NTA Abhyas JEE Main2020MathematicsContinuity and DifferentiabilityPractice
Let f x = l i m n → ∞ x 2 + 2 x + 4 + sin π x n - 1 x 2 + 2 x + 4 + sin π x n + 1 , then
Options
- Af x is continuous and differentiable for all x ∈ R .
- Bf x is continuous but not differentiable for all x ∈ R .
- Cf x is discontinuous at infinite number of points.
- Df x is discontinuous at two points.
Correct answer
A. f x is continuous and differentiable for all x ∈ R .
Step-by-step solution
∵   x 2 + 2 x + 4 + s i n π x = x + 1 2 + 3 + sin ⁡ π x ≥ 2 , ∀ x ∈ R . ∴   f x = l i m n → ∞ 1 - x + 1 2 + 3 + sin ⁡ π x - n 1 + x + 1 2 + 3 + sin ⁡ π x - n = 1 - 0 1 + 0 = 1 Clearly, f x = 1 ,   ∀ x ∈ R Hence, f ( x ) is continuous and differentiable ∀ x ∈ R