JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Application of Derivatives

First, analyze $3x-x^2+2=-(x^2-3x-2)$. The roots are $x=\frac{3\pm \sqrt{17}}{2}$.

In $[-1,2]$, the expression $3x-x^2+2$ is positive except between the roots. So:
$f(x)=\{\begin{array}{ll}{x}^2-4x-2& \text{for }x\in [-1,\frac{3-\sqrt{17}}{2})\\ -x^2+2x+2& \text{for }x\in [\frac{3-\sqrt{17}}{2},2]\end{array}$

Find critical points and evaluate at endpoints and critical points:

• $f(-1)=3$
• $f(\frac{3-\sqrt{17}}{2})=\frac{\sqrt{17}-3}{2}$
• $f(1)=3$ (critical point in the second piece)
• $f(2)=2$

Absolute minimum: $\frac{\sqrt{17}-3}{2}$, absolute maximum: $3$.

Sum: $\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}$.