InterviewSolution
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InterviewSolution
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Let the function `f ; R-{-b}->R-{1}` be defined by `f(x) =(x+a)/(x+b),a!=b,` thenA. f is one-one but not ontoB. f is onto but not one-oneC. f is both one-one and ontoD. `f^(-1)(2) = a - 2b` |
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Answer» 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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