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RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS5 Marks
Question
Show that the function $\text{f}:\text{R}_\ast\rightarrow\text{R}_\ast$ defined by $\text{f(x)}=\frac{1}{\text{x}}$ is one-one and onto, where $\text{R}_\ast$ is the set of all non-zero real numbers. Is the result true, if the domain $\text{R}_\ast$ is replaced by N with co-domain being same as $\text{R}_\ast?$
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Answer

It is given that $\text{f}:\text{R}_\ast\rightarrow\text{R}_\ast$ is defined by $\text{f(x)}=\frac{1}{\text{x}}$
One-one:
f(x) = f(y)
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\text{y}}$
⇒ x = y
$\therefore$ f is one-one.
Onto:
It is clear that for $\text{y}\in\text{R}_\ast,$ there exists$\text{x}=\frac{1}{\text{y}}\in\text{R}_\ast\ (\text{Exists as y}\neq0)$ such that $\text{f(x)}=\frac{1}{\frac{1}{\text{y}}}=\text{y}.$
$\therefore$ f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function $\text{g}:\text{N}\rightarrow\text{R}_\ast$ defined by
$\text{g(x)}=\frac{1}{\text{x}}.$
We have,
$\text{g}(\text{x}_1)=\text{g}(\text{x}_2)\Rightarrow\frac{1}{\text{x}_1}=\frac{1}{\text{x}_2}\Rightarrow\text{x}_1=\text{x}_2$
$\therefore$ g is one-one.
Further, it is clear that g is not onto as for $1.2\in\text{R}_\ast$ there does not exit any x in N such that $\text{g(x)}=\frac{1}{1.2}.$
Hence, function g is one-one but not onto.

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