A) \[x=2,4\]
B) \[x=4,\,3/2\]
C) \[x=2,3/2\]
D) \[x=4\]
Correct Answer: C
Solution :
\[f(x)=\frac{1}{2-x}\] Then, \[f\{f(x)\}=f\left\{ \frac{1}{2-x} \right\}=\frac{1}{2-\left( \frac{1}{2-x} \right)}\] \[(fof)\,(x)=\frac{2-x}{4-2x-1}=\frac{2-x}{3-2x}\] \[(fof)\,(x)=\frac{x-2}{2x-3}\] Here, \[f(x)\] is discontinuous at \[x=2\] and \[f\,(f(x))\] is discontinuous at \[x=\frac{3}{2}.\] \[\because \] For the continuity of \[f\,\{(x)\},\,\,f(x)\] should be continuous. \[\therefore \] The discontinuous point of \[f\{f(x)\}=2,\,\,\frac{3}{2}\]
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