If A=2 sin
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) $[-1,3]$

$A=2 \sin ^{2} x-\cos 2 x$

$=2 \sin ^{2} x-\left(1-2 \sin ^{2} x\right)$

$=4 \sin ^{2} x-1$

$\because 0 \leq \sin ^{2} x \leq 1$

$\Rightarrow 4 \times 0 \leq 4 \times \sin ^{2} x \leq 4 \times 1$

$\Rightarrow 0 \leq 4 \sin ^{2} x \leq 4$

$\Rightarrow 0-1 \leq 4 \sin ^{2} x-1 \leq 4-1$

$\Rightarrow-1 \leq 4 \sin ^{2} x-1 \leq 3$

$\Rightarrow-1 \leq A \leq 3$

$\Rightarrow A \in[-1,3]$

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