Short Answer Typef : R → R is given by f (x) = [x]
Different elements in R can have the Same image {∵ for all x ∈ [0, 1), f (x) = 0 } ∵ f is not one-one Also Rf = set of integers ≠ R
∵ f is not onto.
Consider 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.
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.
Switch
