Question

Let $f\left( x \right)=x{{e}^{x\left( 1-x \right)}},\text{ }x\in R$. Then
(a) ${{f}^{'}}\left( x \right)\ge 0\text{ in }\left[ \dfrac{-1}{2},1 \right]$
(b) ${{f}^{'}}\left( x \right)<0\text{ in }\left[ \dfrac{-1}{2},1 \right]$
(c) ${{f}^{'}}\left( x \right)\ge 0\forall x\in R$
(d) ${{f}^{'}}\left( x \right)<0\forall x\in R$

Answer 1

Hint: If $h(x)$ is a composite function given by $h(x)=f(g(x))$ , then$h'(x)=f'(g(x))\times g'(x)$. Use the fact that if ${{x}_{1}}$ and ${{x}_{2}}$are the roots of any quadratic equation of $x$given by $f(x)\equiv a{{x}^{2}}+bx+c=0$, where $a,b$and $c$ are real and $a>0$, then $f({{x}_{3}})\le 0,\forall {{x}_{3}}\in [{{x}_{1}},{{x}_{2}}]$ and hence , $-f({{x}_{3}})\ge 0,\forall {{x}_{3}}\in [{{x}_{1}},{{x}_{2}}]$

Complete step-by-step answer:
The given function is $f\left( x \right)=x{{e}^{x\left( 1-x \right)}}$
It is of the type $f\left( x \right)=g\left( x \right).h\left( p\left( x \right) \right)$where $g\left( x \right)=x\text{, }h\left( x \right)={{e}^{x}}\text{, }p\left( x \right)=x\left( 1-x \right)$
Now ,first we will determine the derivative of the function $h(p(x))$with respect to $x$ . It will be helpful while applying product rules to determine the derivative of $f\left( x \right)$.
Clearly, we can see ${{e}^{x\left( 1-x \right)}}$is a composite function . So , to determine the value of the derivative of ${{e}^{x\left( 1-x \right)}}$ with respect to $x$, we need to apply chain rule of differentiation. The chain rule of differentiation is given as: “If $h(x)$ is a composite function given by $h(x)=f(g(x))$ , then the derivative of $h(x)$with respect to $x$is $h'(x)=f'(g(x))\times g'(x)$.”
So , $\dfrac{d}{dx}h\left( p\left( x \right) \right)=\dfrac{d}{dx}{{e}^{x\left( 1-x \right)}}$
$={{e}^{x\left( 1-x \right)}}.\dfrac{d}{dx}\left( x\left( 1-x \right) \right)$
$={{e}^{x\left( 1-x \right)}}.\dfrac{d}{dx}p\left( x \right)$
Now to evaluate $\dfrac{d}{dx}p\left( x \right)$, we need to apply product rule.
Now , we know , the product rule of differentiation is given as “If $y$is a function given as $y=f(x).g(x)$, then the derivative of $y$with respect to $x$is given as $y'=\dfrac{d}{dx}(f(x).g(x))=g(x).{{f}^{'}}(x)+f(x).{{g}^{'}}(x)$”.
So , $\dfrac{d}{dx}p\left( x \right)=\dfrac{d}{dx}x\left( 1-x \right)$
$=x\left( -1 \right)+\left( 1-x \right).1$
$=-x+1-x$
$=1-2x$
$\therefore \dfrac{d}{dx}h\left( p\left( x \right) \right)={{e}^{x\left( 1-x \right)}}.\left( 1-2x \right)$
Now , we will evaluate the derivative of $f\left( x \right)$with respect to$x$.
To evaluate the derivative of $f\left( x \right)$with respect to$x$, i.e. $\dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( g\left( x \right).h\left( p\left( x \right) \right) \right)$, we need to apply product rule.
So , ${{f}^{'}}\left( x \right)=\dfrac{d}{dx}h\left( p\left( x \right) \right).g\left( x \right)+h\left( p\left( x \right) \right).\dfrac{d}{dx}\left( g\left( x \right) \right)$
$={{e}^{x\left( 1-x \right)}}.\left( 1-2x \right).x+{{e}^{x\left( 1-x \right)}}.1$
$={{e}^{x\left( 1-x \right)}}.\left( 1+x-2{{x}^{2}} \right)$
$=-2{{e}^{x\left( 1-x \right)}}\left( x-1 \right)\left( x+\dfrac{1}{2} \right)$
Now , we can see ${{f}^{'}}\left( x \right)=0$at $x=-\dfrac{1}{2}$and at $x=1$.
We know , ${{e}^{x\left( 1-x \right)}}\ge 0\forall x\in R$ and $\left( x-1 \right)\left( x+\dfrac{1}{2} \right)$ represents a quadratic in $x$.
Now , if ${{x}_{1}}$ and ${{x}_{2}}$are the roots of any quadratic equation of $x$given by $f(x)\equiv a{{x}^{2}}+bx+c=0$, where $a,b$and $c$ are real and $a>0$, then $f({{x}_{3}})\le 0,\forall {{x}_{3}}\in [{{x}_{1}},{{x}_{2}}]$ and hence , $-f({{x}_{3}})\ge 0,\forall {{x}_{3}}\in [{{x}_{1}},{{x}_{2}}]$
So , ${{f}^{'}}\left( x \right)\ge 0$ when $-\dfrac{1}{2}\le x\le 1$.
So , ${{f}^{'}}\left( x \right)\ge 0$in $\left[ -\dfrac{1}{2},1 \right]$.
OPTION (a) is the correct answer.

Note: ${{e}^{x\left( 1-x \right)}}$is a composite function. Hence , its derivative will be ${{e}^{x\left( 1-x \right)}}.\left( 1-2x \right)$and not ${{e}^{x\left( 1-x \right)}}$. Students generally make this mistake . These details should be taken care of and such mistakes should be avoided as these mistakes result in getting a wrong answer .