JEE Main 2023 — Application of Derivatives Question with Solution

Given, 

$x=2$ be a local minima of the function $f\left(x\right)=2x^4-18x^2+8x+12,x\in \left(-4,4\right)$

Now differentiating the above function we get,

$f^{\text{'}}\left(x\right)=8x^3-36x+8$

$\Rightarrow f^{\text{'}}\left(x\right)=4\left(2x^3-9x+2\right)$

Now equating it to zero we get, $f^{\text{'}}\left(x\right)=0$

$\Rightarrow 2x^3-9x+2=0$

$\Rightarrow \left(x-2\right)\left(2x^2+4x-1\right)=0$

$\Rightarrow x=\frac{-4\pm \sqrt{24}}{4}$

$\Rightarrow x=\frac{-2\pm \sqrt{6}}{2}$

$\therefore x=\frac{\sqrt{6}-2}{2}$ {by checking sign change of $f^{\text{'}}\left(x\right)$ for maxima}

Now rewriting $f\left(x\right)=2x^4-18x^2+8x+12$ we get,

$f\left(x\right)=\left(x^2-2x-\frac{9}{2}\right)\left(2x^2+4x-1\right)+24x+7.5$

Hence, $f\left(\frac{\sqrt{6}-2}{2}\right)=M=12\sqrt{6}-\frac{33}{2}$