Logo
Jeeify
MCQMediumJEE 2025Linear Differential Equations

JEE Mathematics 2025 Question with Solution

Let f:[0,∞)→Rf: [0, \infty) \to \mathbb{R} be a differentiable function such that f(x)=1−2x+∫0xex−tf(t) dtf(x) = 1 - 2x + \int_0^x e^{x-t} f(t) \, dt for all x∈[0,∞)x \in [0, \infty). Then the area of the region bounded by y=f(x)y = f(x) and the coordinate axes is](streamdown:incomplete-link)

  • A

    5\sqrt{5}

  • B

    12\frac{1}{2}

  • C

    2\sqrt{2}

  • D

    22

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

f(x)=1−2x+∫0xex−tf(t) dtf(x)=1-2x+\int_0^x e^{x-t}f(t)\,dt

with f:[0,∞)→Rf:[0,\infty)\to\mathbb{R} differentiable.

Find: The area of the region bounded by y=f(x)y=f(x) and the coordinate axes.](streamdown:incomplete-link)

Differentiate the given relation. Write

∫0xex−tf(t) dt=ex∫0xe−tf(t) dt.\int_0^x e^{x-t}f(t)\,dt=e^x\int_0^x e^{-t}f(t)\,dt.

So,

f′(x)=−2+ex∫0xe−tf(t) dt+f(x).f'(x)=-2+e^x\int_0^x e^{-t}f(t)\,dt+f(x).

From the original equation,

ex∫0xe−tf(t) dt=f(x)−1+2x.e^x\int_0^x e^{-t}f(t)\,dt=f(x)-1+2x.

Substituting this in the differentiated equation gives

f′(x)=−2+(f(x)−1+2x)+f(x)=2x−3+2f(x).f'(x)=-2+\big(f(x)-1+2x\big)+f(x)=2x-3+2f(x).

Hence,

f′(x)−2f(x)=2x−3.f'(x)-2f(x)=2x-3.

Now use the initial condition from the original equation at x=0x=0:

f(0)=1.f(0)=1.

Solve the linear differential equation using integrating factor e−2xe^{-2x}:

ddx(f(x)e−2x)=(2x−3)e−2x.\frac{d}{dx}\big(f(x)e^{-2x}\big)=(2x-3)e^{-2x}.

Integrating,

f(x)e−2x=∫(2x−3)e−2x dx+C.f(x)e^{-2x}=\int (2x-3)e^{-2x}\,dx + C.

Using the working from the solution,

∫(2x−3)e−2x dx=e−2x(1−x)+C,\int (2x-3)e^{-2x}\,dx=e^{-2x}(1-x)+C,

so

f(x)=1−x+Ce2x.f(x)=1-x+Ce^{2x}.

Applying f(0)=1f(0)=1, we get C=0C=0. Therefore,

f(x)=1−x.f(x)=1-x.

The curve y=1−xy=1-x meets the axes at x=1x=1 and y=1y=1. Thus the bounded region is a right triangle with base 11 and height 11.

Area=12×1×1=12.\text{Area}=\frac{1}{2}\times 1\times 1=\frac{1}{2}.

Therefore, the correct option is B.

Differentiate and convert to ODE

Given: f(x)=1−2x+∫0xex−tf(t) dtf(x)=1-2x+\int_0^x e^{x-t}f(t)\,dt.

Find: The area enclosed by the graph of y=f(x)y=f(x) and the coordinate axes.

A useful substitution is

g(x)=∫0xe−tf(t) dt.g(x)=\int_0^x e^{-t}f(t)\,dt.

Then

∫0xex−tf(t) dt=exg(x),\int_0^x e^{x-t}f(t)\,dt=e^x g(x),

and the given equation becomes

f(x)=1−2x+exg(x).f(x)=1-2x+e^x g(x).

Differentiate both sides:

f′(x)=−2+ddx(exg(x))=−2+exg(x)+exg′(x).f'(x)=-2+\frac{d}{dx}\big(e^x g(x)\big)=-2+e^x g(x)+e^x g'(x).

Since

g′(x)=e−xf(x),g'(x)=e^{-x}f(x),

we obtain

f′(x)=−2+exg(x)+f(x).f'(x)=-2+e^x g(x)+f(x).

From

f(x)=1−2x+exg(x),f(x)=1-2x+e^x g(x),

we have

exg(x)=f(x)−1+2x.e^x g(x)=f(x)-1+2x.

Substitute into the derivative equation:

f′(x)=−2+(f(x)−1+2x)+f(x)=2x−3+2f(x).f'(x)=-2+\big(f(x)-1+2x\big)+f(x)=2x-3+2f(x).

Therefore,

f′(x)−2f(x)=2x−3.f'(x)-2f(x)=2x-3.

At x=0x=0, the integral vanishes, so

f(0)=1.f(0)=1.

Now solve

f′−2f=2x−3,f'-2f=2x-3,

with initial condition f(0)=1f(0)=1.

Using integrating factor e−2xe^{-2x},

ddx(f(x)e−2x)=(2x−3)e−2x.\frac{d}{dx}\big(f(x)e^{-2x}\big)=(2x-3)e^{-2x}.

Integrating gives

f(x)e−2x=e−2x(1−x)+C,f(x)e^{-2x}=e^{-2x}(1-x)+C,

which implies

f(x)=1−x+Ce2x.f(x)=1-x+Ce^{2x}.

From f(0)=1f(0)=1, we get C=0C=0, hence

f(x)=1−x.f(x)=1-x.

So the graph is a straight line cutting the axes at  (0,1) \,(0,1)\, and  (1,0) \,(1,0)\,. The enclosed area is

12×1×1=12.\frac{1}{2}\times 1\times 1=\frac{1}{2}.

Hence the area is 12\frac{1}{2}, so the correct option is B.

Common mistakes

  • Differentiating ∫0xex−tf(t) dt\int_0^x e^{x-t}f(t)\,dt incorrectly by ignoring the dependence on both the upper limit and the factor exe^x. This is wrong because Leibniz rule must be applied carefully. Rewrite the integral as ex∫0xe−tf(t) dte^x\int_0^x e^{-t}f(t)\,dt before differentiating.

  • Using the first extracted solution step f′(x)−f(x)=−2f'(x)-f(x)=-2 directly. That misses the substitution from the original equation and leads to an incorrect function. Instead, replace ex∫0xe−tf(t) dte^x\int_0^x e^{-t}f(t)\,dt by f(x)−1+2xf(x)-1+2x to obtain the correct ODE f′(x)−2f(x)=2x−3f'(x)-2f(x)=2x-3.

  • Trying to find area by integrating f(x)f(x) from 00 to ∞\infty. This is wrong because the bounded region with the coordinate axes ends at the positive xx-intercept, not at infinity. First find where f(x)=0f(x)=0, then compute the finite enclosed area.

Practice more Linear Differential Equations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Linear Differential Equations

Related questions