A) \[\frac{1}{4x(2-x)}\]
B) \[\frac{1}{4{{(x-2)}^{2}}}\]
C) \[\frac{1}{2-x}\]
D) \[\frac{1}{2+x}\]
E) \[\frac{1}{{{(x-4)}^{2}}}\]
Correct Answer: A
Solution :
Given, \[f(x)=\frac{x-1}{4}+\frac{{{(x-1)}^{3}}}{4.3}+\frac{{{(x-1)}^{2}}}{4.5}\]\[+\frac{{{(x-1)}^{7}}}{4.7}+....\] \[=\frac{1}{2}\left[ \frac{x-1}{1}+\frac{{{(x-1)}^{3}}}{3}+\frac{{{(x-1)}^{5}}}{4.5} \right.\]\[\left. +\frac{{{(x-1)}^{7}}}{7}+.... \right]\] \[=\frac{1}{2}\left[ \frac{1}{2}\log \left( \frac{1+(x-1)}{1-(x-1)} \right) \right]\] \[\Rightarrow \] \[f(x)=\frac{1}{8}\log \left( \frac{x}{2-x} \right)\] On differentiating w.r.t.\[~x,\]we get \[f(x)=\frac{1}{8}\times \,\frac{1}{\left( \frac{x}{2-x} \right)}\left[ \frac{(2-x)1-x(-1)}{{{(2-x)}^{2}}} \right]\] \[=\frac{2}{8x(2-x)}\,=\frac{1}{4x(2-x)}\]
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