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प्रश्न
Find the maximum and minimum values of each of the following trigonometrical expression:
\[5 \cos x + 3 \sin \left( \frac{\pi}{6} - x \right) + 4\]
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उत्तर
\[\text{ Let } f\left( x \right) = 5 \cos x + 3 \sin\left( \frac{\pi}{6} - x \right) + 4\]
\[\text{ Now } f\left( x \right) = 5\cos x + 3\left( \sin30°\cos x - \cos30°\sin x \right) + 4\]
\[ = 5\cos x + \frac{3}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 4\]
\[ = \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 4\]
\[\text{ We know that }\]
\[ - \sqrt{\left( \frac{13}{2} \right)^2 + \left( - \frac{3\sqrt{3}}{2} \right)^2} \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x \leq \sqrt{\left( \frac{13}{2} \right)^2 + \left( - \frac{3\sqrt{3}}{2} \right)^2} \text{ for all x }\]
\[\text{ Therefore }, \]
\[ - \sqrt{\frac{169 + 27}{4}} \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x \leq \sqrt{\frac{169 + 27}{4}}\]
\[ \Rightarrow - \frac{14}{2} + 4 \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 4 \leq \frac{14}{2} + 4\]
\[ \Rightarrow - 3 \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 4 \leq 11\]
\[\text{ Hence, maximum and minimun values of } f\left( x \right) \text{ are 11 and - 3, respectively } .\]
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