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

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

Answer 1

Answer: (C)

Given limit, $\underset{x\to 0}{\lim }{\left(\frac{a^x+b^x}{2}\right)}^{\frac{2}{x}}$
$=\underset{x\to 0}{\lim }{\left(1+\frac{a^x+b^x-2}{2}\right)}^{\frac{2}{a^x+b^x-2}\underset{x\to 0}{\lim }\left(\frac{a^x-1+b^x-1}{x}\right)}$
$=e^{\log ab}=ab=6$.
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 $4/36=1/9$.