If alpha is the positive root of the equation | vedclass

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(A) Given the equation $p(x) = x^{2} - x - 2 = 0$. Factoring the quadratic,we get $(x - 2)(x + 1) = 0$,so the roots are $x = 2$ and $x = -1$. Since $\

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$\frac{3}{\sqrt{2}}$

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(A) Given the equation $p(x) = x^{2} - x - 2 = 0$. Factoring the quadratic,we get $(x - 2)(x + 1) = 0$,so the roots are $x = 2$ and $x = -1$. Since $\alpha$ is the positive root,$\alpha = 2$. Substituting $\alpha = 2$ into the limit expression,we get $\lim_{x ightarrow 2^{+}} \frac{\sqrt{1 - \cos(x^{2} - x - 2)}}{x + 2 - 4} = \lim_{x ightarrow 2^{+}} \frac{\sqrt{1 - \cos(x^{2} - x - 2)}}{x - 2}$. Using the identity $1 - \cos(	heta) = 2 \sin^{2}(\frac{	heta}{2})$,the expression becomes $\lim_{x ightarrow 2^{+}} \frac{\sqrt{2 \sin^{2}(\frac{x^{2} - x - 2}{2})}}{x - 2}$. Since $x ightarrow 2^{+}$,$\sin(\frac{x^{2} - x - 2}{2})$ is positive,so $\sqrt{\sin^{2}(	heta)} = \sin(	heta)$. This simplifies to $\lim_{x ightarrow 2^{+}} \frac{\sqrt{2} \sin(\frac{(x - 2)(x + 1)}{2})}{x - 2}$. Using the limit $\lim_{u ightarrow 0} \frac{\sin(u)}{u} = 1$,we multiply and divide by $\frac{x + 1}{2}$: $\lim_{x ightarrow 2^{+}} \sqrt{2} \cdot \frac{\sin(\frac{(x - 2)(x + 1)}{2})}{\frac{(x - 2)(x + 1)}{2}} \cdot \frac{x + 1}{2} = \sqrt{2} \cdot 1 \cdot \frac{2 + 1}{2} = \frac{3\sqrt{2}}{2} = \frac{3}{\sqrt{2}}$.

If $\alpha$ is the positive root of the equation $p(x) = x^{2} - x - 2 = 0$,then $\lim_{x \rightarrow \alpha^{+}} \frac{\sqrt{1 - \cos(p(x))}}{x + \alpha - 4}$ is equal to

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