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JEE Main 2019 · 10 Jan · Shift II

JEE Main 2019 Mathematics question on Application of Derivatives

The tangent to the curve, y=xex2y = xe^{x^2} passing through the point (1,e)(1, e) also passes through the point
  1. A.
    (43,2e)\left(\dfrac{4}{3}, 2e\right)
  2. B.
    (3,6e)(3, 6e)
  3. C.
    (2,3e)(2, 3e)
  4. D.
    (53,2e)\left(\dfrac{5}{3}, 2e\right)
Show answer and solution

Correct option: A

Solution

At (1, e): y = 1·e1 = e ✓, so (1, e) is on the curve.

Slope at x=1: dy/dx = e^(x²) + x·2x·e^(x²) = e^(x²)(1 + 2x²). At x=1: slope = e(1+2) = 3e.

Tangent line: y - e = 3e(x - 1) → y = 3ex - 2e.

Check (4/3, 2e): 3e(4/3) - 2e = 4e - 2e = 2e ✓

[JEE Main 2019 · 10 Jan · Shift II]

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