑦 = 3𝑥 + 2,then find the set of rail lines in R related to it.
𝑂𝑅
(b) Let S be the relation defined by
S= {(𝑙1 , 𝑙2 ): 𝑙1 𝑖𝑠 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑙2 }
Check whether the relation S is symmetric and transitive.

37.

A rectangular cake dish is made by cutting out squares from the corners of a 25 cm by 40 cm
rectangle of tin plate, and then folding the metal to form the container.

Based on the given information, answer the following questions.

i. If x cm from each corner is cut out ,then find length ,breadth and height of the dish .
ii. Express the volume(x), in terms of x.
iii. Find the value of x for which the capacity of dish is maximum also find maximum
volume.
OR

Find the interval in which the volume function V(x) is strictly

increasing or strictly decreasing.

38. In an office four employees Ram, Rahim, John, Sital process incoming copies of a certain

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form. Ram processes 30% of the form. Rahim processes 20%, John Processes 10% and Sital
the remaining 40% of the form. Ram has an error rate of 0.06, Rahim has an error rate of 0.03,
John has an error rate of 0.02 and Sital has an error rate of 0.01.

Based on the above situation, answer the following:

(i) Find the total probability of committing an error in processing the form.
(ii) The manager of the company wants to do a quality check. During inspection he
selects a form at random from the days output of processed forms. If the form
selected at random has an error, find the probability that the form is not
processed by Ram.

SAMPLE QUESTION PAPER-3

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 SECTION-A
1. If AB = C then A, B, C is equal to
(a) 𝐴2×3 , 𝐵3×2 , 𝐶2×3 (b) 𝐴3×2 , 𝐵2×3 , 𝐶3×3 (c) 𝐴3×3 , 𝐵2×3 , 𝐶3×3 (d) 𝐴3×2 , 𝐵2×3 , 𝐶3×2
2. If A is a square matrix of order 3 such that det(3A) = 27, then det(A) is
(a) 1 (b) 27 (c) 9 (d) 3
3. If A and B are square matrices of order 2such that 2A + 3B is a null matrix such that det(A) = 9, then
det(B) is
(a) 6 (b) - 4 (c) - 6 (d) 4
2𝑥 + 1 𝑥 < 1
4. If the function f(x) = { 2 is differentiable at x = 1. Then (a, b) is
𝑎𝑥 + 𝑏 𝑥 ≥ 1
(a) (1, 1) (b) (1, 2) (c) (2, 1) (d) (- 1, 1)
5. Equation of a line passing through point (1, 2, 3) and equally inclined to the coordinate axes, is
𝑥 𝑦 𝑧 𝑥 𝑦 𝑧 𝑥−1 𝑦−1 𝑧−1 𝑥−1 𝑦−2 𝑧−3
(𝑎) = = (b) 1 = 1 = 1 (c) 1 = 2 = 3 (d) 1 = 1 = 1
1 2 3
𝑑2 𝑦 𝑑𝑦 2 𝑑2 𝑦
6. The degree of the differential equation + 3 (𝑑𝑥 ) = 𝑥 2 𝑙𝑜𝑔 (𝑑𝑥 2 ) is
𝑑𝑥 2
(a) 2 (b) 3 (c) 1 (d) Not defined
7. The constraints of a linear programming problemare 𝑥 + 2𝑦 ≥ 3, 𝑥 ≥ 10, 𝑦 ≥ 0
Which of the following objective function has an optimal solution with respect to above set of
constraints?
(a) Minimize Z = x + y
(b) Minimize Z = 0.5 x + y
(c) Maximize Z = x + y
(d) Maximize Z = 2x + y
8. If 𝑎⃗. 𝑖̂ = 𝑎⃗. (𝑖̂ + 𝑗̂) = 𝑎⃗. (𝑖̂ + 𝑗̂ + 𝑘̂ )= 1, then 𝑎⃗ is
(𝑎)𝑘̂ (b) 𝑖̂ (c)𝑗̂ (d) 𝑖̂ + 𝑗̂ + 𝑘̂
13 √21−𝑥
9. The value of ∫8 𝑑𝑥 is
√ 𝑥+√21−𝑥
21 5 21
(a) (b) 0 (c) 2 (d) - 2
2
2 0
10. If [ ] = P + Q, where P is a symmetric and Q is a skew symmetric matrix, then P is equal
5 4
to
5 5 5 5
2 0 −2 0 0 −2
2 2
(𝑎) [ 5 ] (b) [ 5 ] (c)[ 5 ] (d) [ 5 ]
4 0 −2 0 4
2 2 2
20 4
11. The corner points of the feasible region of a linear programming problem are (0, 4), (8, 0) and ( 3 , 3).
If Z =30x + 24y is the objective function, then (maximum value of Z – minimum
value of Z ) is equal to
(a) 40 (b) 120 (c) 144 (d) 136
12. If 𝑎⃗, 𝑏⃗⃗ and 𝑐⃗ are position vectors of vertices A, B, C of a parallelogram ABCD, then position vector of
D is
(𝑎)𝑎⃗ − 𝑐⃗ + 𝑏⃗⃗ (b) 𝑎⃗ + 𝑐⃗ − 𝑏⃗⃗ (c)𝑎⃗ − 𝑐⃗ − 𝑏⃗⃗ (d) 𝑐⃗ − 𝑎⃗ + 𝑏⃗⃗
1 1
13. If A = [ ], then 𝐴2021 is
0 1
(𝑎) [2021 2021] (b)[
2021 1
] (c)[
1 2021
] (d) [
1 1
]
0 2021 0 2021 0 1 0 1

Page | 104
 1 1
14. A andB are two students. Their chances of solving a problem correctly are and respectively. If the
3 4
1
probability of their making a common error is 20and they obtain the same answer, then the probability
of their answer to be correct is
1 1 10 13
(𝑎) (b) 40 (c)13 d) 120
12
𝑑𝑦 𝑥 3 −𝑦 𝑛
15. The value of n for which the differential equation 𝑑𝑥 = 𝑥 2 𝑦+𝑦 2𝑥 is homogeneous is
(a) 1 (b) 3 (c)4 (d) 2
16. If 𝑎⃗ × 𝑏⃗⃗ = 𝑖̂ + 𝑗̂ + 𝑘̂and |𝑎⃗| = 2, |𝑏⃗⃗| = 1, then the angle between 𝑎⃗ and 𝑏⃗⃗ is
(a) 300 (b) 600 (c)900 (d)1200
17. The value of 𝑓 / (1)where the function f(x) = |𝑥| + |𝑥 − 2| is
(a) 0, 2 (b) 0 (c) 1 (d) 2
18. If direction ratios of a line are 2, 6, -3 and it makes obtuse angle with y-axis, then its direction cosines
are
2 6 3 2 6 3 2 6 3 2 6 3
a) 7 , 7 , − 7 b) 7 , − 7 , − 7 c) - 7 , − 7 , 7 d) − 7 , − 7 , − 7

ASSERTION-REASON BASED QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of Reason
(R).Choose the correct answer out of the following choices.
(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.
19. Assertion (A) : 𝑓(𝑥) = −2 + |𝑥 − 1| has minimum value at x = 1.
𝑑 𝑑
Reason (R) :When (𝑓(𝑥))< 0 for all x ∈ (𝑎 − ℎ, 𝑎) and (𝑓(𝑥))> 0 for all x ∈ (𝑎, 𝑎 + ℎ) where 'h'
𝑑𝑥 𝑑𝑥
is an infinitesimally small positive quantity, then f(x) has a minimum at x = a. provided f(x) is
continuous at x = a.
|𝑥−1|
20. Assertion (A) :Let f :𝑅 → 𝑅defined by 𝑓(𝑥) = 𝑥−1 is a bijective function.
Reason (R) :A function f :𝐴 → 𝐵is said to be bijective if range of the function is codomain.

