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JEE Main20202 Sep 2020Morning ShiftMathematicsContinuity and DifferentiabilityActual

If a function f x defined by f x = a e x + b e − x , − 1 ≤ x < 1 c x 2 , 1 ≤ x ≤ 3 a x 2 + 2 c x , 3 < x ≤ 4 be continuous for some a , b , c ∈ R and f ' 0 + f ' 2 = e , then the value of a is

Options

  1. A1 e 2 - 3 e + 13
  2. Be e 2 - 3 e - 13
  3. Ce e 2 + 3 e + 13
  4. De e 2 - 3 e + 13

Correct answer

D. e e 2 - 3 e + 13

Step-by-step solution

Continuous at x = 1 , 3 f 1 - = f 1    ⇒ a e + b e - 1 = c   . . . ( 1 ) f 3 = f 3 +   ⇒ 9 c = 9 a + 6 c   ⇒ c = 3 a     . . . 2 From 1 and 2   b = a e ( 3 - e )     . . . 3 f ' x = a e x - b e − x , − 1 < x < 1 2 c x , 1 < x < 3 2 a x + 2 c , 3 < x < 4 f ' 0 = a - b ,   f ' 2 = 4 c Given f ' ( 0 ) + f ' ( 2 ) = e a - b + 4 c = e     . . . 4 By using eq. 1 ,   2 ,   3   &  

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