The number of real solutions of
tan−1√x(x+1)+sin−1√x2+x+1=π2 is
[IIT 1999]
Two
tan−1√x(x+1)+sin−1√x2+x+1=π2
tan−1√x(x+1) is defined when
x(x+1)≥0...........(i)
sin−1√x2+x+1 is defined when
0≤x(x+1)+1≤1 or ≤ x(x+1)≤0.......(ii)
From (i) and (ii), x(x + 1) = 0
or x = 0 and - 1.
Hence number of solution is 2.