SECTION-B
This section comprises of very short answer type-questions (VSA) of 2 marks each
1 2
21. Find the value of 𝑡𝑎𝑛2 (2 𝑠𝑖𝑛−1 3)
OR
𝑥
Is f : R→ R defined by f(x) = 𝑥 2+1 one-one and onto ? Justify
𝜋
22. Find the interval(s) in which the function f(x)= sinx+cosx, x∈ (0, 2 )is strictly increasing or decreasing
23. The total cost C(x) associated with the production x units of an item is given by C(x) = 0.005𝑥 3 -
0.02𝑥 2 + 30x + 5000.Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output.
OR
If the product of two positive numbers is 9, find the numbers so that the sum of their squares is
minimum

Page | 105
 𝑑𝑥
24. Evaluate ∫
√𝑥+𝑎+√𝑥+𝑏
25. Show that the function f(x)=|𝑥 − 2|,x∈R is continuous but not differentiable.
SECTION C
(This section comprises of short answer type questions (SA) of 3 marks each)
𝑑𝑥
26. Find ∫ (1+𝑒 𝑥 )(1−𝑒 −𝑥 )
27. In a dice game, a player pays a stake of Rs 1 for each throw of a die. She receives Re 5 if the die shows
a 3, Rs 2 if the die shows a 1 or 6 and nothing otherwise. What is the player expected profit per throw
over a long series of throws.
𝑠𝑖𝑛−1 √𝑥−𝑐𝑜𝑠−1 √𝑥
28. Evaluate∫ 𝑠𝑖𝑛−1 𝑑𝑥; x𝜖[0,1].
√𝑥+𝑐𝑜𝑠−1 √𝑥

OR
4
Evaluate∫0 [|𝑥| + |𝑥 − 2|]dx.

𝑑𝑦
29. Solve 𝑑𝑥 = cos(x+y) + sin(x+y)
OR
𝑑𝑦
Find the particular solution of the differential equation log(𝑑𝑥 ) = 3x + 4y. given that y(0) = 0 .

30. Solve the following Linear Programming Problem graphically:

Maximize 𝑍 = 7𝑥 + 4𝑦 subject to the constraints
3𝑥 + 2𝑦 ≤ 12, 3𝑥 + 𝑦 ≤ 9, 𝑥, 𝑦 ≥ 0
𝑑𝑦 1−𝑦 2
31. If y√1 − 𝑥 2 +x√1 − 𝑦 2 =1then prove that𝑑𝑥 = -√1−𝑥 2
OR

𝑑2 𝑦 1 𝑑𝑦 2 𝑦
If y = 𝑥 𝑥 , then show that 𝑑𝑥 2 - 𝑦 (𝑑𝑥 ) - 𝑥 = 0 .

SECTION D
(This section comprises of long answer-type questions (LA) of 5 marks each)

32. Find 𝑎rea of the region enclosed by the parabola 4y=3𝑥 2 and the line 2y = 3x+12.

33. A relation R on the set A = {1, 2, 3, 4, 5} is given by R = {(a , b) : |𝑎 − 𝑏| is divisible by 2}. Prove that
R is an equivalence relation. Find equivalence class [0].
OR
2 4𝑥+3
Show that the function f in A = R - {3} defined as f(x) = 6𝑥−4 is one-one and onto.
34. Show that the lines 𝑟⃗ = (𝑖̂ + 𝑗̂ − 𝑘̂) + 𝜆(3𝑖̂ − 𝑗̂) and 𝑟⃗ = (4𝑖̂ − 𝑘̂) + 𝜇(2𝑖̂ + 3𝑘̂) intersect.Also find the
point of intersection .
OR

Page | 106
 The equations of motion of a rocket are : x=2t,y= - 4t,z=4t, where the time t is given in seconds, and the
coordinates of a moving point in km. What is the path of the Rocket? At what distances will the rocket
be from the starting point O(0,0,0) and from the following line in 10 seconds?
𝑟⃗ = 20𝑖̂ − 10𝑗̂ + 40𝑘+ K(10𝑖̂ − 20𝑗̂ + 10𝑘)
1 −1 0 2 2 −4
35. Determine the product of [2 3 4]and [−4 2 −4]then use to solve the system of equations x
0 1 2 2 −1 5
- y = 3, 2 x +3y+4z=17 and y + 2z = 7.
SECTION-E
(This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-
parts.First two case study question have three sub-parts (i),(ii),(iii) of marks 1,1,2
respectively.The third case study question has two sub-parts of 2 marks each).
36. Case-Study1:Nanci walks 4km towards west,then she walk 3km in a direction 30°east of north and
stops.

Based on the above information, answer the following questions

(iv) Find the position vector of A
(v) Find area of triangle OAB.
(vi) Determine Nanci displacement from her initial point to departure.
OR
Find the direction cosines of Nanci displacement from her initial point to departure.
37. Case-Study 2: The government of a state, which has mostly hilly area decided to have adventurous
playground on the top of hill having plane area and space for 10000 persons to sit at a time. After
survey it was decided to have rectangular playground with a semi-circular parking at one end of
playground only as space is less. The total perimeter of the field is measured as 1000m as shown.

(i) Looking at the figure(plan), find the relation between x and y.

(ii) Find the area of sports ground in terms of x.
(iii) Find the value of x for which sports ground has the maximum area.
OR

Page | 107
 Find the maximum area of the sports field
38. Case-Study 3: Three persons A, B, C apply for the job of a manager in a private company. The chances of
their selection is given by the relation 4A = 2B = C. The probability that if selected A, B, C can bring changes to
improve profitability of the company are 0.8, 0.5 and 0.3 respectively.

i) What is the probability that C is selected as a manager?

ii) What is the conditional probability that if changes has taken place, it is due to B?

**** BEST OF LUCK ****

Page | 108
 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):

𝟐𝟎𝟐𝟓 𝟎 𝟎
Q.1. If for a square matrix A, 𝑨. (𝒂𝒅𝒋𝑨) = [ 𝟎 𝟐𝟎𝟐𝟓 𝟎 ], then the value of A  adj A is
𝟎 𝟎 𝟐𝟎𝟐𝟓
equal to:

(C)  2025   45 (D) 2025   2025 

2 2
(A) 1 (B) 2025  1
Q.2. Assume X , Y , Z , W and P are matrices of order 2  n, 3  k , 2  p, n  3 and p  k ,
respectively. Then the restriction on n, k and p so that PY  WY will be defined are:
(A) k  3, p  n (B) k is arbitrary, p  2
(C) p is arbitrary, k  3 (D) k  2, p  3
Q.3. The interval in which the function f defined by 𝑓(𝑥) = 𝑒 𝑥 is strictly increasing, is

(A) [1, ∞) (B)   ,0  (C) (−∞, ∞) (D) (0,∞)

Class-XII/Sample Paper/2024-25/Mathematics/Page 1 of 9
 𝟐
Q.4. If A and B are non-singular matrices of same order with 𝒅𝒆𝒕(𝑨) = 𝟓, then 𝒅𝒆𝒕(𝑩−𝟏 𝑨𝑩) is equal to

2 4
(A) 5 (B) 5 (C) 5 (D) 5 5
𝒅𝒚
Q.5. The value of ' n ' , such that the differential equation 𝒙𝒏 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 𝒙𝟏 𝒚𝟐 is equal to

(A) x2 y1 (B) x1 y1 (C) x2 y2 (D) x1 x2

0 1 c 
Q.7. If A   1 a  b  is a skew-symmetric matrix then the value of a  b  c 
 2 3 0 

