Domain of a 2 − x 2 ( a > 0 ) is (a) ( − a , a ) (b) [ − a , a ] (c) [ 0 , a ] (d) ( − a , 0 ]

This is an Objective Type Questions as classified in NCERT Exemplar L e t f ( x ) = a 2 − x 2 f ( x ) i s d e f i n e d i f a 2 − x 2 ≥ 0 ⇒ x 2 − a 2 ≤ 0 ⇒ x 2 ≤ a 2 ⇒ x ≤ ± a ⇒ − a ≤ x ≤ a ∴ D o m a i n o f f ( x ) = [ − a , a ] H e n c e , t h e c o r r e c t o p t i o n i s ( b ) .

If [ x ] 2 − 5 [ x ] + 6 = 0 , where [.] denotes the greatest integer function, then (a) x ∈ [ 3 , 4 ] (b) x ∈ ( 2 , 3 ] (c) x ∈ [ 2 , 3 ] (d) x ∈ [ 2 , 4 ]

This is an Objective Type Questions as classified in NCERT Exemplar W e h a v e [ x ] 2 − 5 [ x ] + 6 = 0 ⇒ [ x ] 2 − 3 [ x ] − 2 [ x ] + 6 = 0 ⇒ [ x ] ( [ x ] − 3 ) − 2 ( [ x ] − 3 ) = 0 ⇒ ( [ x ] − 3 ) ( [ x ] − 2 ) = 0 ⇒ [ x ] = 2 , 3 S o , x ∈ [ 2 , 3 ] H e n c e , t h e c o r r e c t o p t i o n i s ( c ) .

Range of f ( x ) = 1 1 − 2 c o s x is (a) [ 1 3 , 1 ] (b) [ − 1 , 1 3 ] (c) ( − ∞ , − 1 ] ∪ [ 1 3 , ∞ ) (d) [ − 1 3 , 1 ]

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t : f ( x ) = 1 1 − 2 c o s x W e k n o w t h a t − 1 ≤ c o s x ≤ 1 ⇒ 1 ≥ c o s x ≥ − 1 ⇒ − 1 ≤ − c o s x ≤ 1 ⇒ − 2 ≤ − 2 c o s x ≤ 2 ⇒ − 2 + 1 ≤ 1 − 2 c o s x ≤ 2 + 1 ⇒ − 1 ≤ 1 − 2 c o s x ≤ 3 ⇒ − 1 ≤ 1 1 − 2 c o s x ≤ 1 3 ⇒ − 1 ≤ f ( x ) ≤ 1 3 S o , t h e r a n g e o f f ( x ) = [ − 1 , 1 3 ] H e n c e , t h e c o r r e c t o p t i o n i s ( b ) .

Let f ( x ) = 1 + x 2 then (a) f ( x y ) = f ( x ) ⋅ f ( y ) (b) f ( x y ) ≥ f ( x ) ⋅ f ( y ) (c) f ( x y ) ≤ f ( x ) ⋅ f ( y ) (d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t : f ( x ) = 1 + x 2 ⇒ f ( x y ) = 1 + x 2 y 2 a n d f ( x ) . f ( y ) = 1 + x 2 . 1 + y 2 = 1 + x 2 + y 2 + x 2 y 2 ? 1 + x 2 y 2 ≤ 1 + x 2 + y 2 + x 2 y 2 ⇒ f ( x y ) ≤ f ( x ) . f ( y ) H e n c e , t h e c o r r e c t o p t i o n i s ( c ) .

Let A = {–1, 2, 3} and B = {1, 3}. Determine (i) A × B (ii) B × A (iii) B × B (iv) A × A

This is a short answer type question as classified in NCERT Exemplar G i v e n t h a t : A = { − 1 , 2 , 3 } a n d B = { 1 , 3 } ( i ) A × B = { ( − 1 , 1 ) , ( − 1 , 3 ) , ( 2 , 1 ) , ( 2 , 3 ) , ( 3 , 1 ) , ( 3 , 3 ) } ( i i ) B × A = { ( 1 , − 1 ) , ( 3 , − 1 ) , ( 1 , 2 ) , ( 3 , 2 ) , ( 1 , 3 ) , ( 3 , 3 ) } ( i i i ) B × B = { 1 , 3 } × { 1 , 3 } = { ( 1 , 1 ) , ( 1 , 3 ) , ( 3 , 1 ) , ( 3 , 3 ) } ( i v ) A × A = { − 1 , 2 , 3 } × { − 1 , 2 , 3 } = { ( − 1 , − 1 ) , ( − 1 , 2 ) , ( − 1 , 3 ) , ( 2 , − 1 ) , ( 2 , 2 ) , ( 2 , 3 ) , ( 3 , − 1 ) , ( 3 , 2 ) , ( 3 , 3 ) }

If P = {x: x < 3, x ∈ N}, Q = {x: x ≤ 2, x ∈ W}. Find (P ∪ Q) × (P ∩ Q), where W is the set of whole numbers.

This is a short answer type question as classified in NCERT Exemplar G i v e n t h a t : P = { x : x < 3 , x ∈ N } ⇒ P = { 1 , 2 } Q = { x : x ≤ 2 , x ∈ W } ⇒ Q = { 0 , 1 , 2 } N o w ( P ∪ Q ) = { 0 , 1 , 2 } a n d ( P ∩ Q ) = { 1 , 2 } ∴ ( P ∪ Q ) × ( P ∩ Q ) = { ( 0 , 1 ) , ( 0 , 2 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) , ( 2 , 2 ) }

If A = {x: x ∈ W, x (i) A × (B ∩ C) (ii) A × (B ∪ C)

This is a short answer type question as classified in NCERT Exemplar G i v e n t h a t : A = { x : x ∈ W , x < 2 } ⇒ A = { 0 , 1 } B = { x : x ∈ N , 1 < x < 5 } ⇒ B = { 2 , 3 , 4 } C = { 3 , 5 } ( i ) A × ( B ∩ C ) = { 0 , 1 } × { 3 } = { ( 0 , 3 ) , ( 1 , 3 ) } ( i i ) A × ( B ∪ C ) = { 0 , 1 } × { 2 , 3 , 4 , 5 } = { ( 0 , 2 ) , ( 0 , 3 ) , ( 0 , 4 ) , ( 0 , 5 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 4 ) , ( 1 , 5 ) }

In each of the following cases, find a and b. (i) (2a + b, a – b) = (8, 3) (ii) a 4 , a - b = 0,6 + b a4,a-b=0,6+b

This is a short answer type question as classified in NCERT Exemplar ( i ) G i v e n t h a t : ( 2 a + b , a − b ) = ( 8 , 3 ) C o m p a r i n g t h e d o m a i n s a n d r a n g e s , w e g e t 2 a + b = 8 … ( i ) a − b = 3 … ( i i ) S o l v i n g ( i ) a n d ( i i ) w e g e t a = 1 1 3 a n d b = 2 3 ( i i ) G i v e n t h a t : ( a 4 , a − 2 b ) = ( 0 , 6 + b ) C o m p a r i n g t h e d o m a i n s a n d r a n g e s , w e g e t a 4 = 0 ⇒ a = 0 ⇒ a − 2 b = 6 + b ⇒ a − 3 b = 6 ⇒ 0 − 3 b = 6 ∴ b = − 2 S o , a = 0 , b = − 2 .

Given A = {1, 2, 3, 4, 5}, S = {(x, y): x ∈ A, y ∈ A}. Find the ordered pairs which satisfy the conditions given below: (i) x + y = 5 (ii) x + y (iii) x + y > 8

This is a short answer type question as classified in NCERT Exemplar G i v e n t h a t : A = { 1 , 2 , 3 , 4 , 5 } a n d S = { ( x , y ) : x ∈ A , y ∈ A } ( i ) x + y = 5 , s o , t h e o r d e r e d p a i r s s a t i s f y i n g t h e g i v e n c o n d i t i o n s a r e ( 1 , 4 ) , ( 4 , 1 ) , ( 2 , 3 ) , ( 3 , 2 ) . ( i i ) x + y 8 , s o , t h e o r d e r e d p a i r s s a t i s f y i n g t h e g i v e n c o n d i t i o n s a r e ( 4 , 5 ) , ( 5 , 4 ) , ( 5 , 5 ) .

