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If x is real and (x^2+2x+c)//(x^2+4x+3c)...

If `x` is real and `(x^2+2x+c)//(x^2+4x+3c)` can take all real values, of then show that `0lt=clt=1.`

Text Solution

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The correct Answer is:
` a in ( - (11)/(3), - 2 sqrt(2)]`
Let `y = (x^(2) + 2x + c ) /(x^(2) + 4x + 3c)`
or ` (y - 1 )x^(2) + (4y - 2) x + 3cy - c = 0`
Now , x is real . Hence,
`D = (4y - 2)^(2) - 4 (y - 1) (3cy - c) ge 0 AA y in ` R
or ` (2y - 1 )^(2) - (y - 1)(3cy - c) ge 0 , AA y in ` R
or ` (4 - 3c)y^(2) + (-4 + c + 3c) y + 1 - c ge 0, AA y in ` R.
` or 4-3c gt 0 and (4c - 4)^(2) - 4 (4 - 3c)(1 - c) le 0`
or ` c lt (4)/(3) and 4 (c - 1)^(2) - (4-3c)(1 -c) le 0 `
or ` c lt (4)/(3) and (c - 1)xx (4c - 4 + 4 - 3c)le 0`
or ` c lt (4)/(3) and (c - 1)(c)le 0`
or ` c lt (4)/(3) and - le cle 1`
or ` 0 le c le 1` .
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