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if f:[1,oo)->[2,oo) is given by f(x)=x+1...

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)`

Text Solution

Verified by Experts

The correct Answer is:
A
`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)`
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