Question

Let $f(x)=\frac{x-1}{4}+\frac{(x-1{)}^3}{12}+\frac{(x-1{)}^5}{20}+\frac{(x-1{)}^7}{28}\dots \dots \dots \infty$ 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)

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