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Question : 23 of 86

Marks: +1, -0

if

\;\frac{1}{1+a^{(n-m)}}+\;\frac{1}{1+a^{(m-n)}}=x,

then the value of

x

is:

0
$\;\frac{1}{2}$
1
$a^{m+n}$

Solution:

Concept:

Simplify the sum using the property that

a^{m-n} = \frac{1}{a^{n-m}}

.

Explanation:

Let

t = a^{n-m}

.

Then

a^{m-n} = a^{-(n-m)} = \frac{1}{t}

.

The expression becomes:

\frac{1}{1+t} + \frac{1}{1+\frac{1}{t}}

.

Simplify the second term:

\frac{1}{1+\frac{1}{t}} = \frac{1}{\frac{t+1}{t}} = \frac{t}{1+t}

.

So sum =

\frac{1}{1+t} + \frac{t}{1+t} = \frac{1+t}{1+t} = 1

.

Thus

x = 1

.

Answer:

x = 1

, which corresponds to option C.

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