Question

Let $f:R\to R$ be a continuous function such that $f\left(x\right)-2f\left(\frac{x}{2}\right)+f\left(\frac{x}{4}\right)=x^2$

The equation $f\left(x\right)-x-f\left(0\right)=0$ have exactly :

• A. no solution
• B. one solution
• C. two solutions
• D. Infinite solutions

Answer 1

Answer: (C)

two solutions

$f\left(x\right)=2f\left(\frac{x}{2}\right)+f\left(\frac{x}{4}\right)=x^2$
$\therefore f\left(\frac{x}{2}\right)-2f\left(\frac{x}{4}\right)+f\left(\frac{x}{8}\right)={\left(\frac{x}{2}\right)}^2$
$f\left(\frac{x}{4}\right)-2f\left(\frac{x}{8}\right)+f\left(\frac{x}{16}\right)={\left(\frac{x}{4}\right)}^2$
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$f\left(\frac{x}{2^n}\right)-2f\left(\frac{x}{2^{n+1}}\right)+f\left(\frac{x}{2^{n+2}}\right)={\left(\frac{x}{2^n}\right)}^2$
Adding, we get
$f\left(x\right)-f\left(\frac{x}{2}\right)-f\left(\frac{x}{2^{n+1}}\right)+f\left(\frac{x}{2^{n+2}}\right)$
$=x^2(1+\frac{1}{2^2}+...+\frac{1}{2^{2n}})$
As $n\to \infty$, we get $f\left(x\right)-f\left(\frac{x}{2}\right)=\frac{4x^2}{3}$
Repeating the same procedure again, we get
$f\left(x\right)-f\left(0\right)=\frac{16x^2}{9}$
Hence
$f\left(x\right)-f\left(0\right)-x=0$
$\Rightarrow \frac{16x^2}{9}-x=0\Rightarrow x=0,\frac{9}{16}$
$\therefore$ Two solutions.