Question

If $f\left( x \right) = x{e^{x\left( {1 - x} \right)}}$, then $f\left( x \right)$ is
A. increasing on $R$
B. decreasing on $\left[ { - \left( {\dfrac{1}{2}} \right),1} \right]$
C. increasing on $\left[ { - \left( {\dfrac{1}{2}} \right),1} \right]$
D. decreasing on $R$

Answer 1

Hint: Differentiate the given function $x{e^{x\left( {1 - x} \right)}}$ with respect to $x$, then put the first derivative equal to zero to find the value of $x$ or range of $x$. Now, check whether $f\left( x \right)$ is increasing or decreasing in that range by putting required values of $x$in the function.

Formula Used:
Chain rule –
$\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right) \times \dfrac{d}{{dx}}g\left( x \right)$
Product rule –
$\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)$

Complete step by step solution:
Given that,
$f\left( x \right) = x{e^{x\left( {1 - x} \right)}}$
Differentiate $f\left( x \right)$ with respect to $x$,
$f'\left( x \right) = {e^{x\left( {1 - x} \right)}}\dfrac{d}{{dx}}\left( x \right) + x\dfrac{d}{{dx}}\left( {{e^{x\left( {1 - x} \right)}}} \right)$
$f'\left( x \right) = {e^{x\left( {1 - x} \right)}}\left( 1 \right) + x{e^{x\left( {1 - x} \right)}}\left( {1 - 2x} \right)$
$f'\left( x \right) = {e^{x\left( {1 - x} \right)}} + x{e^{x\left( {1 - x} \right)}} - 2{x^2}{e^{x\left( {1 - x} \right)}}$
$f'\left( x \right) = {e^{x\left( {1 - x} \right)}}\left( {2{x^2} - x - 1} \right)$
$f'\left( x \right) = {e^{x\left( {1 - x} \right)}}\left( {2{x^2} - 2x + x - 1} \right)$
$f'\left( x \right) = {e^{x\left( {1 - x} \right)}}\left( {2x(x - 1) + 1\left( {x - 1} \right)} \right)$
$f'\left( x \right) = {e^{x\left( {1 - x} \right)}}\left( {2x + 1} \right)(x - 1)$
Take, $f'\left( x \right) = 0$
$ \Rightarrow x = - \dfrac{1}{2},1$
Here, $f'\left( x \right) > 0$ when $ - \dfrac{1}{2} < x < 1$
$ \Rightarrow $ $f\left( x \right)$ is increasing on $\left[ { - \left( {\dfrac{1}{2}} \right),1} \right]$

Option ‘C’ is correct

Note: The key concept involved in solving this problem is the good knowledge of chain rule. Students must remember that while doing derivatives using chain rule, do the derivative of the first function and then do the derivative of the function inside the first function.