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Ex 6.2
Ex 6.2,2
Ex 6.2,3 Important
Ex 6.2,4
Ex 6.2, 5 Important
Ex 6.2, 6 (a)
Ex 6.2, 6 (b) Important
Ex 6.2, 6 (c) Important
Ex 6.2, 6 (d)
Ex 6.2, 6 (e) Important
Ex 6.2, 7
Ex 6.2,8 Important
Ex 6.2,9 Important
Ex 6.2,10
Ex 6.2,11
Ex 6.2, 12 (A)
Ex 6.2, 12 (B) Important
Ex 6.2, 12 (C) Important
Ex 6.2, 12 (D)
Ex 6.2, 13 (MCQ) Important
Ex 6.2,14 Important
Ex 6.2,15
Ex 6.2, 16 You are here
Ex 6.2,17 Important
Ex 6.2,18
Ex 6.2,19 (MCQ) Important
Last updated at May 29, 2023 by Teachoo
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(đĽ) = log sin đĽ We need to show that f(đĽ) is strictly increasing on (0 , đ/2) & strictly decreasing on (đ/2 , đ) i.e. We need to show fâ(đ) > 0 for đĽ â (đ , đ /đ) & fâ(đ) < 0 for đĽ â (đ /đ , đ ) Finding fâ(đ) fâ(đĽ) = (đđđ.sinâĄđĽ )â fâ(đĽ) = (1 )/sinâĄđĽ . đ(sinâĄđĽ )/đđĽ fâ (đĽ) = 1/sinâĄđĽ .cosâĄđĽ fâ (đ) = đđ¨đŹâĄđ/đŹđ˘đ§âĄđ Checking sign of fâ (đĽ) on (0 , đ/2) & (đ/2 , đ) For 0 < đ < đ /đ Here, x is in the 1st quadrant â´ cos đĽ > 0 & sin đĽ > 0 Thus, fâ(đĽ) = cosâĄđĽ/sinâĄđĽ > 0 for đ â (đ , đ /đ) Hence, fâ(đ) > 0 for đĽ â (0 , đ/2) Thus f(đĽ) is strictly increasing for đĽ 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(đĽ) = cosâĄđĽ/sinâĄđĽ = ((â))/((+) ) < 0 Hence, fâ(đ) < 0 for đĽ â (đ/2 , đ) Thus f(đĽ) is strictly decreasing for đĽ â (đ/2 , đ)