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JEE Main202122 Jul 2021Morning ShiftMathematicsApplication of DerivativesActual

Let f : R → R be defined as f x = - 4 3 x 3 + 2 x 2 + 3 x , x > 0 3 x e x , x ≤ 0 . Then f is increasing function in the interval

Options

  1. A- 1 2 ,   2
  2. B0 ,   2
  3. C- 1 ,   3 2
  4. D- 3 ,   - 1

Correct answer

C. - 1 ,   3 2

Step-by-step solution

We know that, d d x x n = n x n - 1 and d d x e x = e x Hence, d d x - 4 3 x 3 + 2 x 2 + 3 x = - 4 3 × 3 x 2 + 2 × 2 x + 3 = - 4 x 2 + 4 x + 3 And, using product rule, we get d d x 3 x e x = 3 x d d x e x + 3 e x d d x x = 3 x e x + 3 e x = 3 e x x + 1 ⇒ f ' x = - 4 x 2 + 4 x + 3 , x > 0 3 e x ( 1 + x ) , x ≤ 0 For x > 0 ,   f ' ( x ) = - 4 x 2 + 4 x + 3 = - 2 x + 1 2 x - 3 The sign scheme of f ' x is We know that, if f ' x > 0 then for those value of x , &#160

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