Functions – Summation Series – JEE Main 24 January 2025 Shift 1

Solution:

First, simplify the function $f(x)$:
$$ f(x) = \frac{2^x \cdot 2^2 + 16}{2 \cdot (2^x)^2 + 2^4 \cdot 2^x + 32} $$
$$ f(x) = \frac{4(2^x + 4)}{2((2^x)^2 + 8 \cdot 2^x + 16)} = \frac{4(2^x + 4)}{2(2^x + 4)^2} $$
$$ f(x) = \frac{2}{2^x + 4} $$

Now consider the sum $f(x) + f(4-x)$:
$$ f(4-x) = \frac{2}{2^{4-x} + 4} = \frac{2}{\frac{16}{2^x} + 4} = \frac{2 \cdot 2^x}{16 + 4 \cdot 2^x} $$
$$ f(4-x) = \frac{2 \cdot 2^x}{4(4 + 2^x)} = \frac{2^x}{2(2^x + 4)} $$
$$ f(x) + f(4-x) = \frac{2}{2^x + 4} + \frac{2^x}{2(2^x + 4)} = \frac{4 + 2^x}{2(2^x + 4)} = \frac{1}{2} $$

The given series is $S = \sum_{r=1}^{59} f\left(\frac{r}{15}\right)$.
Notice that for terms $f\left(\frac{r}{15}\right)$ and $f\left(\frac{60-r}{15}\right)$, the arguments sum to $\frac{60}{15} = 4$.
Thus, $f\left(\frac{r}{15}\right) + f\left(\frac{60-r}{15}\right) = f(x) + f(4-x) = \frac{1}{2}$.

The series has 59 terms. We can pair the first 29 terms with the last 29 terms (excluding the middle term):
Pairs: $(r=1, r=59), (r=2, r=58), \dots, (r=29, r=31)$.
Total pairs = 29.
Sum of pairs = $29 \times \frac{1}{2}$.

The middle term is at $r=30$:
$$ \text{Middle term} = f\left(\frac{30}{15}\right) = f(2) = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4} $$

Total Sum $S = \frac{29}{2} + \frac{1}{4} = \frac{58}{4} + \frac{1}{4} = \frac{59}{4}$.

We need to find $8S$:
$$ 8S = 8 \times \frac{59}{4} = 2 \times 59 = 118 $$

Ans. (1)