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Functions Part 1

Section: Mathematics

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Question : 8 of 100

Marks: +1, -0

If

f(x)=\frac{2^{2x}}{2^{2x}+2},\;x\in R

f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+\cdots+f\left(\frac{2022}{2023}\right)

[24-Jan-2023 Shift 2]

2011
1010
2010
1011

Solution:

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

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

\;=\;\frac{4^x}{4^x+2}+\;\frac{4}{4+2(4^x)}

\;=\;\frac{4^x}{4^x+2}+\;\frac{2}{2+4^x}

\;=1

\;\Rightarrow f(x)+f(1-x)=1

\;\;\text{ Now }\; f\left(\frac{1}{2023}\right)+f\left(\frac{2}{2023}\right)+f\left(\frac{3}{2023}\right)+\cdots+

\;\cdots+f\left(1-\frac{3}{2023}\right)+f\left(1-\frac{2}{2023}\right)+f\left(1-\frac{1}{2023}\right)

Now sum of terms equidistant from beginning and end is 1

\;\;\text{ Sum }\;=1+1+1+\cdots+1\;\text{ (1011 times) }

\;=1011

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