The number of real solutions of `tan^(-1)sqrt(x(x+1))+sin^(-1)sqrt(x^2+x+1)=pi/2` is (a) zero b. one c. two d. infinite

Answer

N/a

Explanation

We have, `tan^(-1)sqrt(x(x+1))+sin^(-1)sqrt(x^(2)+x+1) = pi/2"......."(i)`
Let `sin^(-1) sqrt(x^(2)+x+1)=theta`
`rArr sin theta = sqrt((x^(2)+x+1)/(1))`
`rArr tantheta = (sqrt(x^(2)+x+1))/(sqrt(-x^(2)-x))`
`:' theta = tan^(-1)'(sqrt(x^(2)+x+1))/(sqrt(-x-x))`
`=sin^(-1)sqrt(x^(2)+x+1)`

On putting the value of `theta` in Eq. (i), we get
`tan^(-1)sqrt(x(x+1))+tan^(-1)'(sqrt(x^(2)+x+1))/(sqrt(-x^(2)-x)) = (pi)/(2)`
We know that, ` tan^(-1)x+tan^(-1)y=tan^(-1)((x+y)/(1-xy)), xy lt 1`
`:. tan^(-1)[(sqrt(x(x+1))+sqrt((x^(2)+x+1)/(-x^(2)-x)))/(1-sqrt(x(x+1)).sqrt((x^(2)+x+1)/(-x^(2)-x)))]=(pi)/(2)`
`rArr tan^(-1)[(sqrt(x^(2)+x)+sqrt((x^(2)+x+1)/(-1(x^(2)+x))))/(1-sqrt((x^(2)+x).((x^(2)+x+1))/(-1(x^(2)+x))))] = (pi)/(2)`
`rArr (x^(2)+x+sqrt(-(x^(2)+x+1)))/([1-sqrt(-(x^(2)+x+1))]sqrt((x^(2)+x))) = tan'(pi)/2 = 1/0`
`rArr [1-sqrt(-(x^(2)+x+1))]sqrt((x^(2)+x))= 0`
`rArr - (x^(2)+x+1) = 1` or `x^(2) + x = 0`
`rArr -x^(2)-x-1=1` or `x(x+1)=0`
`rArr x^(2)+x+2 = 0` or `x(x+1)=0`
`:. x = (-1+-sqrt(1-4xx 2))/(2)`
`rArr x = 0` or `x = - 1`
For real solution, we have `x = 0, -1`.