Find the domain of each of the following functions given by (i) f ( x ) = 1 1 − c o s x (ii) f ( x ) = 1 x + | x | (iii) f ( x ) = | x | (iv) f ( x ) = x 3 − x + 3 x 2 − 1 (v) f ( x ) = 3 x 2 x − 8

This is a long answer type question as classified in NCERT Exemplar ( i ) G i v e n t h a t : f ( x ) = 1 1 − c o s x W e k n o w t h a t − 1 ≤ c o s x ≤ 1 ⇒ 1 ≥ − c o s x ≥ − 1 ⇒ 1 + 1 ≥ 1 − c o s x ≥ − 1 + 1 ⇒ 2 ≥ 1 − c o s x ≥ 0 ⇒ 0 ≤ 1 − c o s x ≤ 2 F o r r e a l v a l u e o f d o m a i n 1 − c o s x ≠ 0 ⇒ c o s x ≠ 1 ⇒ x ≠ 2 n π ∀ n ∈ Z H e n c e , t h e d o m a i n o f f = R − { 2 n π , n ∈ Z } ( i i ) G i v e n t h a t : f ( x ) = 1 x + | x | ? x + | x | = x + x = 2 x i f x ≥ 0 a n d x + | x | = x − x = 0 i f x < 0 S o f a r x < 0 , f i s n o t d e f i n e d . H e n c e , t h e d o m a i n f = R + . ( i i i ) G i v e n t h a t : f ( x ) = x | x | I t i s c l e a r t h a t f ( x ) i s d e f i n e d f o r a l l x ∈ R . H e n c e , t h e d o m a i n f = R . ( i v ) G i v e n t h a t : f ( x ) = x 3 − x + 3 x 2 − 1 H e r e , f ( x ) i s o n l y d e f i n e d i f x 2 &min

Find the range of the following functions given by (i) f ( x ) = 3 2 − x 2 (ii) f ( x ) = 1 − | x − 2 | (iii) f ( x ) = | x − 3 (iv) f ( x ) = 1 + 3 c o s 2 x (Hint:- 1 ≤ c o s 2 x ≤ 1 ⇒ – 3 ≤ 3 c o s 2 x ≤ 3 ⇒ – 2 ≤ 1 + 3 c o s 2 x ≤ 4 )

This is a long answer type question as classified in NCERT Exemplar ( i ) G i v e n t h a t : f ( x ) = 3 2 − x 2 L e t y = f ( x ) ∴ y = 3 2 − x 2 ⇒ y ( 2 − x 2 ) = 3 ⇒ 2 y − y x 2 = 3 ⇒ y x 2 = 2 y − 3 ⇒ x 2 = 2 y − 3 y ⇒ x = 2 y − 3 y H e r e , x i s r e a l i f 2 y − 3 ≥ 0 a n d y ≥ 0 ⇒ y ≥ 3 2 H e n c e , t h e r a n g e o f f = [ 3 2 , ∞ ) . ( i i ) G i v e n t h a t : f ( x ) = 1 − | x − 2 | W e k n o w t h a t | x − 2 | = − ( x − 2 ) i f x < 2 a n d | x − 2 | = ( x − 2 ) i f x ≥ 2 ∴ − | x − 2 | ≤ 0 ⇒ 1 − | x − 2 | ≤ 1 H e n c e , t h e r a n g e o f f = ( − ∞ , 1 ] . ( i i i ) G i v e n t h a t : f ( x ) = | x − 3 | W e k n o w t h a t | x − 3 | ≥ 0 ⇒ f ( x ) ≥ 0 H e n c e , t h e r a n g e o f f = [ 0 , ∞ ) ( i v ) G i v e n t h a t : f ( x ) = 1 + 3 c o s 2 x W e k n o w t h a t − 1 ≤ c o s 2 x ≤ 1 ⇒ − 3 ≤ 3 c o s 2 x ≤ 3 ⇒ − 3 + 1 ≤ 1 + 3 c o s 2 x ≤ 3 + 1 ⇒ − 2 ≤ 1 + 3 c o s 2 x ≤ 4 ⇒ − 2 ≤ f ( x ) ≤ 4 H e n c e , t h e r a n g e o f f = { − 2 , 4 } .

Redefine the function f ( x ) = | x − 2 | + | 2 + x | , − 3 ≤ x ≤ 3

This is a long answer type question as classified in NCERT Exemplar G i v e n t h a t : f ( x ) = | x − 2 | + | 2 + x | , − 3 ≤ x ≤ 3 Since |x−2|=−(x−2), x<2 a n d | x − 2 | = ( x − 2 ) , x ≥ 2 | 2 + x | = − ( 2 + x ) , x < − 2 a n d | 2 + x | = ( 2 + x ) , x ≥ − 2 N o w f ( x ) = | x − 2 | + | 2 + x | , − 3 ≤ x ≤ 3 = { − ( x − 2 ) − ( 2 + x ) , − 3 ≤ x < − 2 − ( x − 2 ) + ( 2 + x ) , − 2 ≤ x < 2 ( x − 2 ) + ( 2 + x ) , 2 ≤ x ≤ 3 ∴ f ( x ) = { − 2 x , − 3 ≤ x < − 2 4 , − 2 ≤ x < 2 2 x , 2 ≤ x ≤ 3

If f ( x ) = x − 1 x + 1 , then show that (i) f ( 1 x ) = − f ( x ) (ii) f ( 1 − x ) = − 1 f ( x )

This is a long answer type question as classified in NCERT Exemplar G i v e n t h a t : f ( x ) = x − 1 x + 1 ( i ) f ( 1 x ) = 1 x − 1 1 x + 1 = 1 − x 1 + x = − ( x − 1 ) x + 1 = − f ( x ) H e n c e , f ( 1 x ) = − f ( x ) ( i i ) f ( − 1 x ) = − 1 x − 1 − 1 x + 1 = − ( 1 x + 1 ) − ( 1 x − 1 ) = 1 + x 1 − x = 1 1 − x 1 + x = 1 − ( x − 1 x + 1 ) = − 1 f ( x ) H e n c e , f ( − 1 x ) = − 1 f ( x ) .

If P = { 1 , 2 } , then P × P × P = { ( 1 , 1 , 1 ) , ( 2 , 2 , 2 ) , ( 1 , 2 , 2 ) , ( 2 , 1 , 1 ) } .

This is a True or False Type Questions as classified in NCERT Exemplar G i v e n t h a t : P = { 1 , 2 } ∴ P × P = { 1 , 2 } × { 1 , 2 } = { ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) , ( 2 , 2 ) } P × P × P = { ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) , ( 2 , 2 ) } × { 1 , 2 } = { ( 1 , 1 , 1 ) , ( 1 , 1 , 2 ) , ( 1 , 2 , 1 ) , ( 1 , 2 , 2 ) , ( 2 , 1 , 1 ) , ( 2 , 1 , 2 ) , ( 2 , 2 , 1 ) , ( 2 , 2 , 2 ) } H e n c e , t h e s t a t e m e n t i s ' F a l s e ' .

