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Q.
If a function $f : [2, \infty) \to B$ defined by $f(x) = x^2 - 4x + 5$ is a bijection, then B is equal to:
Relations and Functions - Part 2
Solution:
Given $f: [2, \infty ) \to B$ defined as f (x) = $x^2 - 4x + 5$ for all x = 2, 3, ....where f (x) is bijection.
Since, f(x) is bijection (i.e. 1 - 1 and onto)
$\therefore $ Co-domain of f (x) = Range of f (x)
$\Rightarrow $ Range of f (x) = B
Now, Let $y = x^2 - 4x + 5$
$ \Rightarrow \, x^2 = y + 4x - 5$
$ \Rightarrow \, x^2 - 4x + 5 - y = 0$
Since, this is a quadratic equation with constant term = 5 - y
$\therefore \, b^2 - 4ac \ge 0$
i.e. $(4)^2 - 4(1)(5 - y) \ge 0 $
$ \Rightarrow \, 16 - 20 + 4y \ge 0$
$\Rightarrow \, y \ge 1 $
$\Rightarrow \, y \in [1, \infty)$
Hence, $B = [1, \infty).$