if `f:[1,oo)->[2,oo)` is given by `f(x)=x+1/x` then `f^-1(x)` equals to : a) `(x+sqrt(x^2-4))/2` b) `x/(1+x^2)` c) `(x-sqrt(x^2-4))/2` d) `1+sqrt(x^2-4)`

Options

A. `((x+sqrt(x^(2)-4)))/(2)`
B. `(x)/(1+x^(2))`
C. `((x-sqrt(x^(2)-4)))/(2)`
D. `1+sqrt(x^(2)-4)`

Answer

A

Explanation

`f:[1,oo) to [2,oo)`
`f(x)=x+(1)/(x)=y`
`or x^(2)-yx+1=0`
`or x=(y+-sqrt(y^(2)-4))/(2)`
But given ` f:[1,oo) to [2,oo)`
` :. x=(y+sqrt(y^(2)-4))/(2)`
` or f^(-1)(x)=(x+sqrt(x^(2)-4))/(2)`