If f ( x ) = x - 1 x + 1 then show that f 1 x = - f ( x )

Given, f x = x - 1 x + 1 ∴ f 1 x = 1 x - 1 1 x + 1 = 1 - x 1 + x = - x - 1 x + 1 = - f x Hence, proved that f 1 x = - f x .

If f ( x ) = x - 1 x + 1 then show that f - 1 x = - 1 f ( x )

Given f x = x - 1 x + 1 ∴ f - 1 x = - 1 x - 1 - 1 x + 1 = - 1 - x - 1 + x = 1 + x 1 - x = 1 1 - x 1 + x = - 1 x - 1 x + 1 = - 1 f x Hence, proved that f - 1 x = - 1 f x .

If f ( x ) = x 3 - 1 x 3 then show that f ( x ) + f 1 x = 0

Given f x = x 3 - 1 x 3 ∴ f 1 x = 1 x 3 - 1 1 x 3 = 1 x 3 - x 3 ∴ f x + f 1 x = x 3 - 1 x 3 + 1 x 3 - x 3 = 0 Hence, proved.

If f ( x ) = x + 1 x - 1 then show that f { f ( x ) } = x

Given, f x = x + 1 x - 1 ∴ f f x = f x + 1 f x - 1 = x + 1 x - 1 + 1 x + 1 x - 1 - 1 = x + 1 + x - 1 x + 1 - x + 1 = 2 x 2 = x Hence, proved that f f x = x .

If f ( x ) = 1 ( 2 x + 1 ) and x ≠ - 1 2 then prove that f f ( x ) = 2 x + 1 2 x + 3 when it is given that x ≠ - 3 2

Given f ( x ) = 1 ( 2 x + 1 ) , where x ≠ - 1 2 To prove: f { f ( x ) } = 2 x + 1 2 x + 3 , x ≠ - 3 2 Now placing f ( x ) in place of x f { f ( x ) } = 1 2 f x + 1 ⇒ f { f ( x ) } = 1 2 1 2 x + 1 + 1 ⇒ f { f ( x ) } = 1 2 + 2 x + 1 2 x + 1 ⇒ f f x = 2 x + 1 2 x + 3 , where x ≠ - 3 2

If f ( x ) = 1 ( 1 - x ) then show that f f f x = x

Given, f x = 1 1 - x ....(1) ∴ f f x = 1 1 - f x = 1 1 - 1 1 - x = 1 - x 1 - x - 1 (Put value of f x from (1)) = 1 - x - x = x - 1 x Again f f f x = f x - 1 f ( x ) = 1 1 - x - 1 1 1 - x = 1 - 1 + x 1 = x ∴ f f f x = x Hence, proved.

If f ( x ) = 2 x 1 + x 2 then show that f ( tan θ ) = sin 2 θ

Given f x = 2 x 1 + x 2 Replacing x with tan θ in the function ∴ f t a n θ = 2 t a n θ 1 + tan 2 θ = sin 2 θ ∵ sin 2 A = 2 tan A 1 + tan 2 A Hence, proved that f tanθ = sin 2 θ .

If y = f ( x ) = 3 x + 1 5 x - 3 prove that x = f ( y ) .

Given y = f ( x ) = 3 x + 1 5 x - 3 Replacing x b y y in the function, f ( y ) = 3 y + 1 5 y - 3 Now, given in the problem that y = f ( x ) f ( y ) = 3 f ( x ) + 1 5 f ( x ) - 3 ⇒ f ( y ) = 3 3 x + 1 5 x - 3 + 1 5 3 x + 1 5 x - 3 - 3 ⇒ f ( y ) = 9 x + 3 + 5 x - 3 15 x + 5 - 15 x + 9 ⇒ f ( y ) = 14 x 14 = x ⇒ x = f ( y ) Hence proved.

E M B I B E

Senior Secondary School Mathematics FOR CLASS 11>Set and Functions>Chapter 3 - Functions>EXERCISE 3B

R S Aggarwal Solutions for Exercise 2: EXERCISE 3B

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 2: EXERCISE 3B

Attempt the practice questions from Exercise 2: EXERCISE 3B with hints and solutions to strengthen your understanding. Senior Secondary School Mathematics FOR CLASS 11 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal Solutions for Exercise 2: EXERCISE 3B with Hints & Solutions

EASY

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 3 - Functions>EXERCISE 3B>Q 2

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

EASY

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 3 - Functions>EXERCISE 3B>Q 2

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

EASY

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 3 - Functions>EXERCISE 3B>Q 3

If $f (x) = x^{3} - \frac{1}{x^{3}}$ then show that $f (x) + f (\frac{1}{x}) = 0$

EASY

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 3 - Functions>EXERCISE 3B>Q 4

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

EASY

11th CBSE

IMPORTANT

Senior Secondary School Mathematics FOR CLASS 11>Chapter 3 - Functions>EXERCISE 3B>Q 5

If $f (x) = \frac{1}{(2 x + 1)}$ and $x \neq \frac{- 1}{2}$ then prove that $f \{f (x)\} = \frac{2 x + 1}{2 x + 3}$ when it is given that $x \neq \frac{- 3}{2}$