JEE Main 2026 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 5 April 2026, Shift 2, Maths Q19

Question 19 of 75 in this shift, Maths question 19 of 25, Section A.

Let

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

be a differentiable function defined as

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

. Then the value of

f(f(1))

is :

(1) $(1+e^e)$ Official answer
(2) $(1+e)$
(3) $(1+e+e^e)$
(4) $1+2e$

Official answer

Option 1

NTA final key.

Chapter: Differential Equations
Topic: Leibniz Rule
Idea tested: Solving Integral Equations

Same idea in other shifts

Asked 4× in all

4 Apr 2026, Shift 2 · Q25Let $f$ be a twice differentiable function such that $f(x)=\int_0^x\tan(t-x)\,dt-\int_0^x f(t)\tan t\,dt$ ,…HardNumerical value
2 Apr 2025, Shift 2 · Q20Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function. If $10\int_1^x f(t)\,\mathrm{d}t = 5xf(x) - x^5 - 9$ for all…MediumSingle correct
9 Apr 2024, Shift 2 · Q13Let $\int_0^x\sqrt{1-(y'(t))^2}\,dt=\int_0^x y(t)\,dt,\ 0\le x\le 3,\ y\ge0,\ y(0)=0$ . Then at $x=2$ , $y''+y+1$ is equal toHardSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.