JEE MAIN - Mathematics (2024 - 30th January Evening Shift - No. 14)
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
108
392
608
600
Explanation
$$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}+3)^2 \cdot 3(\mathrm{x}-2)^2+(\mathrm{x}-2)^3 2(\mathrm{x}+3) \\ & =5(\mathrm{x}+3)(\mathrm{x}-2)^2(\mathrm{x}+1) \\ & \mathrm{f}^{\prime}(\mathrm{x})=0, \mathrm{x}=-3,-1,2 \end{aligned}$$
_30th_January_Evening_Shift_en_14_1.png)
$$\begin{aligned} & f(-4)=-216 \\ & f(-3)=0, f(4)=49 \times 8=392 \\ & M=392, m=-216 \\ & M-m=392+216=608 \\ & \text { Ans }=\text { '3' } \end{aligned}$$
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