Q92 / 200 · Manipal (MET) 2012 · Mathematics

The function $f(x)=\frac{\ln (1+a x)-\ln (1-b x)}{x}$ is not defined at $x=0$. The value which should be assigned to $f$ at $x=0$ so that it is continuous at $x=0$, is

  • A
    $a-b$
  • Correct answer
    B
    $a+b$
  • C
    $\ln a+\ln b$
  • D
    None of these

Solution

For $f(x)$ to be continuous, we must have $\begin{aligned} & f(0)=\lim _{x \rightarrow 0} f(x) \\ \therefore \quad & \lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x} \\ = & \lim _{x \rightarrow 0} \frac{a \log (1+a x)}{a x}+\frac{b \log (1-b x)}{-b x} \\ = & a \cdot 1+b \cdot 1 \quad\left[\text { using, } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right] \\ = & a+b \\ \therefore & f(0)=a+b \end{aligned}$

Hence, the correct answer is option: 'B'.

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