Application Of Derivatives Question 2
Question 2 - 25 January - Shift 1
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 $M=$
(1) $12 \sqrt{6}-\frac{33}{2}$
(2) $12 \sqrt{6}-\frac{31}{2}$
(3) $18 \sqrt{6}-\frac{33}{2}$
(4) $18 \sqrt{6}-\frac{31}{2}$
Show Answer
Answer: (1)
Solution:
Formula: Maxima and minima of a function
$f^{\prime}(x)=8 x^{3}-36 x+8=4(2 x^{3}-9 x+2)$
$f^{\prime}(x)=0$
$\therefore x=\frac{\sqrt{6}-2}{2}$
Now
$f(x)=(x^{2}-2 x-\frac{9}{2})(2 x^{2}+4 x-1)+24 x+7.5$
$\therefore f(\frac{\sqrt{6}-2}{2})=M=12 \sqrt{6}-\frac{33}{2}$
