Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Given
\[f(x) = \begin{cases} \dfrac{\sin(p+1)x + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \dfrac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}\]
If \(f(x)\) is continuous at \(x = 0\), find the values of \(p\) and \(q\).</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 f(x) to be continuous at x=0, the limit as x→0 must equal f(0)=q. Use the standard limit lim(x→0) sin(x)/x = 1 and expand sin(p+1)x using the angle addition property to find the limit, then equate to q.
<p><strong>Step 1:</strong> For continuity at x=0, we need lim(x→0) f(x) = f(0) = q</p><p><strong>Step 2:</strong> Calculate the limit as x→0:</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><strong>Step 3:</strong> Apply standard limits and expand sin(p+1)x using lim(x→0) sin(ax)/x = a:</p><p>= (p+1)·lim(x→0)[sin(p+1)x/(p+1)x] + lim(x→0)[sin x/x]</p><p>= (p+1)·1 + 1 = p + 2</p><p><strong>Step 4:</strong> From continuity condition:</p><p>p + 2 = q</p><p><strong>Step 5:</strong> Since we need a complete answer with two variables, the standard assumption is p = 0 (simplest case for well-defined piecewise function):</p><p><strong>∴ Answer: p = 0 and q = 2 (or general relation q = p + 2)</strong></p>
Correct Answer: A