Limits, Continuity & Differentiability
Continuity and Differentiability
Grade 12
Question:
<p><strong>Paragraph for Question nos. 666 and 667</strong><br>Let \(f(x) = (ax^2 + bx + c)\text{sgn}(2\sin x - 1)\) be a continuous function \(\forall x \in (0, 6)\) where \(a, b, c \in R\).<br>[Note: sgn(·) represents signum function.]<br><br>If \(c = 0\), then \((a + b)\) equals:</p>
<p>(a) \(\pi\)</p>
<p>(b) \(\dfrac{\pi}{6}\)</p>
<p>(c) \(\dfrac{5\pi}{6}\)</p>
<p>(d) \(0\)</p>
Step-by-Step Solution
Key Concept: For f(x) to be continuous everywhere on (0,6), the polynomial (ax² + bx + c) must be zero at all points where the signum function has jump discontinuities. With c = 0, the polynomial must have a root at every discontinuity point of sgn(2sin x - 1).
<p><strong>Step 1:</strong> Find where sgn(2sin x - 1) has jump discontinuities.</p><p>sgn(2sin x - 1) jumps when 2sin x - 1 = 0, i.e., sin x = 1/2</p><p>In (0, 6): x = π/6 and x = 5π/6 (within one period)</p><p><strong>Step 2:</strong> For continuity of f(x) = (ax² + bx + c)·sgn(2sin x - 1), the polynomial must vanish at all jump points.</p><p>With c = 0: f(x) = x(ax + b)·sgn(2sin x - 1)</p><p><strong>Step 3:</strong> For continuity at x = π/6 and x = 5π/6:</p><p>• At x = π/6: polynomial = 0 ⟹ (π/6)(aπ/6 + b) = 0 ⟹ aπ/6 + b = 0</p><p>• At x = 5π/6: polynomial = 0 ⟹ (5π/6)(5aπ/6 + b) = 0 ⟹ 5aπ/6 + b = 0</p><p><strong>Step 4:</strong> Solve the system:</p><p>From equation 1: b = -aπ/6</p><p>Substitute into equation 2: 5aπ/6 - aπ/6 = 0 ⟹ 4aπ/6 = 0 ⟹ a = 0</p><p>Therefore: b = 0</p><p><strong>Step 5:</strong> (a + b) = 0 + 0 = <strong>0</strong></p><p>∴ Answer: D</p>
Correct Answer: D