The function $\mathrm{f}(\mathrm{t})=\frac{1}{\mathrm{t}^2+\mathrm{t}-2}$ where $\mathrm{t}=\frac{1}{x-1}$…
The function $\mathrm{f}(\mathrm{t})=\frac{1}{\mathrm{t}^2+\mathrm{t}-2}$ where $\mathrm{t}=\frac{1}{x-1}$ is discontinuous at
$-2,1$
$2, \frac{1}{2}$
$\frac{1}{2}, 1$
2,1
Solution
$f(t)=\frac{1}{t^2+t-2}=\frac{1}{(t+2)(t-1)}$
$\mathrm{f}(\mathrm{t})$ is not defined at $\mathrm{t}=-2$ and $\mathrm{t}=1$.
$\begin{aligned}
& \mathrm{t}=-2 \\
& \Rightarrow \frac{1}{x-1}=-2 \\
& \Rightarrow x=\frac{1}{2} \\
& \mathrm{t}=1 \\
& \Rightarrow \frac{1}{x-1}=1 \\
& \Rightarrow x=2
\end{aligned}$
$\therefore \quad$ The function $\mathrm{f}(\mathrm{t})$ is discontinuous at $x=\frac{1}{2}$ and $x=2$.