Question

If the function $f ( x )= axe ^{ bx }{ }^2$ have the maximum value $f (2)=1$, then

• A. $a=\frac{\sqrt{e}}{2}$ and $b=\frac{-1}{8}$
• B. $a=\frac{-\sqrt{ e }}{2}$ and $b =\frac{1}{8}$
• C. $a =\frac{1}{8}$ and $b =\frac{\sqrt{ e }}{2}$
• D. $a=\frac{-\sqrt{e}}{2}$ and $b=\frac{-1}{8}$

Answer 1

Answer: (A)

$f ( x )= axe ^{ bx ^2} ; f (2)=1$ and $f ^{\prime}(2)=0 ; 2 ae e ^{4 b }=1$....(1)
also $ f^{\prime}(x)=a\left[x e^{b x^2} \cdot 2 b x+e^{b x^2}\right]=a x e^{b x^2}\left[2 b x^2+1\right]$
$f^{\prime}(2)=a e^{4 b}(8 b+1)=0 ; a=0$
or $b=-1 / 8$ but $a \neq 0 ; a=e / 2$
hence $a=\frac{\sqrt{ e }}{2}$ and $b =\frac{-1}{8}$