1. Chapter 1 Class 12 Relation and Functions
  2. Serial order wise
  3. Examples
Check sibling questions

Example 26 - Chapter 1 Class 12 Relation and Functions

Last updated at Dec. 16, 2024 by Teachoo

 

Transcript

Example 26 Consider a function f : [0,π/2 ] → R given by f (x) = sin x and g: [0,π/2 ] → R given by g(x) = cos x. Show that f & g are one-one, but f + g is not Checking one-one for f f : [0, π/2 ] → R f (x) = sin x f(x1) = sin x1 f(x2) = sin x2 Putting f(x1) = f(x2) sin x1 = sin x2 So, x1 = x2 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Checking one-one for g g : [0,π/2 ] → R g(x) = cos x g(x1) = cos x1 g(x2) = cos x2 Putting g(x1) = g(x2) cos x1 = cos x2 So, x1 = x2 Hence, if g(x1) = g(x2) , then x1 = x2 ∴ g is one-one Checking one-one for f + g f + g : [0,π/2 ] → R f + g (x) = sin x + cos x But (f + g) (0) = sin 0 + cos 0 = 0 + 1 = 1 &(f + g) [𝛑/𝟐] = sin π/2 + cos π/2 = 1 + 0 = 1 Since, different elements 0, 𝜋/2 have the same image 1, ∴ f + g is not one-one.
  1. Chapter 1 Class 12 Relation and Functions
  2. Serial order wise

    3. Examples

    Miscellaneous →
Miscellaneous →
Chapter 1 Class 12 Relation and Functions
Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo