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Ex 6.2

Ex 6.2, 1

Ex 6.2,2

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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

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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

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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)