Probability
Probability
Allen Star Batch
Grade 12
Question:
If $a$ and $b$ are chosen randomly from the set consisting of numbers $1, 2, 3, 4, 5, 6$ with replacement. If the probability that $\lim_{x \to 0} \left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}} = 6$ is $\frac{p}{q}$ (where $H.C.F(p, q) = 1$) then $q - p = \ldots\ldots\ldots\ldots\ldots\ldots$
Step-by-Step Solution
Key Concept: L'Hôpital's rule converts the limit into a tractable form revealing that the limit equals the geometric mean of $a$ and $b$.
Using L'Hôpital's rule on the indeterminate form $\lim_{x \to 0} \left(\frac{a^x + b^x}{2}\right)^{2/x}$, we take logarithm to get $\ln k = \lim_{x \to 0} \frac{2}{x}\left(\frac{a^x + b^x}{2} - 1\right)$. Applying L'Hôpital's rule yields $\ln k = a^x \ln a + b^x \ln b$ evaluated, giving $k = ab$. The favorable cases where the product equals $ab$ are: $(6,1), (1,6), (3,3), (2,2)$, giving 4 favorable outcomes out of 36 total.
Correct Answer: 8