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Question
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
Options
\[\left[ - 1, 3 \right]\]
\[\left[ 1, 2 \right]\]
\[\left[ - 2, 4 \right]\]
none of these
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Solution
\[\left[ - 1, 3 \right]\]
\[A = 2 \sin^2 x - \cos2x\]
\[ = 2 \sin^2 x - \left( 1 - 2 \sin^2 x \right)\]
\[ = 4 \sin^2 x - 1\]
\[ \because 0 \leq \sin^2 x \leq 1\]
\[ \Rightarrow 4 \times 0 \leq 4 \times \sin^2 x \leq 4 \times 1\]
\[ \Rightarrow 0 \leq 4 \sin^2 x \leq 4\]
\[ \Rightarrow 0 - 1 \leq 4 \sin^2 x - 1 \leq 4 - 1\]
\[ \Rightarrow - 1 \leq 4 \sin^2 x - 1 \leq 3\]
\[ \Rightarrow - 1 \leq A \leq 3\]
\[ \Rightarrow A \in \left[ - 1, 3 \right]\]
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