Question

If $f(x)=\frac{x}{2}-1$ then on the interval $[0,\pi ]$

• A. $\tan [f(x)]$ and $\frac{1}{f(x)}$ are both continuous
• B. $\tan [f(x)]$ and $\frac{1}{f(x)}$ are discontinuous
• C. $\tan [f(x)]$ is continuous but $\frac{1}{f(x)}$ is not continuous
• D. $\tan [f(x)]$ is not continuous but $\frac{1}{f(x)}$ is continuous

Answer 1

Answer: (C)

Given $f(x)=\frac{x}{2}-1$
$\therefore \tan (f(x))=\tan (\frac{x}{2}-1)$
and $\frac{1}{f(x)}=\frac{1}{\frac{x}{2}-1}=\frac{2}{x-2}$
By using graph transformation method we can draw the graph of
$\tan (f(x))$ and $\frac{1}{f(x)}$ as follow :
Graph of $\tan f(x)$

Graph at $\tan [f(x)]$ in $x\in (-\pi +2,\pi +2)$
Graph of $\frac{1}{f(x)}$

From the above graphs of $\tan (f(x))$ and $\frac{1}{f(x)}$, we can easily observe that $\tan (f(x))$ is continuous in $x\in [0,\pi ]$ but $\frac{1}{f(x)}$ is not continuous in $x\in [0,\pi ]$