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Friday, 9 October

JEE Main 2026 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 5 April 2026, Shift 1, Maths Q19

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

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a differentiable function such that f(x+y3)=f(x)+f(y)3f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)}{3} for all x,y∈Rx, y \in \mathbb{R}, and f′(0)=3f'(0) = 3. Then the minimum value of the function g(x)=3+exf(x)g(x) = 3 + e^x f(x), is:
  1. (1)3(e+1e)3\left(\frac{e+1}{e}\right)
  2. (2)3(e−1e)3\left(\frac{e-1}{e}\right)Official answer
  3. (3)3−ee\frac{3-e}{e}
  4. (4)3e3e

Official answer

Option 2

NTA final key.

Chapter
Continuity and Differentiability
Topic
Continuity and Differentiability
Idea tested
Differentiability via Functional Equations

Same idea in other shifts

Asked 3× in all
  1. 2 Apr 2025, Shift 1 · Q14Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a twice differentiable function such that…HardSingle correct
  2. 30 Jan 2024, Shift 2 · Q7Let f:R−{0}→Rf: \mathbb{R} - \{0\} \to \mathbb{R} be a function satisfying f(xy)=f(x)f(y)f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)} for all…MediumSingle 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.