Show that the function f : R* → R* defined by f(x) = 1/x is one-one and onto,where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R* ?
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : R → R given by f(x) = x2
(iv) f : N → N given by f(x) = x3
(v) f : Z → Z given by f(x) = x3
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
Show that the Signum Function f : R → R, given by
is neither one-one nor onto.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
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 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.
Let f : N → N be defined by State whether the function f is bijective. Justify your answer.
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
Let f : R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto.
Let f : R → R be defined as f(x) = 3x. Choose the correct answer.
f : R → R, given by
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or - 1) for the element - 2 in co-domain R, there does not exist any x in domain R such that f(x) = - 2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
