Question : If $\mathrm{K}+\frac{1}{\mathrm{~K}}+2=0$ and $\mathrm{K}<0$, then what is the value of $\mathrm{K}^{10}+\frac{1}{\mathrm{~K}^{11}}$?
Option 1: 1
Option 2: 0
Option 3: –1
Option 4: 2
Correct Answer: 0
Solution :
Given, $\mathrm{K}+\frac{1}{\mathrm{~K}}+2=0$
⇒ $\mathrm{K}^2+1+2\mathrm{K}=0$
⇒ $(\mathrm{K}+1)^2=0$
$\therefore \mathrm{K}=-1$
So, $\mathrm{K}^{10}+\frac{1}{\mathrm{~K}^{11}}=\mathrm{(-1)}^{10}+\frac{1}{\mathrm{(-1)}^{11}}=1-1=0$
Hence, the correct answer is 0.
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