If $K+\frac{1}{K}+2=0$ and K < 0, then what is the value of $K^{10}+\frac{1}{K^{11}}$ ?
Answer & explanation
Correct answer: option 2
$K+\frac{1}{K}+2=0$ and K < 0, then what is the value of $K^{10}+\frac{1}{K^{11}}$
= $K+\frac{1}{K}=-2$
= put k = -1
$K^{10}+\frac{1}{K^{11}}$ = $-1^{10}+\frac{1}{-1^{11}}$ = 0