Example 51
Last updated at May 29, 2018 by Teachoo
Transcript
Example 51 Consider a function f : [0, π2 ] → R given by f (x) = sin x and g: [0, π2 ] → R given by g(x) = cos x. Show that f and g are one-one, but f + g is not Checking one-one for f f : [0, π2 ] → 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 Hence, if f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Checking one-one for g g : [0, π2 ] → 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, π2 ] → R f + g (x) = sin x + cos x But (f + g) (0) = sin 0 + cos 0 = 0 + 1 = 1 &(f + g) 𝛑𝟐 = sin π2 + cos π2 = 1 + 0 = 1 Since, different elements 0, 𝜋2 have the same image 1, ∴ f + g is not one-one.
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Example 51 You are here
Chapter 1 Class 12 Relation and Functions
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.