What is f(x) given the functional equation and f(6)=37?

f(x)=x²+1. The functional equation simplifies to (f(x)−1)(f(1/x)−1)=1, giving f(x)−1=x^n. With f(6)=37: 6^n=36, so n=2.

What is the sum f(1)+f(2)+...+f(10)?

395. Sum = Σ(n²+1) = 385+10 = 395.

f polynomial, log₂(f(x))=log₂(3)·log₃(1+f(x)/f(1/x)), f(6)=37 — find Σf(n) for n=1 to 10

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MathFunctions

QInteger TypeFunctions & Polynomials

Let $f$ be a polynomial function such that $$\log_2(f(x))=\left[\log_2\!\left(2+\dfrac{2}{3}+\dfrac{2}{9}+\cdots\infty\right)\right]\cdot\log_3\!\left(1+\dfrac{f(x)}{f(1/x)}\right),\quad x>0$$ and $f(6)=37$. Then $\displaystyle\sum_{n=1}^{10}f(n)$ is equal to __________.

✅ Correct Answer

395

✓

Solution

Evaluate the geometric series

$2+\dfrac{2}{3}+\dfrac{2}{9}+\cdots=\dfrac{2}{1-\frac{1}{3}}=\dfrac{2}{\frac{2}{3}}=3$

$\therefore\quad\log_2(3)$ is the bracket value.

Simplify the functional equation

$\log_2(f(x))=\log_2(3)\cdot\log_3\!\left(1+\dfrac{f(x)}{f(1/x)}\right)$

Use change of base: $\log_2(3)\cdot\log_3(A)=\log_2(A)$. So:

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

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

$\Rightarrow\quad f(x)\cdot f(1/x)=f(1/x)+f(x)$

$\Rightarrow\quad f(x)\cdot f(1/x)-f(x)-f(1/x)=0$

Adding 1 to both sides: $[f(x)-1][f(1/x)-1]=1$.

Find f(x) as a polynomial

Let $g(x)=f(x)-1$. Then $g(x)\cdot g(1/x)=1$.

For a polynomial $g(x)=x^n$ (monic monomial): $g(x)\cdot g(1/x)=x^n\cdot x^{-n}=1$ ✓ for all $x>0$.

So $f(x)-1=x^n$, i.e., $f(x)=x^n+1$.

Use $f(6)=37$: $6^n+1=37\Rightarrow6^n=36=6^2\Rightarrow n=2$.

$$f(x)=x^2+1$$

Compute the sum

$\displaystyle\sum_{n=1}^{10}f(n)=\sum_{n=1}^{10}(n^2+1)=\sum_{n=1}^{10}n^2+\sum_{n=1}^{10}1$

$=\dfrac{10\cdot11\cdot21}{6}+10=385+10=\boxed{395}$

📘 Functional Equation f(x)·f(1/x)=f(x)+f(1/x)

The relation $(f(x)-1)(f(1/x)-1)=1$ is satisfied by $f(x)-1=x^n$ for integer $n$. This gives $f(x)=x^n+1$. The value of $n$ is determined by the condition $f(6)=37$: $6^n=36$ gives $n=2$.

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Important Concepts

Geometric Series Evaluation$2+2/3+2/9+\cdots=2/(1-1/3)=3$. This is a geometric series with first term $a=2$ and common ratio $r=1/3$. Sum $=a/(1-r)=3$.

Log Identity: log_a(b)·log_b(c)=log_a(c)$\log_2(3)\cdot\log_3(X)=\log_2(X)$ by the chain rule of logarithms. This directly simplifies $\log_2(f(x))=\log_2(1+f(x)/f(1/x))$.

Functional Equation Solution$(f(x)-1)(f(1/x)-1)=1$ for a polynomial means $f(x)-1=x^n$ for some positive integer $n$. This is because $g(x)\cdot g(1/x)=1$ with $g$ polynomial forces $g(x)=\pm x^n$.

Sum of Squares Formula$\sum_{n=1}^{N}n^2=\frac{N(N+1)(2N+1)}{6}$. For $N=10$: $\frac{10\cdot11\cdot21}{6}=385$. Adding $\sum_{n=1}^{10}1=10$ gives $395$.

FAQs

How does log₂(3)·log₃(A)=log₂(A)?

⌄

By change of base: $\log_3(A)=\ln A/\ln3$ and $\log_2(3)=\ln3/\ln2$. Product $=(\ln3/\ln2)\cdot(\ln A/\ln3)=\ln A/\ln2=\log_2(A)$.

How does f(x)·f(1/x)=f(x)+f(1/x) give (f−1)(f(1/x)−1)=1?

⌄

Rearrange: $f(x)f(1/x)-f(x)-f(1/x)=0$. Add 1: $f(x)f(1/x)-f(x)-f(1/x)+1=1$, i.e., $(f(x)-1)(f(1/x)-1)=1$.

Why must g(x)=x^n?

⌄

For a polynomial $g$, $g(x)g(1/x)=1$ for all $x>0$ means $g(1/x)=1/g(x)$. The only monomials satisfying this are $g(x)=x^n$ (with $g(1/x)=x^{-n}=1/x^n=1/g(x)$ ✓).

Verify f(6)=37 with f(x)=x²+1.

⌄

$f(6)=36+1=37$ ✓

Is this from JEE Main 2026?

⌄

Yes, this appeared in JEE Main 2026 Section B.

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