सिद्ध कीजिए कि f(x)=frac{1}{x} द्वारा परिभाषि | vedclass

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Question

It is given that $f : R ^{*} \rightarrow R$. is defined by $f ( x )=\frac{1}{x}$

For one-one:

Let $x, y \in R *$ such that $f(x)=f(y)$

$\Righ

Vedclass · Accepted Answer

Vedclass · Answer

It is given that $f : R ^{*} ightarrow R$. is defined by $f ( x )=\frac{1}{x}$

For one-one:

Let $x, y \in R *$ such that $f(x)=f(y)$

$\Rightarrow \frac{1}{x}=\frac{1}{y}$

$\Rightarrow x=y$

$	herefore f$ is one $-$ one.

For onto:

It is clear that for $y \in R *$, there exists $x=\frac{1}{y} \in R *[	ext { as } y 
eq 0]$ such that

$f(x)=\frac{1}{\left(\frac{1}{y}ight)}=y$

$	herefore f$ is onto.

Thus, the given function $f$ is one $-$ one and onto.

Now, consider function g: $N ightarrow R$. defined by $g ( x )=\frac{1}{x}$

We have, $g\left(x_{1}ight)=g\left(x_{2}ight)$

$\Rightarrow=\frac{1}{x_{1}}=\frac{1}{x_{2}}$

$\Rightarrow x_{1}=x_{2}$

$	herefore g$ is one-one.

Further, it is clear that $g$ is not onto as for $1.2 \in = R_*$. there does not exit any $x$ in $N$ such that $g ( x )$

$=\frac{1}{1.2}$

Hence, function $g$ is one-one but not onto.

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