If A = { 1 , 2 , 3 } , B = { 3 , 4 } , C = { 4 , 5 , 6 } , then ( A × B ) ∪ ( A × C ) = { ( 1 , 3 ) , ( 1 , 4 ) , ( 1 , 5 ) , ( 1 , 6 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 2 , 6 ) , ( 3 , 3 ) , ( 3 , 4 ) , ( 3 , 5 ) , ( 3 , 6 ) }

This is a True or False Type Questions as classified in NCERT Exemplar G i v e n t h a t : A = { 1 , 2 , 3 } , B = { 3 , 4 } a n d C = { 4 , 5 , 6 } ∴ A × B = { ( 1 , 3 ) , ( 1 , 4 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 3 , 3 ) , ( 3 , 4 ) } a n d A × C = { ( 1 , 4 ) , ( 1 , 5 ) , ( 1 , 6 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 2 , 6 ) , ( 3 , 4 ) , ( 3 , 5 ) , ( 3 , 6 ) } ( A × B ) ∪ ( A × C ) = { ( 1 , 3 ) , ( 1 , 4 ) , ( 1 , 5 ) , ( 1 , 6 ) , ( 2 , 3 ) , ( 2 , 4 ) , ( 2 , 5 ) , ( 2 , 6 ) , ( 3 , 3 ) , ( 3 , 4 ) , ( 3 , 5 ) , ( 3 , 6 ) } H e n c e , t h e s t a t e m e n t i s ' T r u e ' .

If ( x − 2 , y + 5 ) = ( − 2 , 1 3 ) are two equal ordered pairs, then x = 4 , y = 1 4 3 .

This is a True or False Type Questions as classified in NCERT Exemplar G i v e n t h a t : ( x − 2 , y + 5 ) = ( − 2 , 1 3 ) ⇒ x − 2 = − 2 ⇒ x = 0 a n d y + 5 = 1 3 ⇒ y = 1 3 − 5 = − 1 4 3 H e n c e , t h e s t a t e m e n t i s ' F a l s e ' .

If A × B = { ( a , x ) , ( a , y ) , ( b , x ) , ( b , y ) } , then A = { a , b } , B = { x , y } .

This is a True or False Type Questions as classified in NCERT Exemplar G i v e n t h a t : A = { a , b } , B = { x , y } ∴ A × B = { ( a , x ) , ( a , y ) , ( b , x ) , ( b , y ) } H e n c e , t h e s t a t e m e n t i s ' T r u e ' .

Let cr denote the binomial coefficient of xr in the expansion of (1 + x)10. If for ∈ R , α, β c1 + 3.2C2 + 5.3C3 + ….upto 10 terms = α × 2 1 1 2 β − 1 ( C 0 + C 1 2 + C 2 3 + . . . . u p t o 1 0 t e r m s ) then the value of a + b is equal to………..

∑ k = 1 1 0 ( 2 k − 1 ) . k . 1 0 C k = 2 ∑ k 2 1 0 C k − ∑ k . 1 0 C k x = 1 ⇒ ∑ k = 0 1 0 k . 1 0 C k = 1 0 . 2 9 x = 1 ⇒ ∑ k 2 1 0 C k = 9 0 . 2 8 + 1 0 . 2 9 = 2 8 ( 9 0 + 2 0 1 ) = 2 8 . 1 1 0 S = 29 . 110 – 10.29 = 29 . 100 S = 29 . 100 2 1 1 . 2 5 2 1 − 1

Let θ be the angle between the vectors a → a n d b → , where | a → | = 4 , | b → | = 3 and θ ∈ ( π 4 , π 3 ) . Then | ( a → − b → ) × ( a → + b → ) | 2 + 4 ( a → . b → ) 2 is equal to…………

| 2 ( a → × b → ) | 2 + 4 ( a → . b → ) 2 4 | a → | 2 | b → | 2 = 4 × 16 × 9 = 576

Let the abscissae of the two points P and Q be the roots of 2x2 – rx + p = 0 and the ordinates of P and Q be the roots of x2 – sx – q = 0. If the equation of the circle described on PQ as diameter is 2(x2 + y2) – 11x – 14y – 22 = 0, then 2r + s – 2q + is equal to…..

P ( x 1 , y 1 ) Q ( x 2 , y 2 ) x 1 + x 2 = r 2 , x 1 x 2 = P 2 y 1 + y 2 = s , y 1 , y 2 = − 9 ( x − x 1 ) ( x − x 2 ) + ( y − y 1 ) ( y − y 2 ) = 0 2 ( x 2 + y 2 ) − r x − 2 s y + p − 2 q = 0 r = 11, s = 7, p – 2q = -22

The number of values of x in the interval ( π 4 , 7 π 4 ) for which 14 cosec2x – 2sin2 x = 21 – 4cos2x holds, is………

Kindly consider the following answer π 4 < x < π 4 1 4 s i n 2 x − 2 s i n 2 x = 2 1 − 4 + 4 s i n 2 x t = − 1 7 ± 2 8 9 + 3 3 6 1 2 = − 1 7 ± 2 5 1 2 = 8 1 2 , − 4 2 1 2 = 2 3 , No. of solutions = 4

If log?(x+?) x² < log?(x+?) (2x+3) then value of x

x ∈ (-3/2, -1) U (-1, 3) - {0}

x ∈ [-3/2, -1) U (1, 3)

) x ∈ [-3/2, -1] U [-1, 3)

If α and β are the roots of the equation 2x(2x + 1) = 1, then β is equal to:

2α(α + 1)

2α(α – 1)

Let f: R → R be a function defined by f(x) = max{x, x²}. Let S denote the set of all points in R, where f is not differentiable. Then:

ϕ (an empty set)

{1}

Let z = x + iy be a non-zero complex number such that z² = i|z|², where i = √-1, then z lies on the:

line, y = x

line, y = -x

imaginary axis

Find minimum distance between the circle x² + (y-3)² = 1 and parabola y² = 4x

√2 - 1

√2

Let θ = π/5 and A = [cosθ sinθ; -sinθ cosθ]. If B = A + A?, then det(B) :

lies in (2,3).

For all twice differentiable functions f: R → R, with f(0) = f(1) = f'(0) = 0,

f''(x) ≠ 0, at every point x ∈ (0,1)

f''(x) = 0, for some x ∈ (0,1)

f''(0) = 0

f''(x) = 0, at every point x ∈ (0,1)

The centre of the circle passing through the point (0,1) and touching the parabola y = x² at the point (2,4) is:

(-16/5, 23/10)

(6/5, 53/10)

Maths NCERT Exemplar Solutions Class 11th Chapter Two: Overview, Questions, Preparation

Updated on Jul 23, 2025 17:36 IST

By Raj Pandey

Table of contents

Relations and Functions Short Answer Type Questions
Relations and Functions Long Answer Type Questions
Relations and Functions Objective Type Questions
Relations and Functions Fill in the blanks Type Questions
Relations and Function True or False Type Questions
JEE Mains 2022

Relations and Functions Long Answer Type Questions

1. Is $g = {(1, 1), (2, 3), (3, 5), (4, 7)}$ a function? Justify. If this is described by the relation, $g (x) = α x + β$ , then what values should be assigned to $α$ and $β$ ?

Sol.

\begin{array}{l} G i v e n t h a t : g = {(1, 1), (2, 3), (3, 5), (4, 7)} \\ Since e v e r y e l e m e n t o f t h e d o m a i n i n t h i s r e l a t i o n s h a s u n i q u e i m a g e, s o g i s a f u n c t i o n . \\ N o w g (x) = α x + β \\ F o r (1, 1) g (1) = α (1) + β = 1 \Rightarrow α + β = 1 \dots (i) \\ F o r (2, 3) g (2) = α (2) + β = 3 \Rightarrow 2 α + β = 3 \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i) w e h a v e \\ α = 2 a n d β = - 1 \\ H e n c e, t h e v a l u e o f α = 2 a n d β = - 1 . \end{array}

2. Find the domain of each of the following functions given by

(i) $f (x) = \frac{1}{\sqrt[]{1 - c o s x}}$

(ii) $f (x) = \frac{1}{\sqrt[]{x + | x |}}$

(iii) $f (x) = | x |$

(iv) $f (x) = \frac{x^{3} - x + 3}{x^{2} - 1}$

(v) $f (x) = \frac{3 x}{2 x - 8}$

Sol.

