A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{5}\]
D) \[\frac{1}{3}\]
Correct Answer: D
Solution :
\[f(x)=\frac{1}{4}\left( (x-1)+\frac{{{(x-1)}^{3}}}{3}+\frac{{{(x-1)}^{5}}}{5}+\frac{{{(x-1)}^{7}}}{7}....... \right)\]\[=\frac{1}{4}\cdot \frac{1}{2}\ln \left( \frac{1+(x-1)}{1-(x-1)} \right)=\frac{1}{8}\ln \left( \frac{x}{2-x} \right)\] \[f'(x)=\frac{1}{1x(2-x)}\Rightarrow f'\left( \frac{3}{2} \right)=\frac{1}{3}\]
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