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(B) Given the equation: $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$. Differentiating both sides with respect to $x$ using Leibniz's rule: $f'(x) + x f(x

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$\lim_{x \rightarrow -\infty} f(x) = -2$

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(B) Given the equation: $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$. Differentiating both sides with respect to $x$ using Leibniz's rule: $f'(x) + x f(x) + 2x = 0$. Rearranging the terms: $f'(x) = -x(f(x) + 2)$. This is a first-order linear differential equation. Separating variables: $\frac{f'(x)}{f(x) + 2} = -x$. Integrating both sides: $\ln|f(x) + 2| = -\frac{x^2}{2} + C$. Thus,$f(x) + 2 = A e^{-x^2/2}$,or $f(x) = A e^{-x^2/2} - 2$. Using the initial condition from the original equation at $x = 0$: $f(0) + \int_{0}^{0} t f(t) dt + 0^2 = 0 \Rightarrow f(0) = 0$. Substituting $x = 0$ into $f(x) = A e^{-x^2/2} - 2$: $0 = A(1) - 2 \Rightarrow A = 2$. So,$f(x) = 2 e^{-x^2/2} - 2$. Now,evaluating the limits: $\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} (2 e^{-x^2/2} - 2) = 0 - 2 = -2$. $\lim_{x \rightarrow -\infty} f(x) = \lim_{x \rightarrow -\infty} (2 e^{-x^2/2} - 2) = 0 - 2 = -2$. Since $f(x) = 2(e^{-x^2/2} - 1)$,$f(x) = f(-x)$,so it is an even function. Comparing with the options,$\lim_{x \rightarrow -\infty} f(x) = -2$ is correct.

Let $f: R \rightarrow R$ be a continuous function satisfying $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$ for all $x \in R$. Then:

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