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JEE Mathematics 2025 Question with Solution

Let f(x)=x−1f(x) = x - 1 and g(x)=exg(x) = e^x for x∈Rx \in \mathbb{R}. If

dydx=(e−2xg(f(f(x)))−yx),  y(0)=0,\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f\left(f(x)\right)\right) - \frac{y}{\sqrt{x}} \right), \; y(0) = 0,

then y(1)y(1) is

  • A

    2e−1e3\frac{2e-1}{e^3}

  • B

    1−e2e4\frac{1-e^2}{e^4}

  • C

    e−1e4\frac{e-1}{e^4}

  • D

    1−e3e4\frac{1-e^3}{e^4}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: f(x)=x−1f(x) = x - 1, g(x)=exg(x) = e^x, and

dydx=e−2xg(f(f(x)))−yx,y(0)=0\frac{dy}{dx} = e^{-2\sqrt{x}} g\left(f\left(f(x)\right)\right) - \frac{y}{\sqrt{x}}, \qquad y(0)=0

Find: y(1)y(1)

First evaluate the nested functions:

f(f(x))=f(x−1)=(x−1)−1=x−2f(f(x)) = f(x-1) = (x-1)-1 = x-2

So,

g(f(f(x)))=g(x−2)=ex−2g(f(f(x))) = g(x-2) = e^{x-2}

Substituting into the differential equation,

dydx+yx=e−2xex−2=ex−2−2x\frac{dy}{dx} + \frac{y}{\sqrt{x}} = e^{-2\sqrt{x}} e^{x-2} = e^{x-2-2\sqrt{x}}

This is a first-order linear differential equation.

The integrating factor is

μ(x)=e∫1x dx=e2x\mu(x) = e^{\int \frac{1}{\sqrt{x}} \, dx} = e^{2\sqrt{x}}

Multiplying the equation by the integrating factor,

e2xdydx+e2xyx=ex−2e^{2\sqrt{x}} \frac{dy}{dx} + e^{2\sqrt{x}} \frac{y}{\sqrt{x}} = e^{x-2}

Hence,

ddx(ye2x)=ex−2\frac{d}{dx}\left(y e^{2\sqrt{x}}\right) = e^{x-2}

Integrating both sides,

ye2x=∫ex−2 dx=ex−2+Cy e^{2\sqrt{x}} = \int e^{x-2} \, dx = e^{x-2} + C

Using the initial condition y(0)=0y(0)=0,

0⋅e0=e−2+C0 \cdot e^0 = e^{-2} + C

So,

C=−e−2C = -e^{-2}

Therefore,

ye2x=ex−2−e−2y e^{2\sqrt{x}} = e^{x-2} - e^{-2}

and hence

y=e−2x(ex−2−e−2)y = e^{-2\sqrt{x}}\left(e^{x-2} - e^{-2}\right)

Now put x=1x=1:

y(1)e2=e−1−e−2y(1)e^2 = e^{-1} - e^{-2}

So,

y(1)=e−1−e−2e2=e−1e4y(1) = \frac{e^{-1} - e^{-2}}{e^2} = \frac{e-1}{e^4}

Therefore, the correct option is C.

Direct evaluation at x = 1

From

ye2x=ex−2−e−2y e^{2\sqrt{x}} = e^{x-2} - e^{-2}

set x=1x=1 directly:

y(1)e2=e−1−e−2=1e−1e2=e−1e2y(1)e^2 = e^{-1} - e^{-2} = \frac{1}{e} - \frac{1}{e^2} = \frac{e-1}{e^2}

Therefore,

y(1)=e−1e2⋅1e2=e−1e4y(1) = \frac{e-1}{e^2} \cdot \frac{1}{e^2} = \frac{e-1}{e^4}

So the required value is e−1e4\frac{e-1}{e^4}.

Common mistakes

  • A common mistake is computing f(f(x))f(f(x)) incorrectly as x−1x-1 instead of x−2x-2. This makes the entire right-hand side wrong. Apply the function twice carefully: first f(x)=x−1f(x)=x-1, then f(x−1)=x−2f(x-1)=x-2.

  • Students often choose the integrating factor incorrectly by integrating 1x\frac{1}{\sqrt{x}} wrongly. Since

    ∫1x dx=2x+C\int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} + C

    the integrating factor is e2xe^{2\sqrt{x}}, not exe^{\sqrt{x}}.

  • Another mistake is forgetting to rewrite the equation in linear form as dydx+yx=Q(x)\frac{dy}{dx} + \frac{y}{\sqrt{x}} = Q(x). If the yy-term is left on the wrong side, the integrating factor method is applied incorrectly.

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