Question

For the given equations $\cos (4y-3x-2)-\cos (4y+3x+2)=2+2\ln (k^4-255)$ and $\cos (4y-3x-2)+\cos (4y+3x+2)=2k+8$ to have real solutions $(x,y)$, the possible value of k is

• A. 4
• B. -4
• C. -5
• D. none of these

Answer 1

The correct option is B -4

$2\sin \left(4y\right)\sin (3x+2)=2+2\ln (k^4-255)$

$1+\ln (k^4-255)=\sin \left(4y\right)\sin (3x+2)\le 1$

$k^4-255\le 1$

$k^4\le 256\Rightarrow -4\le k\le 4$

and $2\cos \left(4y\right)\cos (3x+2)=2k+8$

$k+4\le 1\Rightarrow k\le -3$

So $k=-4$