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If x² + 4y²–8x +12 = 0 is satisfied by real values of x and y then x must lie between

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2, 6

2, 5

–1, 1

–2, –1

answer is A.

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Detailed Solution

4y² = -x² + 8x - 12 $\geq$ 0
$\Rightarrow$ x² - 8x + 12 $\leq$ 0 $\Rightarrow$ x $\in$ (2, 6)

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