Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function such that $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^{2}$ . Then $f(3)$ =? Options: (a)$4$,    (b)$4f(0)$,       (c)$4-f(0)$,        (d)$4+f(0)$,   (e)$16+f(0)$.

CORRECT ANSWER USING REDUCTION

Deep thanks to @martini and @A.S. , soo much respect .

Since we don't exactly know the nature of $f(x)$ we will start of by considering our function as  $f(x)=g(x)+cx^2$ where $g(x)$ represents all the other functions (including constant values) inside $f(x)$ and $cx^{2}$ includes the no. of $x^{2}$ terms. 
Now we have $cx^{2}-2\frac{cx^{2}}{4}+\frac{cx^{2}}{16}=x^{2}$
Comparing coefficients of $x^{2}$ in first equation we have $c=\frac{16}{9}$. 
And 
$g(x)-2g(\frac{x}{2})+g(\frac{x}{4})=0$
Now to solve this recurring equation we substitute $x=\frac{x}{2},\frac{x}{4},\frac{x}{8}......$ and add it up . 
Yielding $g(\frac{x}{2^i})=ig(\frac{x}{2})-(i-1)g(x)$ 
Now as $i\to\infty$. Since $g$ is continuous we will have 
$g(x)=g(x/2)$. 
Now solving the recurring equation $g(x)-g(x/2)=0$ by substituting $x=\frac{x}{2},\frac{x}{4},\frac{x}{8}......$ and adding it up . 
Now as $i\to\infty$. Since $g$ is continuous we will have 
$g(x)=g(0)$
Hence $f(3)=g(3)+(16/9)3^2$. 
Now as $g(3)=g(0)= f(0)-(16/9)0^2$. 
We finally have $f(3)=f(0)+16.$
 A: Let $x = 3\cdot 2^{-n}$, giving 
$$ f(3\cdot 2^{-n}) = 9 \cdot 2^{-2n} + 2 f(3 \cdot 2^{-(n+1)}) - f(3 \cdot 2^{-(n+2)}) $$
Write brevitatis causa $a_n := f(3 \cdot 2^{-n})$, we have
$$ a_{n+2} - 2a_{n+1} + a_n = 9 \cdot 2^{-2n} $$
The general solution of the homogenous recursion 
$$ a_{n+2} - 2a_{n+1} + a_n = 0 $$
is $$ a_n^{\text{hom}} = \alpha n + \beta $$
to find a solution of the inhomogenous equation, we make the ansatz  $\gamma \cdot 2^{-2n}$, giving 
$$ \gamma 2^{-2n}(2^{-4} - 2\cdot 2^{-2} + 1) = 9 \cdot 2^{-2n} \iff \gamma = 16 $$
So we have 
$$ f(3 \cdot 2^{-n}) = a_n = 16\cdot 2^{-2n} + \alpha n + \beta $$
for some $\alpha, \beta \in \mathbf R$. As $f$ is continuous, we must have $a_n \to f(0)$. As $16\cdot 2^{-2n}$ converges, this implies $\alpha = 0$, or 
$$ f(3 \cdot 2^{-n}) = 16 \cdot 2^{-2n} + \beta $$
For $n \to \infty$ this gives $f(0) = \beta$, or 
$$ f(3 \cdot 2^{-n}) = 16 \cdot 2^{-2n} + f(0) $$
for $n=0$ we have
$$ f(3) = 16 + f(0) $$
A: Look for a simple solution first to "homogenize" the equation. $f(x)=cx^2$ works for $c=16/9$. That means that $g(x)=f(x)-cx^2$ satisfies $g(x)-2g(x/2)+g(x/4)=0$. Rewriting $g(x/4)=2g(x/2)-g(x)$ and iterating yields $$g(x/2^i)=ig(x/2)-(i-1)g(x)$$ Take $i\to\infty$. Since $g$ is continuous at $0$, we must have $g(x)=g(x/2)=g(0)$. Hence $f(x)=f(0)+(16/9)x^2$. Hence $$f(3)=f(0)+16.$$
