Differential EquationsmediumPYQ · JEE Main · 5 Apr 2026Free

Differential Equations: Let Defined Value (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

Let

f:[1,\infty)\to\mathbb{R}

be defined as

f(x)=\displaystyle\int_1^x f(t)\,dt+(1-x)(\ln x-1)+e

. Then the value of

f(f(1))

is:

1+e^e

correct

1+e

1+e+e^e

1+2e

Solution

Step 1: Evaluate

f(1)

directly

f(1) = \int_1^1 f(t)\,dt + (1-1)(\ln1-1)+e = 0+0+e = e

Step 2: Differentiate the functional equation to obtain a DE

f'(x) = f(x)+(-1)(\ln x-1)+(1-x)\cdot\frac{1}{x} = f(x)-\ln x+\frac{1}{x}

f'(x)-f(x) = \frac{1}{x}-\ln x

Step 3: Solve using integrating factor

e^{-x}

\frac{d}{dx}\bigl(f(x)e^{-x}\bigr) = e^{-x}\!\left(\frac{1}{x}-\ln x\right) = \frac{d}{dx}(e^{-x}\ln x)

(Verified:

\frac{d}{dx}(e^{-x}\ln x) = e^{-x}(\frac{1}{x}-\ln x)

.) Therefore

f(x)e^{-x} = e^{-x}\ln x + C

. Step 4: Apply

f(1)=e

to determine

C

e\cdot e^{-1} = e^{-1}\cdot0+C \implies C=1 \implies f(x) = \ln x+e^x

Step 5: Compute

f(f(1)) = f(e)

f(e) = \ln e+e^e = 1+e^e

Answer: (1)

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