Let $f(x)=\frac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}$. Then the value of $8\left(\mathrm{f}\left(\frac{1}{15}\right)+\mathrm{f}\left(\frac{2}{15}\right)+\ldots+\mathrm{f}\left(\frac{59}{15}\right)\right)$ is equal to

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Relations And Functions XII

Let

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

. Then the value of

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

is equal to

Answer: (A)

Let

t = 2^{x}

. Then

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

Compute

f (4 - x)

f (4 - x) = \frac{2}{2 ^{4 - x} + 4} = \frac{2}{\frac{16}{2 ^{x}} + 4} = \frac{2 \cdot 2 ^{x}}{16 + 4 \cdot 2 ^{x}} = \frac{2 ^{x}}{2 ( 2 ^{x} + 4 )} .

Adding:

f (x) + f (4 - x) = \frac{2}{2 ^{x} + 4} + \frac{2 ^{x}}{2 ( 2 ^{x} + 4 )} = \frac{4 + 2 ^{x}}{2 ( 2 ^{x} + 4 )} = \frac{1}{2} .

So any two arguments summing to

4

pair up to give

\frac{1}{2}

. Since

\frac{k}{15} + \frac{60 - k}{15} = 4

, pair

k = 1, \dots, 29

with

60 - k = 59, \dots, 31

— that's

29

pairs, each summing to

\frac{1}{2}

— leaving the unpaired middle term

k = 30

f (\frac{30}{15}) = f (2) = \frac{2}{2 ^{2} + 4} = \frac{2}{8} = \frac{1}{4} .

k = 1 \sum 59 f (\frac{k}{15}) = 29 \times \frac{1}{2} + \frac{1}{4} = \frac{29}{2} + \frac{1}{4} = \frac{59}{4} .

8 (\frac{59}{4}) = 118.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option