Question

Let the function $f:R-\{-b\}\to R-\left\{1\right\}$ be defined by $f\left(x\right)=\frac{x+a}{x+b},a\ne b$, then

• A. $f$ is one-one but not onto
• B. $f$ is onto but not one-one
• C. $f$ is both one-one and onto
• D. none of these

Answer 1

Answer: (C)

$f$ is both one-one and onto

Let $x,y$ be two numbers in set $R-\left\{-b\right\}$ such that $f\left(x\right)=f\left(y\right)$
$\therefore \frac{x+a}{x+b}=\frac{y+a}{y+b}$.
Simplifying the equation we get
$ay+bx=ax+by$
$\therefore y\left(a-b\right)=x\left(a-b\right)$
$\therefore y=x$
Hence the function is one-one.
Now for given $y\in R-\left\{1\right\}$, let $x=\frac{a-by}{y-1}$.
Then for this $x$ we have $f\left(x\right)=y$. Hence the function is onto.