JEE Main 12 Jan 2019 Morning Question 19
Question: The maximum value of the expression $ 3\cos \theta +5\sin ( \theta -\frac{\pi }{6} ) $ for any real value of $ \theta $ is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ \frac{\sqrt{79}}{2} $
B) $ \sqrt{31} $
C) $ \sqrt{34} $
D) $ \sqrt{19} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ x=3\cos \theta +5\sin ( \theta -\frac{\pi }{6} ) $ $ =3\cos \theta +5( \frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos \theta ) $ $ =\frac{5\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta $
So, maximum value of x for real $ \theta $ is $ \sqrt{\frac{75}{4}+\frac{1}{4}}=\sqrt{19} $
