Question

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

Answer 1

Hint: At first, we differentiate on both sides with respect to x. Then, we take $\dfrac{1}{4}$ common from each term. The resulting series becomes a geometric series with a common ratio ${{\left( x-1 \right)}^{2}}$ . We then use the formula for an infinite geometric series with first term a and common ratio r as $S=\dfrac{a}{1-r}$ . We then simplify the expression.

Complete step by step answer:
The given function that we have in this problem is,
$f\left( x \right)=\dfrac{x-1}{4}+\dfrac{{{\left( x-1 \right)}^{3}}}{12}+\dfrac{{{\left( x-1 \right)}^{5}}}{20}+\dfrac{{{\left( x-1 \right)}^{7}}}{28}+...$
Taking derivatives on both sides with respect to $x$ , we get,
$\Rightarrow f'\left( x \right)=\dfrac{1}{4}+\dfrac{3{{\left( x-1 \right)}^{2}}}{12}+\dfrac{5{{\left( x-1 \right)}^{4}}}{20}+\dfrac{7{{\left( x-1 \right)}^{6}}}{28}+...$
Simplifying the above expression, we get,
$\Rightarrow f'\left( x \right)=\dfrac{1}{4}+\dfrac{{{\left( x-1 \right)}^{2}}}{4}+\dfrac{{{\left( x-1 \right)}^{4}}}{4}+\dfrac{{{\left( x-1 \right)}^{6}}}{4}+...$
We can see a $\dfrac{1}{4}$ term multiplied with each term of the series. This means that we can take $\dfrac{1}{4}$ from the terms and get,
$\Rightarrow f'\left( x \right)=\dfrac{1}{4}\left[ 1+{{\left( x-1 \right)}^{2}}+{{\left( x-1 \right)}^{4}}+{{\left( x-1 \right)}^{6}}+... \right]$
If we observe the series inside the bracket, we can see that it looks as if an expression has been multiplied with each term to get the next term. This means the series within the brackets is geometric. The common ration of the series will be,
$\dfrac{{{\left( x-1 \right)}^{4}}}{{{\left( x-1 \right)}^{2}}}={{\left( x-1 \right)}^{2}}$
Again, it is given that,
$\begin{align}
  & 0 < x < 2 \\
 & \Rightarrow -1 < x-1 < 1 \\
\end{align}$
We know that the sum of an infinite geometric series with common ratio less than $1$ is, $S=\dfrac{a}{1-r}$ where a is the first term and r is the common ratio. The sum will be,
$\Rightarrow S=\dfrac{1}{1-{{\left( x-1 \right)}^{2}}}=\dfrac{1}{\left( 1+x-1 \right)\left( 1-x+1 \right)}=\dfrac{1}{x\left( 2-x \right)}$
$f'\left( x \right)$ will become,
$\Rightarrow f'\left( x \right)=\dfrac{1}{4}\left[ \dfrac{1}{x\left( 2-x \right)} \right]=\dfrac{1}{4x\left( 2-x \right)}$
Thus, we can conclude that the value of $f'\left( x \right)$ will be $\dfrac{1}{4x\left( 2-x \right)}$

So, the correct answer is “Option A”.

Note: We can also solve the problem in another way. We know that the binomial expansion for ${{\left( 1-x \right)}^{-1}}$ is given as,
${{\left( 1-x \right)}^{-1}}=1+x+{{x}^{2}}+{{x}^{3}}+...$
If we put $x$ as ${{\left( x-1 \right)}^{2}}$ , we get,
$\begin{align}
  & \Rightarrow {{\left( 1-\left( {{\left( x-1 \right)}^{2}} \right) \right)}^{-1}}=1+{{\left( x-1 \right)}^{2}}+{{\left( x-1 \right)}^{4}}+{{\left( x-1 \right)}^{6}}+... \\
 & \Rightarrow \dfrac{1}{\left( 2x-{{x}^{2}} \right)}=1+{{\left( x-1 \right)}^{2}}+{{\left( x-1 \right)}^{4}}+{{\left( x-1 \right)}^{6}}+... \\
\end{align}$
Thus, $f'\left( x \right)$ can be written as,
$\Rightarrow f'\left( x \right)=\dfrac{1}{4}\left[ \dfrac{1}{\left( 2x-{{x}^{2}} \right)} \right]=\dfrac{1}{4x\left( 2-x \right)}$