Let f: [2, ∞) → R be a function defined by f(x) = x2 - 4x + 5. The range of f is : |
R [1, ∞) [4, ∞) [5, ∞) |
[1, ∞) |
Given that, f(x) = x2 - 4x + 5 Let (f(x)) = y $y = x^2-4x+5⇒y=x^2-4x+4+1$ $y=(x-2)^2+1$ $y-1=(x-2)^2⇒(x-2)^2=y-1$ $x-2=\sqrt{y-1}⇒x=2+\sqrt{y-1}$ Now, if f is real valued function then. $⇒ y-1≥0$ $⇒ y≥1$ The range of f is [1, ∞). |