JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Application of Derivatives

With $f(x)=2x^3-3x^2-12x$, $f^{\prime }(x)=6x^2-6x-12=6(x-2)(x+1)$, so $f^{\prime }$ turns from positive to negative at $x=-1$ and from negative to positive at $x=2$. Hence $a=-1$ and $b=2$.
Factorising, $f(x)=x\left(2x^2-3x-12\right)$, and the quadratic factor has roots $\frac{3\pm \sqrt{105}}{4}$, both outside $[-1,2]$. So the only sign change of $f$ on $[-1,2]$ is at $x=0$, with $f>0$ on $(-1,0)$ and $f<0$ on $(0,2)$. Taking $G(x)=\frac{x^4}{2}-x^3-6x^2$ as an antiderivative,

The total area is $A=\frac{9}{2}+24=\frac{57}{2}$, so $4A=114$.