If $f: R-\{0\} \rightarrow R$ is defined by $3 f(x)+4 f\left(\frac{1}{x}\right)=\frac{2-x}{x}$, then find the value of $f(3)$.

Let $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$. Suppose that $f(3)=3$ and $f^{\prime}(0)=11$,then $f^{\prime}(3)$ is given by

Let $a, b, c \in \mathbb{R}$. If $f(x) = ax^2 + bx + c$ is such that $a + b + c = 3$ and $f(x + y) = f(x) + f(y) + xy, \forall x, y \in \mathbb{R}$,then $\sum_{n=1}^{10} f(n)$ is equal to:

$A$ real valued function $y = f(x)$ satisfies the relation $f\left( x - \frac{4}{9} \right) + 2x \le \frac{9}{4}x^2 + \frac{8}{9} \le f\left( x + \frac{4}{9} \right) - 2x$. The value of $f''(2)$ is

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

If $a$ and $b$ are two fixed positive integers such that $f(a + x) = b + [b^3 + 1 - 3b^2f(x) + 3b\{f(x)\}^2 - \{f(x)\}^3]^{1/3}$ for all real $x$,then $f(x)$ is a periodic function with period: