Limits, Continuity & Differentiability
Continuity of functions
Grade 12
Question:
<p>If <span>\(f(x) = \begin{cases} \dfrac{\sin(p+1)x + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \dfrac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}\)</span> is continuous at <span>\(x = 0\)</span>, then find the values of <span>\(p\)</span> and <span>\(q\)</span>.</p>
<p>\(p = \dfrac{3}{2},\ q = \dfrac{1}{2}\)</p>
<p>\(p = -\dfrac{3}{2},\ q = \dfrac{1}{2}\)</p>
<p>\(p = \dfrac{1}{2},\ q = \dfrac{3}{2}\)</p>
<p>\(p = -\dfrac{1}{2},\ q = \dfrac{3}{2}\)</p>
Step-by-Step Solution
Key Concept: For a piecewise function to be continuous at a point, the limit from both sides must equal the function value at that point. Use L'Hôpital's rule or the standard limit sin(u)/u → 1 to evaluate limits of the numerator expressions.
<p><strong>Step 1: Find lim(x→0⁻) f(x)</strong></p><p>For x < 0: f(x) = [sin(p+1)x + sin x]/x</p><p>lim(x→0⁻) [sin(p+1)x + sin x]/x = lim(x→0⁻) [sin(p+1)x]/x + sin x/x</p><p>= (p+1)·lim(u→0) sin u/u + lim(v→0) sin v/v = (p+1)·1 + 1 = p+2</p><p><strong>Step 2: Find lim(x→0⁺) f(x)</strong></p><p>For x > 0: f(x) = (2p sin qx)/x</p><p>lim(x→0⁺) (2p sin qx)/x = 2p·q·lim(w→0) sin w/w = 2pq</p><p><strong>Step 3: Apply continuity at x = 0</strong></p><p>For f to be continuous at x = 0:</p><p>lim(x→0⁻) f(x) = f(0) = lim(x→0⁺) f(x)</p><p>p + 2 = q = 2pq</p><p><strong>Step 4: Solve the system</strong></p><p>From p + 2 = q: substitute into q = 2pq</p><p>p + 2 = 2p(p+2)</p><p>p + 2 = 2p² + 4p</p><p>2p² + 3p - 2 = 0</p><p>(2p - 1)(p + 2) = 0</p><p>p = 1/2 or p = -2</p><p>If p = 1/2: q = 1/2 + 2 = 5/2 ✓</p><p>If p = -2: q = 0, which makes the right side 0 (check: 2(-2)(0) = 0 ✓)</p><p>∴ <strong>p = 1/2, q = 5/2</strong> or <strong>p = -2, q = 0</strong></p>
Correct Answer: B