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(D) Given the equation $\int_0^x f(t) dt = p f(x)$.If $p = 0$,then $\int_0^x f(t) dt = 0$ for all $x \in \mathbb{R}$. Differentiating with respect to

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the whole set $\mathbb{R}$

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(D) Given the equation $\int_0^x f(t) dt = p f(x)$.If $p = 0$,then $\int_0^x f(t) dt = 0$ for all $x \in \mathbb{R}$. Differentiating with respect to $x$,we get $f(x) = 0$ for all $x$. Thus,there is no non-zero function $f$ for $p = 0$.If $p 
eq 0$,we differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus:$f(x) = p f'(x) \implies \frac{f'(x)}{f(x)} = \frac{1}{p}$.Integrating both sides,we get $\ln|f(x)| = \frac{x}{p} + C$,which implies $f(x) = A e^{x/p}$ for some constant $A$.Substituting this back into the original integral equation at $x = 0$:$\int_0^0 f(t) dt = p f(0) \implies 0 = p A e^0 \implies p A = 0$.Since $p 
eq 0$,we must have $A = 0$,which implies $f(x) = 0$ for all $x$.Thus,for any $p \in \mathbb{R}$,there exists no non-zero continuous function $f$ satisfying the given equation.Therefore,$S = \mathbb{R}$.

Let $S$ be the set of real numbers $p$ such that there is no non-zero continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $\int_0^x f(t) dt = p f(x)$ for all $x \in \mathbb{R}$. Then,$S$ is

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