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 $\underset{x\to 0}{lim}{\left[(a^x+b^x)/2\right]}^{2/x}=6$ is

• A. $1/3$
• B. $1/4$
• C. $1/9$
• D. $2/9$

Answer 1

The correct option is C $1/9$

Given limit,
$li{m}_{x\to 0}(\frac{a^x+b^x}{2}{)}^{\frac{2}{x}}$
$=li{m}_{x\to 0}(1+\frac{a^x+b^x-2}{2}{)}^{\frac{2}{a^x+b^x}-2Li{m}_{x\to 0}\left(\frac{a^x-1+b^x-1}{x}\right)}$
$=e^{logab}=ab=6$
Total number of possible ways in a,b can take values is $6X6=36$.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 $4/36=1/9.$