<p>Show \(\displaystyle\int_1^3 e^{x^2-3x}\,dx\le 1\). Find the maximum of \(e^{x^2-3x}\) on \([1,3]\). [JEE Advanced 2016]</p>
Step-by-Step Solution
Key Concept: f(x)=x^2-3x has vertex at x=3/2 (minimum -9/4). At endpoints x=1,3: f=-2. So max of e^f = max(e^(-9/4),e^(-2)) = e^(-2) < 1. Integral \leq 2 \cdot e^(-2) < 1.
<div class='solution'>
<p>$f(x)=x^2-3x=(x-3/2)^2-9/4$. Minimum at $x=3/2$: $f=-9/4$. At $x=1,3$: $f=-2$.</p>
<p>So $e^{f(x)}\le e^{-2}$ on $[1,3]$ (since $-9/4 < -2$ and the max occurs at endpoints).</p>
<p>$$\int_1^3 e^{x^2-3x}dx\le e^{-2}\cdot 2=2e^{-2}<1\qquad(\because e^2>2)$$. ✓</p>
<p>The minimum of $e^{f(x)}$ is $e^{-9/4}$ at $x=3/2$. Maximum of integrand is $e^{-2}$ at endpoints. ✓(B)</p>
Correct Answer: B