If f x = x p + 1 cos 1 x :x ≠ 0 0 :x = 0 , then at x = 0 the function f x is

Question

If f x = x p + 1 cos 1 x :x ≠ 0 0 :x = 0 , then at x = 0 the function f x is, Continuous if p > - 1 and differentiable if p > 0, Continuous if p > 0 and differentiable if p > - 1, Continuous and differentiable if p > - 1, None of these

MARKS App · Accepted Answer

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 + - h p cos - 1 ⁡ 1 h = 0 if p > 0 , R.H.D. = lim h → 0 + h p + 1 cos − 1 1 h − 0 h = l i m h → 0 + h p cos - 1 ⁡ 1 h = 0 if p > 0 ∴ f x is differentiable at x = 0 if p > 0

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