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

$$ \\ $$$$\:\:\:\:\:\begin{cases}{\:\:\:{f}\::\:\:\left[\:\:\mathrm{0}\:\:,\:\:\mathrm{1}\:\right]\:\rightarrow\:\mathbb{R}}\\{\:\:\:\:{f}\:\left({x}\:\right)\:=\:\frac{\:\mathrm{4}^{\:{x}} }{\mathrm{2}\:+\:\mathrm{4}^{\:{x}} }}\end{cases} \\ $$$$\:\:\:\:\:\:{is}\:\:{given}\:.\:\:{find}\:{the}\:{value}\:{of} \\ $$$$\:\:\:\:\:\:{the}\:{following}\:{expression}. \\ $$$$\:\:\:\:\:\:\mathrm{E}\:=\:{f}\:\left(\frac{\mathrm{1}}{\mathrm{20}}\:\right)+\:{f}\left(\frac{\:\mathrm{2}}{\mathrm{20}}\:\right)+…\:+{f}\:\left(\frac{\mathrm{19}}{\mathrm{20}}\right)\:−{f}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:\right)=? \\ $$

Answered by ARUNG_Brandon_MBU last updated on 30/Jan/23

f(x)=(4^x /(2+4^x )) ⇒f(1−x)=(4^(1−x) /(2+4^(1−x) ))=(2/(4^x +2)) f(x)+f(1−x)=(4^x /(2+4^x ))+(2/(2+4^x ))=1 ⇒E=Σ_(k=1) ^9 (f((k/(20)))+f(1−(k/(20))))+f(((10)/(20)))−f((1/2)) =Σ_(k=1) ^9 (1)+f((1/2))−f((1/2))=9

$${f}\left({x}\right)=\frac{\mathrm{4}^{{x}} }{\mathrm{2}+\mathrm{4}^{{x}} }\:\Rightarrow{f}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{4}^{\mathrm{1}−{x}} }{\mathrm{2}+\mathrm{4}^{\mathrm{1}−{x}} }=\frac{\mathrm{2}}{\mathrm{4}^{{x}} +\mathrm{2}} \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{4}^{{x}} }{\mathrm{2}+\mathrm{4}^{{x}} }+\frac{\mathrm{2}}{\mathrm{2}+\mathrm{4}^{{x}} }=\mathrm{1} \\ $$$$\Rightarrow\mathrm{E}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\left({f}\left(\frac{{k}}{\mathrm{20}}\right)+{f}\left(\mathrm{1}−\frac{{k}}{\mathrm{20}}\right)\right)+{f}\left(\frac{\mathrm{10}}{\mathrm{20}}\right)−{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\underset{{k}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\left(\mathrm{1}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{9} \\ $$

Commented by ARUNG_Brandon_MBU last updated on 30/Jan/23

Generally if f(x)=(a^(2x) /(a+a^(2x) )) , then f(x)+f(1−x)=1

$$\mathrm{Generally}\:\mathrm{if}\:{f}\left({x}\right)=\frac{{a}^{\mathrm{2}{x}} }{{a}+{a}^{\mathrm{2}{x}} }\:,\:\mathrm{then}\:{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\mathrm{1} \\ $$

Commented by mnjuly1970 last updated on 30/Jan/23

$${thanks}\:{alot}\:{sir}\:{Arung} \\ $$

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