Question 14 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Last updated at August 10, 2026 by

At x = 5π/6, f (x) = 2 sin 3x + 3 cos 3x is :

(A) maximum 

(B) minimum

(C) zero 

(D) neither maximum or minimum

At x = 5π/6, f (x) = 2 sin3x + 3 cos3x is: - Teachoo Maths [MCQ] - NCERT Exemplar - MCQs

part 2 - Question 14 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Question 14 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Transcript

Question 14 At x = 5๐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 = 5ฯ€/6 So, we will find f โ€ (x) f (x) = 2 sin 3๐‘ฅ + 3 cos 3๐‘ฅ Finding f โ€™ (x) f โ€™ (x) = 6 cos 3๐‘ฅ โˆ’ 9 sin 3๐‘ฅ Finding f โ€™โ€™ (x) fโ€™โ€™ (x) = โˆ’18 sin 3๐‘ฅ โˆ’ 27 cos 3๐‘ฅ At x = ๐Ÿ“๐…/๐Ÿ” fโ€™โ€™ (๐Ÿ“๐…/๐Ÿ”) = โˆ’18 sin (3(5๐œ‹/6))โˆ’ 27 cos (3(5๐œ‹/6)) = โˆ’18 sin (5๐œ‹/2) โˆ’ 27 cos (5๐œ‹/2) = โˆ’18 sin (2๐œ‹+๐œ‹/2) โˆ’ 27 cos (2๐œ‹+๐œ‹/2) = โˆ’18 sin ๐œ‹/2 โˆ’ 27 cos ๐œ‹/2 = โˆ’ 18 (1) โˆ’ 27 (0) = โˆ’18 < 0 Since fโ€™โ€™(x) < 0 at x = 5๐œ‹/6 โˆด f has maximum at x = 5๐œ‹/6 So, the correct answer is (B)

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