Let S be the set of all real numbers x that satisfy the nested absolute value inequality |… — GRE General Test

Let $S$ be the set of all real numbers $x$ that satisfy the nested absolute value inequality $|3 - |2x - 1|| \le 4$ . Which of the following inequalities MUST be satisfied by every value of $x$ in $S$ ? Select all such inequalities.

$|x| \le 4$ Answer
$(x + 3)(x - 4) \le 0$ Answer
C
$|2x - 1| \le 4$
D
$-1 \le x \le 7$
E
$x \ge -2$

Answer

The inequalities that must be satisfied by every value of

x

S

are

|x| \le 4

and

(x + 3)(x - 4) \le 0

The solution set to the nested inequality is

S = [-3, 4]

. For any value of

x

[-3, 4]

, the absolute value

|x|

ranges from

0

4

, so the inequality stating that the magnitude of

x

is at most 4 is satisfied. Additionally, the quadratic expression factored as

(x + 3)(x - 4)

has roots at

x = -3

and

x = 4

and opens upward, meaning it takes on values less than or equal to zero for all

x

between

-3

and

4

Step-by-Step Solution

Unfold the outer absolute value inequality

-4 \le 3 - |2x - 1| \le 4

By definition,

|u| \le k

(with

k \ge 0

) is equivalent to

-k \le u \le k

Isolate the inner absolute value expression by subtracting 3 and multiplying by -1

-1 \le |2x - 1| \le 7

Subtracting 3 yields

-7 \le -|2x - 1| \le 1

. Multiplying by

-1

flips the inequality signs, giving

-1 \le |2x - 1| \le 7

Simplify the compound absolute value bound and solve for

x

x \in [-3, 4]

Since an absolute value is non-negative,

|2x - 1| \ge -1

is satisfied for all real

x

. Thus, we only need

|2x - 1| \le 7

, which gives

-7 \le 2x - 1 \le 7 \implies -6 \le 2x \le 8 \implies -3 \le x \le 4

Test the solution set

S = [-3, 4]

against each given statement

Statements

|x| \le 4

and

(x + 3)(x - 4) \le 0

hold for all

x \in [-3, 4]

For

x \in [-3, 4]

, the extreme values of

x

yield

|x| \le 4

. Furthermore, a quadratic with roots at

-3

and

4

is non-positive on

[-3, 4]

Key Concept

Solving nested absolute value inequalities by systematic expansion and isolating valid intervals.

Question

Answer

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