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Friday, 9 October

JEE Main 2025 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 2 April 2025, Shift 1, Maths Q14

Question 14 of 75 in this shift, Maths question 14 of 25, Section A.

Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a twice differentiable function such that (sin⁡xcos⁡y)(f(2x+2y)−f(2x−2y))=(cos⁡xsin⁡y)(f(2x+2y)+f(2x−2y))(\sin x\cos y)(f(2x+2y)-f(2x-2y))=(\cos x\sin y)(f(2x+2y)+f(2x-2y)), for all x,y∈Rx, y\in\mathbf{R}. If f′(0)=12f'(0)=\frac{1}{2}, then the value of 24f′′(5π3)24f''\left(\frac{5\pi}{3}\right) is :
  1. (1)3
  2. (2)−3-3Official answer
  3. (3)2
  4. (4)−2-2

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. 5 Apr 2026, Shift 1 · Q19Let 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…MediumSingle 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.