Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\log (1+2ax)-\log (1-bx)}{x},& x\ne 0\\ k,& x=0\end{array}$ is continuous at $x=0$, then the value of $k$ is

• A. $b+a$
• B. $b-2a$
• C. $2a-b$
• D. $2a+b$

Answer 1

The correct option is D $2a+b$

Given, $f\left(x\right)=\{\begin{array}{cc}\frac{\log (1+2ax)-\log (1-bx)}{x},& x\ne 0\\ k,& x=0\end{array}$ is continuous at $x=0$.
$\therefore f\left(0\right)=\underset{x\to 0}{lim}\frac{\log (1+2ax)-\log (1-bx)}{x}\left(\frac{0}{0}form\right)$
$\Rightarrow k=\underset{x\to 0}{lim}\frac{\frac{1}{2ax+1}\left(2a\right)-\frac{1}{1-bx}(-b)}{+1}$ (by 'L' Hospital's rule)
$\Rightarrow k=\frac{2a}{0+1}+\frac{b}{1-0}=2a+b$