Let f(x)=2^x+2+162^2 x+1+2^x+4+32.

Question

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

Accepted Answer

$$\begin{aligned}
& f(x)=\frac{42^x+16}{2.2^{2 x}+16.2^x+32} \
& f(x)=\frac{2\left(2^x+4\right)}{2^{2 x}+8.2^x+16} \
& f(x)=\frac{2}{2^x+4} \
& f(4-x)=\frac{2^x}{2\left(2^x+4\right)} \
& f(x)+f(4-x)=\frac{1}{2}
\end{aligned}$$

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

Solutions · 1