Let $f: [1, \infty) \to \mathbb{R}$ be a differentiable function defined as $f(x) = \int_1^x f(t) dt + (1-x)(\log_e x - 1) + e$. Then the value of $f(f(1))$ is :
- $(1 + e^e)$
- $(1 + e)$
- $(1 + e + e^e)$
- $1 + 2e$
The Correct Option is A
Solution and Explanation
Convert the integral equation into a differential equation using the Leibniz rule, solve the ODE, and find the value.
Step 2: Detailed Explanation:
Differentiating both sides w.r.t $x$:
$f'(x) = f(x) + [ -1(\log x - 1) + (1-x) \cdot \frac{1}{x} ] = f(x) - \log x + 1 + \frac{1}{x} - 1$.
$f'(x) - f(x) = \frac{1}{x} - \log x$.
Integrating factor $I.F. = e^{\int -1 dx} = e^{-x}$.
$f(x) e^{-x} = \int e^{-x} (\frac{1}{x} - \log x) dx$.
Using Integration by parts on $\int e^{-x} \log x dx$: $\int \log x d(-e^{-x}) = -e^{-x} \log x + \int e^{-x} \frac{1}{x} dx$.
So, $f(x) e^{-x} = e^{-x} \log x + C \implies f(x) = \log x + C e^x$.
From original equation, at $x=1$, $f(1) = 0 + (0) + e = e$.
$e = \log 1 + C e^1 \implies C = 1$.
$f(x) = \log x + e^x$.
We need $f(f(1)) = f(e)$.
$f(e) = \log e + e^e = 1 + e^e$.
Step 3: Final Answer:
The value is $1 + e^e$.
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