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JEE Main202125 Jul 2021Evening ShiftMathematicsContinuity and DifferentiabilityActual

If f x = ∫ 0 x 5 + 1 - t d t , x > 2 5 x + 1 , x ≤ 2 , then

Options

  1. Af x is not continuous at x = 2
  2. Bf x is everywhere differentiable
  3. Cf x is continuous but not differentiable at x = 2
  4. Df x is not differentiable at x = 1

Correct answer

C. f x is continuous but not differentiable at x = 2

Step-by-step solution

We have, f x = ∫ 0 x 5 + 1 - t d t , x > 2 5 x + 1 , x ≤ 2 Therefore, f x = ∫ 0 1 5 + 1 - t d t + ∫ 1 x 5 + t - 1 d t = 6 - 1 2 + 4 t + t 2 2 | 1 x = 11 2 + 4 x + x 2 2 - 4 - 1 2 = x 2 2 + 4 x + 1 Now, f 2 + = 2 + 8 + 1 = 11 and f 2 = f 2 - = 5 × 2 + 1 = 11 Here, f 2 + = f 2 - So, f x is continuous at x = 2 Clearly, f x is differentiable at x = 1 Now, LHD   f ' 2 - = 5 and RHD   f ' 2 + = 6 Therefore, f x is not differentiable at x = 2

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