Question

If $f\left(x\right)=\frac{\left(x-1\right)}{4}+\frac{{\left(x-1\right)}^3}{12}+\frac{{\left(x-1\right)}^5}{20}+\frac{{\left(x-1\right)}^7}{28}+......$where $0<x<2$ then $f\text{'}\left(x\right)$ is equal to.

• A. $\frac{1}{4x\left(2-x\right)}$
• B. $\frac{1}{4{\left(x-2\right)}^2}$
• C. $\frac{1}{2-x}$
• D. $\frac{1}{2+x}$

Answer 1

The correct option is A

$\frac{1}{4x\left(2-x\right)}$

Explanation for the correct option:

Finding the value of $f\text{'}\left(x\right)$:

Given that,

$f\left(x\right)=\frac{\left(x-1\right)}{4}+\frac{{\left(x-1\right)}^3}{12}+\frac{{\left(x-1\right)}^5}{20}+\frac{{\left(x-1\right)}^7}{28}+......$

Differentiate the given series with respect to $x$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{4}+\frac{3{\left(x-1\right)}^2}{12}+\frac{5{\left(x-1\right)}^4}{20}+\frac{7{\left(x-1\right)}^6}{28}+......\\ & =& \frac{1}{4}+\frac{{\left(x-1\right)}^2}{4}+\frac{{\left(x-1\right)}^4}{4}+\frac{{\left(x-1\right)}^6}{4}+......\\ & =& \frac{1}{4}\left[1+{\left(x-1\right)}^2+{\left(x-4\right)}^4+{\left(x-6\right)}^2+......\right]\end{array}$

This differentiate series is in GP then,

$a=1$ and $r={\left(x-1\right)}^2$

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{4}\left[\frac{1}{1-{\left(x-1\right)}^2}\right]\left[\because S_n=\frac{a}{1-r}\right]\\ & =& \frac{1}{4}\left[\frac{1}{1-x^2-1+2x}\right]\\ & =& \frac{1}{4}\left[\frac{1}{-x^2+2x}\right]\\ & =& \frac{1}{4x\left(2-x\right)}\end{array}$

Hence, the correct option is A.