At x = 5Ο/6, f (x) = 2 sin 3x + 3 cos 3x is :
(A) maximumΒ
(B) minimum
(C) zeroΒ
(D) neither maximum or minimum
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Last updated at Dec. 4, 2021 by Teachoo
Question 27 At x = 5π/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 = 5Ο/6 So, we will find f β (x) f (x) = 2 sin 3π₯ + 3 cos 3π₯ Finding f β (x) f β (x) = 6 cos 3π₯ β 9 sin 3π₯ Finding f ββ (x) fββ (x) = β18 sin 3π₯ β 27 cos 3π₯ At x = ππ /π fββ (ππ /π) = β18 sin (3(5π/6))β 27 cos (3(5π/6)) = β18 sin (5π/2) β 27 cos (5π/2) = β18 sin (2π+π/2) β 27 cos (2π+π/2) = β18 sin π/2 β 27 cos π/2 = β 18 (1) β 27 (0) = β18 < 0 Since fββ(x) < 0 at x = 5π/6 β΄ f has maximum at x = 5π/6 So, the correct answer is (B)