Let $f$ be a function such that $f(x) + 3f\left(\frac{24}{x}\right) = 4x$,where $x \neq 0$. Then $f(3) + f(8)$ is equal to:

How many functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ are there such that $f(x+y)=f(x)+f(y)$ for all $x, y \in \mathbb{Z}$?

If a function $f$ satisfies $f(x+1)+f(x-1)=\sqrt{2} f(x)$,then $f(x+2)+f(x-2)=$

$A$ real-valued function $f(x)$ satisfies the functional equation $f(x - y) = f(x)f(y) - f(a - x)f(a + y)$,where $a$ is a given constant and $f(0) = 1$. Then $f(2a - x)$ is equal to:

If $f: R \rightarrow R$ is a differentiable function such that $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$ and if $f^{\prime}(4)=24$ and $f^{\prime}(0)=3$,then $f(4)=$

If $3 f(x) - f\left(\frac{1}{x}\right) = 8 \log_2 x^3$ for $x > 0$,then $f(2), f(4), f(8)$ are in