If f(x) = 2^x2^x + 2, x R, then k=1^81 f ( k82) is equal to: — Solution | JEE Mathematics

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If $f(x) = \frac{2^x}{2^x + \sqrt{2}}$ , $x \in \mathbb{R}$ , then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ is equal to:

Answer

Correct answer:B

Standard Method

Given: $f(x) = \frac{2^x}{2^x + \sqrt{2}}$ and

S = \sum_{k=1}^{81} f\left(\frac{k}{82}\right)

Find: the value of $S$ .

Use the symmetry relation between $f(x)$ and $f(1-x)$ .

First, compute

f(1-x) = \frac{2^{1-x}}{2^{1-x} + \sqrt{2}} = \frac{2/2^x}{2/2^x + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2} + 2^x}

Therefore,

f(x) + f(1-x) = \frac{2^x}{2^x + \sqrt{2}} + \frac{\sqrt{2}}{\sqrt{2} + 2^x} = 1

Now pair the terms of the sum as

f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right) = f\left(\frac{k}{82}\right) + f\left(1 - \frac{k}{82}\right) = 1

for $k = 1,2,\dots,40$ .

So,

S = \sum_{k=1}^{40} \left[f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right)\right] + f\left(\frac{41}{82}\right)

Hence,

S = 40 + f\left(\frac{1}{2}\right)

Now evaluate the middle term:

f\left(\frac{1}{2}\right) = \frac{2^{1/2}}{2^{1/2} + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2} + \sqrt{2}} = \frac{1}{2}

Thus,

S = 40 + \frac{1}{2} = \frac{81}{2}

Therefore, the correct option is B.

Pairing Argument

Given: $f(x) = \frac{2^x}{2^x + \sqrt{2}}$ with $x \in \mathbb{R}$ .

Find:

\sum_{k=1}^{81} f\left(\frac{k}{82}\right)

Observe that the arguments $\frac{k}{82}$ and $\frac{82-k}{82}$ add up to $1$ . So the key step is to prove

f(x) + f(1-x) = 1

Now,

f(1-x) = \frac{2^{1-x}}{2^{1-x} + \sqrt{2}} = \frac{2/2^x}{2/2^x + \sqrt{2}}

Multiplying numerator and denominator by $2^x$ gives

f(1-x) = \frac{2}{2 + \sqrt{2}\,2^x} = \frac{\sqrt{2}}{\sqrt{2} + 2^x}

Hence,

\begin{aligned} f(x) + f(1-x) &= \frac{2^x}{2^x + \sqrt{2}} + \frac{\sqrt{2}}{\sqrt{2} + 2^x} \\ &= 1 \end{aligned}

Therefore each pair contributes $1$ :

f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right) = 1

There are $40$ such pairs among the $81$ terms, and the remaining term is at $k=41$ :

\frac{41}{82} = \frac{1}{2}

So,

S = 40 + f\left(\frac{1}{2}\right)

Now,

f\left(\frac{1}{2}\right) = \frac{\sqrt{2}}{\sqrt{2}+\sqrt{2}} = \frac{1}{2}

Thus,

S = 40 + \frac{1}{2} = \frac{81}{2}

Therefore, the value of the sum is $\frac{81}{2}$ and the correct option is B.

Pairing $f\left(\frac{k}{82}\right)$ with the wrong complementary term. The correct partner is $f\left(\frac{82-k}{82}\right) = f\left(1-\frac{k}{82}\right)$ , because the identity uses $f(x)+f(1-x)$ . Always match arguments whose sum is $1$ .
Forgetting the unpaired middle term when the total number of terms is odd. Here there are $81$ terms, so after forming $40$ pairs, the term at $k=41$ remains. It must be evaluated separately as $f\left(\frac{1}{2}\right)$ .
Algebraic error while simplifying $f(1-x)$ . If the transformation from $2^{1-x}$ is mishandled, the identity $f(x)+f(1-x)=1$ may not appear. Rewrite $2^{1-x}$ as $\frac{2}{2^x}$ and simplify carefully.

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