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) = x2
3 < x ≤ 10, f(x) = 3x
Also, at x = 3, f(x) = 32 = 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) = 22 = 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.
