Let $\mathrm{f}:[1, \infty) \rightarrow \mathrm{R}$ be a differentiable function defined as $f(x)=\int_{1}^{x} f(t) d t+(1-x)\left(\log _{e} x-1\right)+e$. Then the value of $\mathrm{f}(\mathrm{f}(1))$ is:

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Definite Integrals

Let

f : [1, \infty) \to R

be a differentiable function defined as

f (x) = \int_{1}^{x} f (t) d t + (1 - x) (lo g_{e} x - 1) + e

. Then the value of

f (f (1))

is:

Answer: (A)

Putting

x = 1

gives

f (1) = e

. Differentiating

f (x) = \int_{1}^{x} f (t) d t + (1 - x) (ln x - 1) + e

gives

f^{'} (x) = f (x) - ln x + \frac{1}{x}

, a linear equation

f^{'} - f = \frac{1}{x} - ln x

with integrating factor

e^{- x}

. Since

\frac{d}{d x} (e^{- x} ln x) = e^{- x} (\frac{1}{x} - ln x)

, integration gives

e^{- x} f (x) = e^{- x} ln x + C

, i.e.

f (x) = ln x + C e^{x}

;

f (1) = e

forces

C = 1

, so

f (x) = e^{x} + ln x

. Hence

f (f (1)) = f (e) = e^{e} + ln e = 1 + e^{e}

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option