Question : If $x^2-4 x-1=0$, then find the value of $x^2+\frac{1}{x^2}-5$.
Option 1: 17
Option 2: 12
Option 3: 15
Option 4: 13
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Correct Answer: 13
Solution : $x^2-4 x-1=0$ $x-\frac{1}{x} = 4$ Squaring on both sides, $ x^{2} + \frac{1}{x^{2}} - 2×x×\frac{1}{x} = 16$ $ x^{2} + \frac{1}{x^{2}} = 16+2 = 18$ -----------(i) Now, $x^2+\frac{1}{x^2}-5$ = 18 – 5 = 13 Hence, the correct answer is 13.
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Question : If $x+\frac{1}{x}=2$, then find the value of $x^5+\frac{1}{x^5}$.
Option 1: 1
Option 2: 3
Option 3: 4
Option 4: 2
Question : If $x+\frac{1}{x}=0$, then the value of $x^{5}+\frac{1}{x^{5}}$ is:
Option 1: 2
Option 2: –1
Option 3: 1
Option 4: 0
Question : If $x^4+\frac{1}{x^4}=194, x>0$, then find the value of $x^3+\frac{1}{x^3}+x+\frac{1}{x}$
Option 1: 76
Option 2: 66
Option 3: 56
Option 4: 46
Question : If $x+\frac{1}{x}=3$, then the value of $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ is:
Option 1: $\frac{4}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{5}{3}$
Question : If $x+\frac{1}{x}=2$, then find the value of $x^{1823}+\frac{1}{x^{1929}}$.
Option 2: 1
Option 3: 0
Option 4: –1
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