question_answer

If \[f(x)\] satisfies \[x+|f(x)|=2f(x),\] then \[{{f}^{-1}}(x)\] satisfies

A) \[3x+|{{f}^{-1}}(x)|=2{{f}^{-1}}(x)\]  

B) \[x+|{{f}^{-1}}(x)|=2{{f}^{-1}}(x)\]

C) \[{{f}^{-1}}(x)-|x|=2x\]

D) \[3x-|{{f}^{-1}}(x)|=2{{f}^{-1}}(x)\]

Correct Answer: D

Solution :

[d] \[x+|f(x)|=2f(x)\] \[\Rightarrow \,\,\,\,\,\,\,3x-|{{f}^{-1}}(x)|=2{{f}^{-1}}(x)\]