Let f:[0, ∞) arrow ℝ be differentiable function such that f(x)=1-2 x+∫_0^x e^ x-t f(t) dt for all x ∈[0, ∞) . Then the area of the region bounded by y=f(x) and the coordinate axes is

Correct answer: B. (1)/(2) aligned & y=1-2 x+e^x ∫_0^x e^ -t f(t) d t & (d y)/(d x)=-2+e^ -x · e^x f(x)+e^x ∫_0^x e^ -t f(t) d t & (d y)/(d x)=-2+y+y+2 x-1 & (d y)/(d x)-2 y=(2 x-3) & y e^ -2 x =∫(2 x-3) d x · e^ -2 x & y e^ -2 x =(-(2 x-3))/(2) e^ -2 x +∫ e^ -2 x d x & y e^ -2 x =(-(2 x-3))/(2) e^ -2 x -(1)/(2) e^ -2 x +c & f(0)=1 c=1-(3)/(2)+(1)/(2)=0 aligned aligned & y=-((2 x-3))/(2)-(1)/(2) & y=-x+1 & x+y=1 & area =(1)/(2)(1)(1)=(1)/(2) aligned

JEE Main 20254 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Differential Equations

Let f:[0, ∞) arrow ℝ be differentiable function such that f(x)=1-2...

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

Options

A $\sqrt{5}$
B $\frac{1}{2}$
C $\sqrt{2}$
D2

Correct answer

B. $\frac{1}{2}$

Step-by-step solution

\begin{aligned} & y=1-2 x+e^x \int_0^x e^{-t} f(t) d t \\ & \frac{d y}{d x}=-2+e^{-x} \cdot e^x f(x)+e^x \int_0^x e^{-t} f(t) d t \\ & \frac{d y}{d x}=-2+y+y+2 x-1 \\ & \frac{d y}{d x}-2 y=(2 x-3) \\ & y e^{-2 x}=\int(2 x-3) d x \cdot e^{-2 x} \\ & y e^{-2 x}=\frac{-(2 x-3)}{2} e^{-2 x}+\int e^{-2 x} d x \\ & y e^{-2 x}=\frac{-(2 x-3)}{2} e^{-2 x}-\frac{1}{2} e^{-2 x}+c \\ & f(0)=1 \Rightarrow c=1-\frac{3}{2}+\frac{1}{2}=0\end{aligned}

\begin{aligned} & y=-\frac{(2 x-3)}{2}-\frac{1}{2} \\ & y=-x+1 \\ & x+y=1 \\ & \text{ area }=\frac{1}{2}(1)(1)=\frac{1}{2}\end{aligned}

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