Question

Let $x=2$ be a local minima of the function $f(x)=2 x^4-18 x^2+8 x+12$, $x \in(-4,4)$. If $M$ is local maximum value of the function $f$ in $(-4,4)$, then $M =$

• A. $18 \sqrt{6}-\frac{33}{2}$
• B. $12 \sqrt{6}-\frac{33}{2}$
• C. $12 \sqrt{6}-\frac{31}{2}$
• D. $18 \sqrt{6}-\frac{31}{2}$

Answer 1

Answer: (B)

$f^{\prime}(x)=8 x^3-36 x+8=4\left(2 x^3-9 x+2\right) $
$ f^{\prime}(x)=0 $
$ \therefore x=\frac{\sqrt{6}-2}{2}$
Now
$f(x)=\left(x^2-2 x-\frac{9}{2}\right)\left(2 x^2+4 x-1\right)+24 x+7.5 $
$ \therefore f\left(\frac{\sqrt{6}-2}{2}\right)=M=12 \sqrt{6}-\frac{33}{2}$