Relations and Functions — Class 12 Maths Solution

$f:{R_ + } \to [4,\infty )$ and $f(x) = {x^2} + 4$

Injectivity

Consider ${x_1},{x_2} \in R$

Now, $f({x_1}) = x_1^2 + 4$ and $f({x_2}) = x_2^2 + 4$
if $f({x_1}) = f({x_2}) \Rightarrow x_1^2 + 4 = x_2^2 + 4 \Rightarrow {x_1} = {x_2}$
( both ${x_1},{x_2} > 0$)

$\Rightarrow f(x)$is one-one.

Surjectivity

$f:{R_ + } \to [4,\infty ]$ be given, let $y \in [4,\infty )$ (co-domain of f), then there exist an element $x \in R$ . (domain of f) such that f(x) $=$ y

Now, $f(x) = y$ $\Rightarrow$ $y = {x^2} + 4 \Rightarrow x = \sqrt {y - 4}$

for f(x) $f(\sqrt {y - 4} ) = {(\sqrt {y - 4} )^2} + 4 = y - 4 + 4 = y$
$\Rightarrow$ ${f^{ - 1}}(y) = \sqrt {y - 4}$ or ${f^{ - 1}}(x) = \sqrt {x - 4}$