FunctionHard

Question

The function f : [2, ∞) → Y, defined by f(x) = x2 - 4x + 5 is both one-one and onto if

Options

A.Y = R
B.Y = [1, ∞)
C.Y = [4, ∞)
D.Y = [5, ∞)

Solution

  
f : [2, ∞) → Y
f(x) = x2 - 4x + 5
f(x) = (x - 2)2 + 1
For given domain by graph range is [1, ∞)
For function to be onto codomain y = [1, ∞)

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