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(C) Using the formula for the sum of an infinite geometric series $\sum_{i=1}^{\infty} ar^{i-1} = \frac{a}{1-r}$ for $|r| < 1$: $\sum_{i=1}^{\infty} x

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$2$

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(C) Using the formula for the sum of an infinite geometric series $\sum_{i=1}^{\infty} ar^{i-1} = \frac{a}{1-r}$ for $|r| $\sum_{i=1}^{\infty} x^{i+1} = x^2 + x^3 + \dots = \frac{x^2}{1-x}$ $\sum_{i=1}^{\infty} (\frac{x}{2})^i = \frac{x/2}{1-x/2} = \frac{x}{2-x}$ $\sum_{i=1}^{\infty} (-\frac{x}{2})^i = \frac{-x/2}{1+x/2} = \frac{-x}{2+x}$ $\sum_{i=1}^{\infty} (-x)^i = \frac{-x}{1+x}$ Given $\sin^{-1}(A) = \frac{\pi}{2} - \cos^{-1}(B)$,we know $\sin^{-1}(A) + \cos^{-1}(B) = \frac{\pi}{2}$. Since $\sin^{-1}(B) + \cos^{-1}(B) = \frac{\pi}{2}$,this implies $A = B$. $\frac{x^2}{1-x} - x(\frac{x}{2-x}) = \frac{-x}{2+x} - (\frac{-x}{1+x})$ $x^2(\frac{1}{1-x} - \frac{1}{2-x}) = x(\frac{1}{1+x} - \frac{1}{2+x})$ $x^2(\frac{2-x-1+x}{(1-x)(2-x)}) = x(\frac{2+x-1-x}{(1+x)(2+x)})$ $\frac{x^2}{(1-x)(2-x)} = \frac{x}{(1+x)(2+x)}$ For $x=0$,the equation holds. For $x 
eq 0$,$x(1+x)(2+x) = (1-x)(2-x) \implies x(x^2+3x+2) = x^2-3x+2 \implies x^3+3x^2+2x = x^2-3x+2 \implies x^3+2x^2+5x-2=0$. Let $f(x) = x^3+2x^2+5x-2$. $f(0) = -2$ and $f(1/2) = 1/8 + 2/4 + 5/2 - 2 = 1.125 > 0$. Since $f'(x) = 3x^2+4x+5$,the discriminant $D = 16 - 60 Thus,there is exactly one root in $(0, 1/2)$. Including $x=0$,there are $2$ solutions.

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