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)

$tan [f(x)]$ is continuous but $\frac{1}{f(x)}$ is not continuous

Answer 2

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 $\frac{1}{f(x)}$ is continuous in $x \in [0, \pi]$ but $\frac{1}{f(x)}$ is not continuous in $x \in [0, \pi]$