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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Functions

If f(x)= 2^x 2^x+ 2 , x ∈ ℝ , then ∑_ k=1 ^ 81 f( k 82 ) is equal to

If f(x)=2x2x+2,x∈Rf(x)=\frac{2^x}{2^x+\sqrt{2}}, \mathrm{x} \in \mathbb{R}, then $\sum_{\mathrm{k}=1}^{81} f\left(\frac{\mathrm{k}}{82}\right) is equal to

Options

  1. A1.812_{1 .} 81 \sqrt{2}
  2. B41
  3. C82
  4. D812\frac{81}{2}

Correct answer

D. 812\frac{81}{2}

Step-by-step solution

f(x)=2x2x+2f(x)+f(1−x)=2x2x+2+21−x21−x+2=2x2x+2+22+22x=2x+22x+2=1\begin{aligned} & f(x)=\frac{2^x}{2^x+\sqrt{2}} \\ & f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}} \\ & =\frac{2^x}{2^x+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^x}=\frac{2^x+\sqrt{2}}{2^x+\sqrt{2}}=1\end{aligned} Now, ∑k=181f(k82)=f(182)+f(282)+……+f(8182)=f(182)+f(182)+……+f(1−282)+f(1−182)\begin{aligned} & \text{ Now, } \sum_{\mathrm{k}=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(\frac{81}{82}\right) \\ & =\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{1}{82}\right)+\ldots \ldots+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\end{aligned} [f(182)+f(1−182)]+[f(282)+f(1−282)]+….40 cases +f(4182)(1+1+….+1)40 times +21/221/2+21/240+12=812\begin{aligned} & {\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots .40 \text{ cases }+\mathrm{f}\left(\frac{41}{82}\right)} \\ & (1+1+\ldots .+1) 40 \text{ times }+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}} \\ & 40+\frac{1}{2}=\frac{81}{2}\end{aligned}$
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