Short Answer TypeConsider the identity function 1N : N → N defined as lN(x) = x ∀ x ∈ N. Show that although IN is onto but IN + IN : N → N defined as (IN + IN) (x) = IN(x) + IN(x) = x + x = 2 x is not onto.
f : N → N is given by f (1) = f (2) = 1 and f (x) = x – 1.
Since f (1) = f (2)
∴ f is not one-one Given any y ∈ N, y ≠ 1, we can choose x as y + 1 such that f(y +1) = y + 1 – 1 = y Also f (1) = 1
∴ f is onto.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f (x) = 3 – 4 x
(ii) f : R → R defined by f (x) = 1 + x2.
Let f : N – {1} → N defined by f (n) = the highest prime factor of n. Show that f is neither one-to-one nor onto. Find the range of f.
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