Answer 1

The function f is defined by

\[f\left( x \right) = \begin{cases}x^2 & 0 \leqslant x \leqslant 3 \\ 3x & 3 \leqslant x \leqslant 10\end{cases}\] 

It is observed that for 0 ≤ x < 3, f (x) = x² .
3 <  x ≤ 10, f (x) = 3x
Also, at x = 3, f(x) = 3² = 9. And
f (x) = 3 × 3 = 9.
That is, at x = 3, f (x) = 9.
Therefore, for 0 ≤ x ≤ 10, the images of f (x) are unique.
Thus, the given relation is a function.
Again,
the relation g is defined as

\[g\left( x \right) = \begin{cases}x^2 , & 0 \leqslant x \leqslant 2 \\ 3x, & 2 \leqslant x \leqslant 10\end{cases}\]

It can be observed that for x = 2, g(x) = 2² = 4 and also, 
g(x) = 3 × 2 = 6.
Hence, 2 in the domain of the relation g corresponds to two different images, i.e. 4 and 6.
Hence, this relation is not a function.

Hence proved.