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

Q.8. For any two events A and B , if P A    1

2
  2
3
1
, P B  and P  A  B   , then 𝑃 (𝐴⁄ ̅ ) equals:
4
̅
𝐵

3 8 5 1
(A) (B) (C) (D)
8 9 8 4
Q.9. The value of 𝛼 if the angle between 𝑝⃗ = 2𝛼 2 𝑖̂ − 3𝛼𝑗̂ + 𝑘̂ and 𝑞⃗ = 𝑖̂ + 𝑗̂ + 𝛼𝑘̂ is obtuse, is

(A) 𝑅 − [0, 1] (B) (0, 1) (C) [0, ∞) (D) [1, ∞)

Q.10. If |𝑎⃗| = 3, |𝑏⃗⃗| = 4 and |𝑎⃗ + 𝑏⃗⃗| =5, then |𝑎⃗ − 𝑏⃗⃗| =

(A) 3 (B) 4 (C) 5 (D) 8

Q.11. For the linear programming problem (LPP), the objective function is Z  4 x  3 y and the feasible
region determined by a set of constraints is shown in the graph:

Class-XII/Sample Paper/2024-25/Mathematics/Page 2 of 9
 (Note: The figure is not to scale.)

Which of the following statements is true?

(A) Maximum value of Z is at R  40,0  .

(B) Maximum value of Z is at Q  30, 20  .

(C) Value of Z at R  40,0  is less than the value at P  0,40  .

(D) The value of Z at Q  30, 20  is less than the value at R  40,0  .

𝒅𝒙
Q.12. ∫ 𝟏 equals
𝟑 𝟒
𝒙 (𝟏+𝒙 )𝟐
1 1
(A) − 2𝑥 2 √1 + 𝑥 4 + 𝑐 (B) 2𝑥 √1 + 𝑥 4 + 𝑐

1 1
(C) − 4𝑥 √1 + 𝑥 4 + 𝑐 (D) 4𝑥 2 √1 + 𝑥 4 + 𝑐

𝟐𝝅
Q.13. ∫𝟎 𝒄𝒐𝒔𝒆𝒄𝟕 𝒙 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.

Class-XII/Sample Paper/2024-25/Mathematics/Page 3 of 9
 The corresponding objective function is: Z  18 x  10 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 x  10 y  134 does not have any point in common with the feasible
region.
(C) exists as the inequality 18 x  10 y  134 has infinitely many points in common with the feasible region.
(D) exists as the inequality 18 x  10 y  134 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 x  2.5 but not differentiable at x  2.5

(B) Not Continuous at x  2.5 but differentiable at x  2.5
(C) Not Continuous at x  2.5 and not differentiable at x  2.5
(D) Continuous as well as differentiable at x  2.5

Q.18. A student observes an open-air Honeybee nest on the branch of a tree, whose plane figure is parabolic
shape given by x 2  4 y . Then the area (in sq units) of the region bounded by parabola x 2  4 y and the line
y  4 is

32 64 128 256
(A) 3 (B) 3 (C) 3 (D) 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.)

Class-XII/Sample Paper/2024-25/Mathematics/Page 4 of 9
 (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 x  0 and x  1 .

Reason (R): Suppose f be defined and continuous on  a , b  and c   a , b  , then f  x  is not

𝑓(𝑐+ℎ)−𝑓(𝑐) 𝑓(𝑐+ℎ)−𝑓(𝑐)
differentiable at 𝑥 = 𝑐 if lim− ≠ lim+ .
ℎ→0 ℎ ℎ→0 ℎ
𝜋
Q.20. Assertion (A): The function 𝑓: 𝑅 − {(2𝑛 + 1) 2 : 𝑛 ∈ 𝑍 } → (−∞, −1] ∪ [1, ∞) defined by

f  x   sec x is not one-one function in its domain.

Reason (R): The line y  2 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.)
𝜋
Q.21. If 𝑐𝑜𝑡 −1 (3𝑥 + 5) > 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) = 0.00013𝑥 3 + 0.002 𝑥 2 + 5𝑥 + 2200

Find the marginal cost when 150 items are produced.

Q.23. (a) Find the derivative of tan 1 x with respect to log x ; (where x   1,   ).

OR
𝜋
Q.23. (b) Differentiate the following function with respect to x : (𝑐𝑜𝑠 𝑥)𝑥 ; (where𝑥 ∈ (0, 2 )).

Q.24. (a) If vectors 𝑎⃗ = 2ı̂ + 2ȷ̂ + 3k̂, 𝑏⃗⃗ = − ı̂ + 2ȷ̂ + k̂ and 𝑐⃗ = 3ı̂ + ȷ̂ are such that 𝑏⃗⃗ + λ𝑐⃗ is perpendicular
to 𝑎⃗ , 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

⃗⃗⃗⃗⃗⃗ makes with the x,y and z axes.

𝐵𝐴

Q.25. The two co-initial adjacent sides of a parallelogram are 2ı̂ − 4ȷ̂ − 5k̂ and 2ı̂ + 2ȷ̂ + 3k̂. Find its
diagonals and use them to find the area of the parallelogram.
Class-XII/Sample Paper/2024-25/Mathematics/Page 5 of 9
 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
A t, where t is the age in years, t   5,18 . Show that the rate of increase of the ability to
3
understand spatial concepts decreases with age in between 5 and 18.
Q.28. (a) An ant is moving along the vector ⃗⃗⃗
𝑙1 = 𝑖̂ − 2𝑗̂ + 3𝑘̂. Few sugar crystals are kept along the vector
⃗𝑙⃗⃗⃗2 = 3𝑖̂ − 2𝑗̂ + 𝑘̂ 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 𝑟⃗ = 2ı̂ + ȷ̂ − 3k̂ + λ(ı̂ + 2ȷ̂ + 5k̂) and 𝑟⃗ = 3ı̂ + 3ȷ̂ − 7k̂ + μ(3ı̂ − 2ȷ̂ + 5k̂).

𝟏 𝟏
Q.29. (a) Evaluate: ∫ {𝒍𝒐𝒈 𝒙 − (𝒍𝒐𝒈 𝒙)𝟐 } 𝒅𝒙; (where𝒙 > 𝟏).

OR
𝟏
Q.29. (b) Evaluate : ∫𝟎 𝒙(𝟏 − 𝒙)𝒏 𝒅𝒙; (𝐰𝐡𝐞𝐫𝐞 𝒏 ∈ 𝑵).
Q.30. Consider the following Linear Programming Problem:
Minimise Z  x  2 y
Subject to 2 x  y  3, x  2 y  6, x , y  0.
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 P1  P4  P2  P3 .  2 Marks 

(ii) Calculate the probability of raining tomorrow. 1 Mark 

OR

Q.31. (b) A random variable X can take all non – negative integral values and the probability that X takes

Class-XII/Sample Paper/2024-25/Mathematics/Page 6 of 9
 the value r is proportional to 5−𝑟 . Find P  X  3  .

