Question

If $f\left(x\right)=x^2{e}^{-x^2/a^2}$ is an increasing function then (for $a>0$), x lies in the interval

• A. $[a,2a]$
• B. $(-\infty ,-a)\cup (0,a)$
• C. $(-a,0)$
• D. None of these

Answer 1

The correct option is B $(-\infty ,-a)\cup (0,a)$

For a function $f\left(x\right)$ to be increasing, $f^{{}^{\prime }}\left(x\right)>0$

$f\left(x\right)=x^2{e}^{-x^2/a^2}$

$ln\left(f\right(x\left)\right)=2lnx-\frac{x^2}{a^2}$

$\therefore \frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}=\frac{2}{x}-\frac{2x}{a^2}$

$\therefore f^{{}^{\prime }}\left(x\right)=\frac{2f\left(x\right)}{a^2}\left(\frac{a^2-x^2}{x}\right)$

$\left(f\right(x)>0$ $\forall x$

$\therefore f^{{}^{\prime }}\left(x\right)⟹\frac{a^2-x^2}{x}>0$

$x\in (-\infty ,-a)\cup (0,a)$