Question

Let $f(x)=\frac{x-1}{4}+\frac{(x-1)^3}{12}+\frac{(x-1)^5}{20}+\frac{(x-1)^7}{28} \ldots \ldots \ldots \infty$ 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} \ldots \ldots . . .\right) $
$= \frac{1}{4} \cdot \frac{1}{2} \ln \left(\frac{1+(x-1)}{1-(x-1)}\right)=\frac{1}{8} \ln \left(\frac{x}{2-x}\right)$
$f^{\prime}(x) =\frac{1}{4 x(2-x)} \Rightarrow f^{\prime}\left(\frac{3}{2}\right)=\frac{1}{3}$