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Obaid Hora
Asked: 2022-11-08T08:37:03+05:30 In:

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

Let the function `f ; R-{-b}->R-{1}` be defined by `f(x) =(x+a)/(x+b),a!=b,` then
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`
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  1. Correct Answer – C::D
    `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`

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