Ex 6.2, 16 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
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(π₯) = log sin π₯ We need to show that f(π₯) is strictly increasing on (0 , π/2) & strictly decreasing on (π/2 , π) i.e. We need to show fβ(π) > 0 for π₯ β (π , π /π) & fβ(π) < 0 for π₯ β (π /π , π ) Finding fβ(π) fβ(π₯) = (πππ.sinβ‘π₯ )β fβ(π₯) = (1 )/sinβ‘π₯ . π(sinβ‘π₯ )/ππ₯ fβ (π₯) = 1/sinβ‘π₯ .cosβ‘π₯ fβ (π) = ππ¨π¬β‘π/π¬π’π§β‘π Checking sign of fβ (π₯) on (0 , π/2) & (π/2 , π) For 0 < π < π /π Here, x is in the 1st quadrant β΄ cos π₯ > 0 & sin π₯ > 0 Thus, fβ(π₯) = cosβ‘π₯/sinβ‘π₯ > 0 for π β (π , π /π) Hence, fβ(π) > 0 for π₯ β (0 , π/2) Thus f(π₯) is strictly increasing for π₯ β (0 , π/2) For π /π < π < Ο Here, x is in the 2nd quadrant β΄ cos π < 0 & sin π > 0 Now, fβ(π₯) = cosβ‘π₯/sinβ‘π₯ = ((β))/((+) ) < 0 Hence, fβ(π) < 0 for π₯ β (π/2 , π) Thus f(π₯) is strictly decreasing for π₯ β (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 and Computer Science at Teachoo