NTA Abhyas JEE Main2020MathematicsContinuity and DifferentiabilityPractice
Given f ( x ) = x 2 e 2 ( x − 1 ) , 0 ≤ x ≤ 1 . a sgn ( x + 1 ) cos ( 2 x − 2 ) + b x 2 , 1 < x ≤ 2 If f ( x ) is differentiable at x = 1, then the value of | a − b | is
Correct answer
3
Step-by-step solution
f 1 - = 1 = f 1 and f 1 + = a + b For continuity at x = 1 , a + b = 1 … … i In 0 < x < 1 , f ' x = 2 x e 2 x - 1 + 2 x 2 e 2 x - 1 ∴ f ' 1 - = 4 In 1 < x < 2 , f ' x = - 2 a sin 2 x - 2 + 2 b x ∴ f ' 1 + = 2 b For differentiability at x = 1 , 2 b = 4 . This with ( i ) gives the values a = - 1 , b = 2 then | b − a | = 3