Application of Derivatives 2 Question 2
2. 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 $h(x)=(f \circ g)(x)$ is increasing, is
(2019 Main, 10 April II)
(a) $0, \frac{1}{2} \cup[1, \infty)$
(b) $-1, \frac{-1}{2} \cup \frac{1}{2}, \infty$
(c) $[0, \infty)$
(d) $\frac{-1}{2}, 0 \cup[1, \infty)$
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Solution:
- The given functions are
$$ \text { and } \quad g(x)=x^{2}-x, \forall x \in R $$
Then, $h(x)=(f \circ g)(x)=f(g(x))$
Now, $h^{\prime}(x)=f^{\prime}(g(x)) \cdot g^{\prime}(x)$
$$ \begin{aligned} & =\left(e^{g(x)}-1\right) \cdot(2 x-1)=\left(e^{\left(x^{2}-x\right)}-1\right)(2 x-1) \\ & =\left(e^{x(x-1)}-1\right)(2 x-1) \end{aligned} $$
$\because$ It is given that $h(x)$ is an increasing function, so $h^{\prime}(x) \geq 0$
$$ \Rightarrow \quad\left(e^{x(x-1)}-1\right)(2 x-1) \geq 0 $$
Case I $(2 x-1) \geq 0$ and $\left(e^{x(x-1)}-1\right) \geq 0$
$\Rightarrow \quad x \geq \frac{1}{2}$ and $x(x-1) \geq 0$
$\Rightarrow x \in[1 / 2, \infty)$ and $x \in(-\infty, 0] \cup[1, \infty)$, so $x \in[1, \infty)$
Case II $(2 x-1) \leq 0$ and $\left[e^{x(x-1)}-1\right] \leq 0$
$\Rightarrow \quad x \leq \frac{1}{2}$ and $x(x-1) \leq 0 \Rightarrow x \in-\infty, \frac{1}{2}$ and $x \in[0,1]$
So, $\quad x \in 0, \frac{1}{2}$
From, the above cases, $x \in 0, \frac{1}{2} \cup[1, \infty)$.
