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# If $f(x)=\large\frac{x}{2}$$-1, then on the interval [0,\pi] \begin{array}{1 1}(a)\;\tan[f(x)]\;and \;1/f(x)\;are\;both\;continuous\\(b)\;\tan[f(x)]\;and \;1/f(x)\;are\;both\;discontinuous\\(c)\;\tan[f(x)]\;and \;f^{-1}(x)\;are\;both\;continuous\\(d)\;\tan[f(x)]\;is\; continuous\;but\;1/f(x)\;is\;not\end{array} Can you answer this question? ## 1 Answer 0 votes We have f(x)=\large\frac{x}{2}$$-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.