Question

The cube of a number is less than five times the square of the number. For what set of numbers is this true?

(-0,5)
(5.00)
(-0.0) U (0.5)
​

Answer 1

C

Explanation:

Answer 2

`(-\infty,0)\cup (0,5)`.

Explanation:

Note: In the given options it should be negative infinity instead of -0 because -0 does not make any sense.

It is given that cube of a number is less than five times the square of the number.

Let the unknown number be x.

`x^3<5x^2`

`x^3-5x^2<0`

`x^2(x-5)<0`

This is possible when both factors
`x^2\text{ and }(x-5)` have different sign.

We know that
`x^2` is always greater than or equal to 0. So,

`x-5<0`

`x<5`

`x\in (-\infty,0)\cup (0,5)`

Because
`x^2(x-5)<0` is not true for x=0, so, 0 is not included in the solution set.

Therefore, the solution set is
`(-\infty,0)\cup (0,5)`.