\begin{array}{l} (i) G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{1 - c o s x}} \\ W e k n o w t h a t - 1 \leq c o s x \leq 1 \\ \Rightarrow 1 \geq - c o s x \geq - 1 \\ \Rightarrow 1 + 1 \geq 1 - c o s x \geq - 1 + 1 \\ \Rightarrow 2 \geq 1 - c o s x \geq 0 \\ \Rightarrow 0 \leq 1 - c o s x \leq 2 \\ F o r r e a l v a l u e o f d o m a i n \\ 1 - c o s x \neq 0 \Rightarrow c o s x \neq 1 \\ \Rightarrow x \neq 2 n π \forall n \in Z \\ H e n c e, t h e d o m a i n o f f = R - {2 n π, n \in Z} \\ (i i) G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{x + | x |}} \\ ∵ x + | x | = x + x = 2 x i f x \geq 0 \\ a n d x + | x | = x - x = 0 i f x < 0 \\ S o f a r x < 0, f i s n o t d e f i n e d . \\ H e n c e, t h e d o m a i n f = R^{+} . \\ (i i i) G i v e n t h a t : f (x) = x | x | \\ I t i s c l e a r t h a t f (x) i s d e f i n e d f o r a l l x \in R . \\ H e n c e, t h e d o m a i n f = R . \\ (i v) G i v e n t h a t : f (x) = \frac{x^{3} - x + 3}{x^{2} - 1} \\ H e r e, f (x) i s o n l y d e f i n e d i f x^{2} - 1 \neq 0 \\ (x - 1) (x + 1) \neq 0 \\ ∴ x \neq 1, x \neq - 1 \\ H e n c e, t h e d o m a i n f = R - {- 1, 1} \\ (v) G i v e n t h a t : f (x) = \frac{3 x}{28 - x} \\ H e r e, f (x) i s o n l y d e f i n e d i f 28 - x \neq 0 \Rightarrow x \neq 28 \\ H e n c e, t h e d o m a i n f = R - {28} . \end{array}

Commonly asked questions

Is $g = {(1, 1), (2, 3), (3, 5), (4, 7)}$ a function? Justify. If this is described by the relation, $g (x) = α x + β$ , then what values should be assigned to $α$ and $β$ ?

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : g = {(1, 1), (2, 3), (3, 5), (4, 7)} \\ Since e v e r y e l e m e n t o f t h e d o m a i n i n t h i s r e l a t i o n s h a s u n i q u e i m a g e, s o g i s a f u n c t i o n . \\ N o w g (x) = α x + β \\ F o r (1, 1) g (1) = α (1) + β = 1 \Rightarrow α + β = 1 \dots (i) \\ F o r (2, 3) g (2) = α (2) + β = 3 \Rightarrow 2 α + β = 3 \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i) w e h a v e \\ α = 2 a n d β = - 1 \\ H e n c e, t h e v a l u e o f α = 2 a n d β = - 1 . \end{array}$

Find the domain of each of the following functions given by

(i) $f (x) = \frac{1}{\sqrt[]{1 - c o s x}}$

(ii) $f (x) = \frac{1}{\sqrt[]{x + | x |}}$

(iii) $f (x) = | x |$

(iv) $f (x) = \frac{x^{3} - x + 3}{x^{2} - 1}$

(v) $f (x) = \frac{3 x}{2 x - 8}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{1 - c o s x}} \\ W e k n o w t h a t - 1 \leq c o s x \leq 1 \\ \Rightarrow 1 \geq - c o s x \geq - 1 \\ \Rightarrow 1 + 1 \geq 1 - c o s x \geq - 1 + 1 \\ \Rightarrow 2 \geq 1 - c o s x \geq 0 \\ \Rightarrow 0 \leq 1 - c o s x \leq 2 \\ F o r r e a l v a l u e o f d o m a i n \\ 1 - c o s x \neq 0 \Rightarrow c o s x \neq 1 \\ \Rightarrow x \neq 2 n π \forall n \in Z \\ H e n c e, t h e d o m a i n o f f = R - {2 n π, n \in Z} \\ (i i) G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{x + | x |}} \\ ? x + | x | = x + x = 2 x i f x \geq 0 \\ a n d x + | x | = x - x = 0 i f x < 0 \\ S o f a r x < 0, f i s n o t d e f i n e d . \\ H e n c e, t h e d o m a i n f = R^{+} . \\ (i i i) G i v e n t h a t : f (x) = x | x | \\ I t i s c l e a r t h a t f (x) i s d e f i n e d f o r a l l x \in R . \\ H e n c e, t h e d o m a i n f = R . \\ (i v) G i v e n t h a t : f (x) = \frac{x^{3} - x + 3}{x^{2} - 1} \\ H e r e, f (x) i s o n l y d e f i n e d i f x^{2} &min \end{array}$

Find the range of the following functions given by

(i) $f (x) = \frac{3}{2 - x^{2}}$

(ii) $f (x) = 1 - | x - 2 |$

(iii) $f (x) = | x - 3$

(iv) $f (x) = 1 + 3 c o s 2 x$

(Hint:- $1 \leq c o s 2 x \leq 1 \Rightarrow - 3 \leq 3 c o s 2 x \leq 3 \Rightarrow - 2 \leq 1 + 3 c o s 2 x \leq 4)$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} (i) G i v e n t h a t : f (x) = \frac{3}{2 - x^{2}} \\ L e t y = f (x) ∴ y = \frac{3}{2 - x^{2}} \\ \Rightarrow y (2 - x^{2}) = 3 \Rightarrow 2 y - y x^{2} = 3 \\ \Rightarrow y x^{2} = 2 y - 3 \\ \Rightarrow x^{2} = \frac{2 y - 3}{y} \Rightarrow x = \sqrt[]{\frac{2 y - 3}{y}} \\ H e r e, x i s r e a l i f 2 y - 3 \geq 0 a n d y \geq 0 \\ \Rightarrow y \geq \frac{3}{2} \\ H e n c e, t h e r a n g e o f f = [\frac{3}{2}, \infty) . \\ (i i) G i v e n t h a t : f (x) = 1 - | x - 2 | \\ W e k n o w t h a t | x - 2 | = - (x - 2) i f x < 2 \\ a n d | x - 2 | = (x - 2) i f x \geq 2 \\ ∴ - | x - 2 | \leq 0 \Rightarrow 1 - | x - 2 | \leq 1 \\ H e n c e, t h e r a n g e o f f = (- \infty, 1] . \\ (i i i) G i v e n t h a t : f (x) = | x - 3 | \\ W e k n o w t h a t | x - 3 | \geq 0 \Rightarrow f (x) \geq 0 \\ H e n c e, t h e r a n g e o f f = [0, \infty) \\ (i v) G i v e n t h a t : f (x) = 1 + 3 c o s 2 x \\ W e k n o w t h a t - 1 \leq c o s 2 x \leq 1 \\ \Rightarrow - 3 \leq 3 c o s 2 x \leq 3 \Rightarrow - 3 + 1 \leq 1 + 3 c o s 2 x \leq 3 + 1 \\ \Rightarrow - 2 \leq 1 + 3 c o s 2 x \leq 4 \Rightarrow - 2 \leq f (x) \leq 4 \\ H e n c e, t h e r a n g e o f f = {- 2, 4} . \end{array}$

Redefine the function $f (x) = | x - 2 | + | 2 + x |, - 3 \leq x \leq 3$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = | x - 2 | + | 2 + x |, - 3 \leq x \leq 3 \\ Since | x - 2 | = - (x - 2), x < 2 \\ a n d | x - 2 | = (x - 2), x \geq 2 \\ | 2 + x | = - (2 + x), x < - 2 \\ a n d | 2 + x | = (2 + x), x \geq - 2 \\ N o w f (x) = | x - 2 | + | 2 + x |, - 3 \leq x \leq 3 \\ = {\begin{cases} - (x - 2) - (2 + x), - 3 \leq x < - 2 \\ - (x - 2) + (2 + x), - 2 \leq x < 2 \\ (x - 2) + (2 + x), 2 \leq x \leq 3 \end{cases} \\ ∴ f (x) = {\begin{cases} - 2 x, - 3 \leq x < - 2 \\ 4, - 2 \leq x < 2 \\ 2 x, 2 \leq x \leq 3 \end{cases} \end{array}$

