JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The sum of the infinite geometric progression is given by: S = 2 + 2 3 + 2 9 + = 2 1 - 1 3 = 3 Substituting this into the given equation: _2(f(x)) = _2(3) _3 (1 + f(x) f(1/x) ) Using the base change property _a(b) _b(c) = _a(c): _2(f(x)) = _2 (1 + f(x) f(1/x) ) Equating the arguments: f(x) = 1 + f(x) f(1/x) f(x)f(1/x) - f(1/x) = f(x) f(x)f(1/x) = f(x) + f(1/x) The only polynomial functions satisfying this relation are of the form f(x) = 1 x^n. Given f(6) = 37: 1 6^n = 37 Taking the positive sign, 6^n = 36 n = 2. Thus, the polynomial is f(x) = x^2 + 1. The required sum is: _ n=1 ^ 10 f(n) = _ n=1 ^ 10 (n^2 + 1) _ n=1 ^ 10 f(n) = _ n=1 ^ 10 n^2 + _ n=1 ^ 10 1 _ n=1 ^ 10 f(n) = 10 11 21 6 + 10 _ n=1 ^ 10 f(n) = 385 + 10 = 395 Answer: 395