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Let A=R-{3},B=R-{1}.Let $f \;:\;A \rightarrow B$ be defined by $f(x)=\Large{\frac{x-2}{x-3}}\normalsize \forall x \in A$.Then show that f is bijective.

1 Answer

  • A function$f :A \to B$ is bijective if it is one-one ie $f(x)=f(y)=>x=y \qquad x \in A$ and onto ie for every $y \in B$ then exists $x \in A$ such that $f(x)=y$
Given :Let $A=R-\{3\}\qquad B=R -\{1\}$
$f:A \to B \qquad f(x)=\frac{x-2}{x-3} \qquad x \in A$
Step1: Injective or One-One function:
Let $f(x)=f(y) \qquad x,y \neq 3$
Hence f is one -one
Step 2: Surjective or On-to function:
Let $y \in B$ such that $y \neq 1$
$ yx-x=3y-2$
$x=\frac {3y-2}{y-1}$
$ x \in A$ since $y \neq 1$
Hence there exists $x \in A$ for every $y \in B$ such that $f(x)=y$
f is onto
Solution:Hence f is bijective



answered Mar 4, 2013 by meena.p
edited Mar 27, 2013 by meena.p

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