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

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

Answer 1

Answer: (C)

Given limit, $\lim\limits _{x \rightarrow 0}\left(\frac{a^{x}+b^{x}}{2}\right)^{\frac{2}{x}}$
$=\lim\limits _{x \rightarrow 0}\left(1+\frac{a^{x}+b^{x}-2}{2}\right)^{\frac{2}{a^{x}+b^{x}-2} \lim\limits _{x \rightarrow 0}\left(\frac{a^{x}-1+b^{x}-1}{x}\right)}$
$=e^{\log a b}=a b=6$.
Total number of possible ways in which $a, b$ 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 $4 / 36=1 / 9$.