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