JEE Main 2026 Apr 8 Shift 2, Mathematics Q25: Summation of Series

JEE Main2026Apr 8, Shift 2Mathematics

Let $f$ be a polynomial function such that

$lo g_{2} (f (x)) = (lo g_{2} (2 + \frac{2}{3} + \frac{2}{9} + \dots \infty)) \cdot lo g_{3} (1 + \frac{f ( x )}{f ( 1/ x )})$ , $x > 0$ and $f (6) = 37$ . Then $n = 1 \sum 10 f (n)$ is equal to _______.

Your answer

Solution

Simplify the geometric series: $2 + \frac{2}{3} + \frac{2}{9} + \dots = \frac{2}{1 - 1/3} = 3$ . So $lo g_{2} (3) = lo g_{2} 3$ , and the equation becomes:

$lo g_{2} f (x) = lo g_{2} 3 \cdot lo g_{3} (1 + \frac{f ( x )}{f ( 1/ x )}) = lo g_{2} (1 + \frac{f ( x )}{f ( 1/ x )})$

Hence $f (x) = 1 + \frac{f ( x )}{f ( 1/ x )}$ , which rearranges to the symmetric form:

$f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$

Identify the polynomial: A standard result for this functional equation among polynomials is $f (x) = 1 \pm x^{n}$ .

Using $f (6) = 37$ : $1 + 6^{n} = 37 ⟹ 6^{n} = 36 ⟹ n = 2$ . So $f (x) = 1 + x^{2}$ .

Compute the sum:

$n = 1 \sum 10 (1 + n^{2}) = 10 + \frac{10 \cdot 11 \cdot 21}{6} = 10 + 385 = 395$ .

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q25: Summation of Series

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