If f and g are two real functions defined by f x = x - 2 and g x = x 2 - 1 , then find f g .

Given, f x = x - 2 and g x = x 2 - 1 We note that, f x has domain [ 2 , ∞ ) and g x has domain x ≥ 1 or x ≤ - 1 . Therefore, the common part of the domain is [ 2 , ∞ ) . Now, f g = f x g x , x ∈ [ 2 , ∞ ) f x g x = x - 2 x 2 - 1 , x ∈ [ 2 , ∞ ) ∴ f x g x = x - 2 x 2 - 1 , x ∈ [ 2 , ∞ )

If f and g are two real functions defined by f x = x - 2 and g x = x 2 - 1 , then find 2 f 2 + 3 g .

Given, f x = x - 2 and g x = x 2 - 1 We note that, f x has domain [ 2 , ∞ ) and g x has domain x ≥ 1 or x ≤ - 1 . Therefore, the common part of the domain is [ 2 , ∞ ) . Now, 2 f 2 + 3 g = 2 f x 2 + 3 g x , x ∈ [ 2 , ∞ ) 2 f x 2 + 3 g x = 2 x - 2 2 + 3 x 2 - 1 , x ∈ [ 2 , ∞ ) ∴ 2 f x 2 + 3 g x = 2 x - 2 + 3 x 2 - 1 , x ∈ [ 2 , ∞ ) .

If f and g are two real functions defined by f x = x - 2 and g x = x 2 - 1 , then find 1 f .

Given, f x = x - 2 and g x = x 2 - 1 We note that, f x has domain [ 2 , ∞ ) and g x has domain x ≥ 1 or x ≤ - 1 . Therefore, the common part of the domain is [ 2 , ∞ ) . Now, 1 f = 1 f x , x ∈ ( 2 , ∞ ) as we know that denominator can not be zero. ∴ 1 f x = 1 x - 2 , x ∈ ( 2 , ∞ ) .

If f and g are two real functions defined by f x = x - 2 and g x = x 2 - 1 , then find g f .

Given, f x = x - 2 and g x = x 2 - 1 We note that, f x has domain [ 2 , ∞ ) and g x has domain x ≥ 1 or x ≤ - 1 . Therefore, the common part of the domain is [ 2 , ∞ ) . Now, g f = g x f x , x ∈ 2 , ∞ (as denominator can not be zero) g x f x = x 2 - 1 x - 2 , x ∈ 2 , ∞ ∴ g x f x = x 2 - 1 x - 2 , x ∈ 2 , ∞

If f x = x 3 - 1 x 3 , prove that f x + f 1 x = 0 .

Given, f x = x 3 - 1 x 3 . . . . 1 , D f = R - { 0 } ∴ f 1 x = 1 x 3 - 1 1 x 3 = 1 x 3 - x 3 . . . . 2 (Replace x by 1 x ) Now, from LHS we have, = f x + f 1 x = x 3 - 1 x 3 + 1 x 3 - x 3 (From 1 and 2 ) = 0 = RHS Hence, proved.

If f x = x + 1 x , prove that ( f ( x ) ) 3 = f x 3 + 3 f 1 x .

Given, f x = x + 1 x ⇒ f x 3 = x 3 + 1 x 3 → 1 , D f = R - { 0 } ∴ f 1 x = 1 x + 1 1 x = 1 x + x → 2 (Replace x by 1 x ) To prove ( f ( x ) ) 3 = f x 3 + 3 f 1 x We have, LHS = ( f ( x ) ) 3 = x + 1 x 3 = x 3 + 1 x 3 + 3 x · 1 x x + 1 x = x 3 + 1 x 3 + 3 x + 1 x = x 3 + 1 x 3 + 3 1 x + x = f x 3 + 3 f 1 x (From 1 and 2 ) = RHS Hence, proved.

If y = f x = 6 x - 5 5 x - 6 , prove that f ( y ) = x , x ≠ 6 5 .

Given y = f x = 6 x - 5 5 x - 6 → 1 , x ≠ 6 5 . ⇒ f y = 6 y - 5 5 y - 6 (Replace x with y ) = 6 6 x - 5 5 x - 6 - 5 5 6 x - 5 5 x - 6 - 6 (From 1 ) = 6 ( 6 x - 5 ) - 5 ( 5 x - 6 ) 5 ( 6 x - 5 ) - 6 ( 5 x - 6 ) = 36 x - 30 - 25 x + 30 30 x - 25 - 30 x + 36 = 11 x 11 = x . Hence proved.

E M B I B E

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.7

M L Aggarwal Solutions for Chapter: Relations and Functions, Exercise 7: EXERCISE 2.7

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Relations and Functions, Exercise 7: EXERCISE 2.7

Attempt the practice questions on Chapter 2: Relations and Functions, Exercise 7: EXERCISE 2.7 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 11 Volume 1 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Relations and Functions, Exercise 7: EXERCISE 2.7 with Hints & Solutions

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11th ICSE

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Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.7>Q 1

If $f$ and $g$ are real functions defined by $f (x) = x^{2} + 7$ and $g (x) = 3 x + 5$ , then find the value of $\frac{f (t) - f (5)}{t - 5}, t \neq 5$ .