Let the function `f ; R-{-b}->R-{1}` be defined by `f(x) =(x+a)/(x+b),a!=b,` then

Options

A. f is one-one but not onto
B. f is onto but not one-one
C. f is both one-one and onto
D. `f^(-1)(2) = a - 2b`

Answer

C, D

Explanation

`f:R-{-b}rarrR-{1}`
`f(x)=(x+a)/(x+b)" "[aneb]`
Let `x_(1), x_(2) in D_(f)`
`f(x_(1)) = f(x_(2))`
implies `(x_(1)+a)/(x_(1)+b)=(x_(2)+a)/(x_(2)+b)`
`impliesx_(1)x_(2)+bx_(1)+ax_(2)+ab=x_(1)x_(2)+ax_(1)+bx_(2)+ab`
`impliesb(x_(1)-x_(2))=a(x_(1)-x_(2))`
`implies (x_(1)-x_(2))(b-a)=0`
`implies x_(1)=x_(2)" "[because aneb]`
`therefore` f is one-one function.
Now, let `y=(x+a)/(x+b)`
`xy+by=x+a`
`x(y-1)=a-by`
`x=(a-by)/(y-1)andf^(-1)(y)=(a-by)/(y-1)`
`because y inR-{1}`
`therefore` x is defined, `AAy inR-{1}`
`f^(-1)(2)=(a-2b)/(2-1)=a-2b`