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
Correct Answer: B
Solution :
| We have, \[f(x)=\frac{1}{2}x-1\] for \[0\le x\le \pi \] |
| \[\therefore \,\,\,[f(x)]=\left\{ \begin{align} & -1,0\le x |
| \[\Rightarrow \,\,\tan [f(x)]\,=\left\{ \begin{matrix} \tan (-1), & 0\le x |
| \[\therefore \,\,\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\,\,\tan [f(x)]=-\tan 1\] and \[\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\,\,\tan [f(x)]=0\] |
| So, \[\tan \,f(x)\] is not continuous at \[x=2\] |
| Now \[f(x)=\frac{1}{2}x-1\Rightarrow f(x)=\frac{x-2}{2}\] \[\Rightarrow \,\frac{1}{f(x)}=\frac{2}{x-2}\] |
| Clearly, \[\frac{1}{f(x)}\] is not continuous at\[x=2\]. |
| So, \[\tan \,[f(x)]\] and \[\left[ \frac{1}{f(x)} \right]\] are both discontinuous at \[x=2.\] |
You need to login to perform this action.
You will be redirected in
3 sec
