JEE Main 2026 — Mathematics Algebra

The sum of the infinite geometric progression is given by:

$$S = 2 + \dfrac{2}{3} + \dfrac{2}{9} + \ldots \infty = \dfrac{2}{1 - \dfrac{1}{3}} = 3$$S=2+32​+92​+…∞=1−31​2​=3

Substituting this into the given equation:

$$\log_2(f(x)) = \log_2(3) \cdot \log_3\left(1 + \dfrac{f(x)}{f(1/x)}\right)$$log2​(f(x))=log2​(3)⋅log3​(1+f(1/x)f(x)​)

Using the base change property $\log_a(b) \cdot \log_b(c) = \log_a(c)$loga​(b)⋅logb​(c)=loga​(c):

$$\log_2(f(x)) = \log_2\left(1 + \dfrac{f(x)}{f(1/x)}\right)$$log2​(f(x))=log2​(1+f(1/x)f(x)​)

Equating the arguments:

$$f(x) = 1 + \dfrac{f(x)}{f(1/x)}$$f(x)=1+f(1/x)f(x)​

$$f(x)f(1/x) - f(1/x) = f(x)$$f(x)f(1/x)−f(1/x)=f(x)

$$f(x)f(1/x) = f(x) + f(1/x)$$f(x)f(1/x)=f(x)+f(1/x)

The only polynomial functions satisfying this relation are of the form $f(x) = 1 \pm x^n$f(x)=1±xn.

Given $f(6) = 37$f(6)=37:

$$1 \pm 6^n = 37$$1±6n=37

Taking the positive sign, $6^n = 36 \Rightarrow n = 2$6n=36⇒n=2.

Thus, the polynomial is $f(x) = x^2 + 1$f(x)=x2+1.

The required sum is:

$$\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} (n^2 + 1)$$n=1∑10​f(n)=n=1∑10​(n2+1)

$$\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} n^2 + \sum_{n=1}^{10} 1$$n=1∑10​f(n)=n=1∑10​n2+n=1∑10​1

$$\sum_{n=1}^{10} f(n) = \dfrac{10 \times 11 \times 21}{6} + 10$$n=1∑10​f(n)=610×11×21​+10

$$\sum_{n=1}^{10} f(n) = 385 + 10 = 395$$n=1∑10​f(n)=385+10=395

Answer: $395$395