Question

If $f(x)=x-{{x}^{2}}+{{x}^{3}}-{{x}^{4}}+...to~\infty for \left| x \right|<1$, then ${{f}^{-1}}(x)$=
A. $\dfrac{x}{1+x}$
B. $\dfrac{x}{1-x}$
C. $\dfrac{1-x}{x}$
D. $\dfrac{1}{x}$

Answer 1

Hint: In this case, we are provided with an infinite geometric progression. First, determine the first term and the common difference of the G.P. If the absolute value of its common ratio is smaller than 1 can its sum be determined using the formula of the sum of an infinite geometric series when $\left| x \right|<1$. Do possible substitution and simplification you will get the result.

Formula Used: The sum of an infinite geometric progression is given by
$S_\infty=\dfrac{a}{1-r}$
Where a is the first term and r is the common ratio.

Complete step by step solution: We have given $f(x)=x-{{x}^{2}}+{{x}^{3}}-{{x}^{4}}+...to~\infty$ which is an infinite G.P.
Therefore,
the first term is $a=x$ and,
the common ratio $r= \dfrac{a_1}{a}$
$\rightarrow r=-\dfrac{x^2}{x}\\
\rightarrow r=-x$
The sum of an infinite geometric progression is given by
$S_\infty=\dfrac{a}{1-r}$
Thus,
$\Rightarrow y=\dfrac{x}{1-(-x)} \\
\Rightarrow y=\dfrac{x}{1+x} \\
\Rightarrow y(1+x)=x \\
\Rightarrow y+y x=x \\
\Rightarrow y=x-y x \\
\Rightarrow y=x(1-y) \\
\Rightarrow x=\dfrac{y}{1-y}$
Now,
$\Rightarrow {{f}^{-1}}(y)=\dfrac{y}{1-y}\\
\Rightarrow {{f}^{1}}(x)=\dfrac{x}{1-x}$

So, option B is correct.


Note: In an infinite series, there would be no last term. An infinite geometric series' ability to converge depends on the size of its common ratio. A set of infinite geometric series when $|r| < 1$, it converges, and we may use the formula $\dfrac{a}{1-r}$ to find its sum otherwise it diverges when $|r| > 1$, hence in this situation we are unable to find its total. Remember if the geometric series is finite, we can always determine its sum. But if it is infinite, only when the absolute value of its common ratio is smaller than 1 can its sum be determined.