The maximum and minimum values of frac{x^2 + | vedclass

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Question

(B) Let $y = \frac{x^2 + x + 1}{x^2 - x + 1}$.Then $y(x^2 - x + 1) = x^2 + x + 1$.$yx^2 - yx + y = x^2 + x + 1$.$x^2(y - 1) - x(y + 1) + (y - 1) = 0$.

Vedclass · Accepted Answer

$3, 1/3$

Vedclass · Answer

(B) Let $y = \frac{x^2 + x + 1}{x^2 - x + 1}$.Then $y(x^2 - x + 1) = x^2 + x + 1$.$yx^2 - yx + y = x^2 + x + 1$.$x^2(y - 1) - x(y + 1) + (y - 1) = 0$.Since $x \in \mathbb{R}$,the discriminant $D \geq 0$.$D = [-(y + 1)]^2 - 4(y - 1)(y - 1) \geq 0$.$(y + 1)^2 - 4(y - 1)^2 \geq 0$.$(y + 1 - 2(y - 1))(y + 1 + 2(y - 1)) \geq 0$.$(y + 1 - 2y + 2)(y + 1 + 2y - 2) \geq 0$.$(3 - y)(3y - 1) \geq 0$.Multiplying by $-1$,we get $(y - 3)(3y - 1) \leq 0$.This implies $\frac{1}{3} \leq y \leq 3$.Therefore,the maximum value is $3$ and the minimum value is $\frac{1}{3}$.

The maximum and minimum values of $\frac{x^2 + x + 1}{x^2 - x + 1}$ are respectively ...........

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Statement-$I$: If the roots $\alpha, \beta$ of the equation $x^2 + 2(a - 3)x + 9 = 0$,$a \in R$ satisfy $\alpha < 6 < \beta$,then $a < -3/4$.
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