JEE Main 2026 (Online) 8th April Evening ShiftQuestion 1
Let $f$ be a polynomial function such that
$\log_2(f(x))=\left(\log_2\left(2+\frac{2}{3}+\frac{2}{9}+\ldots\infty\right)\right)\cdot\log_3\left(1+\frac{f(x)}{f\left(\frac{1}{x}\right)}\right)$, $x>0$ and $f(6)=37$. Then $\sum_{n=1}^{10}f(n)$ is equal to $\_\_\_\_\_\_$ .
Functions
Answer$395$
Solution
$2+\frac23+\dots=3$, so $\log_2 3\cdot\log_3(1+\frac{f(x)}{f(1/x)})=\log_2 f(x)$, i.e. $f(x)f(1/x)=f(x)+f(1/x)$, giving $f(x)=1\pm x^n$. $f(6)=37\Rightarrow f(x)=1+x^2$. $\sum_{n=1}^{10}(1+n^2)=10+385=395$.