# Ex 6.2,16

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.2,16 Prove that the function f given by f (x) = log sin x is strictly increasing on 0, 2 and strictly decreasing on 2 , f = log sin We need to show that f is strictly increasing on 0 , 2 & strictly decreasing on 2 , i.e. we need to show f > 0 for 0 , 2 & f < 0 for 2 , Finding f f = log sin f = . f = 1 sin . f = 1 sin . cos f = Case 1: For 0 < < We know that cos > 0 for 0 , 2 & sin > 0 for 0 , 2 f = cos sin = + + for 0 , 2 > 0 for 0 , 2 Hence f > 0 for 0 , 2 Thus f is strictly increasing for 0 , 2 Case 2: For < < We know that cos < 0 for 2 , & sin > 0 for 2 , Now, f = cos sin = + < 0 f < 0 for 2 , f (x) is strictly decreasing for ,

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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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.