Question

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

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

Answer 1

Answer: (B)

$h\left(x\right) = f\left(g\left(x\right)\right) $
$ \Rightarrow h' \left(x\right) = f'\left(g\left(x\right)\right). g'\left(x\right) $ and $ f'\left(x\right) = e^{x} -1$
$ \Rightarrow h'\left(x\right) = \left(e^{g\left(x\right)} -1\right) $
$ \Rightarrow h'\left(x\right) = \left(e^{x^2-x} -1\right) $ and $ \left(2x-1 \right)\ge0 $
Case-I $ e^{x^2-x}\ge1 $ and $ 2x-1 \ge0$
$ \Rightarrow x \in\left[1,\infty\right] $ ...(1)
Case-II $ e^{x^2-x} \le1 $ and $2x -1 \le 0$
$ \Rightarrow x\in\left[0, \frac{1}{2}\right]$ ...(2)
Hence, $ x \in \left[ 0, \frac{1}{2}\right] \cup\left[1, \infty\right]$