JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Hard

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Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^P}\sin \frac{1}{x},x \ne 0}\\ {0,x = 0} \end{array}} \right.

then

f\left( x \right)

is continuous but not differentiable at

x = 0

, if

View answer and explanation

Correct option: A

0 < p \le 1

Option A (

0 < p \le 1

) is correct because it ensures continuity but not differentiability at

x=0

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