Question

Let $f\left(x\right)=e^x-x$ and $g\left(x\right)=x^2-x,\forall x\in R$. Then the set of all $x\in R$, where the function $h\left(x\right)=(f\circ g)\left(x\right)$ is increasing, is :

• A. $[-\frac{1}{2},0]\cup [1,\infty )$
• B. $[0,\frac{1}{2}]\cup [1,\infty )$
• C. $[-1,-\frac{1}{2}]\cup [\frac{1}{2},\infty )$
• D. $[0,\infty )$

Answer 1

The correct option is B $[0,\frac{1}{2}]\cup [1,\infty )$

$h\left(x\right)=(f\circ g)\left(x\right)$
$\Rightarrow h\left(x\right)=e^{x^2-x}-x^2+x\Rightarrow h^{\prime }\left(x\right)=e^{x^2-x}\times (2x-1)-2x+1=(e^{x^2-x}-1)(2x-1)$

For a function to be increasing,
$h^{\prime }\left(x\right)\ge 0$
$\Rightarrow (e^{x^2-x}-1)(2x-1)\ge 0$

$\Rightarrow$ Critical points are $0,\frac{1}{2},1$


$\therefore x\in [0,\frac{1}{2}]\cup [1,\infty )$

Alternate Solution:
$\Rightarrow (e^{x^2-x}-1)(2x-1)\ge 0$
Case 1:
$(e^{x^2-x}-1)\ge 0$ and $(2x-1)\ge 0$
$\Rightarrow x\in [1,\infty )$

Case 2:
$(e^{x^2-x}-1)\le 0$ and $(2x-1)\le 0$
$\Rightarrow x\in [0,\frac{1}{2}]$

$\therefore x\in [0,\frac{1}{2}]\cup [1,\infty )$