Question : If $x+\frac{1}{x}=-2$, then what is the value of $x^7+x^{-7}+x^2+x^{-2} ?(\mathrm{x}<0)$
Option 1: 4
Option 2: 2
Option 3: 1
Option 4: 0
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Correct Answer: 0
Solution : $x+\frac{1}{x}=-2$ $⇒x^2+1=-2x$ $⇒x^2+2x+1=0$ $⇒(x+1)^2 = 0$ $\therefore x = -1$ By putting $x=-1$, we get, $x^7+x^{-7}+x^2+x^{-2}=(-1)^7+(-1)^{-7}+(-1)^2+(-1)^{-2} = -2+2 = 0$ Hence, the correct answer is 0.
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Question : What is the value of $\frac{1+\mathrm{x}}{1-\mathrm{x}^2} \div \frac{1+\mathrm{x}}{1-\mathrm{x}^4}-\frac{1-\mathrm{x}^4}{1-\mathrm{x}} \times \frac{1+\mathrm{x}}{1-\mathrm{x}^2}$?
Option 1: $\frac{2 \mathrm{x}\left(1+\mathrm{x}^2\right)}{(1-\mathrm{x})}$
Option 2: $\frac{2 \mathrm{x}\left(1+\mathrm{x}^2\right)}{(\mathrm{x}-1)}$
Option 3: $(1-\mathrm{x})^2$
Option 4: $\left(1+\mathrm{x}^2\right)$
Question : If $\mathrm{m}-\frac{1}{\mathrm{~m}}=7$, then what is the value of $\mathrm{m}^2+\frac{1}{\mathrm{~m}^2}$?
Option 1: 0
Option 2: 51
Option 3: 53
Option 4: 49
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 $x+\frac{2}{x}=1$, then the value of $\frac{x^2+7x+2}{x^2+13x+2}$ is:
Option 1: $\frac{5}{7}$
Option 2: $\frac{3}{7}$
Option 3: $\frac{4}{7}$
Option 4: $\frac{2}{7}$
Question : If $\frac{x}{y}=\frac{4}{5}$, then the value of $(\frac{4}{7}+\frac{2y–x}{2y+x})$ is:
Option 1: $\frac{3}{7}$
Option 2: $1\frac{1}{7}$
Option 3: $1$
Option 4: $2$
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