# Ex 6.2,17 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 6.2,17 Prove that the function f given by f (𝑥) = log cos 𝑥 is strictly decreasing on 0,𝜋2 and strictly increasing on𝜋2,𝜋 f𝑥 = log cos 𝑥 We need to show f𝑥 is strictly decreasing on 0 , 𝜋2 & Strictly increasing on 𝜋2 , 𝜋 i.e. We need to show f’𝑥 < 0 for 𝑥 ∈ 0 , 𝜋2 & f’ 𝑥 > 0 for 𝑥 ∈ 𝜋2 , 𝜋 Finding f’𝑥 f𝑥=log cos 𝑥 f’𝑥 = 1cos𝑥 . 𝑑𝑑𝑥 cos𝑥 f’𝑥 = 1cos𝑥 × −sin𝑥 f’ 𝑥 = −sin𝑥cos𝑥 Case 1: For 0 < 𝒙 < 𝝅𝟐 We know that cos 𝑥 > 0 for 𝑥 ∈ 0 , 𝜋2 & sin 𝑥 > 0 for 𝑥 ∈ 0 , 𝜋2 f’𝑥 =−sin𝑥cos𝑥 = (−)++ for 𝑥 ∈ 0 , 𝜋2 < 0 for 𝑥 ∈ 0 , 𝜋2 Hence f’𝑥 < 0 for 𝑥 ∈ 0 , 𝜋2 Thus f𝑥 is strictly decreasing for 𝒙 ∈ 𝟎 , 𝝅𝟐 Case 2: For 𝝅𝟐 < 𝒙 < π We know that cos 𝑥 < 0 for 𝑥 ∈ 𝜋2 , 𝜋 & sin 𝑥 > 0 for 𝑥 ∈ 𝜋2 , 𝜋 Now, f’𝑥 =−sin𝑥cos𝑥= −(+)− > 0 ⇒ f’𝑥 > 0 for 𝑥 ∈ 𝜋2 , 𝜋 ∴ f’(x) is strictly increasing for 𝒙 ∈ 𝝅𝟐 , 𝝅

Chapter 6 Class 12 Application of Derivatives

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.