Question : If $x+\frac{1}{x}=c+\frac{1}{c}$, then the value of $x$ is:
Option 1: $c,\frac{1}{c}$
Option 2: $c,c^{2}$
Option 3: $c,2c$
Option 4: $0, 1$
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Correct Answer: $c,\frac{1}{c}$
Solution :
Given: $x+\frac{1}{x}=c+\frac{1}{c}$
⇒ $x-c=\frac{1}{c}–\frac{1}{x}$
⇒ $x-c=\frac{(x–c)}{xc}$
⇒ $(x-c)-\frac{(x–c)}{xc}=0$
⇒ $(x-c)[1-\frac{1}{xc}]=0$
Taking, $(x-c)=0$
⇒ $x=c$
Now, taking $[1-\frac{1}{xc}]=0$
⇒ $\frac{1}{xc}=1$
⇒ $x=\frac{1}{c}$
Hence, the correct answer is $c,\frac{1}{c}$.
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