JEE PYQ: Trigonometric Equations And Inequations Question 19
Question 19 - 2019 (12 Jan Shift 2)
The maximum value of $3\cos \theta + 5\sin\left(\theta - \frac{\pi}{6}\right)$ for any real value of $\theta$ is:
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(1) $\sqrt{19}$
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(2) $\frac{\sqrt{79}}{2}$
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(3) $\sqrt{34}$
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(4) $\sqrt{31}$
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Answer:
Solution
$f(\theta) = \frac{1}{2}\cos\theta + \frac{5\sqrt{3}}{2}\sin\theta$
$\max f(\theta) = \sqrt{\frac{1}{4} + \frac{75}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19}$
BETA
