Explanation

(d)
Let $f(x) = 2x^{3} - 3x^{2} - 12 + 5$
$\therefore\ \ \ \ \ \ \ \ f^{'}(x) = 6x^{2} - 6x - 12$
For maxima or minima, put $f^{'}(x) = 0$
$\therefore\ \ \ \ \ \ \ 6x^{2} - 6x - 12 = 0$
$\Rightarrow \ \ \ \ \ \ x = - 1,\ 2$
Now, $f^{''}(x) = 12x - 6$
$f^{''}( - 1) = - 12 - 6 = - 18 < 0$, maxima
$\therefore\ \ \ f(x)$ is maximum at $x = - 1$
At $x = - 1,\ \ f(x) = - 2 - 3 + 12 + 5 = 12$
At $x\text{=}4,\ f(x) = 128 - 48 - 48 + 5 = 37$
Hence, largest value of $f(x)$ in the given interval is 4