Let a and b respectively be the points of local maximum and local...

.${f}^{'} ( x ) = 6{x}^{2} - 6x - 12 = 6\left({x}^{2} - x - 2\right)$

$= 6 ( x - 2 ) ( x + 1 )$

Clearly, ${f}^{'}\left(- {1}^{-}\right) > 0 \text{ and } {f}^{'}\left(- {1}^{+}\right) < 0$

Hence, $x = - 1$ is the point of local maximum.

Also, ${f}^{'}\left({2}^{-}\right) 0$

Hence, $x = 2$ is the point of local minimum.

$∴ a = - 1 \text{ and } b = 2$

Points=\left(2, - 20\right) \text{ and } \left(- 1, 7\right)$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/1d7af5e9-10e6-4b8c-b088-9cb76e1a6eaa-image.png" style="width: 295px; height: 250px;" /></p><p>$A = {∫}_{- 1}^{0}\left(2{x}^{3} - 3{x}^{2} - 12x\right)dx - {∫}_{0}^{2}\left(2{x}^{3} - 3{x}^{2} - 12x\right)dx$</p><p>$= {\left(\dfrac{{x}^{4}}{2} - {x}^{3} - 6{x}^{2}\right)}_{- 1}^{0} - {\left(\dfrac{{x}^{4}}{2} - {x}^{3} - 6{x}^{2}\right)}_{0}^{2}$</p><p>$= \dfrac{9}{2} + 24 = \dfrac{57}{2} ⇒ 4A = 114