Question

If \( f(x) \) satisfies the equation \( f(x) = \int_1^x f(t) \, dt + (1-x)(\log_e x - 1) + e \), then \( f(f(1)) \) is equal to:

• A. \( 1 - e^{2+e} \)
• B. \( 1 + e^{2+e} \)
• C. \( 1 + e^{e} \)
• D. \( 1 - e^{2+e} \)

Hint:

For equations involving $\int_a^x f(t) dt$, always evaluate at $x=a$ first to find an initial condition, then differentiate to get a differential equation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To solve an integral equation of this form, we differentiate both sides with respect to $x$ using the Leibniz Rule. This converts the integral equation into a linear differential equation.

Step 2: Key Formula or Approach:
1. Differentiate: $f'(x) = f(x) + \frac{d}{dx}[(1-x)(\ln x - 1)]$. 2. Solve the resulting first-order linear differential equation.

Step 3: Detailed Explanation:
1. Find $f(1)$: Substitute $x=1$ into the original equation: \[ f(1) = 0 + (1-1)(0-1) + e = e \] 2. Differentiate the equation: \[ f'(x) = f(x) + [(-1)(\ln x - 1) + (1-x)(\frac{1}{x})] \] \[ f'(x) - f(x) = 1 - \ln x + \frac{1}{x} - 1 = \frac{1}{x} - \ln x \] 3. This is a linear D.E. with I.F. $= e^{\int -1 dx} = e^{-x}$. \[ e^{-x} f(x) = \int e^{-x} (\frac{1}{x} - \ln x) dx \] Using the property $\int e^x (g(x) + g'(x)) dx = e^x g(x)$, we note that $\frac{d}{dx}(-\ln x) = -1/x$. \[ e^{-x} f(x) = e^{-x} \ln x + C \implies f(x) = \ln x + C e^x \] 4. Use $f(1) = e$: $e = \ln 1 + Ce^1 \implies Ce = e \implies C = 1$. \[ f(x) = \ln x + e^x \] 5. Calculate $f(f(1)) = f(e)$: \[ f(e) = \ln e + e^e = 1 + e^e \]

Step 4: Final Answer:
The value is $1 + e^e$ (Note: Options provided may vary slightly due to typos in standard papers).