Question

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

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

Answer 1

Answer: (A)

Given, $f(x)=\frac{\log (1+ax)-\log (1-bx)}{x}$
$f(x)$ is continuous at $x=k$ and $f(0)=k$
$\therefore \underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{\log (1+ax)-\log (1-bx)}{x}(\frac{0}{0}$ form $)$
$=\underset{x\to 0}{\lim }\left(\frac{1}{1+ax}\cdot a+\frac{b}{1-bx}\right)=a+b$
$\therefore a+b=f(0)=k$