Question

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

• A. $-\frac{1}{2}\le \alpha \le \frac{1}{2}$
• B. $-3\le \alpha \le 1$
• C. $-\frac{3}{2}\le \alpha \le \frac{1}{2}$
• D. $\frac{1}{2}\le \alpha \le \frac{3}{2}$

Answer 1

Answer: (C)

$-\frac{3}{2}\le \alpha \le \frac{1}{2}$

${\sin }^{4}x+{\cos }^{4}x+\sin 2x+\alpha =0$
$({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x+\sin 2x+\alpha =0$
$\Rightarrow {\sin }^{2}2x-2\sin 2x-2\alpha -2=0$
$\Rightarrow \sin 2x=1\pm \sqrt{3+2\alpha }$
We have
$-1\le \sin 2x\le 1$
$-1\le 1\pm \sqrt{3+2\alpha }\le 1$
$-2\le \pm \sqrt{3+2\alpha }\le 0$
$0\le 3+2\alpha \le 4$
$-3/2\le \alpha \le 1/2$