If the equation `sin^(-1)(x^2+x+1)+cos^(-1)(lambda x+1)=pi/2` has exactly two solutions for `lambda in [a , b]` , then the value of `a+b` is

Answer

1

Explanation

`sin^(-1) (x^(2) + x -1) + cos^(-1) (lamda x + 1) = (pi)/(2)`
`rArr x^(2) + x + 1 = lamda x + 1`
`rArr x^(2) + (1- lamda) x = 0`
`rArr x = 0 " or " x = lamda -1`
Also, `sin^(-1) (x^(2) + x + 1)` will exist if `-1 le x^(2) + x + 1 le 1`
`rArr x^(2) + x le 0 rArr -1 le x lt 0`
For exactly two solution, we have
`-1 le lamda -1 lt 0`
`rArr 0 le lamda lt 1`
`:. a = 0 and b =1`