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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.` |
Answer» Correct Answer - ` 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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