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

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Differential Equations

Let

f : [0, \infty) \to R

be differentiable function such that

f (x) = 1 - 2 x + \int_{0}^{x} e^{x - t} f (t) dt

for all

x \in [0, \infty)

. Then the area of the region bounded by

y = f (x)

and the coordinate axes is

Answer: (B)

Writing the relation as

f (x) = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t

and differentiating gives

f^{'} (x) = - 2 + f (x) + e^{x} \int_{0}^{x} e^{- t} f (t) d t

. Since

e^{x} \int_{0}^{x} e^{- t} f (t) d t = f (x) - 1 + 2 x

, this becomes

f^{'} (x) = 2 f (x) + 2 x - 3

. Solving

f^{'} - 2 f = 2 x - 3

with

f (0) = 1

gives

f (x) = 1 - x

. The line

y = 1 - x

with the coordinate axes bounds a triangle of area

\int_{0}^{1} (1 - x) d x = \frac{1}{2}

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option

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