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JEE MAIN 2026 · Sequence and Series

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 25

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

Let ff be a polynomial function such that

log⁡2(f(x))=(log⁡2(2+23+29+…∞))⋅log⁡3(1+f(x)f(1x)),x>0\log_2\big(f(x)\big) = \left(\log_2\left(2 + \frac23 + \frac29 + \dots\infty\right)\right)\cdot\log_3\left(1 + \frac{f(x)}{f\left(\frac1x\right)}\right),\quad x > 0

and f(6)=37f(6) = 37. Then ∑n=110f(n)\displaystyle\sum_{n=1}^{10} f(n) is equal to ______.

Your answer

Solution

Solution

Sum the GP. 2+23+29+⋯=21−1/3=32 + \frac23 + \frac29 + \dots = \frac{2}{1 - 1/3} = 3.

Simplify the logs. log⁡23⋅log⁡3Y=log⁡2Y\log_2 3 \cdot \log_3 Y = \log_2 Y, so

log⁡2f(x)=log⁡2(1+f(x)f(1/x))  ⇒  f(x)=1+f(x)f(1/x)  ⇒  f(x)f(1x)=f(x)+f(1x).\log_2 f(x) = \log_2\left(1 + \frac{f(x)}{f(1/x)}\right) \;\Rightarrow\; f(x) = 1 + \frac{f(x)}{f(1/x)} \;\Rightarrow\; f(x)f\left(\tfrac1x\right) = f(x) + f\left(\tfrac1x\right).

The polynomials satisfying this are f(x)=1±xnf(x) = 1 \pm x^n. From f(6)=37=1+62f(6) = 37 = 1 + 6^2, f(x)=1+x2f(x) = 1 + x^2.

∑n=110(1+n2)=10+10⋅11⋅216=10+385=395.\sum_{n=1}^{10}(1 + n^2) = 10 + \frac{10 \cdot 11 \cdot 21}{6} = 10 + 385 = 395.

Answer: 395.

Answer checked against the official answer key. The solution is our own working.

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