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
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Correct Answer: 56
Solution : Given: $x^4+\frac{1}{x^4}=194$ $⇒x^4+\frac{1}{x^4}+2=194+2$ $⇒(x^2+\frac{1}{x^2})^2=196$ $⇒x^2+\frac{1}{x^2}=\sqrt{196}$ $⇒x^2+\frac{1}{x^2}=14$ $⇒x^2+\frac{1}{x^2}+2=14+2$ $⇒(x+\frac{1}{x})^2=16$ $⇒x+\frac{1}{x}=4$ -----------------------------(1) Now, $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3×x×\frac{1}{x}(x+\frac{1}{x})$ $⇒x^3+\frac{1}{x^3}=4^3-3 ×4$ $⇒x^3+\frac{1}{x^3}=52$--------------------------------(2) Therefore, $x^3+\frac{1}{x^3}+x+\frac{1}{x}$ = $52+4 = 56$ (from equations (1) and (2)) Hence, the correct answer is 56.
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Question : If $\frac{x^2+1}{x}=5$, then find the value of $x^4+\frac{1}{x^4}-36$.
Option 1: 491
Option 2: 149
Option 3: 419
Option 4: 194
Question : If $x+\frac{1}{x}=2$, then find the value of $x^{1823}+\frac{1}{x^{1929}}$.
Option 1: 2
Option 2: 1
Option 3: 0
Option 4: –1
Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:
Option 1: $\sqrt{41}$
Option 2: $\sqrt{29}$
Option 3: $\sqrt{23}$
Option 4: $\sqrt{43}$
Question : If $x^2-\sqrt{9.76} x+1=0$ and $x>1$, then the value of $\left(x^3-\frac{1}{x^3}\right)$ is:
Option 1: 21.042
Option 2: 24.024
Option 3: 21.024
Option 4: 24.042
Question : If $x>1$ and $x+\frac{1}{x}=2\frac{1}{12}$, then the value of $x^{4}-\frac{1}{x^{4}}$ is:
Option 1: $\frac{58975}{20736}$
Option 2: $\frac{59825}{20736}$
Option 3: $\frac{57985}{20736}$
Option 4: $\frac{57895}{20736}$
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