(i) f : R → R defined by f(x) = 3 – 4x.
f(x1) = 3 – 4x1, f(x2) = 3 – 4x2
(a) f(x1) = f(x2) ⇒ 3 – 4x1 = 3 – 4x2
∴ x1 = x2
This shows that f is one-one.
(b) f(x) = y = 3 – 4x
∴ x = \(\frac{3-y}{4}\)
For every value of y belonging to its codomain, there is a pre-image in its domain.
⇒ f is onto.
Hence, f is one-one and onto.
(ii) f : R → R given by f(x) = 1 + x2
(a) f(1) = 1 + 1 = 2, f(- 1) = 1 + 1 = 2.
∴ f(- 1) = f(1) = 2
i. e., – 1 and 1 have the same image 2.
⇒ f is not one-one.
(b) No negative number belonging to its codomain has its pre-image in its domain.
⇒ f is not onto.
Thus, f is neither one-one nor onto.
