JEE Main202227 Jul 2022Morning ShiftMathematicsContinuity and DifferentiabilityActual
Let a function f : ℝ → ℝ be defined as: f x = ∫ 0 x 5 - t - 3 d t , x > 4 x 2 + b x , x ≤ 4 where b ∈ ℝ . If f is continuous at x = 4 , then which of the following statements is NOT true?
Options
- Af is not differentiable at x = 4
- Bf ' 3 + f ' 5 = 35 4
- Cf is increasing in - ∞ , 1 8 ∪ 8 , ∞
- Df has a local minima at x = 1 8
Correct answer
C. f is increasing in - ∞ , 1 8 ∪ 8 , ∞
Step-by-step solution
Given f x = ∫ 0 x 5 - t - 3 d t , x > 4 x 2 + b x , x ≤ 4 Also given, f x is continuous at x = 4 So lim x → 4 - f x = lim x → 4 + f x = f 4 So 16 + 4 b = ∫ 0 3 2 - t d t + ∫ 3 4 8 - t d t ⇒ 16 + 4 b = 2 t - t 2 2 0 3 + 8 t - t 2 2 3 4 ⇒ 16 + 4 b = 15 So b = - 1 4 Now differentiating f x we get, f ' x = 5 - x - 3 , x > 4 2 x + b , x < 4 Now at x = 4 LHD = 2 x + b = 2 × 4 - 1 4 = 31 4 RHD = 5 - x - 3 = 4 LHD ≠ RHD So, option A is true And f &#