The real function f (x) = 2x3 – 3x2 – 36x + 7 is:

(a) Strictly increasing in (−∞,−2) and strictly decreasing in ( −2, ∞)
(b) Strictly decreasing in (−2, 3)
(c) Strictly decreasing in (−∞, 3) and strictly increasing in (3, ∞)
(d) Strictly decreasing in (−∞,−2) ∪ (3, ∞)

 

This question is inspired from Ex 6.2, 5 - Chapter 6 Class 12 - Application of Derivatives

Ques 26 (MCQ) - The real function f(x) = 2x3 – 3x2 – 36x + 7 is: - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

part 2 - Question 26 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 26 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 26 The real function f (x) = 2x3 – 3x2 – 36x + 7 is: (a) Strictly increasing in (−∞,−2) and strictly decreasing in ( −2, ∞) (b) Strictly decreasing in (−2, 3) (c) Strictly decreasing in (−∞, 3) and strictly increasing in (3, ∞) (d) Strictly decreasing in (−∞,−2) ∪ (3, ∞) f(𝑥) = 2𝑥3 – 3𝑥2 – 36𝑥 + 7 Calculating f’(𝒙) f’(𝑥) = 6𝑥2 – 6𝑥 – 36 + 0 f’(𝑥) = 6 (𝑥2 – 𝑥 – 6 ) f’(𝑥) = 6(𝑥^2 – 3𝑥 + 2𝑥 – 6) f’(𝑥) = 6(𝑥(𝑥 − 3) + 2 (𝑥 − 3)) f’(𝒙) = 6(𝒙 – 3) (𝒙 + 2) Putting f’(x) = 0 6(𝑥+2)(𝑥 –3)=0 (𝑥+2)(𝑥 –3)=0 So, x = −2 and x = 3 Plotting points on number line Hence, f is strictly increasing in (−∞ ,−𝟐) & (𝟑 ,∞) f is strictly decreasing in (−𝟐, 𝟑) So, the correct answer is (D)

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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