Question

The equation ${\sin }^{4}x+{\cos }^{4}x+\sin 2x+k=0$ must have real solutions if

• A. $k=0$
• B. $|k|\le \frac{1}{2}$
• C. $-\frac{3}{2}\le k\le \frac{1}{2}$
• D. $-\frac{1}{2}\le k\le \frac{3}{2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $k=0$
B $|k|\le \frac{1}{2}$
C $-\frac{3}{2}\le k\le \frac{1}{2}$

${\sin }^{4}x+{\cos }^{4}x=({\sin }^{2}x{)}^2+({\cos }^{2}x{)}^2=1-\frac{1}{2}{\sin }^{2}2x$
So, the equation becomes
$1-\frac{1}{2}{\sin }^{2}2x+\sin 2x+k=0\Rightarrow 2k+2={\sin }^{2}2x-2\sin 2x\Rightarrow 2k+3=(\sin 2x-1{)}^2$
Range of $(\sin 2x-1{)}^2$ is $[0,4]$
For the equation to have solution,
$0\le 2k+3\le 4$
$\Rightarrow -\frac{3}{2}\le k\le \frac{1}{2}$