Continuity
General
Grade 12
Question:
If f(x) = <span class="math-inline">\(\frac{1}{(x-1)(x-2)}\)</span> and g(x) = <span class="math-inline">\(\frac{1}{x^2}\)</span>, then set of points in domain of fog(x) at which fog(x) is discontinuous.
<span class="math-inline">\{ -1, 0, 1, \(\frac{1}{\sqrt{2}}\) \}</span>
<span class="math-inline">∅</span>
<span class="math-inline">\{0, 1\}</span>
<span class="math-inline">\{0, 1, \(\frac{1}{\sqrt{2}}\)\}</span>
Step-by-Step Solution
Key Concept: To find discontinuities of fog(x), we must first determine its domain (where g(x) is defined AND f(g(x)) is defined), then check continuity only at points within that domain. A composite function is continuous wherever it's defined if both component functions are continuous at their respective points.
<p><strong>Step 1: Find the domain of g(x)</strong></p><p>g(x) = 1/x² is defined for all x ≠ 0</p><p><strong>Step 2: Find where f(g(x)) is defined</strong></p><p>We need f(g(x)) = 1/((1/x² - 1)(1/x² - 2)) to be defined.</p><p>This requires:</p><p>• g(x) ≠ 1, so 1/x² ≠ 1 ⟹ x² ≠ 1 ⟹ x ≠ ±1</p><p>• g(x) ≠ 2, so 1/x² ≠ 2 ⟹ x² ≠ 1/2 ⟹ x ≠ ±1/√2</p><p><strong>Step 3: Determine domain of fog(x)</strong></p><p>Domain of fog(x) = {x ∈ ℝ : x ≠ 0, x ≠ ±1, x ≠ ±1/√2}</p><p><strong>Step 4: Check continuity within the domain</strong></p><p>Both f and g are rational functions. At any point in the domain of fog(x):</p><p>• g is continuous (as x ≠ 0)</p><p>• f is continuous at g(x) (as g(x) ≠ 1 and g(x) ≠ 2)</p><p>Therefore, fog(x) = f(g(x)) is continuous at every point in its domain (by the composition theorem for continuous functions).</p><p><strong>Step 5: Conclusion</strong></p><p>There are no points in the domain of fog(x) where it is discontinuous. The points where f or g are undefined are simply excluded from the domain—they don't represent discontinuities.</p><p><strong>∴ Answer: B (empty set)</strong></p>
Correct Answer: B