JEE Main 2025 Apr 4 Shift 1, Mathematics Q8: Methods of Solving First Order, First Degree Differential Equation

JEE Main2025Apr 4, Shift 1Mathematics

Let $f : [0, \infty) \to R$ be differentiable function such that $f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) d t$ for all $x \in [0, \infty)$ . Then the area of the region bounded by $y = f (x)$ and the coordinate axes is

A
$5$
B
$\frac{1}{2}$
C
$2$
D
$2$

Solution

Rewrite the integral term: $y = f (x) = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t$ .

Differentiate both sides: $y^{'} = - 2 + e^{x} \int_{0}^{x} e^{- t} f (t) d t + e^{x} \cdot e^{- x} f (x) = - 2 + (y - (1 - 2 x)) + f (x) = 2 y + 2 x - 3$ .

This gives the linear ODE $y^{'} - 2 y = 2 x - 3$ . Integrating factor $e^{- 2 x}$ :

$(y e^{- 2 x})^{'} = (2 x - 3) e^{- 2 x}$ , hence $y e^{- 2 x} = \int (2 x - 3) e^{- 2 x} d x = - \frac{( 2 x - 3 )}{2} e^{- 2 x} - \frac{1}{2} e^{- 2 x} + C$ .

So $y = - \frac{2 x - 3}{2} - \frac{1}{2} + C e^{2 x} = - x + 1 + C e^{2 x}$ . Using $f (0) = 1$ gives $C = 0$ , so $y = 1 - x$ .

The line $y = 1 - x$ together with both axes bounds a right triangle with legs of length 1. Its area is $\frac{1}{2}$ .

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q8: Methods of Solving First Order, First Degree Differential Equation

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