JEE Main 2026 Apr 5 Shift 2, Mathematics Q19: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Apr 5, Shift 2Mathematics

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:

A
$1 + e^{e}$
B
$1 + e$
C
$1 + e + e^{e}$
D
$1 + 2 e$

Solution

At $x = 1$ : $f (1) = 0 + 0 + e = e$ .

Differentiate both sides with respect to $x$ . Using $\frac{d}{d x} [(1 - x) (ln x - 1)] = - (ln x - 1) + (1 - x) / x = - ln x + 1/ x$ , we get:

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

This is linear in $f$ : integrating factor $e^{- x}$ . So $\frac{d}{d x} (e^{- x} f (x)) = e^{- x} (\frac{1}{x} - ln x)$ .

Note $\frac{d}{d x} (e^{- x} ln x) = - e^{- x} ln x + \frac{e ^{- x}}{x}$ , so the right side is exactly $\frac{d}{d x} (e^{- x} ln x)$ .

Hence $e^{- x} f (x) = e^{- x} ln x + C$ . Using $f (1) = e$ : $e^{- 1} \cdot e = 0 + C \Rightarrow C = 1$ .

$f (x) = e^{x} + ln x$ . Therefore $f (f (1)) = f (e) = e^{e} + 1$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q19: Methods of Solving First Order, First Degree Differential Equation

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