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
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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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