Question

The relation $x^2(x^2-1)\ge0$ is true only for:

Here $x\ge a$ means that $x$ can take on all values greater than $a$ and the value equal to $a,$ while $x\le a$ has a corresponding meaning with “less than.”

$x\ge1$

$-1\le x\le1$

$x=0,$ $x=1,$ $x=-1$

$x=0,$ $x\le-1,$ $x\ge1$

$x\ge0$

Answer: D

Concepts:inequality factoring zero product property

Difficulty rating: 1360

Small Hint:

The factor $x^2$ is positive except at $x=0$

Big Hint:

Away from zero, the sign is controlled by $x^2-1$

Solution:

At $x=0,$ the product is zero. For $x\ne0,$ the factor $x^2$ is positive, so the product is nonnegative exactly when $x^2-1\ge0,$ or $x\le-1$ or $x\ge1.$ Including the isolated value $x=0$ gives the set in choice D.

Thus, the correct answer is D.

1957 AMC 12 Problem 29

Answer: D

Difficulty rating: 1360

Small Hint:

Big Hint:

Solution:

Problem 29 in Other Years