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 $$\mathop {\lim }\limits_{x \to 0} {\left[ {\frac{{\left( {{a^x} + {b^x}} \right)}}{2}} \right]^{\frac{2}{x}}} = 6$$ is :
A.
$$\frac{1}{3}$$
B.
$$\frac{1}{4}$$
C.
$$\frac{1}{9}$$
D.
$$\frac{2}{9}$$
Answer :
$$\frac{1}{9}$$
Solution :
Given limit
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{a^x} + {b^x}}}{2}} \right)^{\frac{2}{x}}} \cr
& = \mathop {\lim }\limits_{x \to 0} {\left( {1 + \frac{{{a^x} + {b^x} - 2}}{2}} \right)^{\frac{2}{{{a^x} + {b^x} - 2}}\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} - 1 + {b^x} - 1}}{x}} \right)}} \cr
& = {e^{\log \,ab}} \cr
& = ab \cr
& = 6 \cr} $$
Total number of possible ways in which $$a,\,b$$ can take values is $$6 \times 6 = 36.$$
Total possible ways are $$\left( {1,\,6} \right),\,\left( {6,\,1} \right),\,\left( {2,\,3} \right),\,\left( {3,\,2} \right).$$
The total number of possible ways is $$4.$$
Hence, the required probability is $$\frac{4}{{36}} = \frac{1}{9}.$$