A) \[a+b\]
B) \[a-b\]
C) \[a\]
D) \[b\]
E) \[b-a\]
Correct Answer: A
Solution :
\[f(x)=\frac{\log (1+ax)-\log (1-bx)}{x}\] \[f(x)\]is continuous at\[x=k\]and\[f(0)=k\]. \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (1+ax)-\log (1-bx)}{x}\] \[\left( \frac{0}{0}form \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{1+ax}.a+\frac{b}{1-bx} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{a}{1+ax}+\frac{b}{1-bx} \right)=a+b\] \[\therefore \] \[a+b=f(0)=k\]You need to login to perform this action.
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