Question:
If $A=2 \sin ^{2} x-\cos 2 x$, then $A$ lies in the interval
(a) $[-1,3]$
(b) $[1,2]$
(c) $[-2,4]$
(d) none of these
Solution:
(a) $[-1,3]$
$A=2 \sin ^{2} x-\cos 2 x$
$=2 \sin ^{2} x-\left(1-2 \sin ^{2} x\right)$
$=4 \sin ^{2} x-1$
$\because 0 \leq \sin ^{2} x \leq 1$
$\Rightarrow 4 \times 0 \leq 4 \times \sin ^{2} x \leq 4 \times 1$
$\Rightarrow 0 \leq 4 \sin ^{2} x \leq 4$
$\Rightarrow 0-1 \leq 4 \sin ^{2} x-1 \leq 4-1$
$\Rightarrow-1 \leq 4 \sin ^{2} x-1 \leq 3$
$\Rightarrow-1 \leq A \leq 3$
$\Rightarrow A \in[-1,3]$