Question-38

How many points of discontinuity does the curve \(\displaystyle y=\frac{1}{u^{2} +u-2}\) have if \(\displaystyle u=\frac{1}{x-1}\).

Answer

\(3\)

Solution

We have \(\displaystyle x=1\) as a point of discontinuity given the way it is defined. Additionally:

\[ \begin{equation*} \begin{aligned} \frac{1}{u^{2} +u-2} & =\frac{1}{( u+2)( u-1)} \end{aligned} \end{equation*} \]

We see that \(\displaystyle u=-2\) and \(\displaystyle u=1\) are two additional points of discontinuity. These correspond to \(\displaystyle x=\frac{1}{2}\) and \(\displaystyle x=2\). Thus the curve has \(\displaystyle 3\) points of discontinuity.