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 :
- A \(600\)
- B \(392\)
- C \(608\)
- D \(108\)
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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