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(B) Given the functional equation $f(x) + f(2x) = 0$ for all $x \in R$. Substituting $x = 0$,we get $f(0) + f(0) = 0$,which implies $2f(0) = 0$,so $f(

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$1$

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(B) Given the functional equation $f(x) + f(2x) = 0$ for all $x \in R$. Substituting $x = 0$,we get $f(0) + f(0) = 0$,which implies $2f(0) = 0$,so $f(0) = 0$. From the given equation,$f(x) = -f(x/2)$. By iterating this relation,we get $f(x) = -f(x/2) = f(x/4) = -f(x/8) = \dots = (-1)^n f(x/2^n)$. Since $f$ is continuous at $x = 0$,we have $\lim_{n 	o \infty} f(x/2^n) = f(\lim_{n 	o \infty} x/2^n) = f(0) = 0$. Thus,$f(x) = \lim_{n 	o \infty} (-1)^n f(x/2^n) = 0$. Therefore,the only continuous function satisfying the condition is the zero function $f(x) = 0$. Hence,there is only $1$ such function.

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

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