Question:
Let $f(x)=e^{x}-x$ and $g(x)=x^{2}-x, \forall x \in \mathbf{R}$. Then the set of all $x \in \mathbf{R}$, where the function $h(x)=(f \circ g)(x)$ is increasing, is :
Correct Option: , 2
Solution:
Given functions are, $f(x)=e^{x}-x$ and $g(x)=x^{2}-x$
$f(g(x))=e^{\left(x^{2}-x\right)}-\left(x^{2}-x\right)$
Given $f(g(x))$ is increasing function.
$\therefore(f(g(x)))^{\prime}=e^{\left(x^{2}-x\right)} \times(2 x-1)-2 x+1$
$=(2 x-1) e^{\left(x^{2}-x\right)}+1-2 x=(2 x-1)\left[e^{\left(x^{2}-x\right)}-1\right] \geq 0$
For $(f(g(x)))^{\prime} \geq 0$
$(2 x-1) \&\left[e^{\left(x^{2}-x\right)}-1\right]$ are either both positive or
negative
$x \in\left[0, \frac{1}{2}\right] \cup[1, \infty)$