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Ex 6.2,16 - Ex 6.2

Ex 6.2,16 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,16 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.2,16 - Chapter 6 Class 12 Application of Derivatives - Part 4


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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œ‹/2) and strictly decreasing on (πœ‹/2,πœ‹) f(π‘₯) = log sin π‘₯ We need to show that f(π‘₯) is strictly increasing on (0 , πœ‹/2) & strictly decreasing 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‘”.sin⁑π‘₯ )’ f’(π‘₯) = (1 )/sin⁑π‘₯ . 𝑑(sin⁑π‘₯ )/𝑑π‘₯ f’ (π‘₯) = 1/sin⁑π‘₯ .cos⁑π‘₯ f’ (𝒙) = πœπ¨π¬β‘π’™/𝐬𝐒𝐧⁑𝒙 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 Thus, f’(π‘₯) = cos⁑π‘₯/sin⁑π‘₯ > 0 for 𝒙 ∈ (𝟎 , 𝝅/𝟐) Hence, f’(𝒙) > 0 for π‘₯ ∈ (0 , πœ‹/2) Thus f(π‘₯) is strictly increasing 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‘₯) = cos⁑π‘₯/sin⁑π‘₯ = ((βˆ’))/((+) ) < 0 Hence, f’(𝒙) < 0 for π‘₯ ∈ (πœ‹/2 , πœ‹) Thus f(π‘₯) is strictly decreasing 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.