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JEE Mains · Maths · STD 12 - 6. Application of derivatives

Let \(f(x)=(x+3)^2(x-2)^3, x \in[-4,4]\). If \(M\) and \(m\) are the maximum and minimum values of \(f\), respectively in \([-4,4]\), then the value of \(M-m\) is :

  1. A \(600\)
  2. B \(392\)
  3. C \(608\)
  4. D \(108\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(608\)

Step-by-step Solution

Detailed explanation

\(\mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}+3)^2 \cdot 3(\mathrm{x}-2)^2+(\mathrm{x}-2)^3 2(\mathrm{x}+3) \) \( =5(\mathrm{x}+3)(\mathrm{x}-2)^2(\mathrm{x}+1) \) \(\mathrm{f}^{\prime}(\mathrm{x})=0, \mathrm{x}=-3,-1,2\) \( f(-4)=-216 \) \( f(-3)=0, f(4)=49 \times 8=392 \)…
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