# 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.