JEE Main 2023 — Mathematics Calculus

Given,

$x=2$x=2 be a local minima of the function $f(x)=2{x}^{4}-18{x}^{2}+8x+12,x\in (-4,4)$f(x)=2x4−18x2+8x+12,x∈(−4,4)

Now differentiating the above function we get,

$${f}^{'}(x)=8{x}^{3}-36x+8$$f′(x)=8x3−36x+8

$$\Rightarrow {f}^{'}(x)=4(2{x}^{3}-9x+2)$$⇒f′(x)=4(2x3−9x+2)

Now equating it to zero we get, ${f}^{'}(x)=0$f′(x)=0

$$\Rightarrow 2{x}^{3}-9x+2=0$$⇒2x3−9x+2=0

$$\Rightarrow (x-2)(2{x}^{2}+4x-1)=0$$⇒(x−2)(2x2+4x−1)=0

$$\Rightarrow x=\frac{-4\pm \sqrt{24}}{4}$$⇒x=4−4±24​​

$$\Rightarrow x=\frac{-2\pm \sqrt{6}}{2}$$⇒x=2−2±6​​

$\therefore x=\frac{\sqrt{6}-2}{2}$∴x=26​−2​ {by checking sign change of ${f}^{'}(x)$f′(x) for maxima}

Now rewriting $f(x)=2{x}^{4}-18{x}^{2}+8x+12$f(x)=2x4−18x2+8x+12 we get,

$$f(x)=({x}^{2}-2x-\frac{9}{2})(2{x}^{2}+4x-1)+24x+7.5$$f(x)=(x2−2x−29​)(2x2+4x−1)+24x+7.5

Hence, $f(\frac{\sqrt{6}-2}{2})=M=12\sqrt{6}-\frac{33}{2}$f(26​−2​)=M=126​−233​