Question

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

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

Answer 1

Answer: (A)

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