Question

If the function $f(x)=ax{e}^{bx}{}^2$ 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)

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