Question:

Step 1: Differentiate the given functional equation with respect to $x$. We start with the equation: $$f(x) + \int_0^x t f(t)\, dt + x^2 = 0$$ Differentiating both sides with respect to $x$ using the Leibniz rule: $$f'(x) + xf(x) + 2x = 0$$ This simplifies to the linear first-order ODE: $$f'(x) + xf(x) = -2x$$ Step 2: Solve the ODE using an integrating factor. The integrating factor is: $$\mu(x) = e^{\int x\, dx} = e^{x^2/2}$$ Multiply both sides of the ODE by $\mu(x)$: $$e^{x^2/2} f'(x) + x e^{x^2/2} f(x) = -2x e^{x^2/2}$$ The left side is the derivative of a product: $$\frac{d}{dx}\left[f(x) e^{x^2/2}\right] = -2x e^{x^2/2}$$ Step 3: Integrate both sides to find the general solution. Integrating both sides: $$f(x) e^{x^2/2} = \int -2x e^{x^2/2}\, dx$$ Using substitution $u = x^2/2$, we get $du = x\, dx$: $$f(x) e^{x^2/2} = -2 e^{x^2/2} + C$$ Solving for $f(x)$: $$f(x) = -2 + C e^{-x^2/2}$$ Step 4: Determine the constant $C$ using the initial condition. Substituting $x = 0$ into the original equation: $$f(0) + \int_0^0 t f(t)\, dt + 0^2 = 0$$ $$f(0) = 0$$ Using this in our general solution: $$0 = -2 + C e^0$$ $$C = 2$$ Therefore: $$f(x) = -2 + 2e^{-x^2/2} = 2\left(e^{-x^2/2} - 1\right)$$ Step 5: Verify each option. **Option 1:** Check if $f(x)$ has more than one point in common with the $x$-axis. Setting $f(x) = 0$: $$2\left(e^{-x^2/2} - 1\right) = 0 \implies e^{-x^2/2} = 1 \implies x = 0$$ Only one intersection point. **FALSE** **Option 2:** Check if $f(x)$ is an odd function. $$f(-x) = 2\left(e^{-(-x)^2/2} - 1\right) = 2\left(e^{-x^2/2} - 1\right) = f(x)$$ Since $f(-x) = f(x)$, the function is **even**, not odd. **FALSE** **Option 3:** Check $\lim_{x \to \infty} f(x)$. $$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} 2\left(e^{-x^2/2} - 1\right) = 2(0 - 1) = -2 \neq 2$$ **FALSE** **Option 4:** Check $\lim_{x \to -\infty} f(x)$. $$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} 2\left(e^{-x^2/2} - 1\right) = 2(0 - 1) = -2$$ **TRUE** The correct answer is **Option 4**.