Limits, Continuity & Differentiability
General
Grade None
Question:
<p>Let BC be the diameter of a circle centred at O. Point A is variable on circumference. If
BC = 1, then limA→B
BM
(Area of sector OAB)2 is:</p>
Step-by-Step Solution
Key Concept: General
<p><strong>1</strong>: Let radius r = 1/2. Let \angleAOB = \theta. As A \to B, \theta \to 0.</p><p><strong>2</strong>: In \triangleOAM, OM = r cos \theta =\Rightarrow BM = r -r cos \theta = r(1 -cos \theta) \approxr\theta2</p> 2 .<p><strong>3</strong>: Area of sector OAB = 1</p> 2r2\theta.<p><strong>4</strong>: Ratio = lim\theta\to 0</p> r\theta2/2 (r2\theta/2)2 = lim\theta\to 0 r\theta2/2 r4\theta2/4 = 2 r3 .<p><strong>5</strong>: For r = 1/2, ratio =</p> 2 (1/8) = 16.
Correct Answer: 4