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JEE Main20205 Sep 2020Evening ShiftMathematicsApplication of DerivativesActual

If x = 1 is a critical point of the function f ( x ) = 3 x 2 + a x - 2 - a e x , then

Options

  1. Ax = 1 and x = - 2 3 are local minima of f
  2. Bx = 1 and x = - 2 3 is a local maxima of f
  3. Cx = 1 is a local maxima and x = - 2 2 is a local minima of f
  4. Dx = 1 is a local minima and x = - 2 3 are local maxima of f

Correct answer

D. x = 1 is a local minima and x = - 2 3 are local maxima of f

Step-by-step solution

f ( x ) = 3 x 2 + a x - 2 - a e x f ' ( x ) = 3 x 2 + a x - 2 - a e x + e x ( 6 x + a ) = e x 3 x 2 + ( a + 6 ) x - 2 ∵    x = 1 is a critical point    ∴    f ' ( 1 ) = 0 ∴ 3 + a + 6 - 2 = 0 a = - 7 ∴ f ' ( x ) = e x 3 x 2 - x - 2 = e x 3 x 2 - 3 x + 2 x - 2 = e x ( 3 x + 2 ) ( x - 1 ) ∴ maxima at x = - 2   3 ∴ minima at x = 1

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