UPSEE Solved Model Paper 2

Putting y = (x+1)2 – 1, we have x + 1 = √y+1 , i.e., x = √y+1 - 1
Therefore, f−1 (x) = √x+1 - 1
Hence, S = {x : f (x) = f−1 (x)} = {x : (x+1)2 - 1 = √x+1 - 1}
= {x : (x+1)2 = √x+1}
= {x : (x+1)4 = x + 1}
= {x : x4+4x3+6x2+3x = 0}
= {x : x (x + 1) (x2+3x+3) = 0} = {0 , - 1}