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To find: g o f

Formula used: g o f = g(f(x))

Given: (i) f = {(1, 2), (3, 5), (4, 1)}

(ii) g = {(1, 3), (2, 3), (5, 1)}

We have,

gof(1) = g(f(1)) = g(2) = 3

gof(3) = g(f(3)) = g(5) = 1

gof(4) = g(f(4)) = g(1) = 3

g o f = {(1, 3), (3, 1), (4, 3)}

(i) f: R → R such that f(x) = x² 

Since the value of x is squared, f(x) will always be equal or greater than 0. 

∴ the range is [0, ∞) 

(ii) g: R → R such that g(x) = x² + 1 

Since, the value of x is squared and also adding with 1, g(x) will always be equal or greater than 1. 

∴ Range of g(x) = [1, ∞) 

(iii) h: R → R such that h(x) = sin x 

We know that, sin (x) always lies between -1 to 1 

∴ Range of h(x) = (-1, 1)

To find: f o f

Formula used: f o f = f(f(x))

Given: (i) f = {(1, 4), (2, 1) (3, 3), (4, 2)}

We have,

fof(1) = f(f(1)) = f(4) = 2

fof(2) = f(f(2)) = f(1) = 4

fof(3) = f(f(3)) = f(3) = 3

fof(4) = f(f(4)) = f(2) = 1

f o f = {(1, 2), (2, 4), (3, 3), (4, 1)}

It is given that f : R → R and g : C → C 

Thus, Domain (f) = R and Domain (g) = C 

We know that, Real numbers ≠ Complex Number 

∵, Domain (f) ≠ Domain (g) 

∴ f(x) and g(x) are not equal functions 

∴ f ≠ g

Answer: (b) = one - one but not onto 

 ∀  (x, y) ∈ (0, ∞) 

f (x) = f (y) ⇒ \(\frac{x}{1+x}\) = \(\frac{y}{1+y}\) 

⇒ x + xy = y + xy 

⇒ x = y 

⇒ f is one-one 

Let z = f (x) = \(\frac{x}{1+x}\) 

⇒ z + zx = x

⇒ z = x (1 – z) ⇒ x = \(\frac{z}{1-z}\) 

⇒ x is not defined for z = 1 and 1 ∈ (0, ∞) 

∴  1 ∈ (0, ∞) has no pre-image in (0, ∞) 

⇒ f is not onto

To find: f{f(x)}

Formula used: (i) f o f = f(f(x))

Given: (i) f : R → R : f(x) = 3x + 2

We have,

f{f(x)} = f(f(x)) = f(3x + 2)

f o f =3(3x + 2) + 2

= 9x + 6 + 2

= 9x + 8

f{f(x)} = 9x + 8

(i) g o f

To find: g o f

Formula used: g o f = g(f(x))

Given: f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3),(4, 9), (5, 9)}

Solution: We have,

gof(3) = g(f(3)) = g(1) = 3

gof(9) = g(f(9)) = g(3) = 3

gof(12) = g(f(12)) = g(4) = 9

g o f = {(3, 3), (9, 3), (12, 9)}

(ii) f o g

To find: f o g

Formula used: f o g = f(g(x))

Given: f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3),(4, 9), (5, 9)}

Solution: We have,

fog(1) = f(g(1)) = f(3) = 1

fog(3) = f(g(3)) = f(3) = 1

fog(4) = f(g(4)) = f(9) = 3

fog(5) = f(g(5)) = f(9) = 3

f o g = {(1, 1), (3, 1), (4, 3), (5, 3)}

Answer: (c) = f is one - one into 

 ∀ (x, y) ∈ R, f (x) = f (y) 

⇒ \(\frac{x-m}{x-n}\) = \(\frac{y-m}{y-n}\) 

⇒ (x – m) (y – n) = (y – m) (x – n) 

⇒ xy – my – nx + mn = yx – mx – ny + mn 

⇒ mx – nx = my – ny 

⇒ (m – n) x = (m – n) y 

⇒ x = y 

⇒ f is one-one.

Let z = f (x) = \(\frac{x-m}{x-n}\)  

⇒ zx – zn = x – m 

⇒ zx – x = zn – m 

⇒ x (z – 1) = zn – m

⇒ x = \(\frac{zn-m}{z-1} = \frac{m-zn}{1-z}\)  

x is not defined for z = 1 

⇒ for z = 1, there exists no pre-image in R 

⇒ f is not onto. 

∴  f is one-one, into function.

To find: f o f

Formula used: (i) f o f = f(f(x))

Given: (i) f : R → R : f(x) = (3 - x³)^1/3

We have,

f o f = f(f(x)) =(3 - x³)^1/3

f o f ={3 -{(3 - x³)^1/3}]^1/3

= [3 - (3 - x³)]^1/3

= [3 - 3 + x³]^1/3

= [x³]^1/3

= x

f o f (x) = x

To prove: f o f = f 

Formula used: f o f = f(f(x)) 

Given: (i) f : R → R : f(x) = |x| 

Solution: We have, 

f o f = f(f(x)) = f(|x|) =||x||= |x| = f(x) 

Clearly f o f = f. 

