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

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.