Let f(x)=ex-x and g(x)=x2-x
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 :

  1. (1) $\left[-1, \frac{-1}{2}\right] \cup\left[\frac{1}{2}, \infty\right)$

  2. (2) $\left[0, \frac{1}{2}\right] \cup[1, \infty)$

  3. (3) $[0, \infty)$

  4. (4) $\left[\frac{-1}{2}, 0\right] \cup[1, \infty)$


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

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

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