Maximum value of $2 sin x + 4 cos x + 3$ is

2 \sqrt{5} + 3

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2 \sqrt{5} - 3

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\sqrt{5} + 3

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2 \sqrt{5}

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Solution

The correct option is A $2 \sqrt{5} + 3$
$- \sqrt{2^{2} + 4^{2}} \leq 2 sin x + 4 cos x \leq \sqrt{2^{2} + 4^{2}}$
$\Rightarrow - 2 \sqrt{5} \leq 2 sin x + 4 cos x \leq 2 \sqrt{5}$ ; so max. value of $2 sin x + 4 cos x + 3 = 2 \sqrt{5} + 3$
Option A

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