SECTION D
 5  4  20
(This section comprises of 4 long answer (LA) type questions of 5 marks each)

𝜋 𝜋
Q.32. Draw the rough sketch of the curve 𝑦 = 20 𝑐𝑜𝑠 2 𝑥; (where 6 ≤ 𝑥 ≤ 3 ).
Using integration, find the area of the region bounded by the curve y = 20 cos2x from the ordinates
𝜋 𝜋
𝑥= to 𝑥 = and the x  axis.
6 3

Q.33. The equation of the path traversed by the ball headed by the footballer is
𝑦 = 𝑎𝑥 2 + 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
3
𝑑𝑦 2 2
[1+( ) ]
𝑑𝑥
Q.34. (b) If (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑐 2 , for some c  0, prove that 𝑑2 𝑦
is a constant independent
𝑑𝑥2

of a and b .
Q.35. (a) Find the shortest distance between the lines l1 and l2 whose vector equations are

𝑟⃗ = (−𝑖̂ − 𝑗̂ − 𝑘̂) + 𝜆(7𝑖̂ − 6𝑗̂ + 𝑘̂) and 𝑟⃗ = (3𝑖̂ + 5𝑗̂ + 7𝑘̂) + 𝜇(𝑖̂ − 2𝑗̂ + 𝑘̂)
where  and  are parameters.
OR
Q.35. (b) Find the image of the point 1,2, 1 with respect to the line
𝑥−3 𝑦+1 𝑧−1
= = . Also find the
1 2 3

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 cm by 40 cm to make container packets without top. Let x cm 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.

Class-XII/Sample Paper/2024-25/Mathematics/Page 7 of 9
 (i) Express the volume (V) of each container as function of x only. 1 Mark 
1 Mark 
𝑑𝑉
(ii) Find 𝑑𝑥

(iii) (a) For what value of x , the volume of each container is maximum?  2 Marks 
OR

 2 Marks 
65
(iii) (b) Check whether V has a point of inflection at x = or not?
6

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 B  b1 , b2 , b3  , G   g1 , g2  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?
1 Mark 
(ii) Write the smallest equivalence relation on G. 1 Mark 
(iii) (a) Ravi defines a relation from B to B as 𝑹𝟏 = {(𝒃𝟏 , 𝒃𝟐 ), (𝒃𝟐 , 𝒃𝟏 )}. Write the minimum ordered
pairs to be added in 𝑹𝟏 so that it becomes (A) reflexive but not symmetric, (B) reflexive and
symmetric but not transitive.  2 Marks 
OR
(iii) (b) If the track of the final race (for the biker b1 ) follows the curve

𝑥 2 = 4𝑦; (where0 ≤ 𝑥 ≤ 20√2&0 ≤ 𝑦 ≤ 200), then state whether the track represents a
one-one and onto function or not. (Justify).  2 Marks 

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:-

Class-XII/Sample Paper/2024-25/Mathematics/Page 8 of 9
(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.  2 Marks 
(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?  2 Marks 

*******************************************************************************

Class-XII/Sample Paper/2024-25/Mathematics/Page 9 of 9
 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 A.  adj A   A I n , where I n is the identity matrix of
order n  n.
 2025 0 0 

So, A.  adj A    0 2025 0   2025 I 3  A  2025 & adj A  A   2025 
3 1 2

 0 0 2025 

 A  adj A  2025   2025  .

2

2. (A)

3. (C) 𝑑𝑦
𝑦 = 𝑒𝑥 = > = 𝑒𝑥
𝑑𝑥
𝑑𝑦
In the domain (R) of the function, > 0 , hence the function is strictly increasing in (−∞, ∞)
𝑑𝑥

 
2
4. (B) 2
A  5, B 1 AB  B 1 A B  A  52 .
2

5. (B) dy
A differential equation of the form  f  x , y  is said to be homogeneous, if f  x , y  is a
dx
homogeneous function of degree 0.

dy  y  dy y   y 
Now, x n  y  log e  log e e    n  log e e .     f  x , y  ;  Let  . f  x , y  will be a
dx  x  dx x   x 
homogeneous function of degree 0, if n  1.
6. (A) Method 1: ( Short cut)

When the points  x1 , y1  ,  x2 , y2  and  x1  x2 , y1  y2  are collinear in the Cartesian plane then

x1  x2 y1  y2 x1  x2 y1  y2
 0    x1 y2  x2 y2  x2 y1  x2 y2   0
x1   x1  x2  y1   y1  y2   x2  y2

 x2 y1  x1 y2 .

Page 1 of 15
 Method 2:
When the points  x1 , y1  ,  x2 , y2  and  x1  x2 , y1  y2  are collinear in the Cartesian plane then

x1 y1 1
x2 y2 1 0
x1  x2 y1  y2 1
 1.  x2 y1  x2 y2  x1 y2  x2 y2   1 x1 y1  x1 y2  x1 y1  x2 y1    x1 y 2  x2 y 1   0
 x2 y1  x1 y2 .
7. (A) 0 1 c 
A   1 a  b 
 2 3 0 
When the matrix A is skew symmetric then AT   A  aij  a ji ;

 c  2; a  0 and b  3
So , a  b  c  0  3  2  1.

   
8. (C) 1 2 1
P A  ;P B  ;P  A  B 
2 3 4
1 1
 P  A  ; P  B  
2 3
1 1 1 7
Wehave, P  A  B   P  A   P  B   P  A  B     
2 3 4 12

 
7
 A P A B P  A  B  1  P  A  B  1  12 5
P       .
 B P B P B  P B   2 8  
3
9. (B) For obtuse angle, cos 𝜃 < 0 => 𝑝⃗. 𝑞⃗ < 0
𝟐𝜶𝟐 − 𝟑𝜶 + 𝜶 < 𝟎 => 𝟐𝜶𝟐 − 𝟐𝜶 < 𝟎 => 𝜶 ∈ (𝟎, 𝟏)
10. (C) a  3, b  4, a  b  5

2
We have , a  b  a  b  2 a  b
2
 2 2
  2 9  16  50  a  b  5.
11. (B) Corner point Value of the objective function Z  4 x  3 y

1. O  0,0  z0

2. R  40,0  z  160

3. Q  30, 20  z  120  60  180

4. P  0,40  z  120

Since , the feasible region is bounded so the maximum value of the objective function z  180 is at

Q  30,20 .

Page 2 of 15
12. (A) 𝑑𝑥 𝑑𝑥
∫ 1 =∫ 1
𝑥 3 (1 + 𝑥 4 ) 2
𝑥 5 (1 +
1 2
)
𝑥4
1 4 𝑑𝑥 1
( Let 1 + 𝑥 −4 = 1 + = 𝑡, 𝑑𝑡 = −4𝑥 −5 𝑑𝑥 = − 𝑑𝑥 ⇒ = − 𝑑𝑡 )
𝑥4 𝑥5 𝑥5 4
1 𝑑𝑡 1
= −4∫ 1 = − 4 × 2 × √𝑡 + 𝑐, where ' c ' denotes any arbitrary constant of integration.
𝑡2

1 1 1
= − 2 √1 + 𝑥 4 + 𝑐 = −
2𝑥 2
√1 + 𝑥 4 + 𝑐

13. (A) We know, 0 f  x  dx  0, if f  2a  x    f  x 

2a

Let f  x   cos ec7 x .

Now, f  2  x   cos ec7  2  x    cos ec 7 x   f  x 

2

  cos ec 7 x dx  0; Using the property 0 f  x  dx  0, if f  2a  x    f  x  .

2a

14. (B) ′ 𝑑𝑦
The given differential equation 𝑒 𝑦 = 𝑥 => 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 y  cos 1 x whose domain is   1,1 and range is  0,   .
16. (D) Since the inequality Z  18 x  10 y  134 has no point in common with the feasible region hence

the minimum value of the objective function Z  18 x  10 y is 134 at P  3,8  .

17. (D) The graph of the function 𝑓: 𝑅 → 𝑅 defined by f  x    x  ;  where  . denotes G . I .F  is a straight
line  x   2.5  h,2.5  h , ' h ' is an infinitesimally small positive quantity. Hence, the function is
continuous and differentiable at x  2.5 .

18. (B) The required region is symmetric about the y  axis.

