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Question : 78 of 86

Marks: +1, -0

a b c=2

\;\frac{1}{1+a+2 b^{-1}}+\;\frac{1}{1+\frac{b}{2}+c^{-1}}+\;\frac{1}{1+a^{-1}+c}=

Solution:

Concept:

Given

abc = 2

, the sum simplifies to 1 by rewriting each term with a common denominator.

Explanation:

Let

S = \frac{1}{1+a+\frac{2}{b}} + \frac{1}{1+\frac{b}{2}+\frac{1}{c}} + \frac{1}{1+\frac{1}{a}+c}

Multiply the first term by

\frac{b}{b}

\frac{b}{b+ab+2}

Multiply the second term by

\frac{2}{2}

\frac{2}{2+b+\frac{2}{c}}

Since

abc=2

, we have

\frac{2}{c}=ab

, so the second term becomes

\frac{2}{2+b+ab} = \frac{2}{b+ab+2}

For the third term, multiply by

\frac{a}{a}

\frac{a}{a+1+ac}

. Using

abc=2

ac = \frac{2}{b}

, so

a+1+ac = a+1+\frac{2}{b}

. Multiply by

\frac{b}{b}

\frac{ab}{ab+b+2} = \frac{ab}{b+ab+2}

Now all three terms have the same denominator

b+ab+2

S = \frac{b}{b+ab+2} + \frac{2}{b+ab+2} + \frac{ab}{b+ab+2} = \frac{b+2+ab}{b+ab+2} = 1

Answer:

Option A:

1

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