If $$y = \frac{1}{{{t^2} + t - 2}}$$ where $$t = \frac{1}{{x - 1}},$$ then find the number of points of discontinuities of $$y = f\left( x \right),$$
A.
1
B.
2
C.
3
D.
4
Answer :
3
Solution :
$$t = \frac{1}{{x - 1}}$$ is discontinuous at $$x = 1$$
Also $$y = \frac{1}{{{t^2} + t - 2}}$$ is discontinuous at $$t = - 2$$ and $$t = 1$$
When $$t = - 2,\,\frac{1}{{x - 1}} = - 2 \Rightarrow x = \frac{1}{2}$$
When $$t = 1,\,\frac{1}{{x - 1}} \Rightarrow x = 2$$
So, $$y = f\left( x \right)$$ is discontinuous at three points $$x = 1,\,\frac{1}{2},\,2$$
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 $$x$$
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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-