If $(f(x))^2 = f(x^2) + f(1)$ holds good,then find $f(x)$.

Let $f$ be a function defined by $f(xy) = \frac{f(x)}{y}$ for all positive real numbers $x$ and $y$. If $f(30) = 20$,then $f(40) = $

Let $f: R \rightarrow R$ be defined by $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x$ and $y$. If $f^{\prime}(0)$ exists and equals $-1$ and $f(0)=1$, then $f(2)=$

If $f(x) + 2f(1/x) = 3x$ for $x \neq 0$ and $S = \{x \in R : f(x) = f(-x)\}$,then $S$:

$A$ 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:

Let $f(x+y)=f(x)f(y)$ for all $x, y$ where $f(0) \neq 0$. If $f(5) = 2$ and $f'(0) = 3$,then $f'(5)$ is equal to