Question : If $\mathrm{N}+\frac{1}{\mathrm{~N}}=\sqrt{3}$, then find the value of $\mathrm{N}^6+\frac{1}{\mathrm{~N}^6}+11$.
Option 1: –11
Option 2: –2
Option 3: 11
Option 4: 9
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Correct Answer: 9
Solution :
Given: $\mathrm{N}+\frac{1}{\mathrm{~N}}=\sqrt{3}$
Squaring both sides, we get:
⇒ $\mathrm{N}^2+2+\frac{1}{\mathrm{~N}^2}=3$
⇒ $\mathrm{N}^2+\frac{1}{\mathrm{~N}^2}=1$
Now, cubing both sides, we get:
⇒ $(\mathrm{N}^2+\frac{1}{\mathrm{~N}^2})^3=1^3$
⇒ $(\mathrm{N}^2)^3 + (\frac{1}{\mathrm{~N}^2})^3 + 3\times\mathrm{N}^2\times\frac{1}{\mathrm{~N}^2}(\mathrm{N}^2+\frac{1}{\mathrm{~N}^2})=1$
⇒ $ \mathrm{N}^6 + \frac{1}{\mathrm{~N}^6} + 3\times1=1$
⇒ $\mathrm{N}^6+\frac{1}{\mathrm{~N}^6}=-2$
Adding 11 on both sides, we get,
$\therefore\mathrm{N}^6+\frac{1}{\mathrm{~N}^6}+11=9$
Hence, the correct answer is 9.
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