If $f (x) = \frac{x - 1}{x + 1}$ , then show that

(i) $f (\frac{1}{x}) = - f (x)$

(ii) $f (\frac{1}{- x}) = \frac{- 1}{f (x)}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{x - 1}{x + 1} \\ (i) f (\frac{1}{x}) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1} = \frac{1 - x}{1 + x} = \frac{- (x - 1)}{x + 1} = - f (x) \\ H e n c e, f (\frac{1}{x}) = - f (x) \\ (i i) f (\frac{- 1}{x}) = \frac{- \frac{1}{x} - 1}{- \frac{1}{x} + 1} = \frac{- (\frac{1}{x} + 1)}{- (\frac{1}{x} - 1)} = \frac{1 + x}{1 - x} = \frac{1}{\frac{1 - x}{1 + x}} \\ = \frac{1}{- (\frac{x - 1}{x + 1})} = \frac{- 1}{f (x)} \\ H e n c e, f (- \frac{1}{x}) = \frac{- 1}{f (x)} . \end{array}$

Let $f (x) = \sqrt[]{x}$ and $g (x) = x$ be two functions defined in the domain $?^{+} \cup {0}$ .

Find

(i) $(f + g) (x)$

(ii) $(f - g) (x)$

(iii) $(f g) (x)$

(iv) $(\frac{f}{g}) (x)$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \sqrt[]{x} a n d g (x) = x b e t w o f u n c t i o n s d e f i n e d i n t h e d o m a i n R^{+} \cup {0} \\ (i) (f + g) (x) = f (x) + g (x) = \sqrt[]{x} + x \\ (i i) (f - g) (x) = f (x) - g (x) = \sqrt[]{x} - x \\ (i i i) (f g) (x) = f (x) . g (x) = \sqrt[]{x} . x = x^{3 / 2} \\ (i v) (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\sqrt[]{x}}{x} = \frac{1}{\sqrt[]{x}} . \end{array}$

Find the domain and range of the function $f (x) = \frac{1}{\sqrt[]{x - 5}}$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{x - 5}} \\ H e r e, i t i s c l e a r t h a t f (x) i s r e a l w h e n x - 5 > 0 \Rightarrow x > 5 \\ H e n c e, t h e d o m a i n = (5, \infty) \\ N o w t o f i n d t h e r a n g e p u t \\ f (x) = y = \frac{1}{\sqrt[]{x - 5}} \\ \Rightarrow \sqrt[]{x - 5} = \frac{1}{y} \Rightarrow x - 5 = \frac{1}{y^{2}} \\ \Rightarrow x = \frac{1}{y^{2}} + 5 \\ F o r x \in (5, \infty), y \in R^{+} . \\ H e n c e, t h e r a n g e o f f = R^{+} . \end{array}$

If $f (x) = y = \frac{a x - b}{c x - a}$ , then prove that $f (y) = x$ .

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = y = \frac{a x - b}{c x - a} \\ f (y) = \frac{a y - b}{c y - a} = \frac{a [\frac{a x - b}{c x - a}] - b}{c [\frac{a x - b}{c x - a}] - a} \\ \Rightarrow f (y) = \frac{a^{2} x - a b - b c x + a b}{c a x - b c - c a x + a^{2}} = \frac{a^{2} x - b c x}{a^{2} - b c} \\ \Rightarrow f (y) = \frac{(a^{2} - b c)}{a^{2} - b c} = x \\ H e n c e, f (y) = x . \end{array}$

Relations and Functions Objective Type Questions

1. Let $n (A) = m$ , and $n (B) = n$ . Then the total number of non-empty relations that can be defined from $A$ to $B$ is:

(a) $m^{n}$

(b) $n^{m} - 1$

(c) $m^{n} - 1$

(d) $2^{m n} - 1$

Sol.

\begin{array}{l} G i v e n t h a t : n (A) = m a n d n (B) = n \\ ∴ n (A \times B) = n (A) . n (B) = m n \\ S o, t h e t o t a l n u m b e r o f r e l a t i o n s f r o m A t o B = 2^{m n} - 1 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}

2. If ${[x]}^{2} - 5 [x] + 6 = 0$ , where [.] denotes the greatest integer function, then

(a) $x \in [3, 4]$

(b) $x \in (2, 3]$

(c) $x \in [2, 3]$

(d) $x \in [2, 4]$

Sol. $\begin{array}{l} W e h a v e {[x]}^{2} - 5 [x] + 6 = 0 \\ \Rightarrow {[x]}^{2} - 3 [x] - 2 [x] + 6 = 0 \\ \Rightarrow [x] ([x] - 3) - 2 ([x] - 3) = 0 \\ \Rightarrow ([x] - 3) ([x] - 2) = 0 \Rightarrow [x] = 2, 3 \\ S o, x \in [2, 3] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Commonly asked questions

Domain of $\sqrt[]{a^{2} - x^{2}}$ ( $a > 0$ ) is

(a) $(- a, a)$

(b) $[- a, a]$

(c) $[0, a]$

(d) $(- a, 0]$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t f (x) = \sqrt[]{a^{2} - x^{2}} \\ f (x) i s d e f i n e d i f a^{2} - x^{2} \geq 0 \\ \Rightarrow x^{2} - a^{2} \leq 0 \Rightarrow x^{2} \leq a^{2} \\ \Rightarrow x \leq \pm a \Rightarrow - a \leq x \leq a \\ ∴ D o m a i n o f f (x) = [- a, a] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Let $n (A) = m$ , and $n (B) = n$ . Then the total number of non-empty relations that can be defined from $A$ to $B$ is:

(a) $m^{n}$

(b) $n^{m} - 1$

(c) $m^{n} - 1$

(d) $2^{m n} - 1$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : n (A) = m a n d n (B) = n \\ ∴ n (A \times B) = n (A) . n (B) = m n \\ S o, t h e t o t a l n u m b e r o f r e l a t i o n s f r o m A t o B = 2^{m n} - 1 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

If ${[x]}^{2} - 5 [x] + 6 = 0$ , where [.] denotes the greatest integer function, then

(a) $x \in [3, 4]$

(b) $x \in (2, 3]$

(c) $x \in [2, 3]$

(d) $x \in [2, 4]$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e {[x]}^{2} - 5 [x] + 6 = 0 \\ \Rightarrow {[x]}^{2} - 3 [x] - 2 [x] + 6 = 0 \\ \Rightarrow [x] ([x] - 3) - 2 ([x] - 3) = 0 \\ \Rightarrow ([x] - 3) ([x] - 2) = 0 \Rightarrow [x] = 2, 3 \\ S o, x \in [2, 3] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Range of $f (x) = \frac{1}{1 - 2 c o s x}$ is

(a) $[\frac{1}{3}, 1]$

(b) $[- 1, \frac{1}{3}]$

(c) $(- \infty, - 1] \cup [\frac{1}{3}, \infty)$

(d) $[- \frac{1}{3}, 1]$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{1}{1 - 2 c o s x} \\ W e k n o w t h a t - 1 \leq c o s x \leq 1 \\ \Rightarrow 1 \geq c o s x \geq - 1 \Rightarrow - 1 \leq - c o s x \leq 1 \\ \Rightarrow - 2 \leq - 2 c o s x \leq 2 \Rightarrow - 2 + 1 \leq 1 - 2 c o s x \leq 2 + 1 \\ \Rightarrow - 1 \leq 1 - 2 c o s x \leq 3 \Rightarrow - 1 \leq \frac{1}{1 - 2 c o s x} \leq \frac{1}{3} \\ \Rightarrow - 1 \leq f (x) \leq \frac{1}{3} \\ S o, t h e r a n g e o f f (x) = [- 1, \frac{1}{3}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Let $f (x) = \sqrt[]{1 + x^{2}}$ then

