Question

Let $f\left(x\right)=\frac{x-1}{4}+\frac{{(x-1)}^3}{12}+\frac{{(x-1)}^5}{20}+\frac{{(x-1)}^7}{28}+.....\infty$ for $x\in (0,2)$, then $f^{\prime }\left(\frac{3}{2}\right)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{1}{3}$

Answer 1

Answer: (D)

$\frac{1}{3}$

$f\left(x\right)=\frac{1}{4}((x-1)+\frac{{(x-1)}^3}{3}+\frac{{(x-1)}^5}{5}+\frac{{(x-1)}^7}{7}+.....)=\frac{1}{4}.\frac{1}{2}\ln \left(\frac{1+(x-1)}{1-(x-1)}\right)=\frac{1}{8}\ln \left(\frac{x}{2-x}\right)$

$f^{\prime }\left(x\right)=\frac{1}{4\times (2-x)}\Rightarrow f^{\prime }\left(\frac{3}{2}\right)=\frac{1}{3}$