Question : If $\frac{a}{1 - 2 a} + \frac{b}{1 - 2 b} + \frac{c}{1 - 2 c} = \frac{1}{2}$ , then the value of $\frac{1}{1 - 2 a} + \frac{1}{1 - 2 b} + \frac{1}{1 - 2 c}$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4

Question

Question : The value of $x$ which satisfies the equation $\frac{x + a^{2} + 2 c^{2}}{b + c} + \frac{x + b^{2} + 2 a^{2}}{c + a} + \frac{x + c^{2} + 2 b^{2}}{a + b} = 0$ is:

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Answer 1

Correct Answer: 4

Solution : Given: $\frac{a}{1 - 2 a} + \frac{b}{1 - 2 b} + \frac{c}{1 - 2 c} = \frac{1}{2}$
Multiplying by 2 on both sides
⇒ $\frac{2 a}{1 - 2 a} + \frac{2 b}{1 - 2 b} + \frac{2 c}{1 - 2 c} = 1$
Adding 3 on both sides
⇒ $1 + \frac{2 a}{1 - 2 a} + 1 + \frac{2 b}{1 - 2 b} + 1 + \frac{2 c}{1 - 2 c} = 1 + 3$
⇒ $\frac{1}{1 - 2 a} + \frac{1}{1 - 2 b} + \frac{1}{1 - 2 c} = 4$
Hence, the correct answer is 4.

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Question : If a1−2a+b1−2b+c1−2c=12, then the value of 11−2a+11−2b+11−2c is:Option 1: 1Option 2: 2Option 3: 3Option 4: 4

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Question : If $\frac{a}{1 - 2 a} + \frac{b}{1 - 2 b} + \frac{c}{1 - 2 c} = \frac{1}{2}$ , then the value of $\frac{1}{1 - 2 a} + \frac{1}{1 - 2 b} + \frac{1}{1 - 2 c}$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4