Question

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

$f^{\prime }\left(0\right)$ is equal to :

• A. $0$
• B. $1$
• C. $f\left(0\right)$
• D. $-f\left(0\right)$

Answer 1

Answer: (A)

$0$

$f\left(x\right)=2f\left(\frac{x}{2}\right)+f\left(\frac{x}{4}\right)=x^2$
$\therefore f\left(\frac{x}{2}\right)-2f\left(\frac{x}{4}\right)+f\left(\frac{x}{8}\right)={\left(\frac{x}{2}\right)}^2$
$f\left(\frac{x}{4}\right)-2f\left(\frac{x}{8}\right)+f\left(\frac{x}{16}\right)={\left(\frac{x}{4}\right)}^2$
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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
$f^{\prime }\left(x\right)=\frac{32x}{9}$
$\therefore f^{\prime }\left(0\right)=0$