QJEEM25Mabbe6: JEE Main 2025 Mathematics - Class 11 JEEM25Mabbe6 Solution: Let . Then the value of is equal to

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

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