If the equation $x^{2} + 9 y^{2} - 4 x + 3 = 0$ is satisfied for real values of $x$ and $y,$ then

1 \leq x \leq 3

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2 \leq x \leq 3

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\frac{- 1}{3} \leq y \leq \frac{1}{3}

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\frac{1}{3} \leq x \leq 1

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Solution

The correct option is A $1 \leq x \leq 3$
$\begin{matrix} 9 y^{2} + (x^{2} - 4 x + 3) = 0 9 y^{2} = - [x^{2} - 4 x + 3] \Rightarrow (x^{2} - 4 x + 3) \leq 0 x \in [1, 3] O p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$

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