The maximum and minimum values of the functio | vedclass

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Question

(A) Given $f(x) = 5 \cos x + 3 \cos \left(x + \frac{\pi}{3}\right) + 8$. Expanding the term $\cos(x + \frac{\pi}{3})$: $f(x) = 5 \cos x + 3 \left[ \co

Vedclass · Accepted Answer

$15, 1$

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(A) Given $f(x) = 5 \cos x + 3 \cos \left(x + \frac{\pi}{3}\right) + 8$. Expanding the term $\cos(x + \frac{\pi}{3})$: $f(x) = 5 \cos x + 3 \left[ \cos x \cdot \cos \frac{\pi}{3} - \sin x \cdot \sin \frac{\pi}{3} \right] + 8$ $f(x) = 5 \cos x + 3 \left[ \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x \right] + 8$ $f(x) = \frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x + 8$. For any expression of the form $A \cos x + B \sin x$,the range is $[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]$. Here,$A = \frac{13}{2}$ and $B = -\frac{3\sqrt{3}}{2}$. Calculating $\sqrt{A^2 + B^2} = \sqrt{\left(\frac{13}{2}\right)^2 + \left(-\frac{3\sqrt{3}}{2}\right)^2} = \sqrt{\frac{169}{4} + \frac{27}{4}} = \sqrt{\frac{196}{4}} = \sqrt{49} = 7$. Thus,the range of $\frac{13}{2} \cos x - \frac{3\sqrt{3}}{2} \sin x$ is $[-7, 7]$. Adding $8$ to the range,we get $f(x) \in [-7 + 8, 7 + 8]$,which is $[1, 15]$. Therefore,the maximum value is $15$ and the minimum value is $1$.

The maximum and minimum values of the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 5 \cos x + 3 \cos \left(x + \frac{\pi}{3}\right) + 8$ for all $x \in \mathbb{R}$,are respectively.

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