JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Application of Derivatives

On $[-1,3]$ the quadratic $x^2-5x+6=(x-2)(x-3)$ is non-negative on $[-1,2]$ and non-positive on $[2,3]$. Thus $f(x)=x^2-8x+8$ on $[-1,2]$ and $f(x)=-x^2+2x-4$ on $[2,3]$. The first branch has vertex at $x=4$, so it decreases across $[-1,2]$ with $f(-1)=17$ and $f(2)=-4$; the second branch has vertex at $x=1$, so it decreases across $[2,3]$ with $f(2)=-4$ and $f(3)=-7$. The greatest value attained is $17$ and the least is $-7$, so the sum of the absolute maximum and minimum is $10$.