The maximum value of 12 sin x - 5 cos x + 3 i | vedclass

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(C) Let $f(x) = 12 \sin x - 5 \cos x + 3$. We know that for any expression of the form $a \sin x + b \cos x$,the range is $[-\sqrt{a^2 + b^2}, \sqrt{a

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Let $f(x) = 12 \sin x - 5 \cos x + 3$. We know that for any expression of the form $a \sin x + b \cos x$,the range is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$. Here,$a = 12$ and $b = -5$. So,$\sqrt{a^2 + b^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$. Thus,$-13 \leq 12 \sin x - 5 \cos x \leq 13$. Adding $3$ to all parts,we get: $-13 + 3 \leq 12 \sin x - 5 \cos x + 3 \leq 13 + 3$. $-10 \leq f(x) \leq 16$. Therefore,the maximum value of $f(x)$ is $16$.

The maximum value of $12 \sin x - 5 \cos x + 3$ is

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