Question

$f \left(x\right) = | x - 1 |, f : R^{+} \to R$ and $g \left(x\right) = e^{x}, g : [-1, \infty) \to R$. If the function fog $\left(x\right)$ is defined, then its domain and range respectively are

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

Answer 1

Answer: (B)

$f (x) | x 1| = $ $ \begin{cases} 1 -x, & \text{ $0 < x < 1 $} \\[2ex] x-1, & \text{ $x \ge 1$ } \end{cases}$
$g (x) = ex, x ^3 \ge -1$
$(fog)(x) = \begin{cases} 1 -g(x), & \text{ $0 < g(x) < 1 \,i.e. \,-1 \le x < 0$} \\[2ex] g(x)-1, & \text{ $g\left(x\right) \ge 1 \,i.e.\, 0 \le x$ } \end{cases}$

$ = \begin{cases} 1-e^x , & \text{$-1 \ge x < 0$} \\[2ex] e^x - 1, & \text{$x \ge 0$ } \end{cases}$
$\therefore $ domain $= [-1, \infty)$
fog is decreasing in $[-1, 0)$ and increasing in $[0, \infty)$
$fog\left(-1\right) = 1 - \frac{1}{e}$ and $\quad fog \left(0\right) = 0$
As $x \to \infty, fog \left(x\right) \to \infty,$
$\therefore \quad$ range $= [0, \infty)$
$\therefore \quad x = \frac{1}{2}log_{e} \left(\frac{y}{2-y}\right)$