If $f:$
If $f:$
Official Solution
VVidaara Team
✓ Verified solution
NCERT & Exemplar
It is given that,, $f(x) = {x^2} - 4x + 5$
Let $y = {x^2} - 4x + 5$
$\Rightarrow$ $y = {x^2} - 4x + 4 + 1 = {(x - 2)^2} + 1$
$\Rightarrow$ ${(x - 2)^2} = y - 1 \Rightarrow x - 2 = \sqrt {y - 1}$
$\Rightarrow$ $x = 2 + \sqrt {y - 1}$
$\therefore y - 1 \ge 0,y \ge 1$
Range $= [1,\infty )$
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