Let f:[0,∞)→ℝ be a continuous function such that f(x) = 1/x + ∫₀ˣ (x-t)f(t)dt for all x∈[0,∞). Then, which of the following statement(s) is (are) TRUE?

f(x) = 1/x + ∫₀ˣ (x-t)f(t)dt ⇒ e⁻ˣf(x) = e⁻ˣ(1/x) + ∫₀ˣ e⁻ᵗf(t)dt
Differentiate w.r.t. x.
-e⁻ˣf(x) + e⁻ˣf'(x) = -e⁻ˣ(1/x) + e⁻ˣ(1/x²) + e⁻ˣf(x) ⇒ -f(x) + f'(x) = -(1/x) + f(x) ⇒ f'(x) - 2f(x) = -1/x
Integrating factor = e⁻²ˣ
f(x)⋅e⁻²ˣ = ∫e⁻²ˣ(-1/x)dx
This integral is difficult to solve analytically. Let's try a different approach.
Let's differentiate the original equation:
f'(x) = -1/x² + ∫₀ˣ f(t)dt + (x-x)f(x) = -1/x² + ∫₀ˣ f(t)dt
Let's try to solve the integral equation differently. We have:
f(x) = 1/x + ∫₀ˣ (x-t)f(t)dt
Let F(x) = ∫₀ˣ f(t)dt. Then F'(x) = f(x). The equation becomes:
F'(x) = 1/x + ∫₀ˣ xf(t)dt - ∫₀ˣ tf(t)dt = 1/x + xF(x) - ∫₀ˣ tf(t)dt
This is still complicated. Let's go back to the original differential equation:
f'(x) - 2f(x) = -1/x
The integrating factor is e⁻²ˣ. Multiplying both sides by the integrating factor:
e⁻²ˣf'(x) - 2e⁻²ˣf(x) = -e⁻²ˣ/x
d/dx(e⁻²ˣf(x)) = -e⁻²ˣ/x
This integral is difficult to solve in closed form. Let's re-examine the problem statement. There must be a simpler approach.
Let's assume a solution of the form f(x) = a/x + b. Then f'(x) = -a/x².
Substituting into f'(x) - 2f(x) = -1/x:
-a/x² - 2(a/x + b) = -1/x
-a/x² - 2a/x - 2b = -1/x
Comparing coefficients, we have -2a = -1 => a = 1/2, and -a/x² - 2b = 0. Since -a/x² must be zero, this is not possible.
Let's assume the solution is f(x) = 1-x. Then f'(x) = -1.
Substituting into f'(x) - 2f(x) = -1/x:
-1 - 2(1-x) = -1/x
-1 - 2 + 2x = -1/x
2x - 3 = -1/x
2x² - 3x + 1 = 0
(2x-1)(x-1) = 0
x = 1/2 or x = 1. This doesn't work.
Let's assume f(x) = 1 - x. Then f(0) is undefined. Let's assume f(x) = 2x -1
If f(x) = 2x - 1, then f'(x) = 2. Substituting into f'(x) - 2f(x) = -1/x:
2 - 2(2x - 1) = -1/x
2 - 4x + 2 = -1/x
4 - 4x = -1/x
4x - 4x² = 1
4x² - 4x + 1 = 0
(2x - 1)² = 0
x = 1/2. This is still not working.
If f(x) = 1-x, then the area = π/4