Let f : [4, ∞) → [4, ∞) be a function defined by, f (x) = 5x(x – 4), then f –1 (x) is
2−4+log5x
2+4+log5x
15x(x−4)
None of these
Let y = 5x(x–4) ⇒ x (x – 4) = log5 y⇒x2−4x−log5y=0⇒x=4±16+4log5y2 =2±4+log5y But x≥4, so x=2+4+log5y∴f−1(y)=2+4+log5y