The function *f* is defined by \[f\left( x \right) = \begin{cases}x^2 , & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10\end{cases}\] The relation *g* is defined by \[g\left( x \right) = \begin{cases}x^2 , & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10\end{cases}\]

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

Advertisement Remove all ads

#### Solution

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 ≤

3 <

Also, at

That is, at

Therefore, for 0 ≤

Thus, the given relation is a function.

Again,

the relation

*x*< 3,*f*(*x*) =*x*^{2}.3 <

*x*≤ 10,*f*(*x*) = 3*x*Also, at

*x*= 3,*f*(*x*) = 3^{2}= 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

Hence, 2 in the domain of the relation

Hence, this relation is not a function.

Hence proved.

*x*= 2,*g*(*x*) = 2^{2}= 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.

Concept: Concept of Functions

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads