# The relation $f$ is defined by $f(x)=left{egin{array}{l}x^{2}, 0 leq x leq 3 \ 3 x, 3 leq x leq 10end{array} ight.$

**Question:**

The relation $f$ is defined by $f(x)=\left\{\begin{array}{l}x^{2}, 0 \leq x \leq 3 \\ 3 x, 3 \leq x \leq 10\end{array}\right.$

The relation $g$ is defined by $g(x)=\left\{\begin{array}{l}x^{2}, 0 \leq x \leq 2 \\ 3 x, 2 \leq x \leq 10\end{array}\right.$

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

**Solution:**

The relation $f$ is defined as $f(x)=\left\{\begin{array}{l}x^{2}, 0 \leq x \leq 3 \\ 3 x, 3 \leq x \leq 10\end{array}\right.$

It is observed that for

$0 \leq x<3, f(x)=x^{2}$

$3

Also, at $x=3, f(x)=3^{2}=9$ or $f(x)=3 \times 3=9$

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

Therefore, for $0 \leq x \leq 10$, the images of $f(x)$ are unique.

Thus, the given relation is a function.

The relation $g$ is defined as $g(x)=\left\{\begin{array}{l}x^{2}, 0 \leq x \leq 2 \\ 3 x, 2 \leq x \leq 10\end{array}\right.$

It can be observed that for $x=2, g(x)=2^{2}=4$ and $g(x)=3 \times 2=6$

Hence, element 2 of the domain of the relation $g$ corresponds to two different images i.e., 4 and 6 . Hence, this relation is not a function.