A) 0
B) -1
C) 1
D) 2
Correct Answer: B
Solution :
[b] \[0\le {{x}^{2}}+x+1\le 1\]and \[0\le {{x}^{2}}+x+\le 1\] \[\therefore x=-1,0\] For \[x=-1\] L.H.S.\[=2{{\sin }^{-1}}1+{{\cos }^{-1}}0=\frac{3\pi }{2}\] \[\therefore x=-1\] is a solutions. For \[x=0,\]L.H.S.\[=2{{\sin }^{-1}}1+{{\cos }^{-1}}0=\frac{3\pi }{2}\] Therefore, \[x=0\] is a solution and sum of the solutions\[=-1\].You need to login to perform this action.
You will be redirected in
3 sec