MCQ
If $\text{A}=2\sin^2\text{x}-\cos2\text{x},$ then A lies in the interval:
- A$[-1,3]$
- B$[1,2]$
- C$[-2,4]$
- DNone of these
Solution:
$\text{A}=2\sin^2\text{x}-\cos2\text{x}$
$=2\sin^2\text{x}-(1-2\sin^2\text{x})$
$=4\sin^2\text{x}-1$
$\therefore0\leq\sin^2\text{x}\leq1$
$\Rightarrow4\times0\leq4\times\sin^2\text{x}\leq4\times1$
$\Rightarrow0\leq4\sin^2\text{x}\leq4$
$\Rightarrow0-1\leq4\sin^2\text{x}-1\leq4-1$
$\Rightarrow-1\leq4\sin^2\text{x}-1\leq3$
$\Rightarrow-1\leq\text{A}\leq3$
$\Rightarrow\text{A}\in[-1,3]$
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