A) \[a=1\]
B) \[a=0\]
C) \[a=e\]
D) \[a=\frac{1}{e}\]
Correct Answer: A
Solution :
Given, \[\underset{x\to a}{\mathop{\lim }}\,\frac{{{a}^{x}}-{{x}^{a}}}{{{x}^{x}}-{{a}^{a}}}=-1\] \[\left( \text{form}\frac{0}{0} \right)\] Using L' Hospital's rule \[\underset{x\to a}{\mathop{\lim }}\,\frac{{{a}^{x}}\log a-a{{x}^{a-1}}}{{{x}^{x}}(1+\log x)}=-1\] \[\Rightarrow \] \[\frac{{{a}^{a}}\log a-{{a}^{a}}}{{{a}^{a}}(1+\log a)}=-1\] \[\Rightarrow \] \[\frac{\log a-1}{1+\log a}=-1\] \[\Rightarrow \] \[\log a=0\Rightarrow a=1\]
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