What's the concept ??

Algebra Level 4

Let $f(x)$ be an invertible function and $g(x)$ be its inverse function . It is given that $f(x)=x$ for only one value of $x$ in the range of $g(x).$

What is the number of values of $x$ in the domain of $f(x)$ for which $f(x)-g(x)=n-\lfloor \frac{2015}{4} \rfloor$ ? ,where $n$ is the only natural number for which the polynomial $1+x+x^2+x^3+...+x^{n-1}$ divides the polynomial $1+x^2+x^4+...+x^{2010}.$

The answer is 1.

1 solution

Ritwik Rudra
Jun 5, 2015

1+x^2+x^4+...x^2010=(1-x^2012)/(1-x^2)=(1-x^1006)(1+x^1006)/(1-x)(1+x)
=(1+x^1006){(1-x^503)/(1-x)}{(1+x^503)/(1+x)} =(1+x^1006)(1-x+x^2-x^3+...x^502)(1+x+x^2+...x^502); This is divisible by 1+x+x^2+...x^(n-1) if n-1=502 => n=503 => n-[2015/4]=0; Now as G(x)is inverse of F(x) therefore range of G(x) is same as domain of F(x). Therefore F(x)=x has only one solution in the domain of F(x). As two functions which are inverse of each other intersect only at y=x. Hence, F(x)=G(x) has only one solution in the domain of F(x). AND SO IS THE SOLUTION

What's the concept ??

The answer is 1.

1 solution

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