The relation *f* is defined by f(x) = `{(x^2,0<=x<=3),(3x,3<=x<=10):}`

The relation* g* is defined by g(x) = `{(x^2, 0 <= x <= 2),(3x,2<= x <= 10):}`

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

#### Solution

The relation *f* is defined as f(x) = `{(x^2,0<=x<=3),(3x,3<=x<=10):}`

It is observed that for

0 ≤ *x* < 3, *f*(*x*) = *x*^{2}

3 < *x* ≤ 10, *f*(*x*) = 3*x*

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 0 ≤ *x* ≤ 10, the images of *f*(*x*) are unique.

Thus, the given relation is a function.

The relation* g* is defined as g(f) = `{(x^2, 0 <= x <= 2),(3x,2<= x <= 10):}`

It can be observed that for *x* = 2, *g*(*x*) = 2^{2} = 4 and *g*(*x*) = 3 × 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.