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(B) The function $u(x) = \frac{1}{x - 1}$ is discontinuous at $x = 1$.The function $f(u) = \frac{1}{u^2 + u - 2} = \frac{1}{(u + 2)(u - 1)}$ is discon

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$3$

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(B) The function $u(x) = \frac{1}{x - 1}$ is discontinuous at $x = 1$.The function $f(u) = \frac{1}{u^2 + u - 2} = \frac{1}{(u + 2)(u - 1)}$ is discontinuous at $u = -2$ and $u = 1$.For the composite function $f(u(x))$,we must consider the points where $u(x)$ is undefined and where $u(x)$ takes the values that make $f(u)$ undefined.$1$. $u(x)$ is discontinuous at $x = 1$.$2$. When $u(x) = -2$,we have $\frac{1}{x - 1} = -2$,which implies $x - 1 = -\frac{1}{2}$,so $x = \frac{1}{2}$.$3$. When $u(x) = 1$,we have $\frac{1}{x - 1} = 1$,which implies $x - 1 = 1$,so $x = 2$.Thus,the composite function $f(u(x))$ is discontinuous at $x = 1, \frac{1}{2}, 2$. There are $3$ such points.

Let $f$ be a composite function of $x$ defined by $f(u) = \frac{1}{u^2 + u - 2}$ and $u(x) = \frac{1}{x - 1}$. Then the number of points $x$ where $f$ is discontinuous is

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