First slide

Extreme values and Periodicity of Trigonometric functions

Question

The equation $\sin^{4} x + \cos^{4} x + \sin 2 x + α = 0$ is solvable for

Difficult

A

$\frac{- 5}{2} \leq α \leq \frac{1}{2}$
B

$- 3 \leq α \leq 1$
C

$\frac{- 3}{2} \leq α \leq \frac{1}{2}$
D

$- 1 \leq α \leq 1$

Solution

$\begin{array}{l} W e h a v e {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x + \sin 2 x + α = 0 \\ 1 - \frac{\sin^{2} 2 x}{2} + \sin 2 x + α = 0 \\ \Rightarrow \sin^{2} 2 x - 2 \sin 2 x - 2 - 2 α = 0 \\ \Rightarrow \sin 2 x = \frac{2 \pm \sqrt{4 - 4 (- 2 - 2 α)}}{2} \\ \Rightarrow \sin 2 x = 1 \pm \sqrt{3 + 2 α} \\ \Rightarrow - 1 \leq \sin 2 x \leq 1 \\ \Rightarrow - 1 \leq 1 \pm \sqrt{3 + 2 α} ≤1 \\ \Rightarrow - 2 \leq \pm \sqrt{3 + 2 α} \leq 0 \\ n e g e c t i n g p o s i t i v e s i g n \\ - 2 \leq - \sqrt{3 + 2 α} \leq 0 \\ 4 \geq 3 + 2 α \geq 0 \\ \frac{- 3}{2} \leq α \leq \frac{1}{2} \end{array}$

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