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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Given limit,
${lim}_{x\to 0}{\left[\frac{(a^x+b^x)}{2}\right]}^{\frac{2}{x}}={lim}_{x\to 0}{[1+\frac{a^x+b^x-2}{2}]}^{\frac{2}{a^x+b^x-2}{lim}_{x\to 0}\left(\frac{a^x-1+b^x-1}{x}\right)}=e^{logab}=ab=6,requiredprobability=\frac{4}{36}$