Question

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

• A. $f$ is one-one
• B. $f$ is onto
• C. $f$ is bijective
• D. none of these

Answer 1

The correct option is A $f$ is one-one

$A=R-\left\{2\right\}$

$B=R-\left\{1\right\}$

$F:A\to B$

$f\left(x\right)=\frac{x-3}{x-2}$

$f\left(x_1\right)=\frac{x_1-3}{x_1-2}$

$f\left(x_2\right)=\frac{x_2-3}{x_2-2}$

$\frac{x_1-3}{x_1-2}=\frac{x_2-3}{x_2-2}$

$(x_1-3)(x_2-2)=(x_1-2)(x_2-3)$

$x_1{x}_2-2x_1-3x_2+6=x_1{x}_2-3x_1-2x_2+6$

$-2x_1-3x_2=-3x_1-2x_2$

$-3x_2+2x_2=-3x_1+2x_1$

$-x_2=-x_1$

$x_1=x_2$

So the function is one-one.

$f\left(x\right)=\frac{x-3}{x-2}$

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

Swapping $y$ and $x$

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

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

$xy-2x=y-3$

$xy-y=-3+2x$

$y(x-1)=-3+2x$

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

So, it is not onto.