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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Question : If $\frac{a}{1-2a}+\frac{b}{1-2b}+\frac{c}{1-2c}=\frac{1}{2}$, then the value of $\frac{1}{1-2a}+\frac{1}{1-2b}+\frac{1}{1-2c}$ is:
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