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

Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \forall x, y \in R$. If $f(x)$ is differentiable at $x=0$,then which of the following statements are true?

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:

The values of $b$ and $c$ for which the identity $f(x + 1) - f(x) = 8x + 3$ is satisfied,where $f(x) = bx^2 + cx + d$,are:

$f: R \rightarrow R$ is defined by $f(x+y)=f(x)+12y, \forall x, y \in R$. If $f(1)=6$,then $\sum_{r=1}^n f(r)=$

$A$ 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