A) 1/3
B) 1/4
C) 1/9
D) 2/9
Correct Answer: C
Solution :
[c] Given limit, \[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{{{a}^{x}}+{{b}^{x}}}{2} \right)}^{\frac{2}{x}}}\] \[\underset{x\to 0}{\mathop{\lim }}\,{{\left( 1+\frac{{{a}^{x}}+{{b}^{x}}-2}{2} \right)}^{\frac{2}{{{a}^{x}}+{{b}^{x}}-2}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{a}^{x}}-1+{{b}^{x}}-1}{x} \right)}}\] \[={{e}^{\log ab}}=ab=6.\] 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 \[4/36=1/9.\]
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