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,∞)→Rf:[1,\infty)\to\mathbb{R} be defined as f(x)=∫1xf(t) dt+(1−x)(ln⁡x−1)+ef(x)=\displaystyle\int_1^x f(t)\,dt+(1-x)(\ln x-1)+e. Then the value of f(f(1))f(f(1)) is:
A1+ee1+e^ecorrect
B1+e1+e
C1+e+ee1+e+e^e
D1+2e1+2e
Solution
Step 1: Evaluate f(1)f(1) directly
f(1)=∫11f(t) dt+(1−1)(ln⁡1−1)+e=0+0+e=ef(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)⋅1x=f(x)−ln⁡x+1xf'(x) = f(x)+(-1)(\ln x-1)+(1-x)\cdot\frac{1}{x} = f(x)-\ln x+\frac{1}{x}
f′(x)−f(x)=1x−ln⁡xf'(x)-f(x) = \frac{1}{x}-\ln x
 Step 3: Solve using integrating factor e−xe^{-x}
ddx(f(x)e−x)=e−x ⁣(1x−ln⁡x)=ddx(e−xln⁡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: ddx(e−xln⁡x)=e−x(1x−ln⁡x)\frac{d}{dx}(e^{-x}\ln x) = e^{-x}(\frac{1}{x}-\ln x).) Therefore f(x)e−x=e−xln⁡x+Cf(x)e^{-x} = e^{-x}\ln x + C. Step 4: Apply f(1)=ef(1)=e to determine CC
e⋅e−1=e−1⋅0+C  ⟹  C=1  ⟹  f(x)=ln⁡x+exe\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(f(1)) = f(e)
f(e)=ln⁡e+ee=1+eef(e) = \ln e+e^e = 1+e^e
 Answer: (1)
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