Question

Let $f\left( x \right)={{e}^{x}}-x$and $g\left( x \right)={{x}^{2}},\forall x\in R$. Then the set of all $x\in R$where the function $h\left( x \right)=\left( fog \right)\left( x \right)$ is increasing is,\[\]
A.$\left[ -1,\dfrac{-1}{2} \right]\bigcup \left[ \dfrac{1}{2},\infty \right)$\[\]
B. $\left[ 0,\dfrac{1}{2} \right]\bigcup \left[ 1,\infty \right)$\[\]
C. $\left[ \dfrac{-1}{2},0 \right]\bigcup \left[ 1,\infty \right)$\[\]
D. $\left[ 0,\infty \right)$\[\]

Answer 1

Hint: We define the function $h\left( x \right)$ using $f\left( x \right)$ and $g\left( x \right)$. We differentiate $h\left( x \right)$ and take the condition${{h}^{'}}\left( x \right)\ge 0$. We factorize ${{h}^{'}}\left( x \right)$ and then investigate each factor for which interval contains the possible values of $x$.\[\]

Complete step by step answer:
We know that the function $f\left( x \right)$ is increasing within the interval $\left[ a,b \right]$ then ${{f}^{'}}\left( x \right)\ge 0$ for all $x\in \left[ a,b \right]$
The given real valued functions are $f\left( x \right)={{e}^{x}}-x$and $g\left( x \right)={{x}^{2}}-x$ . Let us assume $g\left( x \right)=u$.The given composite function can be written with substitution $g\left( x \right)=u$ as
\[h\left( x \right)=fog\left( x \right)=f\left( g\left( x \right) \right)=f\left( u \right)={{e}^{u}}-u\]
We put back $u=g\left( x \right)={{x}^{2}}-x$ above and get ,
\[h\left( x \right)={{e}^{{{x}^{2}}-x}}-{{x}^{2}}-x\]
We now differentiate $h\left( x \right)$ with respect to $x$ and get
\[\begin{align}
  & {{h}^{'}}\left( x \right)=\left( 2x-1 \right){{e}^{{{x}^{2}}-x}}-\left( 2x-1 \right) \\
 & \Rightarrow {{h}^{'}}\left( x \right)=\left( 2x-1 \right)\left( {{e}^{{{x}^{2}}-x}}-1 \right) \\
\end{align}\]
We see that the function $h\left( x \right)$ is increasing in the given domain $R$ then ${{h}^{'}}\left( x \right)>0$. So we have
\[{{h}^{'}}\left( x \right)=\left( 2x-1 \right)\left( {{e}^{{{x}^{2}}-x}}-1 \right)\ge 0\]
We know that the product two numbers is positive if and only if when both of them are positive or both of them are negative. So we get two cases.\[\]
Case-1: If both of them are non-negative , we take the first factor and find the possible values of $x$.
\[\begin{align}
  & 2x-1\ge 0 \\
 & \Rightarrow x\ge \dfrac{1}{2} \\
 & \Rightarrow x\in \left[ \dfrac{1}{2},\infty \right) \\
\end{align}\]
 Now we take the second factor and get,
\[\begin{align}
  & {{e}^{{{x}^{2}}-x}}-1\ge 0 \\
 & \Rightarrow {{e}^{{{x}^{2}}-x}}\ge 1 \\
 & \Rightarrow {{e}^{{{x}^{2}}-x}}\ge {{e}^{0}} \\
\end{align}\]
We know from the identity that for any $a>1$ and $m,n\in R$, ${{a}^{m}}>{{a}^{n}}\Leftrightarrow m>n$. We use it in above and get
\[\begin{align}
  & \Rightarrow {{e}^{{{x}^{2}}-x}}\ge {{e}^{0}} \\
 & \Rightarrow {{x}^{2}}-x\ge 0 \\
 & \Rightarrow x\left( x-1 \right)\ge 0 \\
 & \Rightarrow x\ge 0,x-1\ge 0\text{ or }x\le 0,x-1\le 0 \\
\end{align}\]
So if we consider $x\ge 0,x-1\ge 0$ then the admissible interval for $x$ is $\left[ 0,\infty \right)\bigcap \left[ 1,\infty \right)=\left[ 1,\infty \right)$. If we can also consider $x\le 0,x-1\le 0$ then the admissible interval for $x$ is $\left( -\infty ,0 \right]\bigcap \left( -\infty ,1 \right]=\left( -\infty ,0 \right]$. So all possible of $x$ from the second factor is $\left( -\infty ,0 \right]\bigcup \left[ 1,\infty \right).$
 So if take the intersection of the two intervals obtained from both the factors in case-1 then we get
\[x\in \left[ \dfrac{1}{2},\infty \right)\bigcap \left( \left( -\infty ,0 \right]\bigcup \left[ 1,\infty \right) \right)=\left[ 1,\infty \right)\]
Case-2: If both of them are non-positive , we take the first factor and find the possible values of $x$.
\[\begin{align}
  & 2x-1\le 0 \\
 & \Rightarrow x\le \dfrac{1}{2} \\
 & \Rightarrow x\in \left( -\infty ,\dfrac{1}{2} \right] \\
\end{align}\]
 Now we take the second factor and get,
\[\begin{align}
  & {{e}^{{{x}^{2}}-x}}-1\le 0 \\
 & \Rightarrow {{e}^{{{x}^{2}}-x}}\le 1 \\
 & \Rightarrow {{e}^{{{x}^{2}}-x}}\le {{e}^{0}} \\
 & \Rightarrow {{x}^{2}}-x\le 0 \\
 & \Rightarrow x\left( x-1 \right)\le 0 \\
 & \Rightarrow x\le 0,x\ge 1\text{ or x}\ge \text{0,x}\le \text{1} \\
\end{align}\]

We know that if the So if we consider $x\le 0,x\ge 1$ then the admissible interval for $x$ is $\left( -\infty ,0 \right]\bigcap \left[ 1,\infty \right)=\Phi $. If we consider $x\ge 0,x\le 1$ then the admissible interval for $x$ is $\left[ 0,\infty \right)\bigcap \left( -\infty ,1 \right]=\left[ 0,1 \right]$. So all possible of $x$ from the second factor is $\Phi \bigcup \left[ 0,1 \right]=\left[ 0,1 \right].$
So if take the intersection of the two intervals obtained from both the factors in case-2 then we get
\[x\in \left( -\infty ,\dfrac{1}{2} \right]\bigcap \left[ 0,1 \right]=\left[ 0,\dfrac{1}{2} \right]\]

So we combine all the values from case-1 and case-2 and get ,
\[x\in \left[ 0,\dfrac{1}{2} \right]\bigcup \left[ 1,\infty \right)\]
So the correct option is B. \[\]

Note:
 We can also find the intervals using the wavy-curve method. We take note of the difference between increasing (or non-decreasing) and strictly decreasing function which has the condition ${{h}^{'}}\left( x \right)>0$. The values of $x$ for which ${{h}^{'}}\left( x \right)=0$ are called critical points.