The maximum value of 3cosx + 4sinx + 5 is

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Solution

The correct option is D
10

$∵ a c o s θ + b s i n θ \leq \sqrt{a^{2} + b^{2}} \Rightarrow 3 c o s θ + 4 s i n θ \leq 5 \Rightarrow 3 c o s θ + 4 s i n θ + 5 \leq 10 ∴ M a x i m u m v a l u e = 10$

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