The number of real solutions of the equation
`sin ^(-1)( sum_(i=1)^(oo) x^(i+1) - x sum_(i=1)^(oo) ((x)/(2))^(i)) = (pi)/(2) - cos ^(-1)(sum_(i=1)^(oo) (-(x)/(2))^(i) - sum_(i=1)^(oo)(-x)^(i))` lying in the interval `(-(1)/(2), (1)/(2))` is ............... .
(Here, the inverse trigonometric function `sin ^(-1)x and cos^(-1)x` assume values in `[-(pi)/(2), (pi)/(2)] and [0, pi]`, respectively.)

Answer

2

Explanation

We have ,
` sin ^(-1) ( overset (oo) underset ( i=1)( sum ) x ^(i+ 1 ) - x overset (oo) underset ( i=1) (sum ) ((x)/(2))^(i))`
`" " = (pi)/(2) - cos ^(-1) ( overset (oo) underset (i=1) ( sum) ((-x)/(2)) ^(i) - overset (oo) underset(i=1) (sum) (-x)^(i))`
`rArr sin ^(-1) [ (x^(2)) /(1 - x ) - ( x * (x)/(2)) /( 1 - (x )/(2))] `
`" " = ( pi )/(2 ) - cos ^(-1) [ ((-x)/(2))/( 1 + (x)/(2)) - ((- x ))/( 1 + x )] `
`" " [ because overset (oo) underset (i = 1 ) ( sum ) x ^(i + 1) = x ^(2) + x ^(3) + x ^(4) + ... = (x ^(2))/( 1- x )]`

using sum of infinite terms of GP
`rArr sin ^(-1) [ (x ^(2))/( 1 - x) - (x) /( 2 + x) ]`
`rArr sin ^(-1) [ (x^(2))/( 1 - x ) - ( x^(2))/( 2 - x ) ] = sin ^(-1) (( x) /( 1 + x ) - (x )/( 2 + x))`
`" " [ because sin ^(-1) x = (pi)/( 2) - cos ^(-1) x ]`
`rArr ( x ^(2) ) /(1 - x ) - (x^(2))/( 2 - x ) = ( x)/( 1 + x )- (x )/( 2 + x )`
`rArr x ^(2) (( 2 - x - 1 + x ) /( ( 1- x ) ( 2 -x ))) = x (( 2 + x - 1 - x ))/( ( 1 + x ) (2 + x ))`
` rArr (x)/( 2 - 3x + x ^(2)) = (1 )/( 2+ 3x + x ^(2 ) ) or x = 0 `
`rArr x ^(3) + 3x ^(2) + 2x = x ^(2) - 3x + 2`
`rArr x ^(3) + 2x ^(2) + 5x - 2 = 0 or x - 0 `
Let ` f (x) = x ^(3) + 2x ^(2) + 5x - 2 `
`f ' (x) = 3x ^(2) + 4x + 5`
`f ' (x) gt 0 , AA in R `
` therefore x ^(3) + 2x ^(2) + 5x - 2 ` has only one real roots
Therefore, total number of real solution is 2.