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

• A. $\frac{1}{9}$
• B. $\frac{2}{9}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{1}{9}$

As Limit is of the form $1^{\infty }$
So, $\underset{x\to 0}{lim}f(x{)}^{g\left(x\right)}=e^{\lambda }\text{where}\lambda =\underset{x\to 0}{lim}(f\left(x\right)-1\left)g\right(x)$
$\lambda =\underset{x\to 0}{lim}\left(f\right(x)-1)g\left(x\right)=(\frac{a^x+b^x}{2}-1)\times \frac{2}{x}$
$=\underset{x\to 0}{lim}(\frac{a^x-1}{x}+\frac{b^x-1}{x})$
$=\ln |a|+\ln |b|=\ln |ab|$
Therefore,
$\underset{x\to 0}{lim}{\left(\frac{a^x+b^x}{2}\right)}^{\frac{2}{x}}=e^{\ln |ab|}=ab=6$
Total number of possible ways in 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 $\frac{4}{36}=\frac{1}{9}$