Maths-
General
Easy

Question

Let A equals R minus left curly bracket 3 right curly bracket comma B equals R minus left curly bracket 1 right curly bracket text  Let  end text f colon A not stretchy rightwards arrow B be defined by f left parenthesis x right parenthesis equals fraction numerator x minus 2 over denominator x minus 3 end fraction .Then,

  1.  f is bijective
  2. f is one-one but not onto
  3. f is onto but not one-one

The correct answer is: f is bijective


    Let X and Y be two arbitary elements in A.
    Then, f left parenthesis x right parenthesis equals f left parenthesis y right parenthesis
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow fraction numerator x minus 2 over denominator x minus 3 end fraction equals fraction numerator y minus 2 over denominator y minus 3 end fraction end cell row cell not stretchy rightwards double arrow x y minus 3 x minus 2 y plus 6 equals x y minus 3 y minus 2 x plus 6 end cell row cell not stretchy rightwards double arrow x equals y comma straight for all x comma y element of A end cell end table
    So, f is an injective mapping.
    Again, let Y be an orbitary element in B , then
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell f left parenthesis x right parenthesis equals y end cell row cell not stretchy rightwards double arrow fraction numerator x minus 2 over denominator x minus 3 end fraction equals y end cell row cell not stretchy rightwards double arrow x equals fraction numerator 3 y minus 2 over denominator y minus 1 end fraction end cell end table
    Clearly, straight for all y element of B comma x equals fraction numerator 3 y minus 2 over denominator y minus 1 end fraction element of A, thus for all y element of B there exists x element of A such that
    f left parenthesis x right parenthesis equals f open parentheses fraction numerator 3 y minus 1 over denominator y minus 1 end fraction close parentheses equals fraction numerator fraction numerator 3 y minus 2 over denominator y minus 1 end fraction minus 2 over denominator fraction numerator 3 y minus 2 over denominator y minus 1 end fraction minus 3 end fraction equals y
    Thus, every element in the codomain B has its preimage in  A , so f is a surjection. Hence, f colon A not stretchy rightwards arrow B is bijective.

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