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

Option: 1

tan [f(x)] and \mathrm{~\frac{1}{f(x)}} are both continuous.


Option: 2

tan [f(x)] and \mathrm{\frac{1}{f(x)}} are both discontinuous.


Option: 3

tan [f(x)] and \mathrm{f^{-1}(x)} are both continuous.


Option: 4

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


Answers (1)

best_answer

We have, \mathrm{f(x)=\frac{1}{2} x-1 ~for ~0 \leq x \leq \pi}.
\mathrm{ \therefore[f(x)]= \begin{cases}-1, & \text { for } 0 \leq x<2 \\ 0, & \text { for } 2 \leq x \leq \pi\end{cases} }

\mathrm{ \Rightarrow \tan [f(x)]= \begin{cases}\tan (-1)=-\tan (1), & 0 \leq x<2 \\ \tan 0=0 & , 2 \leq x \leq \pi\end{cases} }

It is evident from the definition of tan [f(x)] that

\mathrm{ \lim _{x \rightarrow 2^{-}} \tan [f(x)]=-\tan 1 ~and ~\lim _{x \rightarrow 2^{+}} \tan [f(x)]=0 }

So, tan [f(x)] is nol continuous at x=2.
Now, \mathrm{ f(x)=\frac{1}{2} x-1 \Rightarrow f(x)=\frac{x-2}{2} \Rightarrow \frac{1}{f(x)}=\frac{2}{x-2} }
Clearly, f(x) is not continuous at x=2. So, tan [f(x)] and \mathrm{ \tan \left[\frac{1}{f(x)}\right] } are both discontinuous at x=2.

Posted by

Gautam harsolia

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