If `f:Rto R` be mapping defined by `f(x)=x^(2)+5`, then `f^(-1)(x)` is equal toA. `(x+5)^(1//3)`B. `(x-5)^(1//3)`C. `(5-x)^(1//3)`D. `5-x`

If `f:Rto R` be mapping defined by `f(x)=x^(2)+5`, then `f^(-1)(x)` is

1.	If `f:Rto R` be mapping defined by `f(x)=x^(2)+5`, then `f^(-1)(x)` is equal toA. `(x+5)^(1//3)`B. `(x-5)^(1//3)`C. `(5-x)^(1//3)`D. `5-x`
Answer» Correct Answer - A Let `y=f(x)=x^(3)+5` `Rightarrow x=(y-5)^(1//3)`

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If `f:Rto R` be mapping defined by `f(x)=x^(2)+5`, then `f^(-1)(x)` is equal toA. `(x+5)^(1//3)`B. `(x-5)^(1//3)`C. `(5-x)^(1//3)`D. `5-x`

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