Ex 6.2, 16 - Prove that f(x) = log sin x is strictly increasing - To show increasing/decreasing in intervals

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.2,16 Prove that the function f given by f (x) = log sin x is strictly increasing on ﷐0,﷐𝜋﷮2﷯﷯ and strictly decreasing on ﷐﷐𝜋﷮2﷯,𝜋﷯ f﷐𝑥﷯ = log sin 𝑥 We need to show that f﷐𝑥﷯ is strictly increasing on ﷐0 , ﷐𝜋﷮2﷯﷯ & strictly decreasing 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 sin 𝑥 f’﷐𝑥﷯ = ﷐𝑙𝑜𝑔.𝑠𝑖𝑛𝑥﷯’ f’﷐𝑥﷯ = ﷐1 ﷮﷐sin﷮𝑥﷯﷯ . ﷐𝑑﷐𝑠𝑖𝑛𝑥﷯﷮𝑑𝑥﷯ f’ ﷐𝑥﷯ = ﷐1﷮﷐sin﷮𝑥﷯﷯ .﷐cos﷮𝑥﷯ f’ ﷐𝑥﷯ = ﷐𝑐𝑜𝑠𝑥﷮𝑠𝑖𝑛𝑥﷯ Case 1: For 0 < 𝒙 < ﷐𝝅﷮𝟐﷯ We know that cos 𝑥 > 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ & sin 𝑥 > 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ f’﷐𝑥﷯ = ﷐﷐cos﷮𝑥﷯﷮﷐sin﷮𝑥﷯﷯ = ﷐﷐+﷯﷮﷐+﷯﷯ for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ > 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ Hence f’﷐𝑥﷯ > 0 for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ Thus f﷐𝑥﷯ is strictly increasing for 𝑥 ∈ ﷐0 , ﷐𝜋﷮2﷯﷯ Case 2: For ﷐𝝅﷮𝟐﷯ < 𝒙 < π We know that cos 𝑥 < 0 for 𝑥 ∈ ﷐﷐𝜋﷮2﷯ , 𝜋﷯ & sin 𝑥 > 0 for 𝑥 ∈ ﷐﷐𝜋﷮2﷯ , 𝜋﷯ Now, f’﷐𝑥﷯ = ﷐﷐cos﷮𝑥﷯﷮﷐sin﷮𝑥﷯﷯ = ﷐﷐−﷯﷮﷐+﷯﷯ < 0 ⇒ f’﷐𝑥﷯ < 0 for 𝑥 ∈ ﷐﷐𝜋﷮2﷯ , 𝜋﷯ ∴ f’(x) is strictly decreasing for 𝒙 ∈ ﷐﷐𝝅﷮𝟐﷯ , 𝝅﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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