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← JEE Main 2026 · 5 Apr · Shift 2
MathematicsSingle correct+4 / -1Question 19 of 75

JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 19

Let f:[1,∞)→Rf:[1,\infty)\to\mathbf R be a differentiable function defined as f(x)=∫1xf(t) dt+(1−x)(log⁡ex−1)+ef(x)=\displaystyle\int_1^xf(t)\,dt+(1-x)(\log_ex-1)+e. Then the value of f(f(1))f(f(1)) is :

Choose one option
Answer and worked solution

Official answer: A

Initial value. f(1)=0+0+e=ef(1)=0+0+e=e.

Differentiate.
f′=f−(ln⁡x−1)+1−xx=f−ln⁡x+1x.f'=f-(\ln x-1)+\frac{1-x}x=f-\ln x+\frac1x.

Solve. Put f=ln⁡x+hf=\ln x+h. Then h′=hh'=h, so h=Cexh=Ce^x, and f(1)=ef(1)=e gives C=1C=1. So f(x)=ln⁡x+exf(x)=\ln x+e^x.

Value: f(f(1))=f(e)=1+eef(f(1))=f(e)=1+e^e.

(A)

Topics:Linear first-order equationsFundamental theorem of calculus
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