The maximum value of the expression frac{x^2+ | vedclass

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Question

(A) Let $y = \frac{x^2+x+1}{2x^2-x+1}$.Then $y(2x^2-x+1) = x^2+x+1$.$(2y-1)x^2 - (y+1)x + (y-1) = 0$.Since $x \in R$,the discriminant $D \ge 0$.$D = (

Vedclass · Accepted Answer

$\frac{7+2\sqrt{7}}{7}$

Vedclass · Answer

(A) Let $y = \frac{x^2+x+1}{2x^2-x+1}$.Then $y(2x^2-x+1) = x^2+x+1$.$(2y-1)x^2 - (y+1)x + (y-1) = 0$.Since $x \in R$,the discriminant $D \ge 0$.$D = (y+1)^2 - 4(2y-1)(y-1) \ge 0$.$y^2+2y+1 - 4(2y^2-3y+1) \ge 0$.$y^2+2y+1 - 8y^2+12y-4 \ge 0$.$-7y^2+14y-3 \ge 0$.$7y^2-14y+3 \le 0$.Solving $7y^2-14y+3 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:$y = \frac{14 \pm \sqrt{196 - 84}}{14} = \frac{14 \pm \sqrt{112}}{14} = \frac{14 \pm 4\sqrt{7}}{14} = \frac{7 \pm 2\sqrt{7}}{7}$.Thus,the range of $y$ is $[\frac{7-2\sqrt{7}}{7}, \frac{7+2\sqrt{7}}{7}]$.The maximum value is $\frac{7+2\sqrt{7}}{7}$.

The maximum value of the expression $\frac{x^2+x+1}{2x^2-x+1}$,for $x \in R$,is

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