Q76 / 170 · Manipal (MET) 2018 · Mathematics

If $f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}$ for $x \neq 0$ and $f(0)=k$ and $f(x)$ is continuous at $x=0$, then $k$ is equal to

  • Correct answer
    A
    a + b
  • B
    a - b
  • C
    a
  • D
    b

Solution

Given, $f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}$ $\begin{aligned} & \quad f(x) \text { is continuous at } x=k \text { and } f(0)=k . \\ & \therefore \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x} \\ & \quad \quad\left(\frac{0}{0} \text { form }\right) \\ & \quad \lim _{x \rightarrow 0}\left(\frac{1}{1+a x} \cdot a+\frac{b}{1-b x}\right)=a+b \\ & \therefore \quad a+b=f(0)=k \end{aligned}$

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

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