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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. |
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Answer» 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. |
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