The least value of $4 cos x + 3 sin x$ is

0

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2

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- 3

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- 5

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Solution

The correct option is C $- 5$
$a c o s x + b s i n x$
Then it has maximum value $\sqrt{a^{2} + b^{2}}$ and minimum value $- \sqrt{a^{2} + b^{2}}$
Now $4 c o s x + 3 s i n x$ has least value $- \sqrt{4^{2} + 3^{2}} = - 5$

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