A) \[\frac{7}{4}\]
B) \[\frac{15}{4}\]
C) \[\frac{9}{4}\]
D) \[\frac{1}{4}\]
Correct Answer: C
Solution :
\[4+\frac{1}{2}{{\sin }^{2}}2x-\frac{1}{2}{{(2{{\cos }^{2}}x)}^{2}}\] \[=4+\frac{1}{2}{{\sin }^{2}}2x-\frac{1}{2}{{(1+\cos 2x)}^{2}}=-{{\cos }^{2}}2x-\cos 2x+4=-\] \[[{{\cos }^{2}}2x+\cos 2x-4]=\frac{17}{4}-{{\left( \cos 2x+\frac{1}{2} \right)}^{2}}\] M = maximum value \[=\frac{17}{4}\] m = minimum value = 2 \[M-m=\frac{17}{4}-2=\frac{9}{4}.\]
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