JEE MAIN - Mathematics (2022 - 25th July Morning Shift - No. 6)
If the absolute maximum value of the function $$f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}$$ in the interval $$[-3,0]$$ is $$f(\alpha)$$, then :
$$\alpha=0$$
$$ \alpha=-3$$
$$\alpha \in(-1,0)$$
$$\alpha \in(-3,-1]$$
Explanation
Given, $f(x)=\underbrace{\left(x^{2}-2 x+7\right)}_{f_{1}(x)} \underbrace{e^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}}_{f_{2}(x)}$
$f_{1}(x)=x^{2}-2 x+7$
$f_{1}^{\prime}(x)=2 x-2$, so $f(x)$ is decreasing in $[-3,0]$ and positive also
$f_{2}(x)=e^{4 x^{3}-12 x^{2}-180 x+31}$
$f_{2}^{\prime}(x)=e^{4 x^{3}-12 x^{2}-180 x+31} \cdot 12 x^{2}-24 x-180$
$=12(x-5)(x+3) e^{4 x^{3}-12 x^{2}-180 x+31}$
So, $f_{2}(x)$ is also decreasing and positive in $\{-3,0\}$
$\therefore$ absolute maximum value of $f(x)$ occurs at $x=-3$
$\therefore \quad \alpha=-3$
$f_{1}(x)=x^{2}-2 x+7$
$f_{1}^{\prime}(x)=2 x-2$, so $f(x)$ is decreasing in $[-3,0]$ and positive also
$f_{2}(x)=e^{4 x^{3}-12 x^{2}-180 x+31}$
$f_{2}^{\prime}(x)=e^{4 x^{3}-12 x^{2}-180 x+31} \cdot 12 x^{2}-24 x-180$
$=12(x-5)(x+3) e^{4 x^{3}-12 x^{2}-180 x+31}$
So, $f_{2}(x)$ is also decreasing and positive in $\{-3,0\}$
$\therefore$ absolute maximum value of $f(x)$ occurs at $x=-3$
$\therefore \quad \alpha=-3$
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