Question

If $0<y<2^{\frac{1}{3}}$ and $x\left(y^3-1\right)=1$ then $\frac{2}{x}+\frac{2}{3x^3}+\frac{2}{5x^5}+.........$is equal to

• A. $\log \left(\frac{y^3}{2-y^3}\right)$
• B. $\log \left(\frac{y^3}{1-y^3}\right)$
• C. $\log \left(\frac{2y^3}{1-y^3}\right)$
• D. $\log \left(\frac{2y^3}{1-2y^3}\right)$

Answer 1

The correct option is A

$\log \left(\frac{y^3}{2-y^3}\right)$

Explanation for correct option

Step-1: Simplify the given data.

Given, $x\left(y^3-1\right)=1$

$\Rightarrow$ $x=\frac{1}{y^3-1}$

Step-2 Apply componendo and Dividendo $\frac{a}{b}=\frac{c}{d}\iff \frac{a+b}{a-b}=\frac{c+d}{c-d}$.

$\Rightarrow$ $\frac{x+1}{x-1}=\frac{1+y^3-1}{1-y^3+1}$

$\Rightarrow$ $\frac{x+1}{x-1}=\frac{y^3}{2-y^3}......\left(i\right)$

Step-3: Apply formula: $\mathit{l}\mathit{o}\mathit{g}\left(\frac{1+x}{1-x}\right)\mathbf{=}\mathbf{2}\left[x+\frac{x^3}{3}+\frac{x^5}{5}+.......\right]$

Given, $\frac{2}{x}+\frac{2}{3x^3}+\frac{2}{5x^5}+.........$

$=$$2\left[\frac{1}{x}+\frac{1}{3x^3}+\frac{1}{5x^5}+.........\right]$

$=$$\log \left(\frac{1+{\displaystyle \frac{1}{x}}}{1-{\displaystyle \frac{1}{x}}}\right)$

$=$$\log \left(\frac{x+1}{x-1}\right)$

$=$$\log \left(\frac{y^3}{2-y^3}\right)$(from equation i )

Hence, correct answer is option $\left(A\right)$.