Question

If the equation ${\sin }^{4}x+{\cos }^{4}x+\sin 2x+k=0$ have real solutions, then range of $k$ is

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

Answer 1

The correct option is B $-\frac{3}{2}\le k\le \frac{1}{2}$

${\sin }^{4}x+{\cos }^{4}x+\sin 2x+k=0$
$\Rightarrow 1-2{\sin }^{2}x{\cos }^{2}x+\sin 2x+k=0$
$\Rightarrow 1-\frac{1}{2}{\sin }^{2}2x+\sin 2x+k=0$
$\Rightarrow 2k={\sin }^{2}2x-2\sin 2x-2$
$\Rightarrow 2k+3=(\sin 2x-1{)}^2$

Now, $0\le (\sin 2x-1{)}^2\le 4$
$\Rightarrow 0\le 2k+3\le 4$
$\Rightarrow -3\le 2k\le 1$
$\Rightarrow -\frac{3}{2}\le k\le \frac{1}{2}$