Hence Proved.

To prove: (g o f) ≠ (f o g)

Formula used: (i) g o f = g(f(x))

(ii) f o g = f(g(x))

Given: (i) f : R → R : f(x) = x²

(ii) g : R → R : g(x) = (x + 1)

Proof: We have,

g o f = g(f(x)) = g(x²) = ( x² + 1 )

f o g = f(g(x)) = g(x+1) = [ (x+1)² + 1 ] = x² + 2x + 2

From the above two equation we can say that (g o f) ≠ (f o g)

Hence Proved

To prove: g o f ≠ f o g

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : R → R : f(x) = x²

(ii)g : R → R g(x) = (x + 1)

We have,

f o g = f(g(x)) = f(x + 7)

f o g = (x + 7)² = x ² + 14x + 49

g o f = g(f(x)) = g(x²)

g o f = (x² + 1) = x² + 1

Clearly g o f ≠ f o g

Hence Proved

Given, 

f : {4, 5, 6} → {p, q, r} such that 

f(4) = p, f (5) = q, f (6) = r 

⇒ f = {4, p), (5, q), (6, r)} (f defined by ordered pairs)

⇒ f is one-one and onto 

⇒ f ^–1 exists 

⇒ f ^–1 = {(p, 4), (q, 5), (r, 6)} 

(∵ components of ordered pains are interchanged in case of inverse functions) 

⇒ (f ^–1) ^–1 = {(4, p), (5, q), (6, r)} 

= f

(i) g o f 

To find: g o f 

Formula used: g o f = g(f(x)) 

Given: (i) f : R → R : f(x) = (x² + 3x + 1) 

(ii) g: R → R : g(x) = (2x - 3) 

Solution: We have, 

g o f = g(f(x)) = g(x² + 3x + 1) = [ 2(x² + 3x + 1) – 3 ] 

⇒ 2x² + 6x + 2 – 3 

⇒ 2x² + 6x – 1 

g o f (x) = 2x² + 6x – 1 

(ii) f o g 

To find: f o g 

Formula used: f o g = f(g(x)) 

Given: (i) f : R → R : f(x) = (x² + 3x + 1) 

(ii) g: R → R : g(x) = (2x - 3) 

Solution: We have, 

f o g = f(g(x)) = f(2x - 3) = [ (2x - 3)² + 3(2x – 3) + 1 ] 

⇒ 4x² - 12x + 9 + 6x – 9 + 1 

⇒ 4x² - 6x + 1 

f o g (x) = 4x² - 6x + 1 

(iii) g o g 

To find: g o g 

Formula used: g o g = g(g(x)) 

Given: (i) g: R → R : g(x) = (2x - 3) 

Solution: We have, 

g o g = g(g(x)) = g(2x - 3) = [ 2(2x – 3) - 3 ] 

⇒ 4x – 6 - 3 

⇒ 4x - 9 

g o g (x) = 4x – 9

To prove: function is one-one and onto

Given: f: R → R : f(x)= 2x

We have,

f(x) = 2x

For, f(x₁) = f(x₂)

⇒ 2x₁ = 2x₂

⇒ x₁ = x₂

When, f(x₁) = f(x₂) then x₁ = x₂

∴ f(x) is one-one

f(x) = 2x

Let f(x) = y such that \(y\in R\)

⇒ y = 2x

\(\Rightarrow x=\frac{y}{2}\)

Since \(y\in R\)

\(\Rightarrow \frac{y}{2}\in R\)

⇒ x will also be a real number, which means that every value of y is associated with some x

∴ f(x) is onto

Hence Proved

Given function f: R → R given by f(x) = 4x + 3

Let us show that the given function is invertible.

Consider injection of f:

Let x and y be any two elements of domain (R),

Such that f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

Therefore, f is one-one.

Now surjection of f:                 

Let y be in the co-domain (R),

Such that f(x) = y.

⇒ 4x + 3 = y 

⇒ 4x = y - 3

⇒ x = (y - 3)/4 in R (domain)

⇒ f is onto.

Therefore, f is a bijection and, hence, is invertible.

