If frac{x^2+ax+3}{x^2+x+1} takes all real val | vedclass

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(D) Let $y = \frac{x^2+ax+3}{x^2+x+1}$.Then $yx^2 + yx + y = x^2 + ax + 3$,which implies $(y-1)x^2 + (y-a)x + (y-3) = 0$.For $x$ to be real,the discri

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(D) Let $y = \frac{x^2+ax+3}{x^2+x+1}$.Then $yx^2 + yx + y = x^2 + ax + 3$,which implies $(y-1)x^2 + (y-a)x + (y-3) = 0$.For $x$ to be real,the discriminant $D \geq 0$.$(y-a)^2 - 4(y-1)(y-3) \geq 0$.$y^2 - 2ay + a^2 - 4(y^2 - 4y + 3) \geq 0$.$-3y^2 + (16-2a)y + (a^2-12) \geq 0$.$3y^2 + (2a-16)y + (12-a^2) \leq 0$.For the range of $y$ to be $(-\infty, \infty)$,this quadratic inequality in $y$ must hold for all $y \in R$,which is impossible for a quadratic expression with a positive leading coefficient $(3 > 0)$.Thus,the expression cannot take all real values for any $a$.

If $\frac{x^2+ax+3}{x^2+x+1}$ takes all real values for all real values of $x$,then $a$ lies in the interval

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