4
 3
 y2 4
64
So, required area (in sq units ) is  2 2 ydy  4    .
3 3
0  
 2 0
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).

Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]

Page 3 of 15
21 𝜋 1
cot −1 (3𝑥 + 5) > = cot −1 1
4 2
1
=>3x + 5 < 1 ( as cot −1 𝑥 is strictly decreasing function in its domain)
2

=> 3x < – 4
4
=> 𝑥 < − 3
4
⸫ 𝑥 ∈ (−∞, − ) 1
3

22. The marginal cost function is C '  x   0.00039 x 2  0.004 x  5 . 1

C '  150   ₹ 14.375 . 1

23.(a) y  tan1 x and z  loge x

.
dy 1 1
Then 
dx 1  x 2 2
dz 1 1
and 
dx x 2
dy
dy dx
 1
dz dz
2
dx
So,
1
 1 x 
2 x 1
.
1 1  x2 2
x
OR Let y  (cos x ) x . Then, y  e x logecosx
23.(b)
dy d
 e x loge cos x ( x log e cos x ) 1
On differentiating both sides with respect to x , we get dx dx
2
dy  d d 
  (cos x ) x  log e cos x ( x )  x (log e cos x )  1
dx  dx dx  2
dy  1  dy
  (cos x ) x  log e cos x  x . (  sin x )    (cos x ) x (log e cos x  x tan x ) . 1
dx  cos x  dx

24.(a) ⃗⃗ + λc⃗ = (−1 + 3λ)î + (2 + λ )ĵ + k̂

We have b 1
2

⃗⃗ + λc⃗) . a⃗⃗ = 0 => 2(−1 + 3λ ) + 2 (2 + λ ) + 3 = 0

(b 1

1
5 2
λ = −8
OR
1
24.(b) ⃗⃗⃗⃗⃗⃗ = 𝑂𝐴
𝐵𝐴 ⃗⃗⃗⃗⃗⃗ = (4𝑖̂ + 3𝑘̂ ) − 𝑘̂ = 4𝑖̂ + 2𝑘̂
⃗⃗⃗⃗⃗⃗ − 𝑂𝐵
2
Page 4 of 15
 4 2 2 1 1
̂ =
𝐵𝐴 𝑖̂ + 𝑘̂ = 𝑖̂ + 𝑘̂
2√5 2√5 √5 √5 2
⃗⃗⃗⃗⃗⃗ with the x , y and the z axes are respectively
So, the angles made by the vector 𝐵𝐴
1
−1 2 𝜋 −1 1
𝑐𝑜𝑠 ( ) , 2 , 𝑐𝑜𝑠 ( ).
√5 √5

25. ⃗⃗⃗⃗⃗
𝑑1 = 𝑎⃗ + 𝑏⃗⃗ = 4𝑖̂ − 2𝑗̂ − 2𝑘̂ , ⃗⃗⃗⃗⃗
𝑑2 = 𝑎⃗ − 𝑏⃗⃗ = −6𝑗̂ − 8𝑘̂ 1
2
1 1
𝑖̂ 𝑗̂ 𝑘̂
Area of the parallelogram = |𝑑1 × 𝑑2 | = ||4 −2 −2|| = 2|𝑖̂ + 8𝑗̂ − 6𝑘̂ |
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 1
2 2
0 −6 −8 1
Area of the parallelogram = 2√101 sq. units. 2

Section –C
[This section comprises of solution short answer type questions (SA) of 3 marks each]

26.

y 1
3 2
3

1
x 𝑥 2 + 32 = 𝑦 2 2

𝑑𝑥 𝑑𝑦
𝑊ℎ𝑒𝑛 𝑦 = 5 𝑡ℎ𝑒𝑛 𝑥 = 4, 𝑛𝑜𝑤 2𝑥 = 2𝑦 1
𝑑𝑡 𝑑𝑡
𝑑𝑦 𝑑𝑦
4 (200) = 5 => = 160 cm/s 1
𝑑𝑡 𝑑𝑡
1
27. 1 𝑑𝐴 1 1 1
𝐴 = 3 √𝑡 ∴ 𝑑𝑡
= 6 𝑡 −2 = 6 𝑡 ; ∀𝑡 ∈ (5,18)
√

𝑑𝐴 1 𝑑2 𝐴 1 1
𝑑𝑡
=6 𝑡
∴ 𝑑𝑡 2
= − 12𝑡
√ √𝑡 1
𝑑2 𝐴 2
So, 𝑑𝑡 2
< 0, ∀𝑡 ∈ (5,18)
This means that the rate of change of the ability to understand spatial concepts decreases 1
2
(slows down) with age.
28(a) ⃗⃗⃗⃗
𝒍 .𝒍⃗⃗⃗⃗ ̂).(3ı̂−2ȷ̂ + 𝑘
(ı̂−2ȷ̂+3k ̂) 1
(i) 𝜽 = 𝐜𝐨𝐬 −𝟏 ( ⃗⃗⃗⃗𝟏 𝟐
) = 𝐜𝐨𝐬 −𝟏 (|(ı̂−2ȷ̂+3k̂)|| (3ı̂−2ȷ̂ + 𝑘̂)|)
|𝒍𝟏 |.|𝒍⃗⃗⃗⃗
𝟐|
1
𝟑+𝟒+𝟑 𝟏𝟎 𝟓 2
= 𝒄𝒐𝒔−𝟏 ( ) = 𝒄𝒐𝒔−𝟏 (𝟏𝟒) = 𝒄𝒐𝒔−𝟏 (𝟕).
√𝟏+𝟒+𝟗√𝟗+𝟒+𝟏

⃗⃗⃗⃗
𝒍𝟏 .𝒍⃗⃗⃗⃗
𝟐
̂).(3ı̂−2ȷ̂ + 𝑘
(ı̂−2ȷ̂+3k ̂) 1
(ii) Scalar projection of ⃗⃗⃗⃗
𝒍𝟏 on ⃗⃗⃗⃗
𝒍𝟐 = =
|𝒍⃗⃗⃗⃗
𝟐 | ̂
| (3ı̂−2ȷ̂ + 𝑘 )| 1
=
3+4+3
=
10
. 2
√9+4+1 √14

Page 5 of 15
28(b) Line perpendicular to the lines

𝑟⃗ = 2ı̂ + ȷ̂ − 3k̂ + λ(ı̂ + 2ȷ̂ + 5k̂) and 𝑟⃗ = 3ı̂ + 3ȷ̂ − 7k̂ + μ(3ı̂ − 2ȷ̂ + 5k̂).
𝑖̂ 𝑗̂ 𝑘̂
has a vector parallel it is given by 𝑏⃗⃗ = ⃗⃗⃗⃗ 𝑏1 × ⃗⃗⃗⃗⃗
𝑏2 = |1 2 5| = 20î + 10ĵ − 8k̂ 1
3 −2 5
⸫ equation of line in vector form is 𝑟⃗ = − ı̂ + 2 ȷ̂ + 7k̂ + a(10ı̂ + 5ȷ̂ − 4k̂)
1
𝑥+1 𝑦−2 𝑧−7
And equation of line in cartesian form is = =
10 5 −4 1

29.(a) 1 1
∫{ − } 𝑑𝑥
𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥)2
𝑑𝑥 1 1 𝑑 1 1
=∫ −∫ 2
𝑑𝑥 = ∫ 𝑑𝑥 − ∫ { ( ) ∫ 𝑑𝑥} 𝑑𝑥 − ∫ 𝑑𝑥
𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥) 𝑙𝑜𝑔𝑒 𝑥 𝑑𝑥 𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥)2 1
𝑥 1 1 1
= +∫ . 𝑥. 𝑑𝑥 − ∫ 𝑑𝑥 1
𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥)2 𝑥 (𝑙𝑜𝑔𝑒 𝑥)2
𝑥 1 𝑑𝑥 𝑥
= +∫ 2
𝑑𝑥 − ∫ 2
= + 𝑐; 1
𝑙𝑜𝑔𝑒 𝑥 (𝑙𝑜𝑔𝑒 𝑥) (𝑙𝑜𝑔𝑒 𝑥) 𝑙𝑜𝑔𝑒 𝑥
where′𝑐′is any arbitary constant of integration.
OR 1
 x  1  x  dx
n

