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

Application of Derivatives PYQ — JEE Main 2022 · 25 Jul · Shift 1 · Q26 with Solution

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Correct answer: B

We need the absolute maximum of $f(x)=\left(x^2-2x+7\right)e^{\left(4x^3-12x^2-180x+31\right)}$ on the interval $[-3,0]$ .

Since $e^{g(x)}>0$ for all $x$ , we can differentiate directly and study critical points.

Using the product rule: $f'(x)=e^{v(x)}\big(u'(x)+u(x)v'(x)\big).$

Now, $u'(x)=2x-2$ and $v'(x)=12x^2-24x-180=12(x^2-2x-15)=12(x-5)(x+3).$

So $f'(x)=e^{v(x)}\left[(2x-2)+(x^2-2x+7)\cdot 12(x-5)(x+3)\right].$

Instead of expanding blindly, note the factorization: $x^2-2x+7=(x-1)^2+6>0$ for all $x$ . Also, on $[-3,0]$ , $x-5<0, \qquad x+3\ge 0,$ so $v'(x)=12(x-5)(x+3)\le 0,$ with equality only at $x=-3$ .

Let us test the sign of $f'(x)$ on $(-3,0]$ . At $x=-2$ ,

\quad v'(-2)=12(-7)(1)=-84, \quad u'(-2)=-6,$$ so $$u'(-2)+u(-2)v'(-2)=-6+15(-84)<0.$$ Thus $f'(x)<0$ there. To confirm globally, observe that on $[-3,0]$, - $u(x)>0$, - $v'(x)<0$ for $x>-3$, - and the large negative term $u(x)v'(x)$ dominates. Hence $$f'(x)<0 \quad \text{for all } x\in(-3,0].$$ So $f$ is strictly decreasing on $[-3,0]$. Therefore, the absolute maximum on $[-3,0]$ occurs at the left endpoint: $$\alpha=-3.$$ --- 4. Check the options: - A: $\alpha=0$ ❌ - B: $\alpha=-3$ ✅ - C: $\alpha\in(-1,0)$ ❌ - D: $\alpha\in(-3,-1]$ ❌ Thus the correct option is $$\boxed{\text{B}}.$$ --- 5. Comparison with stored correct answer: Stored correct answer = B. My derived answer = B. They agree.