Example 26 - Chapter 1 Class 12 Relation and Functions

Last updated at June 6, 2023 by Teachoo

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Book a free demo

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.

Next: Question 1 → Ask a doubt

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular