Question : If $x+\frac{1}{x}=-2$, then what is the value of $x^{17}+x^{-17}+x^{12}+x^{-12} ?(x<0)$
Option 1: –2
Option 2: –1
Option 3: 1
Option 4: 0
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Correct Answer: 0
Solution : Given $x+\frac{1}{x}=-2$ $⇒x^2+1=-2x$ $⇒x^2+2x+1=0$ $⇒(x+1)^2=0$ $⇒x=-1$ Now, putting the value of $x=-1$ in $x^{17}+x^{-17}+x^{12}+x^{-12}$ we get: $=(-1)^{17}+(-1)^{-17}+(-1)^{12}+(-1)^{-12}$ $=-1-1+1+1=0$ Hence, the correct answer is 0.
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Question : If $\mathrm{K}+\frac{1}{\mathrm{~K}}+2=0$ and $\mathrm{K}<0$, then what is the value of $\mathrm{K}^{11}+\frac{1}{\mathrm{~K}^4}$?
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Question : If $x^2+\frac{1}{x^2}=2$, then the value of $x-\frac{1}{x}$ is:
Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
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