Probability
Classical probability with limits
Grade 12
Question:
<p>If \(a\) and \(b\) are chosen randomly by throwing a pair of fair dice, then the probability that \(\lim\limits_{x \to 0}\left(\dfrac{a^x + b^x}{2}\right)^{\frac{2}{x}} = 6\) equals:</p>
<p>(a) \(\dfrac{4}{9}\)</p>
<p>(b) \(\dfrac{2}{9}\)</p>
<p>(c) \(\dfrac{3}{9}\)</p>
<p>(d) \(\dfrac{1}{9}\)</p>
Step-by-Step Solution
Key Concept: Recognize that this limit equals the geometric mean of a and b: $\lim_{x \to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{2}{x}} = \sqrt{ab}$. Therefore, we need $\sqrt{ab} = 6$, which means $ab = 36$.
<p><strong>Step 1: Recognize the limit form.</strong></p><p>Let $L = \lim_{x \to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{2}{x}}$</p><p>Taking natural log: $\ln L = \lim_{x \to 0} \frac{2}{x} \ln\left(\frac{a^x + b^x}{2}\right)$</p><p><strong>Step 2: Apply L'Hôpital's Rule.</strong></p><p>Rewrite as: $\ln L = \lim_{x \to 0} \frac{2\ln\left(\frac{a^x + b^x}{2}\right)}{x}$ (form $\frac{0}{0}$)</p><p>Taking derivative: $\ln L = 2\lim_{x \to 0} \frac{a^x \ln a + b^x \ln b}{a^x + b^x} = 2 \cdot \frac{\ln a + \ln b}{2} = \ln(ab)$</p><p><strong>Step 3: Solve for ab.</strong></p><p>Thus $L = ab$. Given $L = 6$, we need $ab = 36$.</p><p><strong>Step 4: Count favorable outcomes.</strong></p><p>Pairs $(a,b)$ from two dice where $ab = 36$: $(6,6)$ only.</p><p>Total outcomes = 36</p><p>Probability = $\frac{1}{36}$</p><p>∴ Answer: B</p>
Correct Answer: B