Question

The function $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}}$ monotonically increasing if

• A. $\mathrm{x}<0$ only
• B. $\mathrm{x}>2$ only
• C. $0<\mathrm{x}<2$
• D. $\mathrm{x}\in \left(-\infty ,0\right)\cup \left(2,\infty \right)$

Answer 1

The correct option is C

$0<\mathrm{x}<2$

Explanation of correct answer:

Step 1:Find the first order derivative of the given function

If a function increases or decreases across its entire domain, it is said to be monotonic function.

The function $\mathrm{f}\left(\mathrm{x}\right)$ is increasing over an interval if, $\mathrm{f}\text{'}\left(\mathrm{x}\right)>0$, and the function $\mathrm{f}\left(\mathrm{x}\right)$ is decreasing over an interval if, $\mathrm{f}\text{'}\left(\mathrm{x}\right)<0$.

The given function is, $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}}$.

By using exponent rule, we get,

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

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=2{\mathrm{xe}}^{-\mathrm{x}}-{\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{x}}$

$={\mathrm{xe}}^{-\mathrm{x}}\left(2-\mathrm{x}\right)$

Step 2: Apply the condition for function to be monotonically increasing

For $\mathrm{f}\left(\mathrm{x}\right)$ to be monotonic increasing, we get,

$\mathrm{f}\text{'}\left(\mathrm{x}\right)>0$

$\Rightarrow$ ${\mathrm{e}}^{-\mathrm{x}}\mathrm{x}\left(2-\mathrm{x}\right)>0$

$\Rightarrow$ $\mathrm{x}\left(2-\mathrm{x}\right)>0$

$\Rightarrow$ $\mathrm{x}\left(\mathrm{x}-2\right)<0$

$\Rightarrow$ $0<\mathrm{x}<2$

So, the monotone intervals of the given function $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}}$ is $0<\mathrm{x}<2$.

Hence, Option $\left(\mathrm{C}\right)$ is the correct .