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Saturday, 31 October

JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 3 April 2025, Shift 2, Maths Q20

Question 20 of 75 in this shift, Maths question 20 of 25, Section A.

Let y=y(x)y = y(x) be the solution of the differential equation dydx+3(tan⁡2x)y+3y=sec⁡2x, y(0)=13+e3\frac{dy}{dx} + 3\left(\tan^2 x\right)y + 3y = \sec^2 x,\ y(0) = \frac{1}{3} + e^3. Then y(π4)y\left(\frac{\pi}{4}\right) is equal to
  1. (1)23+e3\frac{2}{3} + e^3
  2. (2)43+e3\frac{4}{3} + e^3
  3. (3)23\frac{2}{3}
  4. (4)43\frac{4}{3}Official answer

Official answer

Option 4

NTA final key.

Same idea in other shifts

Asked 7× in all
  1. 2 Apr 2026, Shift 1 · Q20Let y=y(x)y=y(x) be the solution curve of the differential equation (1+sin⁡x)dydx+(y+1)cos⁡x=0(1+\sin x)\frac{dy}{dx}+(y+1)\cos x=0, y(0)=0y(0)=0. If the curve y=y(x)y=y(x)…MediumSingle correct
  2. 2 Apr 2025, Shift 2 · Q25Let y=y(x)y = y(x) be the solution of the differential equation…MediumNumerical value
  3. 29 Jan 2024, Shift 1 · Q13A function y=f(x)y=f(x) satisfies f(x)sin⁡2x+sin⁡x−(1+cos⁡2x)f′(x)=0f(x)\sin 2x+\sin x-(1+\cos^2x)f'(x)=0 with condition f(0)=0f(0)=0. Then, f(π2)f\left(\frac{\pi}{2}\right) is…MediumSingle correct
  4. 31 Jan 2024, Shift 1 · Q11Let y=y(x)y=y(x) be the solution of the differential equation…HardSingle correct
  5. 28 Jul 2022, Shift 1 · Q10If y=y(x), x∈(0,π/2)y=y(x),\ x\in(0,\pi/2) be the solution curve of the differential equation…HardSingle correct
  6. 27 Jun 2022, Shift 2 · Q28Let y=y(x)y=y(x) be the solution of the differential equation (1−x2)dy=(xy+(x3+2)1−x2)dx\left(1-x^2\right)dy=\left(xy+\left(x^3+2\right)\sqrt{1-x^2}\right)dx,…HardNumerical value

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 108 of the 174 JEE Main shifts held from 2017 to 2026.