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 $x \in [0, \pi]$