Question

Let f(x) = { x² , when 0 ≤ x ≤ 2. 2x, when 2 ≤ x ≤ 5.

g(x) = { x² , when 0 ≤ x ≤ 3. 2x, when 3 ≤ x ≤ 5. 

Show that f is a function while g is not a function.

Answer 1

The relation is defined as

f (x) = { x² , when 0 ≤ x ≤ 3 

            2x, when 3 ≤ x ≤ 5 

It is observed that for 

0 ≤ x ≤ 2, f(x) = x² And 2 ≤ x ≤ 5, f(x) = 2x 

Also, at x = 2, f (x) = 2² = 4 or f(x) = 2 × 2 = 4 

i,e at x = 2, f(x) = 4. 

Therefore, for 0≤ x ≤ 5, the image of f (x) are unique Thus, the given relation ‘f’ is a function. The relation g is defined as g(x) = { x² , when 0 ≤ x ≤ 3 2, when 3 ≤ x ≤ 5 It can be observed that for x = 3 g(x) = 3² = 9 and g (x) = 2 × 3 = 6 Hence, element 3 of the domain of relation ‘g’ corresponds to two different images i.e., 9 and 6. Hence, this relation is not a function.