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


Transcript

Ex 6.2, 17 Prove that the function f given by f (𝑥) = log cos 𝑥 is strictly decreasing on (0,𝜋/2) and strictly increasing on(𝜋/2,𝜋) f(𝑥) = log cos 𝑥 We need to show that f(𝑥) is strictly decreasing on (0 , 𝜋/2) & strictly increasing on (𝜋/2 , 𝜋) i.e. We need to show f’(𝒙) < 0 for 𝑥 ∈ (𝟎 , 𝝅/𝟐) & f’(𝒙) > 0 for 𝑥 ∈ (𝝅/𝟐 , 𝝅) Finding f’(𝒙) f’(𝑥) = (𝑙𝑜𝑔.cos⁡𝑥 )’ f’(𝑥) = (1 )/cos⁡𝑥 . 𝑑(cos⁡𝑥 )/𝑑𝑥 f’ (𝑥) = 1/cos⁡𝑥 .〖−sin〗⁡𝑥 f’ (𝒙) = 〖−𝐬𝐢𝐧〗⁡𝒙/𝐜𝒐𝒔⁡𝒙 Checking sign of f’ (𝑥) on (0 , 𝜋/2) & (𝜋/2 , 𝜋) For 0 < 𝒙 < 𝝅/𝟐 Here, x is in the 1st quadrant ∴ cos 𝑥 > 0 & sin 𝑥 > 0 Now, f’(𝑥) =〖−𝐬𝐢𝐧〗⁡𝒙/𝐜𝒐𝒔⁡𝒙 = ((−)(+))/((+) ) < 0 Hence, f’(𝒙) < 0 for 𝑥 ∈ (0 , 𝜋/2) Thus f(𝑥) is strictly decreasing for 𝑥 ∈ (0 , 𝜋/2) For 𝝅/𝟐 < 𝒙 < π Here, x is in the 2nd quadrant ∴ cos 𝒙 < 0 & sin 𝒙 > 0 Now, f’(𝑥) =〖−𝐬𝐢𝐧〗⁡𝒙/𝐜𝒐𝒔⁡𝒙 = ((−)(+))/((−) ) > 0 Hence, f’(𝒙) >0 for 𝑥 ∈ (𝜋/2 , 𝜋) Thus, f(𝑥) is strictly increasing for 𝑥 ∈ (𝜋/2 , 𝜋)

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.