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

$\begin{array}{1 1}(a)\;\tan[f(x)]\;and \;1/f(x)\;are\;both\;continuous\\(b)\;\tan[f(x)]\;and \;1/f(x)\;are\;both\;discontinuous\\(c)\;\tan[f(x)]\;and \;f^{-1}(x)\;are\;both\;continuous\\(d)\;\tan[f(x)]\;is\; continuous\;but\;1/f(x)\;is\;not\end{array}$

1 Answer

We have $f(x)=\large\frac{x}{2}$$-1$
$\therefore [f(x)]=[\large\frac{x}{2}$$-1]=-1\qquad 0\leq x < 2$
$\tan [f(x)]=\tan(-1),0 \leq x < 2$
$\qquad\;\quad\;=0,2\leq x\leq \pi$
$\therefore$ The function $\tan[f(x)]$ is discontinuous at $x=2$
Also the function $\large\frac{1}{f(x)}=\frac{1}{\Large\frac{x}{2}-\normalsize 1}$
$\Rightarrow \large\frac{2}{x-2}$ is discontinuous at $x=2$
Thus both the given functions $\tan[f(x)]$ as well as $\large\frac{1}{f(x)}$ are continuous on the interval $[0,\pi]$
Also $f^{-1}(x)=y$
$\Rightarrow x=f(y)=\large\frac{y}{2}$$-1$
$\Rightarrow y=2(x+1)$
$f^{-1}(x)=2(x+1)$ is continuous on $[0,\pi]$
Hence (b) is the correct answer.
answered Jan 4, 2014 by sreemathi.v

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