If a function f from R to R is defined by f = { ( x , 3 x - 5 ) : x ∈ R } , find the values of a and b given that ( a , 4 ) and ( 1 , b ) belong to f .

Given function f from R to R is defined by f = { ( x , 3 x - 5 ) : x ∈ R } ∴ f x = 3 x - 5 → 1 Since, the ordered pairs a , 4 and 1 , b belongs to the given function ∴ f a = 4 ⇒ 3 a - 5 = 4 From 1 ⇒ a = 3 And f 1 = b ⇒ 3 - 5 = b From 1 ⇒ b = - 2 Hence, a = 3 , b = - 2 .

Show that the function f : Q → Q defined by f ( x ) = 3 x - 2 is One-One function.

Given function, f ( x ) = 3 x - 2 We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ Q , then the function is One-One function. Let x 1 , x 2 ∈ Q be such that f x 1 = f x 2 ⇒ 3 x 1 - 2 = 3 x 2 - 2 ⇒ 3 x 1 = 3 x 2 ⇒ x 1 = x 2 Hence, f is One-One function.

Show that the function f : N → N defined by f ( x ) = 2 x - 1 is One-one function but not Onto.

Given function f : N → N defined by f ( x ) = 2 x - 1 . We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ N , then the function is One-one. Let x 1 , x 2 ∈ N be such that f x 1 = f x 2 ⇒ 2 x 1 - 1 = 2 x 2 - 1 ⇒ 2 x 1 = 2 x 2 ⇒ x 1 = x 2 Hence, f is One-one function. Consider any y ∈ N (co-domain of f ) Let y = f ( x ) = 2 x - 1 ⇒ y + 1 = 2 x ⇒ y + 1 2 = x Thus , for every y ∈ N (Co-domain of f ), there does not exist x ∈ N (Domain of f ) Hence, the f is not Onto.

Show that the function f : N → N defined by f ( n ) = n 2 is injective.

Given function, f : N → N defined by f ( n ) = n 2 . When n = 1 ∈ N , f 1 = 1 ∈ N When n = 2 ∈ N , f 2 = 4 ∈ N When n = 3 ∈ N , f 3 = 9 ∈ N and so on. Therefore, for each n ∈ N , there exists different images. Hence, f is One-one function.

Show that the function f : N → N defined by f ( m ) = m 2 + m + 3 is One-one.

Given function, f : N → N defined by f ( m ) = m 2 + m + 3 . When m = 1 ∈ N , f 1 = 1 + 1 + 3 = 5 ∈ N When m = 2 ∈ N , f 2 = 4 + 2 + 3 = 9 ∈ N When m = 3 ∈ N , f 3 = 9 + 3 + 3 = 15 ∈ N and so on. Therefore, for each n ∈ N , there exists different images. Hence, f is One-one function.

State the function f : R → R defined by f ( x ) = 2 x + 1 is bijective or not. Justify answer.

Given function f : R → R defined by f ( x ) = 2 x + 1 . We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ R , then its a One-one function. Let x 1 , x 2 ∈ R be such that f x 1 = f x 2 ⇒ 2 x 1 + 1 = 2 x 2 + 1 ⇒ 2 x 1 = 2 x 2 ⇒ x 1 = x 2 Hence, f is One-one function. Consider any y ∈ R (co-domain of f ) Let y = f ( x ) = 2 x + 1 ⇒ y - 1 = 2 x ⇒ y - 1 2 = x Thus, for every y ∈ R (Co-domain of f ), there exists x ∈ R (Domain of f ) such that f x = f y - 1 2 = 2 · y - 1 2 + 1 = y Hence, the f is Onto. Therefore, the given function is bijective.

State the function f : R → R defined by f ( x ) = 3 - 4 x is bijective or not. Justify answer.

Given function, f : R → R defined by f ( x ) = 3 - 4 x . We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ R , then its a One-one function. Let x 1 , x 2 ∈ R be such that f x 1 = f x 2 ⇒ 3 - 4 x 1 = 3 - 4 x 2 ⇒ - 4 x 1 = - 4 x 2 ⇒ x 1 = x 2 Hence, f is One-one function. Consider any y ∈ R (co-domain of f ) Let y = f ( x ) = 3 - 4 x ⇒ 3 - y = 4 x ⇒ 3 - y 4 = x Thus, for every y ∈ R (Co-domain of f ), there exists x ∈ R (Domain of f ) such that f x = f 3 - y 2 = 3 - 4 · 3 - y 4 = y Hence, the f is Onto. Therefore, the given function is bijective.

E M B I B E

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.3

M L Aggarwal Solutions for Chapter: Relations and Functions, Exercise 3: EXERCISE 2.3

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Relations and Functions, Exercise 3: EXERCISE 2.3

Attempt the practice questions on Chapter 2: Relations and Functions, Exercise 3: EXERCISE 2.3 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 3: EXERCISE 2.3 with Hints & Solutions

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.3>Q 15

If a function $f$ from $R to R$ is defined by $f = {(x, 3 x - 5) : x \in R}$ , find the values of $a and b$ given that $(a, 4) and (1, b)$ belong to $f$ .