Let us find f ^-1

Let f^-1(x) = y ...(1)

⇒ x = f(y)

⇒ x = 4y + 3

⇒ x − 3 = 4y

⇒ y = (x - 3)/4

Now put these values in 1 we get

Therefore, f^-1(x) = (x - 3)/4 

(i) g o f

To find: g o f

Formula used: g o f = g(f(x))

Given: (i) f : R → R : f(x) = (2x + 1)

(ii) g : R → R : g(x) = (x² - 2)

Solution: We have,

g o f = g(f(x)) = g(2x + 1) = [ (2x + 1)² – 2 ]

⇒ 4x² + 4x + 1 – 2

⇒ 4x² + 4x – 1

g o f (x) = 4x² + 4x – 1

(ii) f o g

To find: f o g

Formula used: f o g = f(g(x))

Given: (i) f : R → R : f(x) = (2x + 1)

(ii) g : R → R : g(x) = (x² - 2)

Solution: We have,

f o g = f(g(x)) = f(x² - 2) = [ 2(x² - 2) + 1 ]

⇒ 2x² - 4 + 1

⇒ 2x² – 3

f o g (x) = 2x² – 3

(iii) f o f

To find: f o f

Formula used: f o f = f(f(x))

Given: (i) f : R → R : f(x) = (2x + 1)

Solution: We have,

f o f = f(f(x)) = f(2x + 1) = [ 2(2x + 1) + 1 ]

⇒ 4x + 2 + 1

⇒ 4x + 3

f o f (x) = 4x+ 3

(iv) g o g

To find: g o g

Formula used: g o g = g(g(x))

Given: (i) g : R → R : g(x) = (x² - 2)

Solution: We have,

g o g = g(g(x)) = g(x² - 2) = [ (x² - 2)² – 2]

⇒ x⁴ -4x² + 4 - 2

⇒ x 4 -4x² + 2

g o g (x) = x⁴ -4x² + 2

To show: h o (g o f ) = (h o g) o f

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : N → N : f(x) = 2x

(ii) g : N → N : g(y) = 3y + 4

(iii) h : N → N : h(z) = sin z

Solution: We have,

LHS = h o (g o f )

⇒ h o (g(f(x))

⇒ h(g(2x))

⇒ h(3(2x) + 4)

⇒ h(6x +4)

⇒ sin(6x + 4)

RHS = (h o g) o f

⇒ (h(g(x))) o f

⇒ (h(3x + 4)) o f

⇒ sin(3x+4) o f

Now let sin(3x+4) be a function u

RHS = u o f

⇒ u(f(x))

⇒ u(2x)

⇒ sin(3(2x) + 4)

⇒ sin(6x + 4) = LHS

Hence Proved

To find: f o g, g o f ,(f o g) (2) and (g o f) (-3)

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : R → R : f(x) = x² + 2

(ii) \(g:R→R :g(x)=\frac{x}{x-1},x\neq1\)

f o g = f(g(x))

\(\Rightarrow f(\frac{x}{x-1})\)

\(\Rightarrow (\frac{x}{x-1})^2+2\)

\(\Rightarrow\frac{(x)^2}{(x-1)^2}+2\)

\(fog(2)=\frac{(2)^2}{(2-1)^2}+2\)

\(=\frac{4}{1}+2\)

= 6

g o f = g(f(x))

⇒ g(x²+2)

\(\Rightarrow \frac{x^2+2}{x^2+2-1}\)

\(\Rightarrow \frac{x^2+2}{x^2+1}\)

\((gof)(-3)=\frac{-3^2+2}{-3^2+1}\)

\(=\frac{9+2}{9+1}\)

\(=\frac{11}{10}\)

To find: (f o g)\((\frac{-3}{2})\)+(g o f)\((\frac{4}{3})\).

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f is a greatest integer function

(ii) g is an absolute value function

f(x) = [x] (greatest integer function)

g(x) = (absolute value function)

\(f(\frac{4}{3})=[\frac{4}{3}]=1.....(i)\)

\(g(\frac{-3}{2})=[\frac{-3}{2}]=1.5......(ii)\)

Now, for (f o g)\((\frac{-3}{2})\)+(g o f)\((\frac{4}{3})\).

Substituting values from (i) and (ii)

\(\Rightarrow f(1.5)+g(1)\)

⇒ [1.5] +[1]

⇒ 1 + 1 = 2

We know that

f (x₁) = f (x₂)

So we get

2x₁ = 2x₂ => x₁ = x₂

Here, f is one-one

Consider y = 2x where x = ½ y

Each y in co domain R there exists ½ y where

f (1/2 y) = (2 × ½ y) = y

Hence, f is onto.

It is given that

f(x) = |x|

We can write it as

f o f = f{f(x)} = f(|x|) = |x| …….. (1)

Same way

f = |x| …….. (2)

From both the equations we know that fof = f.

It is given that

f(x) = (x² + 3x + 1) and g(x) = (2x – 3)

(i) g o f = g o f (x) = g{f(x)} = g(x² + 3x + 1)

So we get

= 2 (x² + 3x + 1) – 3

It can be written as

= 2x² + 6x + 2 – 3

On further calculation

= 2x² + 6x – 1

(ii) f o g = f o g (x) = f{g(x)} = f {2x – 3}

So we get

= (2x – 3)² + 3 (2x – 3) + 1

It can be written as

= 4x² – 12x + 9 + 6x – 9 + 1

On further calculation

= 4x² – 6x + 1

(iii) g o g = g o g (x) = g {g(x)} = g{2x – 3}

So we get

= 2 (2x – 3) – 3

It can be written as

= 4x – 6 – 3

On further calculation

= 4x – 9