Question

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

Answer 1

(a) $\left[-1,3\right]$

$$\begin{gathered} A=2{\sin }^{2}x-\cos 2x \\ =2{\sin }^{2}x-\left(1-2{\sin }^{2}x\right) \\ =4{\sin }^{2}x-1 \\ \because 0\le {\sin }^{2}x\le 1 \\ \Rightarrow 4\times 0\le 4\times {\sin }^{2}x\le 4\times 1 \\ \Rightarrow 0\le 4{\sin }^{2}x\le 4 \\ \Rightarrow 0-1\le 4{\sin }^{2}x-1\le 4-1 \\ \Rightarrow -1\le 4{\sin }^{2}x-1\le 3 \\ \Rightarrow -1\le A\le 3 \\ \Rightarrow A\in \left[-1,3\right] \end{gathered}$$