Answer :

Given A = 2 sin2 x – cos 2x


[ using cos 2x = 1 – 2 sin2 x ]


so A = 2 sin2 x – cos 2x = 2 sin2 x –[ 1 – 2 sin2 x]


= 2 sin2 x -1 + 2 sin2 x]


= 4 sin2 x – 1


Now A = 2 sin2 x – cos 2x = 4 sin2 x – 1


As we know sin x lies between -1 and 1


-1 ≤ sin x ≤ 1


0 ≤ sin2x ≤ 1


Multiplying the inequality by 4


0 ≤ 4 sin2 x ≤ 4


Subtracting 1 from the inequality


-1 ≤ (4 sin2 x – 1) ≤ 3


From the above inequation, we can say that


A = (4 sin2 x – 1) belongs to the closed interval [-1,3]


Hence the answer is A.

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