Applications Of Derivatives Question 84
Question: The minimum value of the function $ 2\cos 2x-\cos 4x $ in $ 0\le x\le \pi $ is
Options:
A) 0
B) 1
C) $ \frac{3}{2} $
D) - 3
Show Answer
Answer:
Correct Answer: D
Solution:
$ y=2\cos 2x-\cos 4x $
$ =2\cos 2x(1-\cos 2x)+1=4\cos 2x{{\sin }^{2}}x+1 $ Obviously, $ {{\sin }^{2}}x\ge 0 $ Therefore to be least value of y, $ \cos 2x $ should be least i.e., $ -1 $ .
Hence least value of y is $ -,4+1=-3 $ .
BETA
