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

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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‹)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo