A) \[\frac{1}{x}{{f}_{3}}\,(x)\]
B) \[{{f}_{2}}\,(x)\]
C) \[{{f}_{3}}\,(x)\]
D) \[{{f}_{1}}\,(x)\]
Correct Answer: C
Solution :
Given \[{{f}_{1}}\,(x)\,=\,\frac{1}{x},\,\,{{f}_{2}}\,(x)\,\,=\,\,1-x,\,\,{{f}_{3}}(x)\,=\,\,\frac{1}{1-x}\]x \[{{f}_{2}}\,(J({{f}_{1}}(x)))\,\,=\,\,{{f}_{3}}\,(x)\] \[{{f}_{2}}\,\left( J\left( \frac{1}{x} \right) \right)\,\,=\,\,\frac{1}{1-x}\] \[1-J\,\left( \frac{1}{x} \right)\,\,=\,\,\frac{1}{1-x}\] \[J\,\left( \frac{1}{x} \right)\,\,=\,1-\,\,\frac{1}{1-x}\] \[J\,\left( \frac{1}{x} \right)\,\,=\,\,\,\frac{-x}{1-x}\] \[J\,\left( \frac{1}{x} \right)\,\,=\,\,\,\frac{-x}{x-1}\] Let \[t\,\,=\,\,\frac{1}{x}\] \[J(t)\,\,=\,\,\frac{1/t}{1/t-1}\] \[J(t)\,\,=\,\,\frac{1}{1-t}\] \[\therefore \,\,\,\,\,\,J(x)\,\,=\,\,\frac{1}{1-x}\] \[=\,\,{{f}_{3}}(x)\]You need to login to perform this action.
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