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JEE Main202118 Mar 2021Morning ShiftMathematicsContinuity and DifferentiabilityActual

If f x = 1 | x | ; | x | ≥ 1 a x 2 + b ; | x | < 1 is differentiable at every point of the domain, then the values of a and b are respectively:

Options

  1. A1 2 , 1 2
  2. B1 2 , - 3 2
  3. C5 2 , - 3 2
  4. D- 1 2 , 3 2

Correct answer

D. - 1 2 , 3 2

Step-by-step solution

f x = 1 | x | , | x | ≥ 1 a x 2 + b , | x | < 1 at x = 1 function must be continuous So, 1 = a + b   … 1 differentiability at x = 1 - 1 x 2 x = 1 = ( 2 · a x ) x = 1 ⇒ - 1 = 2 a ⇒ a = - 1 2 Put in ( 1 ) ⇒ b = 1 + 1 2 = 3 2

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