At x = 5π/6, f (x) = 2 sin 3x + 3 cos 3x is :
(A) maximum
(B) minimum
(C) zero
(D) neither maximum or minimum
Last updated at Dec. 16, 2024 by Teachoo
Question 14 At x = 5๐/6, f (x) = 2 sin 3x + 3 cos 3x is : maximum (B) minimum (C) zero (D) neither maximum or minimum Since, we have to check maximum and minimum value at x = 5ฯ/6 So, we will find f โ (x) f (x) = 2 sin 3๐ฅ + 3 cos 3๐ฅ Finding f โ (x) f โ (x) = 6 cos 3๐ฅ โ 9 sin 3๐ฅ Finding f โโ (x) fโโ (x) = โ18 sin 3๐ฅ โ 27 cos 3๐ฅ At x = ๐๐ /๐ fโโ (๐๐ /๐) = โ18 sin (3(5๐/6))โ 27 cos (3(5๐/6)) = โ18 sin (5๐/2) โ 27 cos (5๐/2) = โ18 sin (2๐+๐/2) โ 27 cos (2๐+๐/2) = โ18 sin ๐/2 โ 27 cos ๐/2 = โ 18 (1) โ 27 (0) = โ18 < 0 Since fโโ(x) < 0 at x = 5๐/6 โด f has maximum at x = 5๐/6 So, the correct answer is (B)
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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