Limits, Continuity & Differentiability
Discontinuity of Composite Functions
Grade 12
Question:
<p>The points of discontinuity of \(y = \frac{1}{u^2 + u - 2}\), where \(u = \frac{1}{x-1}\), are:</p>
<p>(a) \(\frac{1}{2}, 1, \frac{3}{2}\)</p>
<p>(b) \(1, \frac{3}{2}, 2\)</p>
<p>(c) \(\frac{1}{2}, 1, 2\)</p>
<p>(d) \(1, 2, 3\)</p>
Step-by-Step Solution
Key Concept: Find points where the inner function is discontinuous and where the outer function becomes undefined.
<p><strong>Step 1:</strong> The function $u = f(x) = \frac{1}{x-1}$ is discontinuous at $x = 1$.</p><p><strong>Step 2:</strong> The function $\frac{1}{u^2 + u - 2} = \frac{1}{(u+2)(u-1)}$ is discontinuous at $u = -2$ and $u = 1$.</p><p><strong>Step 3:</strong> When $u = -2$: $\frac{1}{x-1} = -2 \Rightarrow x = \frac{1}{2}$</p><p><strong>Step 4:</strong> When $u = 1$: $\frac{1}{x-1} = 1 \Rightarrow x = 2$</p><p><strong>Step 5:</strong> The composite function $y = g[f(x)]$ is discontinuous at three points: $x = \frac{1}{2}, 1, 2$</p><p>∴ Answer is (c).</p>
Correct Answer: C