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Friday, 9 October

JEE Main 2026 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 4 April 2026, Shift 2, Maths Q19

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

Let y=y(x)y=y(x) be the solution of the differential equation: dydx+(6x2+(3x2+2x3+4)e−2x(x3+2)(2+e−2x))y=2+e−2x\frac{dy}{dx}+\left(\frac{6x^2+\left(3x^2+2x^3+4\right)e^{-2x}}{\left(x^3+2\right)\left(2+e^{-2x}\right)}\right)y=2+e^{-2x}, x∈(−1,2)x\in(-1,2), satisfying y(0)=32y(0)=\frac32. If y(1)=α(2+e−2)y(1)=\alpha(2+e^{-2}), then α\alpha is equal to:
  1. (1)138\frac{13}{8}
  2. (2)613\frac{6}{13}
  3. (3)1213\frac{12}{13}
  4. (4)1312\frac{13}{12}Official answer

Official answer

Option 4

NTA final key.