JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q6

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If the absolute maximum value of the function $f(x)=\left(x^2-2x+7\right)e^{\left(4x^3-12x^2-180x+31\right)}$ in the interval $[-3,0]$ is $f(\alpha)$ , then :

\alpha=0

\alpha=-3

\alpha\in(-1,0)

\alpha\in(-3,-1]

Answer and worked solution

Official answer: B

Exponent. $g=4x^3-12x^2-180x+31$ has $g'=12(x-5)(x+3)\le0$ on $[-3,0]$ , so $g$ is largest at $x=-3$ ; $x^2-2x+7$ also decreases on $[-3,0]$ . Both factors peak at $x=-3$ , so $\alpha=-3$ .

(B)

Topics:Maxima and minima Monotonicity

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JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 6