If x,y∈R and 2x2+6xy+5y2=1, then

2x2+6xy+5y2=1 (1)Equation (1) can be rewritten as2x2+(6y)x+5y2−1=0Since x is real, we have36y2−85y2−1≥0or y2≤2or −2≤y≤2Equation (1) can also be rewritten as 5y2+(6x)y+2x2−1=0Since y is real, we have 36x2−202x2−1≥0or 36x2−40x2+20≥0or −4x2≥−20or x2≤5or −5≤x≤5

If x,y∈R and 2x2+6xy+5y2=1, then

2x2+6xy+5y2=1 (1)Equation (1) can be rewritten as2x2+(6y)x+5y2−1=0Since x is real, we have36y2−85y2−1≥0or y2≤2or −2≤y≤2Equation (1) can also be rewritten as 5y2+(6x)y+2x2−1=0Since y is real, we have 36x2−202x2−1≥0or 36x2−40x2+20≥0or −4x2≥−20or x2≤5or −5≤x≤5

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

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$| x | \leq \sqrt{5}$

$| x | \geq \sqrt{5}$

$y^{2} \leq 2$

$y^{2} \leq 4$

answer is A.

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Detailed Solution

$2 x^{2} + 6 xy + 5 y^{2} = 1 (1)$
Equation (1) can be rewritten as
$2 x^{2} + (6 y) x + 5 y^{2} - 1 = 0$
Since x is real, we have
$36 y^{2} - 8 (5 y^{2} - 1) \geq 0$
or    $y^{2} \leq 2$
or    $- \sqrt{2} \leq y \leq \sqrt{2}$
Equation (1) can also be rewritten as
   $5 y^{2} + (6 x) y + 2 x^{2} - 1 = 0$
Since y is real, we have
$36 x^{2} - 20 (2 x^{2} - 1) \geq 0$
or    $36 x^{2} - 40 x^{2} + 20 \geq 0$
or    $- 4 x^{2} \geq - 20$
or    $x^{2} \leq 5$
or $- \sqrt{5} \leq x \leq \sqrt{5}$