29.(b) 0
1 𝑎 𝑎
= ∫ (1 − 𝑥){1 − (1 − 𝑥)}𝑛 𝑑𝑥, (𝑎𝑠, ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥 )
0 0 0 1
1
= ∫ 𝑥 𝑛 (1 − 𝑥)𝑑𝑥
0

1 1 1
= ∫ 𝑥 𝑛 𝑑𝑥 − ∫ 𝑥 𝑛+1 𝑑𝑥 2
0 0
1 1 1
= [𝑥 𝑛+1 ]1 0 − [𝑥 𝑛+2 ]0 1
𝑛+1 𝑛+2
2
1 1 1
= 𝑛+1 − 𝑛+2 = (𝑛+1)(𝑛+2).
1

30. The feasible region determined by the constraints, 2 x  y  3, x  2 y  6, x  0, y  0 is as shown.

Page 6 of 15
 1

The corner points of the unbounded feasible region are A( 6, 0) and B ( 0, 3) .

The values of Z at these corner points are as follows:

Value of the objective function

Corner point Z  x  2y

1
A( 6, 0) 6

B ( 0, 3) 6

We observe the region x  2 y  6 have no points in common with the unbounded feasible region. Hence 1
the minimum value of z  6 . 2
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
x  2 y  6 such as (2,2) on line x  2 y  6, then Z  6 .
1
Thus, the minimum value of Z occurs for more than 2 points, and is equal to 6. 2

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  P1  0.6

Page 7 of 15
 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) P1  P4  P2  P3  0.6  0.3  0.2  0.7  0.04. 1

(ii) Let E1 and E2 be the events that it will rain today and it will not rain today respectively.
1

P  E1   0.4 & P  E2   0.6

𝐴 𝐴 1
A be the event that it will rain tomorrow. 𝑃 (𝐸 ) = 0.8 & 𝑃 (𝐸 ) = 0.7 2
1 2

𝐴 𝐴
We have, 𝑃(𝐴) = 𝑃(𝐸1 )𝑃 (𝐸 ) + 𝑃(𝐸2 )𝑃 (𝐸 ) = 0.4 × 0.8 + 0.6 × 0.7 = 0.74.
1 2
1
The probability of rain tomorrow is 0.74 .
2
1
OR Given 𝑃(𝑋 = 𝑟)𝛼 5𝑟 1
31.(b) 1 2
𝑃(𝑋 = 𝑟) = 𝑘 5𝑟 ( where k is a non-zero constant )
,
1
𝑃(𝑟 = 0) = 𝑘. 0
5
1
𝑃(𝑟 = 1) = 𝑘. 1
5
1 1
𝑃(𝑟 = 2) = 𝑘. 2
5 2
1
𝑃(𝑟 = 3) = 𝑘. 3
5
………………………….
………………………….
We have, 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)+. . . . . . . . . . . . . . . = 1 1
2
Page 8 of 15
 1 1 1
⇒ 𝑘 (1 + + 2 + 3 +. . . . . . . . . . . . . ) = 1
5 5 5
1
1 4
⇒ 𝑘( 1) = 1 ⇒ 𝑘 = 2
1− 5
5
So, 𝑃(𝑋 < 3) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)
4 1 1 4 25 + 5 + 1 124
= (1 + + 2 ) = ( )= . 1
5 5 5 5 25 125
Section –D
[This section comprises of solution of long answer type questions (LA) of 5 marks each]

32.

𝜋 𝜋

Required area = 20 ∫𝜋4 𝑐𝑜𝑠2𝑥 𝑑𝑥 + |20 ∫𝜋3 𝑐𝑜𝑠2𝑥 𝑑𝑥 | 1+1

6 4

𝜋 𝜋
sin 2𝑥 4 sin 2𝑥 3 1
= 20 [ ]𝜋 + |20 [ ] |
2 2 𝜋
6 4

1
√3 √3 √3
= 10 (1 − 2
)+ 10 (1 − 2
) = 20 (1 − 2
) sq. units.

33. y  ax 2  bx  c
15  4a  2b  c 1
25  16a  4b  c
15  196a  14b  c
The set of equations can be represented in the matrix form as AX  B , 1
4 2 1 𝑎 15 4 2 1 𝑎 15 2
where 𝐴 = [ 16 4 1]’ 𝑋 = [𝑏] and 𝐵 = [25] ⇒ [ 16 4 1] [𝑏] = [25].
196 14 1 𝑐 15 196 14 1 𝑐 15 1
1
|𝐴| = 4(4 − 14) − 2(16 − 196) + (224 − 784) = −40 + 360 − 560 = −240 ≠ 0. Hence A 2
exists.

Page 9 of 15
 −10 180 −560 𝑇 −10 12 −2 1
Now,𝑎𝑑𝑗(𝐴) = [ 12 −192 336 ] = [ 180 −192 12 ]
−2 12 −16 −560 336 −16
𝑎 1 −10 12 −2 15 5 −10 12 −2 3 5 24 1
[𝑏 ] = − [ 180 −192 12 ] [25] = − [ 180 −192 12 ] [5] = − [−384]
𝑐 240 240 240
−560 336 −16 15 −560 336 −16 3 −48

1 1
 a   , b  8, c  1 2
2
1 1
So, the equation becomes y   x 2  8 x  1
2 2

34.(a) 𝑥 3 ,if 𝑥 ≥ 0
We have, 𝑓(𝑥) = |𝑥|3 , {
(−𝑥)3 = −𝑥 3 ,if𝑥 < 0 1
𝑓(𝑥)−𝑓(0) −𝑥 3 −0 2
Now, (𝐿𝐻𝐷 𝑎𝑡 𝑥 = 0) = 𝑙𝑖𝑚−
𝑥−0
= 𝑙𝑖𝑚− (
𝑥
) = 𝑙𝑖𝑚−(−𝑥 2 ) = 0
𝑥→0 𝑥→0 𝑥→0
1
𝑓(𝑥)−𝑓(0) 𝑥 3 −0
(𝑅𝐻𝐷 𝑎𝑡𝑥 = 0) 𝑙𝑖𝑚+
𝑥−0
= 𝑙𝑖𝑚+ ( 𝑥
) = 𝑙𝑖𝑚(−𝑥 2 ) = 0 2
𝑥→0 𝑥→0 𝑥→0

1
∴ (𝐿𝐻𝐷 𝑜𝑓 𝑓(𝑥) 𝑎𝑡 𝑥 = 0) = (𝑅𝐻𝐷 𝑜𝑓 𝑓(𝑥) 𝑎𝑡 𝑥 = 0) 2
So, f  x  is differentiable at x  0 and the derivative of f  x  is given by

3𝑥 2 ,if𝑥 ≥ 0 1
𝑓′(𝑥) = {
−3𝑥 2 ,if𝑥 < 0
1
𝑓′(𝑥)−𝑓′(0) −3𝑥 2 −0
Now, (𝐿𝐻𝐷𝑜𝑓𝑓′(𝑥)𝑎𝑡𝑥 = 0) = 𝑙𝑖𝑚− 𝑥−0
= 𝑙𝑖𝑚− ( 𝑥
) = 𝑙𝑖𝑚−(−3𝑥) = 0 2
𝑥→0 𝑥→0 𝑥→0

