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Question: Let ƒ(x) =\(\int_{0}^{x}t\sin\frac{1}{t}dt\). Then the number of points of discontinuity of the func...

Let ƒ(x) =∫0xtsin⁡1tdt\int_{0}^{x}t\sin\frac{1}{t}dt. Then the number of points of discontinuity of the function ƒ(x) in the open interval (0, p) is –

A

0

B

1

C

2

D

Infinite

Answer

0

Explanation

Solution

Since, f¢(x) = x sin 1x\frac{1}{x}

Now, at all points in (0, p), f¢(x) has a definite finite value.

\ f(x) is differentiable finitely in (0, p)

As a finitely differentiable function is also continuous

\ f(x) is continuous in (0, p)