Question : If $a+\frac{1}{a}=1$, then the value of $\frac{a^2-a+1}{a^2+a+1}$ is $(a\neq 0)$:
Option 1: 1
Option 2: –1
Option 3: 0
Option 4: 2
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Correct Answer: 0
Solution : Given: $a+\frac{1}{a}=1$ ⇒ $\frac{a^2+1}{a}=1$ ⇒ $a^2+1=a$ ⇒ $a^2-a+1=0$ Now, $\frac{a^2–a+1}{a^2+a+1}=\frac{0}{a^2+a+1}=0$ Hence, the correct answer is 0.
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Question : If $a+b=2c$, then the value of $\frac{a}{a–c}+\frac{c}{b–c}$ is equal to (where $a\neq b\neq c$):
Option 1: $–1$
Option 2: $1$
Option 3: $0$
Option 4: $\frac{1}{2}$
Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$ where $a \neq b\neq c\neq 0$, then the value of $a^{2}b^{2}c^{2}$ is:
Option 1: –1
Option 2: abc
Option 4: 1
Question : If $(a+\frac{1}{a})=–2$, then the value of $a^{1000}+a^{–1000}$ is:
Option 1: $2$
Option 2: $0$
Option 3: $1$
Question : If $\frac{a}{b}+\frac{b}{a}=-1$ and $a-b=2$, then the value of $a^3-b^3$ is:
Option 1: $0$
Option 2: $\frac{1}{2}$
Option 4: $–1$
Question : If $a+\frac{1}{a-2}=4$, then the value of $(a-2)^{2}+(\frac{1}{a-2})^{2}$ is:
Option 1: 0
Option 2: 2
Option 3: –2
Option 4: 4
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