The maximum and minimum values of the functio | vedclass

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(A) Given,$f(x)=5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+8$ $=5 \cos x+3\left\{\cos x \cos \left(\frac{\pi}{3}\right)-\sin x \sin \left(\frac{\pi}

Vedclass · Accepted Answer

$15, 1$

Vedclass · Answer

(A) Given,$f(x)=5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+8$ $=5 \cos x+3\left\{\cos x \cos \left(\frac{\pi}{3}\right)-\sin x \sin \left(\frac{\pi}{3}\right)\right\}+8$ $=5 \cos x+3\left\{\cos x\left(\frac{1}{2}\right)-\sin x\left(\frac{\sqrt{3}}{2}\right)\right\}+8$ $=5 \cos x+\frac{3}{2} \cos x-\left(\frac{3 \sqrt{3}}{2}\right) \sin x+8$ $=\frac{13}{2} \cos x-\left(\frac{3 \sqrt{3}}{2}\right) \sin x+8$ Now,we know that $A \sin x+B \cos x \in \left[-\sqrt{A^2+B^2}, \sqrt{A^2+B^2}\right]$ Here,$A=-\frac{3 \sqrt{3}}{2}$ and $B=\frac{13}{2}$ Therefore,$-\sqrt{\left(-\frac{3 \sqrt{3}}{2}\right)^2+\left(\frac{13}{2}\right)^2}+8 \leq f(x) \leq \sqrt{\left(-\frac{3 \sqrt{3}}{2}\right)^2+\left(\frac{13}{2}\right)^2}+8$ $\Rightarrow -\sqrt{\frac{27}{4}+\frac{169}{4}}+8 \leq f(x) \leq \sqrt{\frac{27}{4}+\frac{169}{4}}+8$ $\Rightarrow -\sqrt{\frac{196}{4}}+8 \leq f(x) \leq \sqrt{\frac{196}{4}}+8$ $\Rightarrow -7+8 \leq f(x) \leq 7+8$ $\Rightarrow 1 \leq f(x) \leq 15$ Thus,the maximum value is $15$ and the minimum value is $1$.

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

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