The range of the function f(x) = √ {(x-1)(3-x)} is

Question

The range of the function f(x) = \(\sqrt{(x-1)(3-x)}\) is

(a) [0, 1] 

(b) (–1, 1) 

(c) (–3, 3) 

(d) (–3, 1)

Answer 1

Answer : (a) [0, 1]

Let y = f(x) = \(\sqrt{(x-1)(3-x)}\) 

⇒ y² = (x – 1) (3 – x) = – x² + 4x – 3 

⇒ x² – 4x + (3 + y²) = 0 

This is a quadratic in x, so

\(x = \frac{+4\,\pm\sqrt{16-4(3+y^2)}}{2}\)  = \(\frac{4\,\pm\,2\sqrt{1-y^2}}{2}\) 

Since x is real, 1 – y² ≥ 0 

⇒ y² – 1 ≤ 0 

⇒ – 1 ≤ y ≤ 1 

But f (x) attains only non-negative values, 

so range of f = [0, 1]