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(A) Given the integral equation $f(x) = x + \int_0^x f(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule,we get: $f'(x) =

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$1$

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(A) Given the integral equation $f(x) = x + \int_0^x f(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule,we get: $f'(x) = 1 + f(x)$. This is a linear first-order differential equation of the form $f'(x) - f(x) = 1$. The integrating factor is $I.F. = e^{\int -1 dx} = e^{-x}$. Multiplying both sides by $e^{-x}$,we get: $e^{-x} f'(x) - e^{-x} f(x) = e^{-x}$. $\frac{d}{dx} (f(x) e^{-x}) = e^{-x}$. Integrating both sides with respect to $x$: $f(x) e^{-x} = \int e^{-x} dx = -e^{-x} + C$. $f(x) = -1 + C e^x$. From the original equation,at $x=0$,$f(0) = 0 + \int_0^0 f(t) dt = 0$. Substituting $x=0$ into $f(x) = -1 + C e^x$: $0 = -1 + C(e^0) \Rightarrow C = 1$. Thus,$f(x) = e^x - 1$. To find the number of elements in $S = \{x \in R : f(x) = 0\}$,we solve $e^x - 1 = 0$. $e^x = 1 \Rightarrow x = 0$. There is only one solution,$x=0$. Therefore,the number of elements in the set $S$ is $1$.

Let $f: R \rightarrow R$ be a continuous function satisfying $f(x)=x+\int_0^x f(t) dt$,for all $x \in R$. Then,the number of elements in the set $S=\{x \in R: f(x)=0\}$ is

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