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JEE Main201910 Jan 2019Evening ShiftMathematicsApplication of DerivativesActual

The tangent to the curve, y = x e x 2 passing through the point 1 , e also passes through the point:

Options

  1. A4 3 , 2 e
  2. B2 , 3 e
  3. C5 3 , 2 e
  4. D3 , 6 e

Correct answer

A. 4 3 , 2 e

Step-by-step solution

Given y = x e x 2 ⇒ d y d x = 2 x 2 e x 2 + e x 2 ⇒ d y d x at 1 ,   e = 3 e ⇒ Equation of tangent is y - e = 3 e x - 1 , which passes through 4 3 , 2 e .

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