Question

The number of real solutions of the equation
$si{n}^{-1}\left(\sum _{i=1}^{\infty }{x}^{i+1}-x\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i\right)=\frac{\pi }{2}-co{s}^{-1}\left(\sum _{i=1}^{\infty }{\left(-\frac{x}{2}\right)}^i-\sum _{i=1}^{\infty }{\left(-x\right)}^i\right)$
lying in the interval $\left(-\frac{1}{2},\frac{1}{2}\right)$ is_____.
(Here, the inverse trigonometric functions $si{n}^{-1}$ x and $co{s}^{-1}$ assume values in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ and $\left[0,\pi \right]$, respectively.)

Answer 1

Answer: 2

$\sum _{i=1}^{\infty }{X}^{i+1}=\frac{X^2}{1-X}$
$\sum _{i=1}^{\infty }{\left(\frac{x}{2}\right)}^i=\frac{x}{2-x}\sum _{i=1}^{\infty }{\left(-\frac{x}{2}\right)}^i$
$=\frac{-x}{2+x}\sum _{i=1}^{\infty }(-x{)}^i=\frac{-x}{1+x}$
To have real solutions
$\sum _{i=1}^{\infty }{X}^{i+1}-X\sum _{i=1}^{\infty }{\left(\frac{X}{2}\right)}^i=\sum _{i=1}^{\infty }{\left(\frac{-X}{2}\right)}^i-\sum _{i=1}^{\infty }(X{)}^i$
$\frac{X^2}{1-X}-\frac{X^2}{2-X}=\frac{-X}{2+X}+\frac{X}{1+X}$
$X\left(X^3+2X^2+5X-2\right)=0$
$\therefore x=0$ and let $f(x)=X^3+2X^2+5X-2$
$f\left(\frac{1}{2}\right)\cdot f\left(-\frac{1}{2}\right)<0$
Hence two solutions exist