Let $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1),$ where the function $f$ satisfies $f(x + y) = f(x)f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $a$ is

Let $R$ be the set of all real numbers. The number of continuous functions $f: R \rightarrow R$ such that for all real $x$,$f(x) + f(2x) = 0$ is

$A$ 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:

$f: R \rightarrow R$ is a function such that $f(0)=1$ and for all $x, y \in R$,$f(xy+1)=f(x)f(y)-f(y)-x+2$. Then $\frac{df}{dx}$ at $x=e$ is:

Suppose $f : R \rightarrow (0, \infty)$ be a differentiable function such that $5f(x + y) = f(x) \cdot f(y), \forall x, y \in R$. If $f(3) = 320$,then $\sum_{n=0}^5 f(n)$ is equal to:

Let $f$ be a polynomial function such that $\log_2(f(x)) = (\log_2 (2 + \frac{2}{3} + \frac{2}{9} + \dots \infty)) \cdot \log_3 (1 + \frac{f(x)}{f(1/x)}), x > 0$ and $f(6) = 37$. Then $\sum_{n=1}^{10} f(n)$ is equal to: