Let $$f\left( x \right) = \int_0^x {t\,\sin \frac{1}{t}dt} .$$ Then the number of points of discontinuity of the function $$f\left( x \right)$$ in the open interval $$\left( {0,\,\pi } \right)$$ is :
A.
$$0$$
B.
$$1$$
C.
$$2$$
D.
infinite
Answer :
$$0$$
Solution :
$$f'\left( x \right) = x\,\sin \,\frac{1}{x}.$$ At all points in $$\left( {0,\,\pi } \right),\,f'\left( x \right)$$ has a definite finite value.
$$\therefore f\left( x \right)$$ is differentiable finitely in $$\left( {0,\,\pi } \right).$$ As a finitely differentiable function is also continuous, we get $$f\left( x \right)$$ is continuous in $$\left( {0,\,\pi } \right).$$
Releted MCQ Question on Calculus >> Continuity
Releted Question 1
For a real number $$y,$$ let $$\left[ y \right]$$ denotes the greatest integer less
than or equal to $$y:$$ Then the function $$f\left( x \right) = \frac{{\tan \left( {\pi \left[ {x - \pi } \right]} \right)}}{{1 + {{\left[ x \right]}^2}}}$$ is-
A.
discontinuous at some $$x$$
B.
continuous at all $$x,$$ but the derivative $$f'\left( x \right)$$ does not exist for some $$x$$
C.
$$f'\left( x \right)$$ exists for all $$x,$$ but the second derivative $$f'\left( x \right)$$ does not exist for some $$x$$
The function $$f\left( x \right) = \frac{{\ln \left( {1 + ax} \right) - \ln \left( {1 - bx} \right)}}{x}$$ is not defined at $$x = 0.$$ The value which should be assigned to $$f$$ at $$x = 0,$$ so that it is continuous at $$x =0,$$ is-
The function $$f\left( x \right) = \left[ x \right]\cos \left( {\frac{{2x - 1}}{2}} \right)\pi ,\,\left[ . \right]$$ denotes the greatest integer function, is discontinuous at-
The function $$f\left( x \right) = {\left[ x \right]^2} - \left[ {{x^2}} \right]$$ (where $$\left[ y \right]$$ is the greatest integer less than or equal to $$y$$ ), is discontinuous at-