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Question 27 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at March 22, 2023 by

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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Transcript

Question 27 At x = 5πœ‹/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 ” (x) f (x) = 2 sin 3π‘₯ + 3 cos 3π‘₯ Finding f ’ (x) f ’ (x) = 6 cos 3π‘₯ βˆ’ 9 sin 3π‘₯ Finding f ’’ (x) f’’ (x) = βˆ’18 sin 3π‘₯ βˆ’ 27 cos 3π‘₯ At x = πŸ“π…/πŸ” f’’ (πŸ“π…/πŸ”) = βˆ’18 sin (3(5πœ‹/6))βˆ’ 27 cos (3(5πœ‹/6)) = βˆ’18 sin (5πœ‹/2) βˆ’ 27 cos (5πœ‹/2) = βˆ’18 sin (2πœ‹+πœ‹/2) βˆ’ 27 cos (2πœ‹+πœ‹/2) = βˆ’18 sin πœ‹/2 βˆ’ 27 cos πœ‹/2 = βˆ’ 18 (1) βˆ’ 27 (0) = βˆ’18 < 0 Since f’’(x) < 0 at x = 5πœ‹/6 ∴ f has maximum at x = 5πœ‹/6 So, the correct answer is (B)

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