The sum of all the local minimum values of the twice differentiable function $f: \mathbf{R} \rightarrow \mathbf{R}$ defined by $f(\mathrm{x})=\mathrm{x}^{3}-3 \mathrm{x}^{2}-\frac{3 f^{\prime \prime}(2)}{2} \mathrm{x}+f^{\prime \prime}(1)$ is :
Correct Option: , 3
$f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1) \ldots(1)$
$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} f^{\prime \prime}(2) \ldots(2)$
$f^{\prime \prime}(x)=6 x-6$..(3)
Now is $3^{\text {rd }}$ equation
$f^{\prime \prime}(2)=12-6=6$
$f^{\prime \prime}(11=0)$
Use (ii)
$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} f^{\prime \prime}(2)$
$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} \times 6$
$f^{\prime}(x)=3 x^{2}-6 x-9$
$f^{\prime}(x)=0$
$3 x^{2}-6 x-9=0$
$\Rightarrow x=-1 \& 3$
Use (iii)
$f^{\prime \prime}(x)=6 x-6$
$f^{\prime \prime}(-1)=-12<0$ maxima
$f^{\prime}(3)=12>0$ minima.
Use (i)
$f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1)$
$f(x)=x^{3}-3 x^{2}-\frac{3}{2} \times 6 \times x+0$
$f(x)=x^{3}-3 x^{2}-9 x$
$f(3)=27-27-9 \times 3=-27$