Mathematics - Continuity - Differentiability Question with Solution | TestHub

$\mu (x)=\frac{1}{x-1}$, which is discontinous at $x=1$ $$f(u)=\frac{1}{u^2+u-2}=\frac{1}{(u+2)(u-1)},$$ which is discontinous at $u=-2,1$ when $u=-2$, then $\frac{1}{x-1}=-2\Rightarrow x=\frac{1}{2}$ when $u=1$, then $\frac{1}{x-1}=1\Rightarrow x=2$ Hence given composite function is discontinous at three points, $x=1,\frac{1}{2}$ and 2 .