Ex 6.2

Chapter 6 Class 12 Application of Derivatives
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/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₯) = (πππ.sinβ‘π₯ )β fβ(π₯) = (1 )/sinβ‘π₯ . π(sinβ‘π₯ )/ππ₯ fβ (π₯) = 1/sinβ‘π₯ .cosβ‘π₯ fβ (π) = ππ¨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)