NTA Abhyas JEE Main2020MathematicsContinuity and DifferentiabilityPractice
If f x = x p + 1 cos 1 x :x ≠ 0 0 :x = 0 , then at x = 0 the function f x is
Options
- AContinuous if p > - 1 and differentiable if p > 0
- BContinuous if p > 0 and differentiable if p > - 1
- CContinuous and differentiable if p > - 1
- DNone of these
Correct answer
A. Continuous if p > - 1 and differentiable if p > 0
Step-by-step solution
Continuity at x = 0 , L . H . L . = l i m h → 0 + f 0 - h = l i m h → 0 + - h p + 1 cos - 1 ⁡ 1 h = 0   i f p > - 1 R . H . L . = l i m h → 0 + f 0 + h = l i m h → 0 + h p + 1 cos - 1 ⁡ 1 h = 0   i f p > - 1 and f 0 = 0 . ∴ f x is continuous at x = 0  i f  p > - 1 Differentiability at x = 0 : L.H.D. = lim h → 0 + f 0 − h − f 0 − h = l i m h → 0 + - h p + 1 cos - 1 ⁡ 1 h - 0 - h = l i m h → 0