(a) $f (x y) = f (x) \cdot f (y)$

(b) $f (x y) \geq f (x) \cdot f (y)$

(c) $f (x y) \leq f (x) \cdot f (y)$

(d) None of these

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \sqrt[]{1 + x^{2}} \Rightarrow f (x y) = \sqrt[]{1 + x^{2} y^{2}} \\ a n d f (x) . f (y) = \sqrt[]{1 + x^{2}} . \sqrt[]{1 + y^{2}} = \sqrt[]{1 + x^{2} + y^{2} + x^{2} y^{2}} \\ ? \sqrt[]{1 + x^{2} y^{2}} \leq \sqrt[]{1 + x^{2} + y^{2} + x^{2} y^{2}} \\ \Rightarrow f (x y) \leq f (x) . f (y) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

If $f (x) = a x + b$ , where $a$ and $b$ are integers, $f (- 1) = - 5$ and $f (3) = 3$ , then $a$ and $b$ are equal to

(a) $a = - 3, b = - 1$

(b) $a = 2, b = - 3$

(c) $a = 0, b = 2$

(d) $a = 2, b = 3$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = a x + b \\ \Rightarrow f (- 1) = a (- 1) + b \\ \Rightarrow - 5 = - a + b \\ \Rightarrow a - b = 5 \dots (i) \\ \Rightarrow f (3) = 3 a + b \\ \Rightarrow 3 = 3 a + b \\ \Rightarrow 3 a + b = 3 \dots (i i) \\ O n s o l v i n g e q n . (i) a n d (i i), w e g e t a = 2, b = - 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The domain of the function $f$ defined by $f (x) = \sqrt[]{4 - x} + \frac{2}{\sqrt[]{x^{2}} - 1}$ is equal to

(a) $(- \infty, - 1) \cup (1, 4]$

(b) $(- \infty, - 1] \cup (1, 4]$

(c) $(- \infty, - 1) \cup [1, 4]$

(d) $(- \infty, - 1) \cup [1, 4)$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \sqrt[]{4 - x} + \frac{1}{\sqrt[]{x^{2} - 1}} \\ f (x) i s d e f i n e d i f 4 - x \geq 0 o r x^{2} - 1 > 0 \\ \Rightarrow - x \geq - 4 o r (x - 1) (x + 1) > 0 \\ \Rightarrow x \leq 4 o r x < - 1 a n d x > 1 \\ ∴ D o m a i n o f f (x) = (- \infty, - 1) \cup (1, 4] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The domain and range of the real function $f$ defined by $f (x) = \frac{x - 4}{x - 4}$ is given by

(a) Domain = $?$ , Range = {-1, 1}

(b) Domain = $? - {1}$ , Range = $? - {1}$

(c) Domain = $? - {4}$ , Range = {-1}

(d) Domain = $? - {- 4}$ , Range = {-1, 1}.

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{4 - x}{x - 4} \\ W e k n o w t h a t f (x) i s d e f i n e d i f x - 4 \neq 0 \Rightarrow x \neq 4 \\ S o, t h e d o m a i n o f f (x) i s = R - {4} \\ L e t f (x) = y = \frac{4 - x}{x - 4} \\ \Rightarrow y x - 4 y = 4 - x \Rightarrow y x + x = 4 y + 4 \\ \Rightarrow x (y + 1) = 4 y + 4 \Rightarrow x = \frac{4 (1 + y)}{1 + y} \\ I f x i s r e a l n u m b e r, t h e n 1 + y \neq 0 \Rightarrow y \neq - 1 \\ ∴ R a n g e o f f (x) = R - {- 1} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The domain and range of real function $f$ defined by $f (x) = \sqrt[]{x - 1}$ is given by

(a) Domain = $(1, \infty)$ , Range = $(0, \infty)$

(b) Domain = $[1, \infty)$ , Range = $(0, \infty)$

(c) Domain = $[1, \infty)$ , Range = $[0, \infty)$

(d) Domain = $[1, \infty)$ , Range = $[0, \infty)$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \sqrt[]{x - 1} \\ f (x) i s d e f i n e d i f x - 1 \geq 0 \Rightarrow x \geq 1 \\ ∴ D o m a i n o f f (x) = [0, \infty) \\ L e t f (x) = y = \sqrt[]{x - 1} \\ \Rightarrow y^{2} = x - 1 \Rightarrow x = y^{2} + 1 \\ I f x i s r e a l n u m b e r, t h e n y \in R \\ ∴ R a n g e o f f (x) = [0, \infty) \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The domain $f$ of the function given by $f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - x - 6}$ is

(a) $? - {3, - 2}$

(b) $? - {- 3, 2}$

(c) $? - [3, - 2]$

(d) $? - (3, - 2)$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{x^{2} + 2 x + 1}{x^{2} - x - 6} \\ f (x) i s d e f i n e d i f x^{2} - x - 6 \neq 0 \\ \Rightarrow x^{2} - 3 x + 2 x - 6 \neq 0 \\ \Rightarrow (x - 3) (x + 2) \neq 0 \Rightarrow x \neq - 2, x \neq 3 \\ S o, t h e d o m a i n o f f (x) = R - {- 2, 3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The domain and range of the function $f$ given by $f (x) = 2 - \sqrt[]{x - 5}$ is

(a) Domain = $?^{+}$ , Range = $(- \infty, 1]$

(b) Domain = $?,$ Range = $(- \infty, 2]$

(c) Domain = $?,$ Range = $(- \infty, 2)$

(d) Domain = $?^{+}$ , Range = $(- \infty, 2]$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = 2 - | x - 5 | \\ f (x) i s d e f i n e d f o r x \in R \\ ∴ D o m a i n o f f (x) = R \\ N o w, | x - 5 | \geq 0 \Rightarrow - | x - 5 | \leq 0 \\ \Rightarrow 2 - | x - 5 | \leq 2 \Rightarrow f (x) \leq 2 \\ ∴ R a n g e o f f (x) = (- \infty, 2] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The domain for which the functions defined by $f (x) = 3 x^{2} - 1$ and $g (x) = 3 + x$ are equal is

(a) $\{- 1, \frac{4}{3}\}$

(b) $\{- 1, \frac{4}{3}\}$

(c) $\{- 1, \frac{4}{3}\}$

(d) $\{- 1, \frac{4}{3}\}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = 3 x^{2} - 1 a n d g (x) = 3 + x \\ f (x) = g (x) \\ \Rightarrow 3 x^{2} - 1 = 3 + x \\ \Rightarrow 3 x^{2} - x - 4 = 0 \Rightarrow 3 x^{2} - 4 x + 3 x - 4 = 0 \\ \Rightarrow x (3 x - 4) + 1 (3 x - 4) = 0 \Rightarrow (x + 1) (3 x - 4) = 0 \\ \Rightarrow x + 1 = 0 o r 3 x - 4 = 0 \\ \Rightarrow x = - 1 o r x = \frac{4}{3} \\ ∴ D o m a i n = {- 1, \frac{4}{3}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Relations and Functions Fill in the blanks Type Questions

1. Let f and g be two real functions given by $f = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)}$ $g = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)}$ then the domain of $f \circ g$ is given by _________.

