JEE Main 2022 (Online) 25th July Morning ShiftQuestion 1
If the absolute maximum value of the function $f(x)=(x^2-2x+7)e^{(4x^3-12x^2-180x+31)}$ in the interval $[-3,0]$ is $f(\alpha)$, then :
Differentiation and its Application
Solution
Concept. Every term in the bracket of $f'$ is negative on $(-3,0)$, so $f$ is decreasing and the maximum is at the left end. Formulas used. (1) $f'=e^{h}\left[(2x-2)+(x^2-2x+7)\,h'\right]$ with $h'=12x^2-24x-180=12(x-5)(x+3)$Step 1 - Signs. On $(-3,0)$: $2x-2<0$, $x^2-2x+7>0$, and $(x-5)(x+3)<0$, so $h'<0$. Step 2 - Conclude. By (1), $f'<0$, so $f$ decreases on $[-3,0]$ and the maximum is at $\alpha=-3$. Answer: (B)