Let \(\displaystyle f\left( x \right) =\int _{ 0 }^{ x }{ t\sin { \frac { 1 }{ t } }\ dt } \), then the number of points of discontinuity of \(f\left( x \right)\) in \(\left( 0,\pi \right) \) is \(\text{__________} \).

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TopNewestWe have \(f'(x)=x \sin \frac{1}{x}\), which exists and is continuous everywhere in \((0,\pi)\) (Check that \(\lim_{x \to 0}x \sin \frac{1}{x}=0\)). Thus the function \(f(x)\) has no point of discontinuity in the interval \((0,\pi)\). – Abhishek Sinha · 9 months, 3 weeks ago

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@parv mor – Akhilesh Prasad · 9 months, 4 weeks ago

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@Rishabh Cool. I have been trying this one for sometime, – Akhilesh Prasad · 9 months, 4 weeks ago

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