If $f(1)=0$ and $f(n+1)-f(n)=5n$ for all $n \in N$, then $f(n)=$

Let $f: R-\{0\} \rightarrow R$ be a function satisfying $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$ for all $x, y$ where $f(y) \neq 0$. If $f^{\prime}(1)=2024$,then:

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:

Let $f$ and $g$ be functions satisfying $f(x+y)=f(x)f(y)$, $f(1)=7$ and $g(x+y)=g(xy)$, $g(1)=1$ for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then $n$ is equal to:

$f$ is a real-valued function satisfying the relation $f\left(3x + \frac{1}{2x}\right) = 9x^2 + \frac{1}{4x^2}$. If $f\left(x + \frac{1}{x}\right) = 1$, then $x =$

If $f(x)$ is a quadratic function such that $f(x) f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$,then $\sqrt{f\left(\frac{2}{3}\right) + f\left(\frac{3}{2}\right)} = $