Given that y = frac{10}{sin x + sqrt{3} cos x | vedclass

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(C) The given expression is $y = \frac{10}{\sin x + \sqrt{3} \cos x}$.To find the minimum value of $y$,we need to find the maximum value of the denomi

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$5$

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(C) The given expression is $y = \frac{10}{\sin x + \sqrt{3} \cos x}$.To find the minimum value of $y$,we need to find the maximum value of the denominator $t = \sin x + \sqrt{3} \cos x$.The expression $a \sin x + b \cos x$ has a maximum value of $\sqrt{a^2 + b^2}$.Here,$a = 1$ and $b = \sqrt{3}$.So,$t_{\max} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.Therefore,$y_{\min} = \frac{10}{t_{\max}} = \frac{10}{2} = 5$.

Given that $y = \frac{10}{\sin x + \sqrt{3} \cos x}$. The minimum value of $y$ is

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