Question

Let $A=R-$ {$3$}, $B=R-${$1$}. Let $f:A$ $\to$ $B$ be defined by $f\left(x\right)=\frac{(x-2)}{(x-3)}$. The function $f$ is:

• A. Injective
• B. Surjective
• C. Bijective
• D. None of the above

Answer 1

Answer: (A), (B), (C)

Injective
Surjective
Bijective

Given $A=R-\left\{3\right\},B=R-\left\{1\right\}$ and $f\left(x\right)=\frac{x-2}{x-3}$

The function $f$ is injective, then if $x\ne y$ implies $f\left(x\right)\ne f\left(y\right)$. Also $x=y$ implies $f\left(x\right)=f\left(y\right)$ for $x,y\in A$

Taking $f\left(x\right)=f\left(y\right)$,

$\frac{x-2}{x-3}=\frac{y-2}{y-3}$

$(x-2)(y-3)=(y-2)(x-3)$

$xy-3x-2y+6=xy-3y-2x+6$

$⟹y=x$ Hence $f$ is injective.

The function $f$ is surjective if for every $b\in B$ there exists $a\in A$ such that $f\left(a\right)=b$.

Let $y=\frac{x-2}{x-3}$ ($x\in A$)

$xy-3y=x-2$

$x=\frac{3y-2}{y-1}$

So $y\in R-\left\{1\right\}$

$\therefore y\in B$

Hence $f$ is surjective.

As $f$ is both injective and surjective, it is bijective too.