COmplete set of real values of x

Question

The complete set of real values of x for which x(x-1)^2 < (x-1)x^2 is

- all x

- x > 0

- x < q

- 0 < x < 1

- x < 0 or x > 1

Answer 1

Hi Chek,

 

So, first of all, you want to simplify the above inequality:

 

x(x - 1)² < (x - 1)x²

 

Multiply out each side of the inequality.

 

x(x² - 2x + 1) < x³ - x²

 

Then,

 

x³ - 2x² - x < x³ - x²

 

We can then cancel out x3 terms and rearrange to get

 

-x² + x < 0

 

We can multiply through by -1, thus flipping the inequality sign, to get

 

x² - x > 0 

 

or

 

x(x - 1) > 0

 

f(x) = x² - x is a parabola that crosses the x axis at x = 0 and at x = 1. For 0 < x < 1, f(x) is below the x axis. For all other values of x, it is above the axis.

 

So, x² - x > 0 for all real x except for 0 < x < 1.

 

This same statement applies to the original inequality. So, out of your multiple choices, the last one is correct: x < 0 or x > 1

Answer 2

x(x-1)^2 < (x-1)x^2

x(x-1)^2 - (x-1)x^2<0

x(x-1)(x-1-x)<0

x(x-1)>0

x>0, x>1

x<0, x<1 

 

So x<0 or x<1