JEE Main 2025 Jan 24 Shift 1, Mathematics Q5: Some Important Properties Of Functions

Simplify by writing every power in terms of $u=2^x$. Numerator: $2^{x+2}+16=4u+16=4(u+4)$. Denominator: $2^{2x+1}+2^{x+4}+32=2u^2+16u+32=2(u+4{)}^2$.

$f(x)=\frac{4(u+4)}{2(u+4{)}^2}=\frac{2}{2^x+4}$

Now look for a pairing symmetry. Compute

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

Adding:

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

Pair the 59 terms by $k/15$ and $(60-k)/15$ (which sum to 4). The pairs $(1,59),(2,58),\dots ,(29,31)$ give 29 pairs, each summing to $1/2$. The middle term $k=30$ gives $f(2)=\frac{2}{4+4}=\frac{1}{4}$.

Sum $=29\cdot \frac{1}{2}+\frac{1}{4}=\frac{59}{4}$.

Multiply by 8: $8\cdot \frac{59}{4}=118$.

$\therefore$ Option (1).