Let f x = x 2 + x + 1 and g x = sin x . Show that f o g ≠ g o f .

f x = x 2 + x + 1 and g x = sin x f o g = f g x = g x 2 + g x + 1 = sin x 2 + sin x + 1 = sin 2 x + sin x + 1 . . . ( 1 ) and g o f = g f x = sin f x = sin x 2 + x + 1 . . . 2 From 1 and 2 , we can clearly see that f o g ≠ g o f .

Let f x = x 2 + x + 1 and g x = sin x . Show that f o g ≠ g o f .

f x = x 2 + x + 1 and g x = sin x f o g = f g x = g x 2 + g x + 1 = sin x 2 + sin x + 1 = sin 2 x + sin x + 1 . . . ( 1 ) and g o f = g f x = sin f x = sin x 2 + x + 1 . . . 2 From 1 and 2 , we can clearly see that f o g ≠ g o f .

Let f x = x 2 , g x = sin x and h x = x . Then verify that [ f o ( g o h ) ] x and [ ( f o g ) o h ] x are equal.

f x = x 2 , g x = sin x and h x = x Now f o g o h = f g h x = f g x = f sin x = sin x 2 = sin 2 x . . . ( 1 ) and f o g o h = f g x o h = f sin x o h = sin x 2 o h = sin 2 x o h = sin 2 x . . . ( 2 ) From equations 1 and 2 , we can say that [ f o g o h ] x = [ f o g o h ]

Find f o g and g o f , if f x = x 2 + 2 ; g x = 1 - 1 1 - x , x ≠ 1 .

f x = x 2 + 2 ; g x = 1 - 1 1 - x , x ≠ 1 f o g = f g x = f 1 - 1 1 - x = 1 - 1 1 - x 2 + 2 = 1 - x - 1 1 - x 2 + 2 = x x - 1 2 + 2 = 3 x 2 - 4 x + 2 x - 1 2 ; x ≠ 1 g o f = g f x = 1 - 1 1 - x 2 + 2 = 1 - x 2 - 2 - 1 1 - x 2 - 2 = x 2 + 2 x 2 + 1 .

If f x = 1 + x 2 ; x ≤ 1 x + 1 ; 1 < x ≤ 2 and g x = 1 - x ; - 2 ≤ x ≤ 1 , then define the function f o g x .

f x = 1 + x 2 ; x ≤ 1 x + 1 ; 1 < x ≤ 2 and g x = 1 - x ; - 2 ≤ x ≤ 1 f o g x = 1 + g 2 x ; g x ≤ 1 g x + 1 ; 1 < g x ≤ 2 We need to check the interval of x , for which we get the value of g x ≤ 1 and 1 < g x ≤ 2 For 0 ≤ x ≤ 1 ⇒ g x ≤ 1 and - 1 ≤ x < 0 ⇒ 1 < g x ≤ 2 Therefore, f o g x = 1 + 1 - x 2 ; 0 ≤ x ≤ 1 1 - x + 1 ; - 1 ≤ x ≤ 0 = 2 - 2 x + x 2 ; 0 ≤ x ≤ 1 2 - x ; - 1 ≤ x < 0

If f x = x + 2 x + 1 and g x = x - 2 x , then find the domain of fog x .

f x = x + 2 x + 1 and g x = x - 2 x , x ≠ 0 f o g = f g x (for x ≠ 0 , g x , is not defined). = g x + 2 g x + 1 = x - 2 x + 2 x - 2 x + 1 = 3 x - 2 2 x - 2 ; x ≠ 1 . Therefore, domain of f o g is R - 0 , 1 .

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 138

EASY

JEE Main/Advance

IMPORTANT

Earn 100

Let $f (x) = x^{2} + x + 1$ and $g (x) = \sin x .$ Show that $f o g \neq g o f$ .

Important Questions on Functions

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 139

Let

f (x) = x^{2}, g (x) = \sin x

and

h (x) = \sqrt{x}

. Then verify that

[f o (g o h)] (x)

and

[(f o g) o h] (x)

are equal.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 140

Find $f o g$ and $g o f$ , if $f (x) = e^{x}; g (x) = \ln x$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 141

Find $f o g$ and $g o f$ , if $f (x) = |x|; g (x) = \sin x$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 142

Find $f o g$ and $g o f$ , if $f (x) = \sin x; g (x) = x^{2}$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 143

Find $f o g$ and $g o f$ , if $f (x) = x^{2} + 2; g (x) = 1 - \frac{1}{1 - x}, x \neq 1$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 144

If $f (x) = \ln (x^{2} - x + 2); R^{+} \to R$ and $g (x) = \{x\} + 1; [1, 2] \to [1, 2]$ , where $\{x\}$ denotes fractional part of $x$ . Find the domain and range of $f (g (x))$ when defined.