𝑓′(𝑥)−𝑓′(0) 3𝑥 2 −0 1
(𝑅𝐻𝐷 𝑜𝑓𝑓′(𝑥) 𝑎𝑡 𝑥 = 0) = 𝑙𝑖𝑚+ = 𝑙𝑖𝑚+ ( ) = 𝑙𝑖𝑚+(3𝑥) = 0
𝑥→0 𝑥−0 𝑥→0 𝑥−0 𝑥→0 2
∴ (𝐿𝐻𝐷 𝑜𝑓𝑓′(𝑥)𝑎𝑡 𝑥 = 0) = (𝑅𝐻𝐷𝑜𝑓𝑓′(𝑥)𝑎𝑡𝑥 = 0) 1
2

So, 𝑓′(𝑥)is differentiable at x  0. 1

2
6𝑥,if𝑥 ≥ 0
Hence, 𝑓′′(𝑥) = {
−6𝑥,if𝑥 < 0. 1
2

OR Given relation is (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑐 2 , 𝑐 > 0.

34 .(b) 1
Let x  a  c cos  and 𝑦 − 𝑏 = 𝑐 𝑠𝑖𝑛 𝜃.
2
𝑑𝑥 𝑑𝑦
Therefore, 𝑑𝜃 = −𝑐 𝑠𝑖𝑛 𝜃 And = 𝑐 𝑐𝑜𝑠𝜃 1
𝑑𝜃
2
𝑑𝑦
∴ 𝑑𝑥 = − 𝑐𝑜𝑡 𝜃 1

𝑑 𝑑𝑦 𝑑
Differentiate both sides with respect to , we get 𝑑𝜃 (𝑑𝑥 ) = 𝑑𝜃 (− 𝑐𝑜𝑡 𝜃) 1
2

Page 10 of 15
 𝑑 𝑑𝑦 𝑑𝑥
Or, 𝑑𝑥 (𝑑𝑥 ) 𝑑𝜃 = 𝑐𝑜𝑠 𝑒 𝑐 2 𝜃
1
𝑑2 𝑦 2
Or, (−𝑐 𝑠𝑖𝑛 𝜃) = cosec 𝜃 2
𝑑𝑥 2
1
𝑑2𝑦 𝑐𝑜𝑠𝑒𝑐 3 𝜃
= − 2
𝑑𝑥 2 𝑐

3
𝑑𝑦 2 2 3 3 1
[1+( ) ] 𝑐[1+𝑐𝑜𝑡 2 𝜃]2 − 𝑐(𝑐𝑜𝑠 𝑒𝑐 2 𝜃)2
𝑑𝑥
∴ 𝑑2 𝑦
= = = −𝑐,
− 𝑐𝑜𝑠 𝑒𝑐 3 𝜃 cosec 3 𝜃
𝑑𝑥2
1
Which is constant and is independent of a and b . 2

35.(a)

Given that equation of lines are

𝑟⃗ = (−𝑖̂ − 𝑗̂ − 𝑘̂ ) + 𝜆(7𝑖̂ − 6𝑗̂ + 𝑘̂ ). . . . . . . . . . . . . . . . (𝑖) and
𝑟⃗ = (3𝑖̂ + 5𝑗̂ + 7𝑘̂ ) + 𝜇(𝑖̂ − 2𝑗̂ + 𝑘̂ ). . . . . . . . . . . . . . . . . . (𝑖𝑖)
The given lines are non-parallel lines as vectors 7𝑖̂ − 6𝑗̂ + 𝑘̂and 𝑖̂ − 2𝑗̂ + 𝑘̂ 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 line𝑟⃗ = (−𝑖̂ − 𝑗̂ − 𝑘̂ ) + 𝜆(7𝑖̂ − 6𝑗̂ + 𝑘̂ ) where '  ' 1
is a scalar, is (7𝜆 − 1)𝑖̂ − (6𝜆 + 1)𝑗̂ + (𝜆 − 1)𝑘̂ , for some  and the position vector of the point Q 2
1
lying on the line 𝑟⃗ = (3𝑖̂ + 5𝑗̂ + 7𝑘̂ ) + 𝜇(𝑖̂ − 2𝑗̂ + 𝑘̂ )where '  ' is a scalar, is 2
(𝜇 + 3)𝑖̂ + (−2𝜇 + 5)𝑗̂ + (𝜇 + 7)𝑘̂ , for some  . Now, the vector
⃗⃗⃗⃗⃗⃗
𝑃𝑄 = 𝑂𝑄 𝑂𝑃 = (𝜇 + 3 − 7𝜆 + 1)𝑖̂ + (−2𝜇 + 5 + 6𝜆 + 1)𝑗̂ + (𝜇 + 7 − 𝜆 + 1)𝑘̂
⃗⃗⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ = (𝜇 − 7𝜆 + 4)𝑖̂ + (−2𝜇 + 6𝜆 + 6)𝑗̂ + (𝜇 − 𝜆 + 8)𝑘̂ ; (where ' O ' is the origin), is 1
𝑖. 𝑒. , 𝑃𝑄
⃗⃗⃗⃗⃗⃗ is perpendicular to both the vectors 7𝑖̂ − 6𝑗̂ + 𝑘̂ and
perpendicular to both the lines, so the vector 𝑃𝑄
𝑖̂ − 2𝑗̂ + 𝑘̂ .
 (𝜇 − 7𝜆 + 4). 7 + (−2𝜇 + 6𝜆 + 6). (−6) + (𝜇 − 𝜆 + 8). 1 = 0

Page 11 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 1
On solving the above equations, we get     0 1
So, the position vector of the points P and Q are −𝑖̂ − 𝑗̂ − 𝑘̂ and 3𝑖̂ + 5𝑗̂ + 7𝑘̂ respectively. 2
1
𝑃𝑄 = 4𝑖̂ + 6𝑗̂ + 8𝑘̂ and
⃗⃗⃗⃗⃗⃗
2
⃗⃗⃗⃗⃗⃗ | = √42 + 62 + 82 = √116 = 2√29 𝑢𝑛𝑖𝑡𝑠.
|𝑃𝑄 1

OR
35.(b)

Let P 1, 2 , 1 be the given point and L be the foot of the perpendicular from P to the given line AB

(𝑎𝑠 𝑠ℎ𝑜𝑤𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒 𝑎𝑏𝑜𝑣𝑒).

x  3 y 1 z 1
Let’s put     . Then, x    3, y  2  1, z  3  1
1 2 3 1
2
Let the coordinates of the point L be    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)

1
Direction ratios of the given line are 1, 2 and 3, which is perpendicular to PL . Therefore, we have, 2

2
(𝜆 + 2). 1 + (2𝜆 − 3). 2 + 3𝜆. 3 = 0 ⇒ 14𝜆 = 4 ⇒ 𝜆 = 1
7 2
2 23 2 3 2 13
Then, 𝜆 + 3 = 7 + 3 = 7
; 2𝜆 − 1 = 2 (7) − 1 = − 7 ; 3𝜆 + 1 = 3 (7) + 1 = 7 1
23 3 13 2
Therefore, coordinates of the point L are ( 7 , − 7 , 7 ).

