FunctionHard
Question
Let f : R → R be a function such that f(0) = 1 and for any x, y ∈ R, f(xy + 1) = f(x) f(y) - f(y) - x + 2. Then f is
Options
A.one-one and onto
B.one-one but not onto
C.many one but onto
D.many one and into
Solution
f(xy + 1) = f(yx + 1)
f(x) f(y) - f(y) - x + 2 = f(y) f(x) - f(x) - y + 2
f(x) - f(y) = x - y
Putting y = 0
f(x) - 1 = x - 0
f(x) = x + 1
f(x) f(y) - f(y) - x + 2 = f(y) f(x) - f(x) - y + 2
f(x) - f(y) = x - y
Putting y = 0
f(x) - 1 = x - 0
f(x) = x + 1
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