Question

If $f\left(x\right)=\frac{x}{2}-1$, determine which of the following statement(s) is (are) true on the following interval $[0,\pi ].$

• A. $\tan \left(f\right(x\left)\right)$ and $1/f\left(x\right)$ are both continuous
  
• B. $\tan \left(f\right(x\left)\right)$ and $1/f\left(x\right)$ are both discontinuous
  
• C. $\tan \left(f\right(x\left)\right)$ and $f^{-1}\left(x\right)$ are both continuous
  
• D. $\tan \left(f\right(x\left)\right)$ is continuous but $1/f\left(x\right)$ is not.

Answer 1

Answer: (C), (D)

$\tan \left(f\right(x\left)\right)$ is continuous but $1/f\left(x\right)$ is not.

$f\left(x\right)=\frac{x-2}{2}\text{For}x\in [0,\pi ]$
$\frac{1}{f\left(x\right)}=\frac{2}{x-2}$
Clearly, $\frac{1}{f\left(x\right)}$ is not continuous at $x=2$

$\because x\in [0,\pi ]$
$\Rightarrow \frac{x-2}{2}\in [-1,\frac{\pi }{2}-1]$
$\tan \left(f\right(x\left)\right)=\tan \left(\frac{x-2}{2}\right)$
Since,
$\tan x$ is contiuous in $(-\frac{\pi }{2},\frac{\pi }{2})$
$\therefore \tan \left(f\right(x\left)\right)$ is contiuous in $[-1,\frac{\pi }{2}-1]$.

Finding the inverse of the function,
Let
$y=f\left(x\right)=\frac{x}{2}-1\Rightarrow \frac{x}{2}=1+y\Rightarrow x=2+2y\Rightarrow f^{-1}\left(x\right)=2+2x$
$f^{-1}\left(x\right)=2(x+1)$, which is a linear polynomial so its contiuous.