MCQ
The minimum value of the function $2\cos 2x - \cos 4x$ in $0 \le x \le \pi $ is
- A$0$
- B$1$
- C${3 \over 2}$
- ✓$-3$
$ = 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$.
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