If the solve the problem

Question:

Let $f(\mathrm{x})=\mathrm{e}^{\mathrm{x}}-\mathrm{x}$ and $\mathrm{g}(\mathrm{x}) \mathrm{x}^{2}-\mathrm{x}, \forall \mathrm{x} \in \mathrm{R}$.

Then the set of all $x \in R$, where the function $h(x)=(f \circ g)(x)$ is increasing, is :

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

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

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

  4. $[0, \infty)$


Correct Option: , 2

Solution:

$h(x)=f(g(x))$

$\Rightarrow h^{\prime}(x)=f^{\prime}(g(x)) \cdot g^{\prime}(x)$ and $f^{\prime}(x)=e^{x}-1$

$\Rightarrow h^{\prime}(x)=(e g(x)-1) g^{\prime}(x)$

$\Rightarrow \mathrm{h}^{\prime}(\mathrm{x})=\left(\mathrm{e}^{\mathrm{x}^{2}-\mathrm{x}}-1\right)(2 \mathrm{x}-1) \geq 0$

Case-I $\mathrm{e}^{\mathrm{x}^{2}-\mathrm{x}} \geq 1$ and $2 \mathrm{x}-1 \geq 0$

$\Rightarrow x \in[1, \infty) \ldots \ldots(1)$            .....(1)

Case-II $\mathrm{e}^{\mathrm{x}^{2}-x} \leq 1$ and $2 \mathrm{x}-1 \leq 0$

$\Rightarrow x \in\left[0, \frac{1}{2}\right] \ldots .(2)$

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

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