NTA Abhyas JEE Main2020MathematicsContinuity and DifferentiabilityPractice
If f x = a + c o s - 1 x + b : x ≥ 1 - x : x < 1 is differentiable at x = 1 , then the value of b - a is equal to
Options
- A0
- B1
- C- 1
- Dπ 2
Correct answer
D. π 2
Step-by-step solution
Since, f ( x ) is differentiable, it must be continuous at x = 1 ∴ f 1 - = f 1 + = f 1 - 1 = a + c o s - 1 1 + b ... 1 f ' x = − 1 1 − x + b 2 : x ≥ 1 − 1 : x < 1 Now, for f x to be differentiable at x = 1 f ′ 1 − = f ′ 1 + - 1 1 - 1 + b 2 = - 1 ⇒ 1 - 1 + b 2 = 1 ⇒ b = - 1 ... 2 From equation 1 & 2 - 1 = a + π 2 ⇒ a = - π 2 - 1