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

## (A) maximum

## (B) minimum

## (C) zero

## (D) neither maximum or minimum

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

NCERT Exemplar - MCQs

Question 1
Important

Question 2 Important

Question 3 Important

Question 4 Important

Question 5 Important

Question 6

Question 7 Important

Question 8 Important

Question 9 Important

Question 10

Question 11 Important

Question 12 Important

Question 13

Question 14 You are here

Question 15 Important

Question 16 Important

Question 17 Important

Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Deleted for CBSE Board 2024 Exams

Question 7 Important Deleted for CBSE Board 2024 Exams

Question 8 Deleted for CBSE Board 2024 Exams

Question 9 Important Deleted for CBSE Board 2024 Exams

Question 10 Important Deleted for CBSE Board 2024 Exams

Question 11 Deleted for CBSE Board 2024 Exams

Question 12 Deleted for CBSE Board 2024 Exams

Question 13 Deleted for CBSE Board 2024 Exams

Chapter 6 Class 12 Application of Derivatives

Serial order wise

Last updated at May 29, 2023 by Teachoo

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Question 14 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 = 5π/6 So, we will find f ” (x) f (x) = 2 sin 3𝑥 + 3 cos 3𝑥 Finding f ’ (x) f ’ (x) = 6 cos 3𝑥 − 9 sin 3𝑥 Finding f ’’ (x) f’’ (x) = −18 sin 3𝑥 − 27 cos 3𝑥 At x = 𝟓𝝅/𝟔 f’’ (𝟓𝝅/𝟔) = −18 sin (3(5𝜋/6))− 27 cos (3(5𝜋/6)) = −18 sin (5𝜋/2) − 27 cos (5𝜋/2) = −18 sin (2𝜋+𝜋/2) − 27 cos (2𝜋+𝜋/2) = −18 sin 𝜋/2 − 27 cos 𝜋/2 = − 18 (1) − 27 (0) = −18 < 0 Since f’’(x) < 0 at x = 5𝜋/6 ∴ f has maximum at x = 5𝜋/6 So, the correct answer is (B)