The number of real solution of the equation `tan^(-1) sqrt(x^2-3x +2) + cos^(-1) sqrt(4x-x^2 - 3) = pi` is

Answer

2

Explanation

Given equation is
`sin^(-1) (underset(i =1)overset(oo)sum x^(i+1) -x underset(i=1)overset(oo)sum ((x)/(2))^(i) ) + cos^(-1) (underset(i=1)overset(oo)sum (-(x)/(2))^(i) -underset(i=1)overset(oo)sum (-x)^(i)) = (pi)/(2)`
`:. underset(i=1)overset(oo)sum x^(i +1)-x underset(i=1)overset(oo)sum ((x)/(2))^(i) = underset(i=1)overset(oo)sum (-(x)/(2))^(i) - underset(i=1)overset(oo)sum (-x)^(i)`
`rArr (x^(2))/(1-x) -x ((x)/(2))/(1-(x)/(2)) = (x)/(1 +x) + ((- (x)/(2)))/(1 + (x)/(2))` (Each series is infinite G.P.)
`rArr (x^(2))/(1 -x) -(x^(2))/(2 -x) = (x)/(1 + x) - (x)/(2 + x)`
`rArr x = 0 "or " (x)/(1-x) -(1)/(1 + x) = (x)/(2-x) -(1)/(2 + x)`
`rArr x = 0 " or " (x + x^(2) -1 + x)/(1 -x^(2)) = (2x + x^(2) -2 + x)/(4 -x^(2))`
`rArr x = 0 " or " (x^(2) + 2x -1) (4 -x^(2)) = (x^(2) + 3x -2) (1 -x^(2))`
`rArr 4x^(2) + 8x -4 - x^(4) -2x^(3) + x^(2) = x^(2) + 3x -2 -x^(4) - 3x^(3) + 2x^(2)`
`rArr x^(3) + 2x^(2) + 5x -2 = 0`
Let `f(x) = x^(3) + 2x^(2) + 5x -2`
`:. f'(x) = 3x^(2) + 4x + 5 gt 0`
So, `f(x)` is increasing function and it intersects x-axis only once.
`f(0) = -2 and f(1//2) = 9//8`
So, one root of `fs (x) = 0` lies in (0, 1/2)
So, there are two value of x