Q82 / 150 · MHT CET 2012 · Mathematics

If $f(x)=\left\{\begin{array}{cl}\frac{\log (1+2 a x)-\log (1-b x)}{x}, & x \neq 0 \\ k & , x=0\end{array}\right.$ continuous at $x=0$, then value of $k$ is

  • A
    $b+a$
  • B
    $b-2 a$
  • C
    $2 a-b$
  • Correct answer
    D
    $2 a+b$

Solution

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

Hence, the correct answer is option: 'D'.

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