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(B) Given that $\alpha$ and $\beta$ are roots of $\sqrt{3} a \cos x + 2 b \sin x = c$. Since $\alpha$ and $\beta$ are roots,we have: $\sqrt{3} a \cos

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$0.5$

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(B) Given that $\alpha$ and $\beta$ are roots of $\sqrt{3} a \cos x + 2 b \sin x = c$. Since $\alpha$ and $\beta$ are roots,we have: $\sqrt{3} a \cos \alpha + 2 b \sin \alpha = c$ $(i)$ $\sqrt{3} a \cos \beta + 2 b \sin \beta = c$ $(ii)$ Subtracting $(ii)$ from $(i)$,we get: $\sqrt{3} a (\cos \alpha - \cos \beta) + 2 b (\sin \alpha - \sin \beta) = 0$ Using the sum-to-product formulas: $\sqrt{3} a [-2 \sin(\frac{\alpha + \beta}{2}) \sin(\frac{\alpha - \beta}{2})] + 2 b [2 \cos(\frac{\alpha + \beta}{2}) \sin(\frac{\alpha - \beta}{2})] = 0$ Given $\alpha + \beta = \frac{\pi}{3}$,so $\frac{\alpha + \beta}{2} = \frac{\pi}{6}$. Substituting this value: $-\sqrt{3} a [2 \sin(\frac{\pi}{6}) \sin(\frac{\alpha - \beta}{2})] + 4 b [\cos(\frac{\pi}{6}) \sin(\frac{\alpha - \beta}{2})] = 0$ Since $\alpha 
eq \beta$,$\sin(\frac{\alpha - \beta}{2}) 
eq 0$,we can divide by it: $-\sqrt{3} a (2 	imes \frac{1}{2}) + 4 b (\frac{\sqrt{3}}{2}) = 0$ $-\sqrt{3} a + 2 \sqrt{3} b = 0$ $2 \sqrt{3} b = \sqrt{3} a$ $\frac{b}{a} = \frac{\sqrt{3}}{2 \sqrt{3}} = \frac{1}{2} = 0.5$

Let $a, b, c$ be three non-zero real numbers such that the equation $\sqrt{3} a \cos x + 2 b \sin x = c$,$x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ has two distinct real roots $\alpha$ and $\beta$ with $\alpha + \beta = \frac{\pi}{3}$. Then the value of $\frac{b}{a}$ is

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