If x- y - and 2 x2-6 x y-5 y2-1- then

The correct options are A |x| ≤√(5) nbsp; C y^2 ≤ 22x^2 +6xy + 5y^2=1 ⇒ 2x^2 + (6y)x + 5y^2 1 =0 Since, x is real, ∴ 36y^2 8(5y^2 1) ≥ 0 ⇒ y^2 ≤ 2 Also, 5y^ ...

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Byju's Answer

Standard XIII

Mathematics

Quadratic Equation

If x, y ∈ℝ an...

Question

If $x, y \in R$ and $2 x^{2} + 6 x y + 5 y^{2} = 1$ , then

| x | \leq \sqrt{5}

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| x | \geq \sqrt{5}

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y^{2} \leq 2

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y^{2} \geq 2

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Solution

The correct options are
A $| x | \leq \sqrt{5}$
C $y^{2} \leq 2$
$2 x^{2} + 6 x y + 5 y^{2} = 1$
$\Rightarrow 2 x^{2} + (6 y) x + 5 y^{2} - 1 = 0$
Since, $x$ is real,
$∴ 36 y^{2} - 8 (5 y^{2} - 1) \geq 0$
$\Rightarrow y^{2} \leq 2$

Also, $5 y^{2} + (6 x) y + 2 x^{2} - 1 = 0$
Since, $y$ is real,
$∴ 36 x^{2} - 20 (2 x^{2} - 1) \geq 0$
$\Rightarrow x^{2} \leq 5$
$\Rightarrow | x | \leq \sqrt{5}$

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If x,y∈R and 2x2+6xy+5y2=1, then

The correct options are A |x|≤√5 C y2≤22x2+6xy+5y2=1 ⇒2x2+(6y)x+5y2−1=0 Since, x is real, ∴36y2−8(5y2−1)≥0 ⇒y2≤2 Also, 5y2+(6x)y+2x2−1=0 Since, y is real, ∴36x2−20(2x2−1)≥0 ⇒x2≤5 ⇒|x|≤√5

If $x, y \in R$ and $2 x^{2} + 6 x y + 5 y^{2} = 1$ , then