The range of the function y=x2−2x+10 is

x2−2x+10=(x−1)2+9 Here, least value of (x−1)2+9 is 3 when x−1=0 . Also, (x−1)2+9≥9⇒(x−1)2+9≥3 Hence, x2−2x+10∈[3,∞)

The range of the function y=x2−2x+10 is

x2−2x+10=(x−1)2+9 Here, least value of (x−1)2+9 is 3 when x−1=0 . Also, (x−1)2+9≥9⇒(x−1)2+9≥3 Hence, x2−2x+10∈[3,∞)

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

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$[3, \infty]$

$(3, \infty)$

$(- 3, \infty)$

$(- \infty, 3)$

answer is A.

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Detailed Solution

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