Question

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

Answer 1

Hint:- In this question first we need to find the derivative of $x{e^{x(1 - x)}}$ using derivative formulas. Then equate $f'(x) = 0$ to find the local extremum points of given function. After this we have to check between these local extremum points $f'(x)$ is increasing or decreasing.

Complete step-by-step answer:
Given: $f(x) = x{e^{x(1 - x)}}$ ---- eq.1
We know, the derivative formula of $\dfrac{{d(u.v)}}{{dx}} = u.\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$. ---- eq.2
$ \Rightarrow \dfrac{{dx}}{{dx}} = 1$ ----eq.3
And
$ \Rightarrow \dfrac{{d{e^x}}}{{dx}} = {e^x}$ ---eq.4
Now, differentiate $f(x)$with respect to $x$ and use above derivative formulas, we get
$ \Rightarrow \dfrac{{df(x)}}{{dx}} = \dfrac{{d\{ x{e^{x(1 - x)}}\} }}{{dx}}$
$ \Rightarrow f'(x) = {e^{x(1 - x)}} + x{e^{x(1 - x)}}(1 - 2x)$
Now, take ${e^{x(1 - x)}}$ common in RHS, we get
$ \Rightarrow f'(x) = {e^{x(1 - x)}}\{ 1 + x(1 - 2x)\} $
Now, we make a quadratic equation on RHS.
\[ \Rightarrow f'(x) = {e^{x(1 - x)}}\{ 1 + x - 2{x^2}\} \]
Now, take negative sign from \[\{ 1 + x - 2{x^2}\} \] to make coefficient of ${x^2}$ positive, we get
\[ \Rightarrow f'(x) = - {e^{x(1 - x)}}\{ 2{x^2} - x - 1\} \]
Now, factorise the quadratic equation \[\{ 2{x^2} - x - 1\} \], we get
$ \Rightarrow f'(x) = - {e^{x(1 - x)}}\{ 2{x^2} - 2x + x - 1\} \\
   \Rightarrow f'(x) = - {e^{x(1 - x)}}\{ (x - 1)(2x + 1)\} {\text{ ---- eq}}{\text{.5}} \\
$
Now, for finding the increasing or decreasing of function $f(x)$.
We have to put $f'(x) = 0$.
$ \Rightarrow - {e^{x(1 - x)}}\{ (x - 1)(2x + 1)\} = 0$
We know, exponential function (${e^{x(1 - x)}}$) can never be zero. So
$ \Rightarrow \{ (x - 1)(2x + 1)\} = 0$
Above equation gives us the condition $f'(x) = 0$. On equating each factor equal to zero , we get
$ \Rightarrow x = 1,\dfrac{{ - 1}}{2}$
Now, we know if $f'(x)$is positive then $f(x)$is increasing and if $f'(x)$ is negative then $f(x)$is decreasing.
We know, exponential function \[{e^{x(1 - x)}}\] can never be negative. So we have to check for $(x - 1)(2x + 1)$ to when it is greater or less than to zero.
$
  {\text{f'(x) > 0 i}}{\text{.e}}{\text{. f(x) is increasing }} - \dfrac{1}{2} \leqslant {\text{ }}x \leqslant 1 \\
  {\text{f'(x) < 0 i}}{\text{.e}}{\text{. f(x) is decreasing otherwise }} \\
$
Therefore, f(x) is greater than zero i.e. increasing between the range $ - \dfrac{1}{2} \leqslant {\text{ }}x \leqslant 1$ and f(x) is less than zero i.e. decreasing for all x other than range $ - \dfrac{1}{2} \leqslant {\text{ }}x \leqslant 1$. So, $f(x) = x{e^{x(1 - x)}}$ is increasing \[{\text{ }}\left[ { - \dfrac{1}{2},1} \right]\].
Hence, option A is correct.

Note:-Whenever you get this type of question the key concept to solve this is to learn the all basics derivative formulas. Using these formulas find the derivative of a given function ($f'(x)$). And then put $f'(x) = 0$ to get local extreme points. Then check where it is increasing if $f'(x) > 0$ and decreasing $f'(x) < 0$. And remember one more thing that exponential function can never be negative.