NTA Abhyas JEE Main2020MathematicsContinuity and DifferentiabilityPractice
If the function f x = a x + 7 ; 0 ≤ x < 2 b x + 5 ; x ≥ 2 is differentiable ∀ x ≥ 0, then 2 a + 4 b is equal to
Options
- A240 16
- B5
- C85 16
- D250 16
Correct answer
B. 5
Step-by-step solution
f x must be continuous and differentiable at x = 2 For continuity at x = 2 f 2 - = f 2 + = f 2 a 2 + 7 = b 2 + 5 ⇒ 3 a = 2 b + 5 … i For differentiability at x = 2 f ′ x = a 2 x + 7 ; 0 ≤ x < 2 b ; x ≥ 2 f ′ 2 − = f ′ 2 + a 2 2 + 7 = b ⇒ a = 6 b … i i From i ⇒ 18 b = 2 b + 5 ⇒ b = 5 16 So, 2 a + 4 b = 12 b + 4 b = 16 b = 5