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Let $f: R \rightarrow R$ be defined as $f(x)=3 x$. Choose the correct answer.
(a) f is one-one onto
(b) fis many-one onto
(c) f is one-one but not onto
(d) fis neither one-one nor onto
(a) f is one-one onto
(b) fis many-one onto
(c) f is one-one but not onto
(d) fis neither one-one nor onto
Solution:
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Verified Answer
$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ is defined by $\mathrm{f}(\mathrm{x})=3 \mathrm{x}$
(a) $f\left(x_1\right)=3 x_1, f\left(x_2\right)=3 x_2 \Rightarrow f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 3 x_1=3 x_2 \Rightarrow x_1=x_2 \Rightarrow f$ is one-one
(b) for every member $y$ belonging to co-domain has pre-image $x$ in domain of $\mathrm{f}$.
$\because \quad y=3 x \Rightarrow x=\frac{y}{3}, \therefore$ fis onto
$\therefore \mathrm{f}$ is one-one and onto. Option (a) is correct.
(a) $f\left(x_1\right)=3 x_1, f\left(x_2\right)=3 x_2 \Rightarrow f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 3 x_1=3 x_2 \Rightarrow x_1=x_2 \Rightarrow f$ is one-one
(b) for every member $y$ belonging to co-domain has pre-image $x$ in domain of $\mathrm{f}$.
$\because \quad y=3 x \Rightarrow x=\frac{y}{3}, \therefore$ fis onto
$\therefore \mathrm{f}$ is one-one and onto. Option (a) is correct.
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