Question

Two functions are defined as under
$f(x) = \begin{cases} x+1, & \text{ $x \le 1$ } \\[2ex] 2x+1, & \text{ $1 < x \le 2$} \end{cases}$ and $g(x) = \begin{cases} x^2, & \text{ $-1 \le x < 2$} \\[2ex] x+2, & \text{ $2 \le x \le 3$} \end{cases}$
then $(fog)(x)$ equals

• A. $ \begin{cases} x^2+1, & \text{ $|x| \le 1$} \\[2ex] 2x^2+1, & \text{ $1 < x \le \sqrt{2}$} \end{cases}$
• B. $ \begin{cases} x^2+1, & \text{ $|x| \le 1$} \\[2ex] 2x^2+1, & \text{ $1 \le x < \sqrt{2}$ } \end{cases}$
• C. $ \begin{cases} x^2+1, & \text{ $|x| > 1$} \\[2ex] 2x^2+1, & \text{ $|x| \ge \sqrt{2}$} \end{cases}$
• D. None of these

Answer 1

Answer: (A)

$ (f\,o\,g) (x) =f(g(x))= \begin{cases} g(x)+1, & \text{ $g(x) \le 1$} \\[2ex] 2g(x)+1, & \text{ $1 < g(x) \le 2$} \end{cases}$
Now, $g(x) \le 1$
$(i) \,x^2 \le 1$, $-1 \le x < 2$
$\Rightarrow |x| \le 1$ , $- 1$, $\le x < 2 $
$\Rightarrow |x| \le 1$
$\left(ii\right)\,x+2 \le1, 2 \le x \le3 $
$\Rightarrow x \le-1, 2 \le x \le 3$
$\Rightarrow x=\phi$
Now, consider $1 < g\left(x\right) \le 2$
$\left(iii\right)\,1 < x^{2} \le2, -1 \le x < 2$
$\Rightarrow x \in [-\sqrt{2}$, $-1] \cup (1$, $\sqrt{2}]$, $-1 \le x < 2$
$\Rightarrow 1 < x \le \sqrt{2}$
$\left(iv\right)\,1 < x+2 \le2$, $2 \le x \le3$
$\Rightarrow -1 < x \le0$, $2 \le x \le3$
$\Rightarrow x=\phi $
So $(fog)(x) =f(g(x)) = \begin{cases} x^2+1, & \text{ $|x|\le 1$} \\[2ex] 2x^2+1, & \text{ $1 < x \le \sqrt{2}$} \end{cases}$