The range of the function f(x) = \(\sqrt{(x-1)(3-x)}\) is(a) [0, 1]�

1.	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» 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]

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»

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]