Question 13 Marks

A function $f: R \rightarrow R$ defined by $f(x)=\frac{3 x}{5}+2, x \in R$. Show that $f$ is one-one and onto. Hence, find $\mathrm{f}^{-1}$.

Answer

$f: R \rightarrow R$ defined by $f(x)=\frac{3 x}{5}+2$
First we have to prove that $f$ is one-one function for that we have to prove if $f\left(x_1\right)=f\left(x_2\right)$ then $x_1=x_2$
Here $f(x)=\frac{3 x}{5}+2$
Let $f\left(x_1\right)=f\left(x_2\right)$
$
\begin{aligned}
& \therefore \frac{3 x_1}{5}+2=\frac{3 x_2}{5}+2 \\
& \therefore \frac{3 x_1}{5}=\frac{3 x_2}{5} \\
& \therefore \mathrm{x}_1=\mathrm{x}_2
\end{aligned}
$
$\therefore \mathrm{f}$ is a one-one function.
Now, we have to prove that $\mathrm{f}$ is an onto function.
Let $y \in R$ be such that
$
\begin{aligned}
& y=f(x) \\
& \therefore y=\frac{3 x}{5}+2 \\
& \therefore y-2=\frac{3 x}{5} \\
& \therefore x=\frac{5(y-2)}{3} \in R
\end{aligned}
$
$\therefore$ for any $y \in$ co-domain $R$, there exist an element $x=\frac{5(y-2)}{3} \in$ domain $R$ such that $f(x)=$ $\mathrm{y}$
$\therefore \mathrm{f}$ is an onto function.
$\therefore \mathrm{f}$ is one-one onto function.
$\therefore \mathrm{f}^{-1}$ exists.
$
\begin{aligned}
& \therefore \mathrm{f}^{-1}(y)=\frac{5(y-2)}{3} \\
& \therefore f^{-1}(x)=\frac{5(x-2)}{3}
\end{aligned}
$

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