Sol.

\begin{array}{l} G i v e n t h a t : f (x) = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)} \\ a n d g (x) = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)} \\ ∴ d o m a i n o f f = {0, 2, 3, 4, 5} \\ a n d d o m a i n o f g = {1, 2, 3, 4, 5} \\ S o, d o m a i n o f f . g = D o m a i n o f f \cap D o m a i n o f g \\ = {2, 3, 4, 5} \\ H e n c e, t h e f i l l e r i s {2, 3, 4, 5} . \end{array}

2. Let $f = {(2, 4), (5, 6), (8, - 1), (10, - 3)}$ $g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)$ be two real functions. Then Match the following:

(a) $f - g$ (i) ${(2, \frac{4}{5})} (8, \frac{- 1}{4}) (10, \frac{- 3}{13})$

(b) $f + g$ (ii) ${(2, 20), (8, - 4), (10, - 39)}$

(c) $f \cdot g$ (iii) ${(2, - 1), (8, - 5), (10, - 16)}$

(d) $\frac{f}{g}$ (iv) ${(2, 9), (8, 3), (10, 10)}$

Sol.

\begin{array}{l} G i v e n t h a t : f = {(2, 4), (5, 6), (8, - 1), (10, - 3)} \\ a n d g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)} \\ f - g, f + g, f . g, \frac{f}{g} a r e d e f i n e d i n t h e d o m a i n \\ (d o m a i n o f f \cap d o m a i n o f g) \\ \Rightarrow {2, 5, 8, 10} \cap {2, 7, 8, 10, 11} \Rightarrow {2, 8, 10} \\ (i) (f - g) 2 = f (2) - g (2) = 4 - 5 = - 1 \\ (f - g) 8 = f (8) - g (8) = - 1 - 4 = - 5 \\ (f - g) 10 = f (10) - g (10) = - 3 - 13 = - 16 \\ ∴ (f - g) = {(2, - 1), (8, - 5), (10, - 16)} \\ (i i) (f + g) 2 = f (2) + g (2) = 4 + 5 = 9 \\ (f + g) 8 = f (8) + g (8) = - 1 + 4 = 3 \\ (f + g) 10 = f (10) + g (10) = - 3 + 13 = 10 \\ ∴ (f + g) = {(2, 9), (8, 3), (10, 10)} \\ (i i i) (f . g) 2 = f (2) . g (2) = 4.5 = 20 \\ (f . g) 8 = f (8) . g (8) = (- 1) . 4 = - 4 \\ (f . g) 10 = f (10) . g (10) = - 3.13 = - 39 \\ ∴ (f . g) = {(2, 20), (8, - 4), (10, - 39)} \\ (i v) (\frac{f}{g}) (2) = \frac{f (2)}{g (2)} = \frac{4}{5} \\ (\frac{f}{g}) (8) = \frac{f (8)}{g (8)} = \frac{- 1}{4} \\ (\frac{f}{g}) (10) = \frac{f (10)}{g (10)} = \frac{- 3}{13} \\ (\frac{f}{g}) = {(2, \frac{4}{5}), (8, \frac{- 1}{4}), (10, \frac{- 3}{13})} \\ H e n c e, t h e c o r r e c t o p t i o n i s \\ (a) \leftrightarrow (i i i), (b) \leftrightarrow (i v), (c) \leftrightarrow (i i), (d) \leftrightarrow (i) \end{array}

Commonly asked questions

Let f and g be two real functions given by $f = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)}$ $g = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)}$ then the domain of $f ? g$ is given by _________.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)} \\ a n d g (x) = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)} \\ ∴ d o m a i n o f f = {0, 2, 3, 4, 5} \\ a n d d o m a i n o f g = {1, 2, 3, 4, 5} \\ S o, d o m a i n o f f . g = D o m a i n o f f \cap D o m a i n o f g \\ = {2, 3, 4, 5} \\ H e n c e, t h e f i l l e r i s {2, 3, 4, 5} . \end{array}$

Let $f = {(2, 4), (5, 6), (8, - 1), (10, - 3)}$ $g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)$ be two real functions. Then Match the following:

(a) $f - g$ (i) ${(2, \frac{4}{5})} (8, \frac{- 1}{4}) (10, \frac{- 3}{13})$

(b) $f + g$ (ii) ${(2, 20), (8, - 4), (10, - 39)}$

(c) $f \cdot g$ (iii) ${(2, - 1), (8, - 5), (10, - 16)}$

(d) $\frac{f}{g}$ (iv) ${(2, 9), (8, 3), (10, 10)}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f = {(2, 4), (5, 6), (8, - 1), (10, - 3)} \\ a n d g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)} \\ f - g, f + g, f . g, \frac{f}{g} a r e d e f i n e d i n t h e d o m a i n \\ (d o m a i n o f f \cap d o m a i n o f g) \\ \Rightarrow {2, 5, 8, 10} \cap {2, 7, 8, 10, 11} \Rightarrow {2, 8, 10} \\ (i) (f - g) 2 = f (2) - g (2) = 4 - 5 = - 1 \\ (f - g) 8 = f (8) - g (8) = - 1 - 4 = - 5 \\ (f - g) 10 = f (10) - g (10) = - 3 - 13 = - 16 \\ ∴ (f - g) = {(2, - 1), (8, - 5), (10, - 16)} \\ (i i) (f + g) 2 = f (2) + g (2) = 4 + 5 = 9 \\ (f + g) 8 = f (8) + g (8) = - 1 + 4 = 3 \\ (f + g) 10 = f (10) + g (10) = - 3 + 13 = 10 \\ ∴ (f + g) = {(2, 9), (8, 3), (10, 10)} \\ (i i i) (f . g) 2 = f (2) . g (2) = 4.5 = 20 \\ (f . g) 8 = f (8) . g (8) = (- 1) . 4 = - 4 \\ (f . g) 10 = f (10) . g (10) = - 3.13 = - 39 \\ ∴ (f . g) = {(2, 20), (8, - 4), (10, - 39)} \\ (i v) (\frac{f}{g}) (2) = \frac{f (2)}{g (2)} = \frac{4}{5} \\ (\frac{f}{g}) (8) = \frac{f (8)}{g (8)} = \frac{- 1}{4} \\ (\frac{f}{g}) (10) = \frac{f (10)}{g (10)} = \frac{- 3}{13} \\ (\frac{f}{g}) = {(2, \frac{4}{5}), (8, \frac{- 1}{4}), (10, \frac{- 3}{13})} \\ H e n c e, t h e c o r r e c t o p t i o n i s \\ (a) \leftrightarrow (i i i), (b) \leftrightarrow (i v), (c) \leftrightarrow (i i), (d) \leftrightarrow (i) \end{array}$

Relations and Function True or False Type Questions

1. The ordered pair $(5, 2)$ belongs to the relation $R = {(x, y) : y = x - 5, x, y \in ℤ}$ .

Sol.

\begin{array}{l} G i v e n t h a t : R = {(x, y) : y = x - 5, x, y \in Z} \\ F o r (5, 2), y = x - 5 \\ P u t x = 5, y = 5 - 5 = 0 \neq 2 \\ S o, (5, 2) i s n o t t h e o r d e r e d p a i r o f R . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}

2. If $P = {1, 2}$ , then $P \times P \times P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}$ .

Sol.

\begin{array}{l} G i v e n t h a t : P = {1, 2} \\ ∴ P \times P = {1, 2} \times {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} \\ P \times P \times P = {(1, 1), (1, 2), (2, 1), (2, 2)} \times {1, 2} \\ = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)} \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}

Commonly asked questions

The ordered pair $(5, 2)$ belongs to the relation $R = {(x, y) : y = x - 5, x, y \in ?}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : R = {(x, y) : y = x - 5, x, y \in Z} \\ F o r (5, 2), y = x - 5 \\ P u t x = 5, y = 5 - 5 = 0 \neq 2 \\ S o, (5, 2) i s n o t t h e o r d e r e d p a i r o f R . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

If $P = {1, 2}$ , then $P \times P \times P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : P = {1, 2} \\ ∴ P \times P = {1, 2} \times {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)} \\ P \times P \times P = {(1, 1), (1, 2), (2, 1), (2, 2)} \times {1, 2} \\ = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)} \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

If $A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}$ , then $(A \times B) \cup (A \times C)$ $= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = {1, 2, 3}, B = {3, 4} a n d C = {4, 5, 6} \\ ∴ A \times B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} \\ a n d A \times C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} \\ (A \times B) \cup (A \times C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)} \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

If $(x - 2, y + 5) = (- 2, \frac{1}{3})$ are two equal ordered pairs, then $x = 4$ , $y = \frac{14}{3}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : (x - 2, y + 5) = (- 2, \frac{1}{3}) \\ \Rightarrow x - 2 = - 2 \Rightarrow x = 0 \\ a n d y + 5 = \frac{1}{3} \Rightarrow y = \frac{1}{3} - 5 = - \frac{14}{3} \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

If $A \times B = {(a, x), (a, y), (b, x), (b, y)}$ , then $A = {a, b}$ , $B = {x, y}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = {a, b}, B = {x, y} \\ ∴ A \times B = {(a, x), (a, y), (b, x), (b, y)} \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

JEE Mains 2022

Commonly asked questions

Let c_r denote the binomial coefficient of x^r in the expansion of (1 + x)¹⁰. If for $\in R,$ α, β c₁ + 3.2C₂ + 5.3C₃ + ….upto 10 terms = $\frac{α \times 2^{11}}{2^{β} - 1} (C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + . . . . u p t o 10 t e r m s)$ then the value of a + b is equal to………..

