Question

y = {x(x - 3)}² increases for all values of x lying in the interval.

Options

• 0 < x < `3/2`

• 0 < x < ∞

• -∞ < x < 0

• 1 < x < 3

Answer 1

`underline(0 < x < 3/2)`

Explanation:

y = {x(x - 3)}² 

⇒ y = x²(x - 3)²

∴ `dy/dx = 2x(x - 3)^2 + 2(x - 3)x^2`

= 2x(x - 3)[x - 3 + x]

= 2x(x - 3)(2x - 3)

For y to be increasing, `dy/dx > 0`

⇒ 2x(x - 3)(2x - 3) > 0

⇒ x(x - 3)(2x - 3) > 0 ⇒ `x ∈ (0, 3/2)`