A) \[a-b\]
B) \[a+b\]
C) \[a-b\]
D) None of these
Correct Answer: B
Solution :
For\[f(x)\]to be continuous, we must have \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,f(x)\] \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (1+ax)-\log (1-bx)}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{a\log (1+ax)}{ax}+\frac{b\log (1-bx)}{-bx}\] \[=a.1+b.1\,\left[ \text{using}\,\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\log (1+x)}{x}=1 \right]\] \[=a+b\] \[\therefore \,f(0)\,=a+b\]
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