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JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 25

Let ff be a polynomial function such that
log⁡2(f(x))=(log⁡2(2+23+29+…∞))⋅log⁡3(1+f(x)f(1/x)),\log_2(f(x))=\left(\log_2\left(2+\frac23+\frac29+\dots\infty\right)\right)\cdot\log_3\left(1+\frac{f(x)}{f(1/x)}\right),
x>0x>0 and f(6)=37f(6)=37. Then ∑n=110f(n)\displaystyle\sum_{n=1}^{10}f(n) is equal to ________.

Answer and worked solution

Official answer: 395

The series. 2+23+29+⋯=2+1=32+\tfrac23+\tfrac29+\dots=2+1=3.

Equation. log⁡2f=log⁡23⋅log⁡3(1+f(x)f(1/x))=log⁡2(1+f(x)f(1/x))\log_2f=\log_23\cdot\log_3\left(1+\tfrac{f(x)}{f(1/x)}\right)=\log_2\left(1+\tfrac{f(x)}{f(1/x)}\right), so
f(x)f(1/x)=f(x)+f(1/x).f(x)f(1/x)=f(x)+f(1/x).

Polynomial solution. f(x)=1±xnf(x)=1\pm x^n. f(6)=37f(6)=37 gives f(x)=x2+1f(x)=x^2+1.

Sum. ∑n=110(n2+1)=385+10=395\sum_{n=1}^{10}(n^2+1)=385+10=395.

Answer: 395

Topics:Types of functionsGeometric progressionLogarithms
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