If $x^{2} + 9 y^{2} - 4 x + 3 = 0, x, y \in R,$ then x and y respectively lie in the intervals:

Option 1 -

[1, 3] and [1, 3]

Option 2 -

[1, 3] and $[- \frac{1}{3}, \frac{1}{3}]$

Option 3 -

$[- \frac{1}{3}, \frac{1}{3}]$ and [1, 3]

Option 4 -

$[- \frac{1}{3}, \frac{1}{3}] a n d [- \frac{1}{3}, \frac{1}{3}]$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$x^{2} + 9 y^{2} - 4 x + 3 = 0, x, y \in R . . . . . . . . . (i)$

$x^{2} - 4 x + 9 y^{2} + 3 = 0$

Since $x \in R, D \geq 0 i . e . 16 - 4 (9 y^{2} + 3) \geq 0$

or $9 y^{2} - 1 \leq 0$

$\Rightarrow y \in [- \frac{1}{3}, \frac{1}{3}]$

$(i) \Rightarrow 9 y^{2} = - x^{2} + 4 x - 3$

Since L.H.S. $\geq 0 s o R . H . S . \geq 0 i . e . - x^{2} + 4 x - 3 \geq 0$

or $x^{2} - 4 x + 3 \leq 0$

$(x - 3) (x - 1) \leq 0$

$\Rightarrow x \in [1, 3]$

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If $x^{2} + 9 y^{2} - 4 x + 3 = 0, x, y \in R,$ then x and y respectively lie in the intervals:

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If $x^{2} + 9 y^{2} - 4 x + 3 = 0, x, y \in R,$ then x and y respectively lie in the intervals: