If a and b are chosen randomly from the set consisting of number 1, 2, 3, 4, 5, 6 with replacement. Then the probability that `lim_(x to 0)[(a^(x)+b^(x))//2]^(2//x)=6` is

Options

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

Answer

C

Explanation

Given limit,
`underset(x rarr 0)(lim)((a^(x) + b^(x))/(2))^((2)/(x))`
`=(underset(xrarr0)lim(1+(a^(x) + b^(x) - 2)/(2))^((2)/(a^(x) + b^(x) - 2)))^(underset(xrarr0)lim((a^(x) -1 + b^(2)-1)/(x))`
`= e^(log ab) = ab = 6`