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If A = {1, 2, 3, 4), B = {1, 2, 3, 4, 5, 6} are two sets and the function f : A → B is defined by f (x) = x + 2  ∀  x ∈ A, then the function f is

(a) bijective 

(b) one-one 

(c) onto 

(d) many-one

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Answer: (b) = one - one

f (x) = x + 2 and f : A → B 

∴  f (1) = 1 + 2 = 3, f (2) = 2 + 2 = 4 

f (3) = 3 + 2 = 5, f (4) = 4 + 2 = 6 

Thus each element in A has a unique image in B and no two elements in B have the same pre-image in A. 

So f is one-one

But not all the elements of B, i.e., 1 and 2 have a pre-image in A. 

So, the function f is not onto

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