The number of real solutions of the equation lying in the interval is ______. (Here, the inverse trigonometric functions and assume values in and respectively.) [2018]

Question

The number of real solutions of the equation $\sin^{- 1} (\sum_{i = 1}^{\infty} x^{i + 1} - x \sum_{i = 1}^{\infty} {(\frac{x}{2})}^{i}) = \frac{π}{2} - \cos^{- 1} (\sum_{i = 1}^{\infty} {(- \frac{x}{2})}^{i} - \sum_{i = 1}^{\infty} {(- x)}^{i})$ lying in the interval $(- \frac{1}{2}, \frac{1}{2})$ is ______.

(Here, the inverse trigonometric functions $\sin^{- 1} x$ and $\cos^{- 1} x$ assume values in $[- \frac{π}{2}, \frac{π}{2}]$ and $[0, π],$ respectively.) [2018]

Smart Achievers · Accepted Answer

(2)

$\sin^{- 1} (\sum_{i = 1}^{\infty} x^{i + 1} - {x \sum}_{i = 1}^{\infty} {(\frac{x}{2})}^{i})$

$= \frac{π}{2} - \cos^{- 1} [\sum_{i = 1}^{\infty} {(- \frac{x}{2})}^{i} - \sum_{i = 1}^{\infty} {(- x)}^{i}]$

$= \sin^{- 1} (\frac{x^{2}}{1 - x} - x \cdot \frac{\frac{x}{2}}{1 - \frac{x}{2}}) = \sin^{- 1} (\frac{- \frac{x}{2}}{1 + \frac{x}{2}} - \frac{- x}{1 + x})$

$[∵ Sum of infinite terms of a G.P. = \frac{a}{1 - r}, if | r | < 1]$

$\Rightarrow \frac{x^{2}}{1 - x} - \frac{x^{2}}{2 - x} = - \frac{x}{2 + x} + \frac{x}{1 + x}$

$\Rightarrow \frac{x^{2}}{1 - x} - \frac{x}{1 + x} + \frac{x}{2 + x} - \frac{x^{2}}{2 - x} = 0$

$\Rightarrow \frac{x (x^{2} + 2 x - 1)}{1 - x^{2}} + \frac{x (2 - 3 x - x^{2})}{4 - x^{2}} = 0$

$\Rightarrow x (x^{3} + 2 x^{2} + 5 x - 2) = 0$

$\Rightarrow x = 0 or x^{3} + 2 x^{2} + 5 x - 2 = 0 = p (x) (say)$

$We observe that p (0) < 0 and p (\frac{1}{2}) > 0$

$∴$ $One root of p (x) = 0 lies in (0, \frac{1}{2})$

$∴$ $Two solutions lie between - \frac{1}{2} and \frac{1}{2}$

Solution

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Solution

Q. The number of real solutions of the equation sin-1(∑i=1∞x i+1-x∑i=1∞(x2)i)=π2-cos-1(∑i=1∞(-x2)i-∑i=1∞(-x)i) lying in the interval (-12,12) is ______. (Here, the inverse trigonometric functions sin-1x and cos-1x assume values in [-π2,π2] and [0,π], respectively.) [2018]

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