f-0-1-R-f-x-4-x-2-4-x-is-given-find-the-value-of-the-following-expression-E-f-1-20-f-2-20-

Question Number 185983 by mnjuly1970 last updated on 30/Jan/23

{\begin{cases} f : [0, 1] \to R \\ f (x) = \frac{4^{x}}{2 + 4^{x}} \end{cases}

i s g i v e n . f i n d t h e v a l u e o f

t h e f o l l o w i n g e x p r e s s i o n .

E = f (\frac{1}{20}) + f (\frac{2}{20}) + \dots + f (\frac{19}{20}) - f (\frac{1}{2}) = ?

Answered by ARUNG_Brandon_MBU last updated on 30/Jan/23

f (x) = \frac{4^{x}}{2 + 4^{x}} \Rightarrow f (1 - x) = \frac{4^{1 - x}}{2 + 4^{1 - x}} = \frac{2}{4^{x} + 2}

f (x) + f (1 - x) = \frac{4^{x}}{2 + 4^{x}} + \frac{2}{2 + 4^{x}} = 1

\Rightarrow E = \underset{k = 1}{\sum^{9}} (f (\frac{k}{20}) + f (1 - \frac{k}{20})) + f (\frac{10}{20}) - f (\frac{1}{2})

= \underset{k = 1}{\sum^{9}} (1) + f (\frac{1}{2}) - f (\frac{1}{2}) = 9

Commented by ARUNG_Brandon_MBU last updated on 30/Jan/23

Generally if f (x) = \frac{a^{2 x}}{a + a^{2 x}}, then f (x) + f (1 - x) = 1

Commented by mnjuly1970 last updated on 30/Jan/23

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