$\begin{aligned}
& f(x)=\frac{42^x+16}{2.2^{2 x}+16.2^x+32} \
& f(x)=\frac{2\left(2^x+4\right)}{2^{2 x}+8.2^x+16} \
& f(x)=\frac{2}{2^x+4} \
& f(4-x)=\frac{2^x}{2\left(2^x+4\right)} \
& f(x)+f(4-x)=\frac{1}{2}
\end{aligned}So,\quad \mathrm{f}\left(\frac{1}{15}\right)+\mathrm{f}\left(\frac{59}{15}\right)=\frac{1}{2}\begin{aligned}
& \text { Similarly }=f\left(\frac{29}{15}\right)+f\left(\frac{31}{15}\right)=\frac{1}{2} \
& f\left(\frac{30}{15}\right)=f(2)=\frac{2}{2^2+4}=\frac{2}{8}=\frac{1}{4} \
& \Rightarrow 8\left(29 \times \frac{1}{2}+\frac{1}{4}\right)
\end{aligned}$ Ans. 118
Let f(x)= 2^x+2+16 2^2 x+1+2^x+4+32. Then the value of 8 (f ( 1 15 )+f ( 2 15 )+…+f ( 59 15 ) ) is equal to
JEE Main 2025 — Mathematics Algebra
Let f(x)=22x+1+2x+4+322x+2+16. Then the value of 8(f(151)+f(152)+…+f(1559)) is equal to
Held on 24 Jan 2025 · Verified 6 Jul 2026.
Options
- A
92
- B
118
- C
102
- D
108
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