Let 𝑄(𝑥1 , 𝑦1 , 𝑧1 )be the image of P 1, 2 , 1 with respect to the given line. Then, L is the mid-point
1
of PQ.
1+𝑥1 23 2+𝑦1 3 1+𝑧1 13 39 20 19
Therefore, 2
= 7
, 2
= −7, 2
= 7
⇒ 𝑥1 = 7
, 𝑦1 = − 7
, 𝑧1 = 7

Hence, the image of the point P  1, 2,1 with respect to the given line 𝑄 ( , −
39 20 19
, ). 1
7 7 7

The equation of the line joining P  1, 2,1 and 𝑄 ( 7 , −

39 20 19
7
, 7 )is

Page 12 of 15
 𝑥−1 𝑦−2 𝑧−1 𝑥−1 𝑦−2 𝑧−1 1
= = ⇒ = = .
32/7 −34/7 12/7 16 −17 6

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 1

𝑑𝑉
(ii) = 4(3𝑥 − 50)(𝑥 − 5) 1
𝑑𝑥

𝑑𝑉 𝟏⁄
(iii) (a) For extreme values = 4(3𝑥 − 50)(𝑥 − 5) = 0 𝟐
𝑑𝑥
50 𝟏⁄
⇒𝑥= 3
or 𝑥 = 5 𝟐

𝑑2 𝑉
= 24𝑥 − 260 𝟏⁄
𝑑𝑥 2 𝟐
𝑑2 𝑉
∴ 𝑑𝑥 2 at 𝑥 = 5 is − 140 < 0 𝟏⁄
𝟐
∴ 𝑉 is max 𝑤ℎ𝑒𝑛 𝑥 = 5

(iii) OR
𝟏⁄
𝑑𝑉
(b) For extreme values 𝑑𝑥 = 4(3𝑥 2 − 65𝑥 + 250) 𝟐

𝑑2 𝑉
𝟏⁄
= 4(6𝑥 − 65) 𝟐
𝑑𝑥 2

𝑑𝑉 65 𝑑2 𝑉 65
𝑎𝑡 𝑥 = exists and 𝑑𝑥 2 𝑎𝑡 𝑥 = 𝑖𝑠 0.
𝑑𝑥 6 6

𝟏⁄
𝑑2𝑉 65 − 𝑑2𝑉 65 + 𝟐
𝑎𝑡 𝑥 = ( ) is negative and 𝑎𝑡 𝑥 = ( ) is positive
𝑑𝑥 2 6 𝑑𝑥 2 6

65
⸫𝑥= is a point of inflection. 𝟏⁄
6
𝟐

Number of relations is equal to the number of subsets of the set B  G  2  

37. n BG
(i)
n B   n G  1
2  23 2  26
( 𝑾𝒉𝒆𝒓𝒆𝒏(𝑨) 𝒅𝒆𝒏𝒐𝒕𝒆𝒔 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒇𝒊𝒏𝒊𝒕𝒆 𝒔𝒆𝒕 A )
(ii) Smallest Equivalence relation on G is {(𝒈𝟏 , 𝒈𝟏 ), (𝒈𝟐 , 𝒈𝟐 )} 1
(iii) (a) (A) reflexive but not symmetric =
{(𝒃𝟏 , 𝒃𝟐 ), (𝒃𝟐 , 𝒃𝟏 ), (𝒃𝟏 , 𝒃𝟏 ), (𝒃𝟐 , 𝒃𝟐 ), (𝒃𝟑 , 𝒃𝟑 ), (𝒃𝟐 , 𝒃𝟑 )}.

Page 13 of 15
 So the minimum number of elements to be added are
(𝒃𝟏 , 𝒃𝟏 ), (𝒃𝟐 , 𝒃𝟐 ), (𝒃𝟑 , 𝒃𝟑 ), (𝒃𝟐 , 𝒃𝟑 ) 1
{Note : it can be any one of the pair from, (𝒃𝟑 , 𝒃𝟐 ), (𝒃𝟏 , 𝒃𝟑 ), (𝒃𝟑 , 𝒃𝟏 ) in place of
(𝒃𝟐 , 𝒃𝟑 ) 𝒂𝒍𝒔𝒐}
(B) reflexive and symmetric but not transitive =
{(𝒃𝟏 , 𝒃𝟐 ), (𝒃𝟐 , 𝒃𝟏 ), (𝒃𝟏 , 𝒃𝟏 ), (𝒃𝟐 , 𝒃𝟐 ), (𝒃𝟑 , 𝒃𝟑 ), (𝒃𝟐 , 𝒃𝟑 ), (𝒃𝟑 , 𝒃𝟐 ) }.

1
So the minimum number of elements to be added are
(𝒃𝟏 , 𝒃𝟏 ), (𝒃𝟐 , 𝒃𝟐 ), (𝒃𝟑 , 𝒃𝟑 ), (𝒃𝟐 , 𝒃𝟑 ), (𝒃𝟑 , 𝒃𝟐 )

OR (iii) (b) One-one and onto function

𝟐
𝒙𝟐
𝒙 = 𝟒𝒚. let𝒚 = 𝒇(𝒙) =
𝟒
𝒙𝟏 𝟐 𝒙𝟏 𝟐
Let 𝒙𝟏 , 𝒙𝟐 ∈ [𝟎, 𝟐𝟎√𝟐] such that 𝒇(𝒙𝟏 ) = 𝒇(𝒙𝟐 ) ⇒ =
𝟒 𝟒 1
⇒ 𝒙𝟏 𝟐 = 𝒙𝟐 𝟐 ⇒ (𝒙𝟏 − 𝒙𝟐 )(𝒙𝟏 + 𝒙𝟐 ) = 𝟎 ⇒ 𝒙𝟏 = 𝒙𝟐 as 𝒙𝟏 , 𝒙𝟐 ∈ [𝟎, 𝟐𝟎√𝟐]
∴ 𝒇 is one-one function
Now, 𝟎 ≤ 𝒚 ≤ 𝟐𝟎𝟎 hence the value of 𝒚 is non-negative
and 𝒇(𝟐√𝒚) = 𝒚
∴ for any arbitrary 𝒚 ∈ [𝟎, 𝟐𝟎𝟎], the pre-image of 𝒚 exists in [𝟎, 𝟐𝟎√𝟐] 1
hence 𝒇 is onto function.
38. Let E1 be the event that one parrot and one owl flew from cage –I

𝐸2 be the event that two parrots flew from Cage-I

A be the event that the owl is still in cage-I

(i) Total ways for A to happen

From cage I 1 parrot and 1 owl flew and then from Cage-II 1 parrot and 1 owl
flew back + From cage I 1 parrot and 1 owl flew and then from Cage-II 2 parrots
flew back + From cage I 2 parrots flew and then from Cage-II 2 parrots came
1
back.
2
=(5𝐶1 × 1𝐶1 )(7𝐶1 × 1𝐶1 ) + (5𝐶1 × 1𝐶1 )(7𝐶2 ) + (5𝐶2 )(8𝐶2 )
Probability that the owl is still in cage –I = P(𝐸1 ∩ 𝐴) + P(𝐸2 ∩ 𝐴)
(5𝐶1 × 1𝐶1 )(7𝐶1 × 1𝐶1 ) + (5𝐶2 )(8𝐶2 ) 1
(5𝐶1 × 1𝐶1 )(7𝐶1 × 1𝐶1 ) + (5𝐶1 × 1𝐶1 )(7𝐶2 ) + (5𝐶2 )(8𝐶2 )
35 + 280 315 3 1
= = =
35 + 105 + 280 420 4 2

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(i) The probability that one parrot and the owl flew from Cage-I to Cage-II given 1
2
that the owl is still in cage-I is 𝑃 (𝐸1⁄𝐴)
1
𝐸 P(𝐸1 ∩ 𝐴)
𝑃 ( 1⁄𝐴) = P(𝐸1 ∩ 𝐴)+P(𝐸2 ∩ 𝐴)
(by Baye’s Theorem) 2

35
420 1
= 315 = 1
9
420

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