Question

I. If $0<y<2^{1/3}$ and $x(y^3-1)=1$, then $\frac{2}{x}+\frac{2}{3x^3}+\frac{2}{5x^5}+\dots \dots \infty =\log \left(\frac{y^3}{2-y^3}\right)$
II. $\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\dots \dots \infty =1-{\log }_{e}2$
Which of the above statements is true?

• A. Only I
• B. Only II
• C. Both I and II
• D. Neither I or II

Answer 1

The correct option is C Both I and II

Given $x(y^3-1)=1$

$\Rightarrow y^3=1+\frac{1}{x}$
Statement 1:
Consider, $\log \left(\frac{y^3}{2-y^3}\right)$

$=\log y^3-\log (2-y^3)$

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

$=(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-.....)-(-\frac{1}{x}-\frac{1}{2x^2}-\frac{1}{3x^3}-...)$

$=\frac{2}{x}+\frac{2}{3x^3}+\frac{2}{5x^5}+....\infty$

Hence, statement 1 is correct.
Statement 2:
${\log }_e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+....\infty$

$\Rightarrow {\log }_{e}2=(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+.....\infty$

$\Rightarrow 1-{\log }_{e}2=\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+....\infty$

Hence, statement 2 is correct.