Let f = x , 2 x 2 x + 1 : x ∈ R , x ≠ - 1 2 be a function from R into R . Find the range of f .

Given that f = x , 2 x 2 x + 1 : x ∈ R , x ≠ - 1 2 is a function from R into R . We have to find range ( f ) , f ( x ) = 2 x 2 x + 1 is well-defined for all x ∈ R So, domain of f is R Now, for each x ∈ R , we have 0 ≤ 2 x < 2 x + 1 ⇒ 0 ≤ 2 x 2 x + 1 < 1 ⇒ 0 ≤ f ( x ) < 1 We know that range of range ( f ) = { f ( x ) | x ∈ R } Thus, we get range ( f ) = R - { 1 } .

Let f = x , 3 x 2 1 + x 2 : x ∈ R be a function from R into R . Find the range of f .

Given that f = x , 3 x 2 1 + x 2 : x ∈ R is a function from R into R . We have to find range ( f ) f ( x ) = 3 x 2 1 + x 2 is well-defined for all x ∈ R So, domain of f is R Now, for each x ∈ R , we have 0 ≤ 3 x 2 < x 2 + 1 ⇒ 0 ≤ 3 x 2 x 2 + 1 < 1 ⇒ 0 ≤ f ( x ) < 3 We know that range of range ( f ) = { f ( x ) | x ∈ R } Thus, we get range ( f ) = 0 ≤ f ( x ) < 3 .

Find R be a relation from N to N defined by R = { ( a , b ) : a , b ∈ N and a = b 2 } . Is the given true? ( a , a ) ∈ R , for all a ∈ N

Given, R be a relation from N to N defined by R = a , b : a , b ∈ N a n d a = b 2 It can be seen that 2 ∈ N ; however, 2 ≠ 2 2 = 4 ⇒ 2 , 2 ∈ N but not ∉ R as 2 ≠ 2 2 Therefore, the statement " a , a ∈ R , for all a ∈ N " is not true.

Find R be a relation from N to N defined by R = { ( a , b ) : a , b ∈ N and a = b 2 } . Is the given statement true? ( a , b ) ∈ R , implies ( b , a ) ∈ R

Given, R be a relation from N to N defined by R = a , b : a , b ∈ N a n d a = b 2 . It can be seen that 9 , 3 ∈ R because 9 , 3 ∈ N and 9 = 3 2 Now, 3 ≠ 9 2 = 81 ∴ 3 , 9 ∉ R Therefore, the statement " a , b ∈ R ⇒ b , a ∈ R " is not true.

Find R be a relation from N to N defined by R = { ( a , b ) : a , b ∈ N and a = b 2 } . Is the given statement true? ( a , b ) ∈ R and ( b , c ) ∈ R implies ( a , c ) ∈ R .

Given, R be a relation from N to N defined by R = a , b : a , b ∈ N a n d a = b 2 . It can be seen that 16 , 4 ∈ R , 4 , 2 ∈ R because 16 , 4 , 2 ∈ N and 16 = 4 2 and 4 = 2 2 . Now, 16 ≠ 2 2 = 4 ∴ 16 , 2 ∉ R Therefore, the statement " a , b ∈ R , b , c ∈ R ⇒ a , c ∈ R " is not true.

E M B I B E

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(IV)

SURANJAN SAHA and SABITA SAHA Solutions for Chapter: Relations and Functions, Exercise 4: EXERCISE-2(IV)

Author:SURANJAN SAHA & SABITA SAHA

SURANJAN SAHA Mathematics Solutions for Exercise - SURANJAN SAHA and SABITA SAHA Solutions for Chapter: Relations and Functions, Exercise 4: EXERCISE-2(IV)

Attempt the free practice questions on Chapter 2: Relations and Functions, Exercise 4: EXERCISE-2(IV) with hints and solutions to strengthen your understanding. I.S.C MATHEMATICS FOR CLASS XI solutions are prepared by Experienced Embibe Experts.