Probability Question 365
Question: If a and b are chosen randomly from the set consisting of numbers 1, 2, 3, 4, 5, 6, with replacement. Then the probability that $ \underset{x\to 0}{\mathop{\lim }}{{[(a^{x}+b^{x})/2]}^{2/x}}=6 $ is
Options:
A) 1/3
B) 1/4
C) 1/9
D) 2/9
Show Answer
Answer:
Correct Answer: C
Solution:
Given limit, $ \underset{x\to 0}{\mathop{\lim }}{{( \frac{a^{x}+b^{x}}{2} )}^{\frac{2}{x}}} $
$ \underset{x\to 0}{\mathop{\lim }}{{( 1+\frac{a^{x}+b^{x}-2}{2} )}^{\frac{2}{a^{x}+b^{x}-2}\underset{x\to 0}{\mathop{\lim }}( \frac{a^{x}-1+b^{x}-1}{x} )}} $
$ ={e^{\log ab}}=ab=6. $
Total number of possible ways in which a, b can take values is $ 6\times 6=36. $
Total possible ways are $ (1,6),(6,1),(2,3),(3,2). $
The total number of possible ways is 4.
Hence, the required probability is $ 4/36=1/9. $
BETA
