In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) $f:R \to R,$ defned by $f(x) = 3 - 4x$

(ii) $f:R \to R,$ defined by $f(x) = 1 + {x^2}$

(i) $f:R \to R$ defined by $f(x) = 3 - 4x.$

Injectivity
$f({x_1}) = f({x_2}) \Rightarrow 3 - 4x = 3 - 4{x_2}$
$\Rightarrow$ $- 4{x_2} = - 4{x_2} \Rightarrow {x_1} = {x_2}$
Therefore, f is one-one.

Surjectivity

Now, $f:R \to R$ given for everyy$\in$R(co-domain of f), there exists an element x$\in$R (domain of f) such that $f(x) = y$
$\Rightarrow$ $y = 3 - 4x \Rightarrow x = \cfrac{{3 - y}}{4}$

Therefore, $f\left( {\cfrac{{3 - y}}{4}} \right) = 3 - 4\left( {\cfrac{{3 - y}}{4}} \right) = 3 - 3 + y = y$
Hence, f is onto.

Thus, f is one-one and onto or bijective function.

(ii) $f:R \to R$ defined by $f(x) = 1 + {x^2}$
Injectivity

Let ${x_1},\;\;{x_2} \in R,\;\;then\;\;f({x_1}) = 1 + x_1^2\;\;and\;\;f({x_2}) = 1 + x_2^2$
$f({x_1}) = f({x_2}) \Rightarrow x_1^2 = x_2^2 \Rightarrow {x_1} = \pm {x_2}$
Thus, f is not one-one.

Surjectivity

Now, $f:R \to R,$ given for every y$\in$R(co-domain of f), there exists an element x$\in$R(domain of f) such that f(x) $=$ y
$\Rightarrow$

Thus, elements less than 1 has no pre-image.
Therefore, f is not onto.
Hence, f is neither one-one nor onto and hence not bijective.