Limits, Continuity & Differentiability
Discontinuity Types
Grade 12
Question:
<p>If $f(x) = \text{sgn}(\cos 2x - 2\sin x + 3)$, where $\text{sgn}()$ is the signum function, then $f(x)$</p>
<p>(a) is continuous over its domain.</p>
<p>(b) has a missing point discontinuity.</p>
<p>(c) has isolated point discontinuity.</p>
<p>(d) has irremovable discontinuity</p>
Step-by-Step Solution
Key Concept: Analyze the sign of the argument to the signum function. If it never equals zero, the signum function is constant and continuous.
<p><strong>Step 1:</strong> Simplify the argument: $\cos 2x - 2\sin x + 3 = 1 - 2\sin^2 x - 2\sin x + 3 = 4 - 2\sin^2 x - 2\sin x$.</p><p><strong>Step 2:</strong> Let $u = \sin x$ where $u \in [-1, 1]$. Then $g(u) = 4 - 2u^2 - 2u = -2(u^2 + u - 2) = -2(u+2)(u-1)$.</p><p><strong>Step 3:</strong> For $u \in [-1, 1]$: $g(u) = -2(u+2)(u-1)$. Since $u + 2 > 0$ always and $(u-1) \leq 0$ for $u \leq 1$, we have $g(u) \geq 0$ for all $u \in [-1, 1]$. In fact, $g(u) > 0$ for all $u \in [-1, 1)$.</p><p><strong>Step 4:</strong> Since $\cos 2x - 2\sin x + 3 > 0$ for all $x$, the signum function yields $f(x) = +1$ everywhere (or nearly everywhere). The function is continuous.</p><p>∴ Answer is (a).</p>
Correct Answer: A