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

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Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.