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$If \sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2} for 0 < | x | < \sqrt{2},$ then 100x equals

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Solution

$Since \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} for | x | \leq 1 . ∴ x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots = x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots \Rightarrow \frac{x}{1 + \frac{x}{2}} = \frac{x^{2}}{1 + \frac{x^{2}}{2}} (0 < | x | < \sqrt{2}) \Rightarrow \frac{x}{2 + x} = \frac{x^{2}}{2 + x^{2}} \Rightarrow 2 x + x^{3} = 2 x^{2} + x^{3}$
⇒ x² = x ⇒ x = 0, 1
But x ≠ 0, ∴ x = 1.

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