Question : If $\mathrm{K}+\frac{1}{\mathrm{~K}}+2=0$ and $\mathrm{K}<0$, then what is the value of $\mathrm{K}^{11}+\frac{1}{\mathrm{~K}^4}$?
Option 1: 0
Option 2: –2
Option 3: –1
Option 4: –17
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Correct Answer: 0
Solution : Given, $\mathrm{K}+\frac{1}{\mathrm{~K}}+2=0$ ⇒ $K^{2} + 2K + 1 = 0$ ⇒ $(K+1)^{2} = 0$ ⇒ $K+1 = 0$ $\therefore$ $K = -1$ Now, we have to find $K^{11}+\frac{1}{K^4}$ Putting the value of $K$, we get $(-1)^{11} + \frac{1}{(-1)^{4}}$ $= -1 + 1 = 0$ Hence, the correct answer is 0.
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