If f (x) is defined by f(x)=1xloge⁡1+3x1−2x,x≠0kx=0 is continuous on −13,13 then the value of k is

Solution:

For $f (x)$ to be continuous at $x = 0$ , we must have

$\begin{array}{l} lim_{x \to 0} f (x) = f (0) \\ \Rightarrow lim_{x \to 0} \frac{1}{x} \log_{e} (\frac{1 + 3 x}{1 - 2 x}) = k \\ \Rightarrow lim_{x \to 0} \{\frac{\log_{e} (1 + 3 x)}{x} - \frac{\log_{e} (1 - 2 x)}{x}\} = k \\ \Rightarrow 3 lim_{x \to 0} \frac{\log_{e} (1 + 3 x)}{3 x} + 2 lim_{x \to 0} \frac{\log_{e} (1 - 2 x)}{- 2 x} = k \\ \Rightarrow 3 \times 1 + 2 \times 1 = k \Rightarrow k = 5 \end{array}$

If $f (x)$ is defined by
$f (x) = \{\begin{array}{cc} \frac{1}{x} \log_{e} (\frac{1 + 3 x}{1 - 2 x}), & x \neq 0 \\ k & x = 0 \end{array}$ is continuous on $(- \frac{1}{3}, \frac{1}{3})$ then the value of $k$ is

Solution:

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