Let f: R rightarrow R be a function defined b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = \frac{2e^{2x}}{e^{2x} + e}$.Consider $f(x) + f(1-x) = \frac{2e^{2x}}{e^{2x} + e} + \frac{2e^{2(1-x)}}{e^{2(1-x)} + e}$.Multiplying t

Vedclass · Accepted Answer

$99$

Vedclass · Answer

(B) Given $f(x) = \frac{2e^{2x}}{e^{2x} + e}$.Consider $f(x) + f(1-x) = \frac{2e^{2x}}{e^{2x} + e} + \frac{2e^{2(1-x)}}{e^{2(1-x)} + e}$.Multiplying the numerator and denominator of the second term by $e^{2x}$,we get $\frac{2e^{2-2x} \cdot e^{2x}}{e^{2-2x} \cdot e^{2x} + e \cdot e^{2x}} = \frac{2e^2}{e^2 + e^{2x+1}} = \frac{2e^2}{e^2 + e \cdot e^{2x}} = \frac{2e}{e + e^{2x}}$.Thus,$f(x) + f(1-x) = \frac{2e^{2x}}{e^{2x} + e} + \frac{2e}{e^{2x} + e} = \frac{2(e^{2x} + e)}{e^{2x} + e} = 2$.The sum is $S = \sum_{k=1}^{99} f\left(\frac{k}{100}ight)$.Pairing terms $f\left(\frac{k}{100}ight) + f\left(1 - \frac{k}{100}ight) = f\left(\frac{k}{100}ight) + f\left(\frac{100-k}{100}ight) = 2$.There are $49$ such pairs (for $k=1$ to $49$) and the middle term $f\left(\frac{50}{100}ight) = f\left(\frac{1}{2}ight)$.$f\left(\frac{1}{2}ight) = \frac{2e^{2(1/2)}}{e^{2(1/2)} + e} = \frac{2e}{e + e} = 1$.Therefore,$S = 49 	imes 2 + 1 = 98 + 1 = 99$.

Let $f: R \rightarrow R$ be a function defined by $f(x) = \frac{2e^{2x}}{e^{2x} + e}$. Then $f\left(\frac{1}{100}\right) + f\left(\frac{2}{100}\right) + f\left(\frac{3}{100}\right) + \dots + f\left(\frac{99}{100}\right)$ is equal to

Similar Questions

Let $f(x)$ and $g(x)$ be two functions given by $f(x) = \frac{2\sin(\pi x)}{x}$ and $g(x) = f(1 - x) + f(x)$. If $g(x) = k f(\frac{x}{2}) f(\frac{1 - x}{2})$,then the value of $k$ is

Let $f : R \to R$ be a function defined by $f(x) = \frac{4^x}{4^x + 2}$. What is the value of $f(\frac{1}{4}) + 2 f(\frac{1}{2}) + f(\frac{3}{4})$?

The total number of functions $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4, 5, 6\}$ such that $f(1) + f(2) = f(3)$ is equal to:

If $e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$, then $k$ is equal to

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are given by $f(x)=|x|$ and $g(x)=[x]$ for each $x \in R$, then $\{x \in R: g(f(x)) \leq f(g(x))\}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module