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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
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continuous at $x=0$, then value of $k$ is
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The correct answer is:
$2 a+b$
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) $\Rightarrow k=\frac{2 a}{0+1}+\frac{b}{1-0}=2 a+b$
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) $\Rightarrow k=\frac{2 a}{0+1}+\frac{b}{1-0}=2 a+b$
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