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Let f(x)=\frac{x-2}{2}, then in [0, \pi].

Option: 1

\tan (f(x)) and \frac{1}{f(x)}  both are continuous.


Option: 2

\tan f(x) is continuous but f^{-1}(x) is not continuous


Option: 3

\quad \tan (f-1(x)) and f^{-1}(x)are discontinuous


Option: 4

none of these


Answers (1)

best_answer

x \in[0, \pi] \Rightarrow \frac{x-2}{2} \in\left[-1, \frac{\pi}{2}-1\right]

Now \frac{1}{f(x)} =\frac{1}{x-2} which is discontinuations at x=2 . \tan (f(x)) is continuous for x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), f^{-1}(x)=2(x+1)
which is clearly continuous but \tan \left(f^{-1}(x)\right) is not continuous. Hence (B) is the correct answer.

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mansi

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