Check sibling questions

Ex 6.2, 17 - Prove that f (x) = log cos x is strictly decreasing

Ex 6.2,17 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,17 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.2,17 - Chapter 6 Class 12 Application of Derivatives - Part 4

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Ex 6.2, 17 Prove that the function f given by f (π‘₯) = log cos π‘₯ is strictly decreasing on (0,πœ‹/2) and strictly increasing on(πœ‹/2,πœ‹) f(π‘₯) = log cos π‘₯ We need to show that f(π‘₯) is strictly decreasing on (0 , πœ‹/2) & strictly increasing on (πœ‹/2 , πœ‹) i.e. We need to show f’(𝒙) < 0 for π‘₯ ∈ (𝟎 , 𝝅/𝟐) & f’(𝒙) > 0 for π‘₯ ∈ (𝝅/𝟐 , 𝝅) Finding f’(𝒙) f’(π‘₯) = (π‘™π‘œπ‘”.cos⁑π‘₯ )’ f’(π‘₯) = (1 )/cos⁑π‘₯ . 𝑑(cos⁑π‘₯ )/𝑑π‘₯ f’ (π‘₯) = 1/cos⁑π‘₯ .γ€–βˆ’sin〗⁑π‘₯ f’ (𝒙) = γ€–βˆ’π¬π’π§γ€—β‘π’™/πœπ’π’”β‘π’™ Checking sign of f’ (π‘₯) on (0 , πœ‹/2) & (πœ‹/2 , πœ‹) For 0 < 𝒙 < 𝝅/𝟐 Here, x is in the 1st quadrant ∴ cos π‘₯ > 0 & sin π‘₯ > 0 Now, f’(π‘₯) =γ€–βˆ’π¬π’π§γ€—β‘π’™/πœπ’π’”β‘π’™ = ((βˆ’)(+))/((+) ) < 0 Hence, f’(𝒙) < 0 for π‘₯ ∈ (0 , πœ‹/2) Thus f(π‘₯) is strictly decreasing for π‘₯ ∈ (0 , πœ‹/2) For 𝝅/𝟐 < 𝒙 < Ο€ Here, x is in the 2nd quadrant ∴ cos 𝒙 < 0 & sin 𝒙 > 0 Now, f’(π‘₯) =γ€–βˆ’π¬π’π§γ€—β‘π’™/πœπ’π’”β‘π’™ = ((βˆ’)(+))/((βˆ’) ) > 0 Hence, f’(𝒙) >0 for π‘₯ ∈ (πœ‹/2 , πœ‹) Thus, f(π‘₯) is strictly increasing for π‘₯ ∈ (πœ‹/2 , πœ‹)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

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.