Ex 6.2

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### Transcript

Ex 6.2, 16 Prove that the function f given by f (x) = log sin x is strictly increasing on (0,đ/2) and strictly decreasing on (đ/2,đ) f(đĽ) = log sin đĽ We need to show that f(đĽ) is strictly increasing on (0 , đ/2) & strictly decreasing on (đ/2 , đ) i.e. We need to show fâ(đ) > 0 for đĽ â (đ , đ/đ) & fâ(đ) < 0 for đĽ â (đ/đ , đ) Finding fâ(đ) fâ(đĽ) = (đđđ.sinâĄđĽ )â fâ(đĽ) = (1 )/sinâĄđĽ . đ(sinâĄđĽ )/đđĽ fâ (đĽ) = 1/sinâĄđĽ .cosâĄđĽ fâ (đ) = đđ¨đŹâĄđ/đŹđ˘đ§âĄđ Checking sign of fâ (đĽ) on (0 , đ/2) & (đ/2 , đ) For 0 < đ < đ/đ Here, x is in the 1st quadrant â´ cos đĽ > 0 & sin đĽ > 0 Thus, fâ(đĽ) = cosâĄđĽ/sinâĄđĽ > 0 for đ â (đ , đ/đ) Hence, fâ(đ) > 0 for đĽ â (0 , đ/2) Thus f(đĽ) is strictly increasing for đĽ â (0 , đ/2) For đ/đ < đ < Ď Here, x is in the 2nd quadrant â´ cos đ < 0 & sin đ > 0 Now, fâ(đĽ) = cosâĄđĽ/sinâĄđĽ = ((â))/((+) ) < 0 Hence, fâ(đ) < 0 for đĽ â (đ/2 , đ) Thus f(đĽ) is strictly decreasing for đĽ â (đ/2 , đ)