JEE MAIN - Mathematics (2025 - 24th January Morning Shift - No. 5)
Let $f(x)=\frac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}$. Then the value of $8\left(f\left(\frac{1}{15}\right)+f\left(\frac{2}{15}\right)+\ldots+f\left(\frac{59}{15}\right)\right)$ is equal to
108
92
118
102
Explanation
$$\begin{aligned} & f(x)=\frac{42^x+16}{2.2^{2 x}+16.2^x+32} \\ & f(x)=\frac{2\left(2^x+4\right)}{2^{2 x}+8.2^x+16} \\ & f(x)=\frac{2}{2^x+4} \\ & f(4-x)=\frac{2^x}{2\left(2^x+4\right)} \\ & f(x)+f(4-x)=\frac{1}{2} \end{aligned}$$
So, $\quad \mathrm{f}\left(\frac{1}{15}\right)+\mathrm{f}\left(\frac{59}{15}\right)=\frac{1}{2}$
$$\begin{aligned} & \text { Similarly }=f\left(\frac{29}{15}\right)+f\left(\frac{31}{15}\right)=\frac{1}{2} \\ & f\left(\frac{30}{15}\right)=f(2)=\frac{2}{2^2+4}=\frac{2}{8}=\frac{1}{4} \\ & \Rightarrow 8\left(29 \times \frac{1}{2}+\frac{1}{4}\right) \end{aligned}$$
Ans. 118
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