A) 0
B) 1
C) \[\frac{3}{2}\]
D) ? 3
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 \[-\,4+1=-3\].You need to login to perform this action.
You will be redirected in
3 sec