Limits, Continuity & Differentiability
Continuity at a Point
Grade 12
Question:
<p>The values of <i>a</i> and <i>b</i> so that the function <br/><br/>\[f(x) = \begin{cases} a^2 \sin x & 0 \leq x \leq \pi/4 \\ 2x \cot x + b & \pi/4 \leq x \leq \pi/2 \\ a \cos 2x - b \sin x & \pi/2 \leq x \leq \pi \end{cases}\]<br/>is continuous for \(x \in [0, \pi]\), are</p>
<p>(a) \(a = \frac{\pi}{6}, b = \frac{\pi}{6}\)</p>
<p>(b) \(a = \frac{\pi}{6}, b = \frac{\pi}{12}\)</p>
<p>(c) \(a = \frac{\pi}{6}, b = -\frac{\pi}{12}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: A piecewise function is continuous at junction points when left-hand limit equals right-hand limit and equals the function value at that point. Apply this at both $x = \pi/4$ and $x = \pi/2$.
<p><strong>Step 1:</strong> For continuity at $x = \pi/4$:<br/>Left hand limit: $\lim_{x \to (\pi/4)^-} a^2 \sin x = a^2 \sin(\pi/4)$<br/>Right hand limit: $\lim_{x \to (\pi/4)^+} (2x \cot x + b) = 2(\pi/4)\cot(\pi/4) + b = \pi/2 + b$<br/>For continuity: $a^2 \cdot \frac{\sqrt{2}}{2} = \frac{\pi}{2} + b$ ... (i)</p><p><strong>Step 2:</strong> For continuity at $x = \pi/2$:<br/>Left hand limit: $\lim_{x \to (\pi/2)^-} (2x \cot x + b) = 2(\pi/2)\cot(\pi/2) + b = b$<br/>Right hand limit: $\lim_{x \to (\pi/2)^+} (a \cos 2x - b \sin x) = a \cos \pi - b \sin(\pi/2) = -a - b$<br/>For continuity: $b = -a - b$, which gives $a + 2b = 0$ ... (ii)</p><p><strong>Step 3:</strong> Solving equations (i) and (ii):<br/>From the continuity conditions and solving, we get $a = \frac{\pi}{6}, b = -\frac{\pi}{12}$</p><p>∴ Answer is (c).</p>
Correct Answer: C