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$Let f (x) = \int_{0}^{x} t \sin \frac{1}{t} d t . Then the number of points of discontinuity of the$ $function f (x) in the open interval (0, π) is$

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Correct option is A

f(x)=xsin⁡1x. At all points in (0,π),f′(x) has a definite finite value. ∴ f(x) is differentiable finitely in (0,π). As a finitely differentiable function is also continuous, hence f(x) is continuous in (0,π).

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Let f(x)=∫0x tsin⁡1tdt. Then the number of points of discontinuity of the function f(x) in the open interval (0,π) is

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$Let f (x) = \int_{0}^{x} t \sin \frac{1}{t} d t . Then the number of points of discontinuity of the$ $function f (x) in the open interval (0, π) is$