Question

If $f\left(x\right)=\frac{\log (1+ax)-log(1-bx)}{x}$ for $x\ne 0$ and $f\left(0\right)=k$ and $f\left(x\right)$ is continuous at $x=0$, then $k$ is equal to:

• A. $a+b$
• B. $a-b$
• C. $a$
• D. $b$

Answer 1

The correct option is A $a+b$

Given, $f$ is continuous at $x=0$ and $f\left(0\right)=k$, then

$f\left(0\right)=\underset{x\to 0}{lim}\frac{\log (1+ax)-\log (1-bx)}{x}$

$=\underset{x\to 0}{lim}(a\cdot \frac{\log (1+ax)}{ax}+b\cdot \frac{\log (1-bx)}{-bx})=a+b=k$

$\therefore k=a+b$