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 .