Mathematics · 2019
JEE Main · 8 April 2019, Shift 1 · Q61
If f(x)= log_e ((1-x)/(1+x)),|x|<1, then f((2 x)/(1+x^2)) is equal to:
If $\displaystyle f(x)=\log _{\mathrm{e}}\left(\frac{1-x}{1+x}\right),|x|<1$, then $\displaystyle f\left(\frac{2 x}{1+x^2}\right)$ is equal to :
Official answer
(3)
$\displaystyle 2 f(x)$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.