Mathematics · Sequence and Series

JEE Advanced 2020 — Paper 2 — Question 49

Let the function $\mathrm{f}:[0,1] \rightarrow \mathbb{R}$ be defined by

$f(x)=\frac{4^{x}}{4^{x}+2}$ Then the value of $f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+f\left(\frac{3}{40}\right)+\ldots+f\left(\frac{39}{40}\right)-f\left(\frac{1}{2}\right)$

is $\_\_\_\_$

Answer: 19

Numerical answer — enter this value.

Step-by-step solution

$\sum_{r=1}^{39} f\left(\frac{r}{40}\right)=\frac{1}{2} \sum_{r=1}^{39} f\left(\frac{r}{40}\right)+f\left(\frac{40-r}{40}\right)=\frac{39}{2}$

As $(\mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})=1) \& \mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{2}$

So $\sum_{r=1}^{39} f\left(\frac{r}{40}\right)-f\left(\frac{1}{2}\right)=\frac{39}{2}-\frac{1}{2}=19$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Mathematics
Chapter: Sequence and Series
Topic: Introduction to Sequence and Series