Question

The relation $g$ is defined by $g(x)={x_{2},0≤x≤23x,2≤x≤10 $

Show that $f$ is a function and $g$ is not a function.

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It is observed that for $0≤x≤3,$ we have $f(x)=x_{2}$ and for $3≤x≤10,$ we have $f(x)=3x$

Also at $x=3$, $f(x)=3_{2}=9$ or $f(x)=3×3=9$

i.e., at $x=3,f(x)=9$

Therefore for every $x,$ $0≤x≤10$, we have unique image under $f$

Thus, the relation $f$ is a function.

Also, the relation $g$ is defined as $g(x)={x_{2},0≤x≤23x,2≤x≤10 $

Therefore for every $x,$ $0≤x≤10$, we have unique image under $f$

Thus, the relation $f$ is a function.

Also, the relation $g$ is defined as $g(x)={x_{2},0≤x≤23x,2≤x≤10 $

It can be observed that for $x=2,$ we have $g(x)=2_{2}=4$ and $g(x)=3×2=6$

Thus, the element $2$ of the domain of the relation $g$ has two different images i.e., $4$ and $6$

Hence, this relation is but not a function.

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