Mathematics · 2020
JEE Main · 7 January 2020, Shift 2 · Q73
If the function f defined on (-1/3, 1/3) by f(x)= {1/x log_e ((1+3 x)/(1-2 x)), when x ≠ 0; k, when x=0} is continuous, then k is equal to \ \ \ \.
If the function $\displaystyle f$ defined on $\displaystyle \left(-\frac{1}{3}, \frac{1}{3}\right)$ by
$$f(x)= \begin{cases}\frac{1}{x} \log _{\mathrm{e}}\left(\frac{1+3 x}{1-2 x}\right), & \text { when } x \neq 0 \\ \mathrm{k}, & \text { when } x=0\end{cases}
$$
is continuous, then k is equal to \_\_\_\_.
Official answer
5
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.