First slide

Functions (XII)

Question

The range of the function $y = \sqrt{x^{2} - 2 x + 10}$ is

Easy

A

$[3, \infty]$
B

$(3, \infty)$
C

$(- 3, \infty)$
D

$(- \infty, 3)$

Solution

$\begin{array}{l} \sqrt{x^{2} - 2 x + 10} = \sqrt{(x - 1)^{2} + 9} \\ Here, least value of \sqrt{(x - 1)^{2} + 9} is 3 when x - 1 = 0 . \\ Also, (x - 1)^{2} + 9 \geq 9 \Rightarrow \sqrt{(x - 1)^{2} + 9} \geq 3 \\ Hence, \sqrt{x^{2} - 2 x + 10} \in [3, \infty) \end{array}$

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