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

If $f(x)$ is a function satisfying the condition $f(x) = \frac{1}{3}\left[ f(x + 6) + \frac{6}{f(x + 7)} \right]$ and $f(x) \geq 0$ for all $x \in R$. If $\lim_{x \to \infty} f(x) = \sqrt{m}$,then the value of $m$ is:

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}{y}$ for all positive real numbers $x$ and $y.$ If $f(30) = 20,$ then the value of $f(40)$ is-

Let $f: R \rightarrow R$ satisfy $f(x+y)=2^{x} f(y)+4^{y} f(x)$ for all $x, y \in R$. If $f(2)=3$,then $14 \cdot \frac{f^{\prime}(4)}{f^{\prime}(2)}$ is equal to

$f(x)$ is a real-valued function such that $2f(x) + 3f(-x) = 15 - 4x$ for all $x \in R$. Then $f(2) =$

If $f(x + y) = f(x)f(y)$ and $\sum_{x=1}^{\infty} f(x) = 2$,where $x, y \in N$ and $N$ is the set of all natural numbers,then the value of $\frac{f(4)}{f(2)}$ is