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 limx→0 ax+bx/22/x=6 is

Given limit,limx→0 ax+bx22x=limx→0 1+ax+bx−222ax+bx−2limx→0 ax−1+bx−1x=elog⁡ab=ab=6Total 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.