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NDA 2019 Maths · Mathematics · Application of Derivatives · medium

The function f(x)=2x^3-15x^2+36x+7 is strictly decreasing in the interval

Single correct option

The function f(x)=2x3−15x2+36x+7f(x)=2x^3-15x^2+36x+7 is strictly decreasing in the interval
  1. A. (2,3)(2,3)
  2. B. (−∞,2)(-\infty,2)
  3. C. (3,∞)(3,\infty)
  4. D. (0,2)(0,2)
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Answer: (A) (2,3)(2,3)

f′(x)=6x2−30x+36=6(x−2)(x−3)f'(x)=6x^2-30x+36=6(x-2)(x-3). This is negative exactly when 2<x<32<x<3, so ff is strictly decreasing on (2,3)(2,3) and increasing on (−∞,2)(-\infty,2) and (3,∞)(3,\infty). Solving f′(x)=0f'(x)=0 gives the turning points, but the sign between them decides the answer.
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Original question written in the NDA 2019 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake