Question

If $ f(x) = \begin{cases} \frac{log\,\left(1+2ax\right)-log\,\left(1-bx\right)}{x} &, \text{ $ x \ne 0 $ } \\[2ex] k &, \text{ $ x = 0 $ } \end{cases} $
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{cases} \frac{log(1+2ax)-log (1-bx)}{x}, & \quad \text{ } x \neq 0 \\ k & \quad \text{ } x=0 \\ \end{cases} $
is continuous at x = 0
$\therefore f(0)=\displaystyle\lim _{x \rightarrow 0} \frac{\log (1+2 a x)-\log (1-b x)}{x}$
$\left(\frac{0}{0}\right.$ form $)$
$\Rightarrow k=\displaystyle\lim _{x \rightarrow 0} \frac{\frac{1}{2 a x+1}(2 a)-\frac{1}{1-b x}(-b)}{+1}$
(by ‘L’ Hospital’s rule)
$\Rightarrow k=\frac{2a}{0+1}+\frac{b}{1-0}=2 a+b$