If the solve the problem

Solution:

(c) 2

Given : $f(x)=2 x^{3}-3 x^{2}-12 x+5$

$\Rightarrow f^{\prime}(x)=6 x^{2}-6 x-12$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow 6 x^{2}-6 x-12=0$

$\Rightarrow x^{2}-x-2=0$

$\Rightarrow(x-2)(x+1)=0$

$\Rightarrow x=2,-1$

Now,

$f^{\prime \prime}(x)=12 x-6$

$\Rightarrow f^{\prime \prime}(-1)=-12-6=-18<0$

So, $x=1$ is a local maxima.

Also,

$f^{\prime \prime}(2)=24-6=18>0$

So, $x=2$ is a local minima.