$\sum_{k = 1}^{10} (2 k - 1) . k .^{10} C_{k}$

$= 2 \sum {k^{2}}^{10} C_{k} - \sum k .^{10} C_{k}$

$x = 1 \Rightarrow \sum_{k = 0}^{10} k .^{10} C_{k} = 10 . 2^{9}$

$x = 1 \Rightarrow \sum k^{2}^{10} C_{k} = 90 . 2^{8} + 10 . 2^{9} = 2^{8} (90 + 201) = 2^{8} . 110$

S = 2⁹ . 110 – 10.2⁹ = 2⁹ . 100

S = 2⁹ . 100

$\frac{2^{11} . 25}{2^{1} - 1}$

The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is……….

$x_{1} + x_{2} = 6, 13 \to 6 + 6 = 12$

$\begin{array}{l} x_{1} + x_{2} = 4, 11, 18 \to 4 + 8 + 1 = 13 \\ x_{1} + x_{2} = 2, 9, 16 \to 2 + 9 + 3 = 14 \\ x_{1} + x_{2} = 7, 14 \to 7 + 5 = 12 \\ x_{1} + x_{2} = 5, 12 \to 5 + 7 = 12 \end{array}$

=63

Let θ be the angle between the vectors $\vec{a} a n d \vec{b},$ where $| \vec{a} | = 4, | \vec{b} | = 3$ and $θ \in (\frac{π}{4}, \frac{π}{3})$ . Then ${| (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) |}^{2} + 4 {(\vec{a} . \vec{b})}^{2}$ is equal to…………

${| 2 (\vec{a} \times \vec{b}) |}^{2} + 4 {(\vec{a} . \vec{b})}^{2}$

$4 {| \vec{a} |}^{2} {| \vec{b} |}^{2}$

= 4 × 16 × 9 = 576

Let the abscissae of the two points P and Q be the roots of 2x² – rx + p = 0 and the ordinates of P and Q be the roots of x² – sx – q = 0. If the equation of the circle described on PQ as diameter is 2(x² + y²) – 11x – 14y – 22 = 0, then 2r + s – 2q + is equal to…..

$P (x_{1}, y_{1})$

$Q (x_{2}, y_{2})$

$x_{1} + x_{2} = \frac{r}{2}, x_{1} x_{2} = \frac{P}{2}$

$y_{1} + y_{2} = s, y_{1}, y_{2} = - 9$

$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

$2 (x^{2} + y^{2}) - r x - 2 s y + p - 2 q = 0$

r = 11, s = 7, p – 2q = -22

The number of values of x in the interval $(\frac{π}{4}, \frac{7 π}{4})$ for which 14 cosec²x – 2sin² x = 21 – 4cos²x holds, is………

Kindly consider the following answer

$\frac{π}{4} < x < \frac{π}{4}$

$\frac{14}{s i n^{2} x} - 2 s i n^{2} x = 21 - 4 + 4 s i n^{2} x$

$t = \frac{- 17 \pm \sqrt[]{289 + 336}}{12}$

$= \frac{- 17 \pm 25}{12} = \frac{8}{12}, \frac{- 42}{12}$

$= \frac{2}{3},$

No. of solutions = 4

For a natural number n, let a_n = 19ⁿ – 12ⁿ. Then, the value of $\frac{31 α_{9} - α_{10}}{57 α_{8}} i s . . . . . . .$

$α_{n} = 1 9^{n} - 1 2^{n}$

$\frac{31 α_{9} - α_{10}}{57 . α_{2}} = \frac{31 (1 9^{9} - 1 2^{9}) - (1 9^{10} - 1 2^{10})}{57 (1 9^{8} - 1 2^{8})}$

$= \frac{1 9^{9} (31 - 19) - 1 2^{9} (31 - 12)}{57 (1 9^{8} - 1 2^{8})}$

$= \frac{1 9^{9} . 12 - 1 2^{9} . 19}{57 (1 9^{8} - 1 2^{8})} = 4$

Let f : R -> R be a function defined by f(x) $\Rightarrow {(2 (1 - \frac{x^{25}}{2}) (2 + x^{25}))}^{\frac{1}{50}} .$ If the function g(x) = $f (f (f (x))) + f (f (x)),$ then the greatest integer less than or equal to g(1) is……..

g(1) = ?

$= f (f (f (1))) + f (f (1))$

$f (1) = {(2 (1 - \frac{1}{2}) (2 + 1))}^{\frac{1}{50}} = 3^{\frac{1}{50}}$

$3^{\frac{1}{50}} + 1 3^{\frac{1}{50}} > 1^{\frac{1}{50}}$

3 > 1

$3^{\frac{1}{50}} > 2$ $2 > 3^{\frac{1}{50}} > 1$

$3 < 2^{50}$ $[3^{\frac{1}{50}} + 1] = 1 + 1 = 2$

Let the lines

L₁: $\vec{r} = λ (\vec{i} + 2 \hat{j} + 3 \hat{k}), λ \in R$

L₂ : $\vec{r} = (\hat{i} + 3 \hat{j} + \hat{k}) + μ (\hat{i} + \hat{j} + \hat{5}); μ \in R,$

intersects at the point. S. If a plane ax + by – z + d = 0 passes through S and is parallel to both lines L₁ and L₂, then the value of a + b + d is equal to…………

$λ = 1 + μ$

$\begin{array}{l} \underline{2 λ = 3 λ μ} \\ λ = 2 \to S (2, 4, 6) \end{array}$

$μ = 1$

$\vec{n} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 1 & 5 \end{matrix} | = (7, - 2, - 1)$

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

Let A be a 3 × 3 matrix having entries from the set {-1, 0, 1}. The number of all such matrices A having sum of all the entries equal to 5, is……….

$[\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] a_{1}, b_{1}, c_{1} t {- 1, 0, 1}$

$\underset{9 t i m e s}{\underset{?}{a_{1} + a_{2} + a_{3} + . . . .}} = 5$

Total = 414

The greatest integer less than or equal to the sum of first 100 terms of the sequence

$\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, . . . . . . . . . . . . i s e q u a l t o . . . . . . . . . . . .$

$t_{n} = \frac{3^{n} - 2^{n}}{3^{n}} = 1 - {(\frac{2}{3})}^{n}$

$S_{100} = 100 - \frac{\frac{2}{3} ({(\frac{2}{3})}^{100} - 1)}{\frac{2}{3} - 1} = 100 + 2 . \frac{(2^{100} - 3^{100})}{3^{100}}$

$= 100 - 2 + \frac{2^{101}}{3^{100}} = 98 + {(\frac{2}{3})}^{100} . 2 < 98 + 1$

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exam

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Maths NCERT Exemplar Solutions Class 11th Chapter Two: Overview, Questions, Preparation

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