Question

The sum of all the local minimum values of the twice differentiable function $f: R \rightarrow R$ defined by $f(x)=x^{3}-3 x^{2}-\frac{3 f^{\prime \prime}(2)}{2} x+f^{\prime \prime}(1)$ is :

• A. -22
• B. 5
• C. -27
• D. 0

Answer 1

Answer: (C)

$f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1)$
$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} f^{\prime \prime}(2) \ldots \ldots$ (ii)
$f^{\prime \prime}(x)=6 x-6 \ldots \ldots$ (iii)
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
$f^{\prime \prime}(3)=12>0$ minima.
Use (i)
$f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1)$
$f(x)=x^{3}-3 x^{2}-\frac{3}{2} \times 68 x+0$
$f(x)=x^{3}-3 x^{2}-9 x$
$f(3)=27-27-9 \times 3=-27$