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(A) The given quadratic equation is $x^2-2x+4=0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we get $x = \frac{2 \pm \sqrt{4-

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$2^{n+1}$

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(A) The given quadratic equation is $x^2-2x+4=0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we get $x = \frac{2 \pm \sqrt{4-16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$. Let $\alpha = 1+i\sqrt{3}$ and $\beta = 1-i\sqrt{3}$. Converting to polar form, $\alpha = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i\pi/3}$ and $\beta = 2(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3}) = 2e^{-i\pi/3}$. Then $\alpha^n + \beta^n = (2e^{i\pi/3})^n + (2e^{-i\pi/3})^n = 2^n(e^{in\pi/3} + e^{-in\pi/3})$. Using Euler's formula $e^{i	heta} + e^{-i	heta} = 2 \cos 	heta$, we get $\alpha^n + \beta^n = 2^n(2 \cos \frac{n\pi}{3}) = 2^{n+1} \cos \frac{n\pi}{3}$. Comparing this with the given expression $k \cos \frac{n\pi}{3}$, we find $k = 2^{n+1}$.

If $\alpha, \beta$ are the roots of the equation $x^2-2x+4=0$ and for any $n \in N, \alpha^n+\beta^n=k \cos \frac{n \pi}{3}$, then $k=$

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