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

Given limit,

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

$\begin{array}{c}={\left(\underset{x\to 0}{lim} {\left(1+\frac{a^x+b^x-2}{2}\right)}^{\frac{2}{a^x+b^x-2}}\right)}^{\underset{x\to 0}{lim} \left(\frac{a^x-1+b^x-1}{x}\right)}\\ =e^{\log ab}=ab=6\end{array}$

Total number of possible ways in which a, b can take values is 6 x 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.