Question

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

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

Answer 1

Answer: (D)

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