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

Let $R$ denote the set of all real numbers. Let $a_i, b_i \in R$ for $i \in \{1, 2, 3\}$. Define the functions $f: R \rightarrow R$,$g: R \rightarrow R$,and $h: R \rightarrow R$ by $f(x) = a_1 + 10x + a_2x^2 + a_3x^3 + x^4$ and $g(x) = b_1 + 3x + b_2x^2 + b_3x^3 + x^4$. Let $h(x) = f(x+1) - g(x+2)$. If $f(x) \neq g(x)$ for every $x \in R$,then the coefficient of $x^3$ in $h(x)$ is:

If $f: R \rightarrow R$ is defined by $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=7$,then $\sum_{r=1}^n f(r)=$

If $f(x)$ is a function such that $f(x+y)=f(x)+f(y)$ and $f(1)=7$,then $\sum_{r=1}^n f(r)=$

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 $f(2)=$

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