Question : If $\frac{1}{(x-2)}+x=8$, then what is the value of $\frac{1}{(x-2)^2}+(x-2)^2$?
Option 1: 38
Option 2: 36
Option 3: 40
Option 4: 34
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Correct Answer: 34
Solution : Given: $\frac{1}{(x-2)}+x=8$ Subtracting 2 from both sides, we get: ⇒ $\frac{1}{(x-2)}+x-2=6$ Squaring both sides, we get, ⇒ $\frac{1}{(x-2)^2}+(x-2)^2+2\times \frac{1}{(x-2)} \times(x-2) = 36$ ⇒ $\frac{1}{(x-2)^2}+(x-2)^2=36-2$ ⇒ $\frac{1}{(x-2)^2}+(x-2)^2=34$ Hence, the correct answer is 34.
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Question : If $2 x+\frac{2}{x}=5$, then the value of $\left(x^3+\frac{1}{x^3}+2\right)$ will be:
Option 1: $\frac{81}{11}$
Option 2: $\frac{81}{7}$
Option 3: $\frac{71}{8}$
Option 4: $\frac{81}{8}$
Question : If $x+\frac{1}{x}=\mathrm{\frac{K}{2}}$, then what is the value of $\frac{x^8+1}{x^4} ?$
Option 1: $\frac{\mathrm{K}^4-16 \mathrm{~K}^2+32}{16}$
Option 2: $\frac{\mathrm{K}^4-8 \mathrm{~K}^2-36}{32}$
Option 3: $\frac{\mathrm{K}^4-8 \mathrm{~K}^2-32}{16}$
Option 4: $\frac{\mathrm{K}^4-8 \mathrm{~K}^2+32}{16}$
Question : If $x^2+\frac{1}{x^2}=7$, then what is the value of $x^3+\frac{1}{x^3}$?
Option 1: 9
Option 2: 18
Option 3: 27
Option 4: 36
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 $\frac{x^2-1}{x}=8$, then what is the value of $\frac{x^6-1}{x^3}=?$
Option 1: 496
Option 2: 536
Option 3: 512
Option 4: 488
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