Question

Let $f:R\to R$ be defined as $f\left(x\right)=3x$, Choose the correct answer.
(A) $f$ is one-one onto.
(B) $f$ is many-one onto
(C) $f$ is one-one but not onto
(D) $f$ is neither one-one nor onto

Answer 1

Solve for one-one.
$f\left(x\right)=3x$
For one-one,
$f\left(x_1\right)=f\left(x_2\right)$
$3x_1=3x_2$
$x_1=x_2$
Hence, if $f\left(x_1\right)=f\left(x_2\right),$ then $x_1=x_2$
$\therefore$ function $f$ is one-one.

Solve for onto.
$f\left(x\right)=3x$
Let $f\left(x\right)=y,$ such that $y\in R$
$3x=y$
$x=\frac{y}{3}$
Now, for $y=f\left(x\right)$
Putting value of $x$ in $f\left(x\right)$
$f\left(x\right)=f\left(\frac{y}{3}\right)$
$\Rightarrow f\left(x\right)=3\left(\frac{y}{3}\right)$
$\Rightarrow f\left(x\right)=y$
Thus, for every $y\in R$, there exists $x\in R$ such that
$f\left(x\right)=y$
Hence, $f$ is onto
So, $f$ is one-one and onto.
$\therefore A$ is the correct option.