If frac{1}{2} leq frac{x^2+x+a}{x^2-x+a} leq | vedclass

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(C) Let $f(x) = \frac{x^2+x+a}{x^2-x+a}$. The given inequality is $\frac{1}{2} \leq f(x) \leq 2$. For $f(x) \leq 2$: $\frac{x^2+x+a}{x^2-x+a} \leq 2 \

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(C) Let $f(x) = \frac{x^2+x+a}{x^2-x+a}$. The given inequality is $\frac{1}{2} \leq f(x) \leq 2$. For $f(x) \leq 2$: $\frac{x^2+x+a}{x^2-x+a} \leq 2 \implies x^2+x+a \leq 2x^2-2x+2a$ (assuming $x^2-x+a > 0$) $\implies x^2-3x+a \geq 0$. For this to hold for all $x$,the discriminant $D \leq 0$: $(-3)^2 - 4(1)(a) \leq 0 \implies 9 - 4a \leq 0 \implies a \geq \frac{9}{4}$. For $f(x) \geq \frac{1}{2}$: $\frac{x^2+x+a}{x^2-x+a} \geq \frac{1}{2} \implies 2x^2+2x+2a \geq x^2-x+a$ $\implies x^2+3x+a \geq 0$. For this to hold for all $x$,the discriminant $D \leq 0$: $(3)^2 - 4(1)(a) \leq 0 \implies 9 - 4a \leq 0 \implies a \geq \frac{9}{4}$. However,checking the boundary condition $a = \frac{9}{4}$,we find that the expression becomes a perfect square,satisfying the inequality. Thus,$a = \frac{9}{4}$.

If $\frac{1}{2} \leq \frac{x^2+x+a}{x^2-x+a} \leq 2$ for all $x \in R$,then $a=$

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