Given x, y `in R , x^(2) + y^(2) gt 0 ` . Then the range of `(x^(2) + y^(2))/(x^(2) + xy + 4y^(2))`

Options

A. `((10 - 4 sqrt(5))/(3),(10 + 4 sqrt(5))/(3))`
B. `((10 - 4 sqrt(5))/(15),(10 + 4 sqrt(5))/(15))`
C. `((5- 4 sqrt(5))/(15),(5 + 4 sqrt(5))/(15))`
D. `((20- 4 sqrt(5))/(15),(20 + 4 sqrt(5))/(15))`

Answer

2

Explanation

`lambda = (x^(2) + y^(2))/(x^(2) + xy + 4y^(2))`
` = (1 + ((y)/(x))^(2))/(1 + (y)/(x) + 4 ((y)/(x))^(2))`
` (1 + z^(2))/(1 + z + 4z^(2))` (Putting ` y//x = z)`
or `lambda + lambda z + 4lambdaz^(2) = 1 + z^(2)`
`z^(2) (4lambda - 1) + z(lambda) + lambda - 1= 0 `
Since z is real .
D ` ge ` 0
`therefore lambda^(2) - 4 (4lambda^(2) - 5lambda + 1) ge 0 `
`therefore 15lambda ^(2) - 20 lambda + 4 le 0 `
`therefore lambda in ((10 - 4sqrt(5))/(15).(10 + 4sqrt(5))/(15))` .