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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If ${x}^2+\frac{1}{{x}^2}=2$ then the value of ${x}^{256}+\frac{1}{{x}^{256}}$
Solution
Given \(x^2+\dfrac{1}{x^2}=2\). Let \(t=x+\dfrac{1}{x}\). Then \(x^2+\dfrac{1}{x^2}=t^2-2\), so \(t^2-2=2 \Rightarrow t^2=4 \Rightarrow t=\pm2\).
From \(x+\dfrac{1}{x}=2 \Rightarrow x=1\) and from \(x+\dfrac{1}{x}=-2 \Rightarrow x=-1\).
Hence \(x^{256}+\dfrac{1}{x^{256}}= \begin{cases} 1+1=2,& x=1\\ (-1)^{256}+(-1)^{-256}=1+1=2,& x=-1 \end{cases}\).
Final Answer: \(2\).
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