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Functions Part 2

Section: Mathematics

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Question : 30 of 73

Marks: +1, -0

Let

f(x)=\;\frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}

. Then the value of

8\left(f\left(\;\frac{1}{15}\right)+f\left(\;\frac{2}{15}\right)+\dots+f\left(\;\frac{59}{15}\right)\right)

[24 Jan 2025 Shift 1]

118
92
102
108

Solution: 👈: Video Solution

\;f(x)=\;\frac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}=\;\frac{4(2^x+4)}{2(2^{2x})+16(2^x)+32}

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

Now,

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

\;=\;\frac{2}{2^x+4}+\;\frac{2 \times 2^x}{16+4 \times 2^x}=\;\frac{2}{2^x+4}+\;\frac{2x}{2(4+2^x)}

\;=\;\frac{1}{2}

So,

f\left(\;\frac{1}{15}\right)+f\left(\;\frac{59}{15}\right)=\;\frac{1}{2}

\;f\left(\;\frac{2}{15}\right)+f\left(\;\frac{58}{15}\right)=\;\frac{1}{2}

\;\; \text{ and } \; f\left(\;\frac{30}{15}\right)=\;\frac{2}{4+4}=\;\frac{1}{2}

\; \therefore 8\left(f\left(\;\frac{1}{15}\right)+f\left(\;\frac{2}{15}\right)+\dots f\left(\;\frac{59}{15}\right)\right)

\;=8\left(\;\frac{29}{2}+\;\frac{1}{4}\right)=118

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