Question

The function f(x) = `|x - 1|/x^2` is monotonically decreasing on ______.

Options

• (0, 1) ∪ (2, ∞)

• (0, ∞)

• (–∞, 1) ∪ (2, ∞)

• (–∞, ∞)

Answer 1

The function f(x) = `|x - 1|/x^2` is monotonically decreasing on (0, 1) ∪ (2, ∞).

Explanation:

f(x) = `|x - 1|/x^2`

= `{{:((x - 1)/x^2,;,x ≥ 1),((1 - x)/x^2,;, x < 1):}`

f(x) = `|:(1/x - 1/x^2",", x ≥ 1),(1/x^2 - 1/x",", x < 1):}`

f'(x) = `{:((-1)/x^2 + 2/x^3",", x ≥ 1),((-2)/x^3 + 1/x^2",", x < 1):}`

for f'(x) < 0,

`x ≥ 1, 2/x^3 - 1/x^2` = 0

⇒ 2 – x < 0

⇒ x > 2

For `x < 1, 1/x^2 < 2/x^3`

`2/x < 0`

⇒ `(x - 2)/x < 0`

x∈(0, 2)

∴ Finally x∈(0, 1) ∪ (2, ∞)