Let S be the set of all real numbers x that satisfy the inequality x^2 - 6x \leq 0. How ma… — GMAT

Let $S$ be the set of all real numbers $x$ that satisfy the inequality $x^2 - 6x \leq 0$ . How many integer values of $k$ are there such that the equation $||x - 4| - |x + 2|| = k$ has at least one solution $x \in S$ ?

Answer: 7

Answer

The correct answer is 7.

Solving the quadratic inequality

x^2 - 6x \leq 0

gives the domain

S = [0, 6]

. Over this closed interval, the function

g(x) = ||x - 4| - |x + 2||

is continuous and attains its minimum value of 0 at

x = 1

and its maximum value of 6 at

x = 4

(and throughout

[4, 6]

). By the Intermediate Value Theorem,

g(x)

takes on all real values in the interval

[0, 6]

. The integer values of

k

for which

g(x) = k

has a solution in

S

are

0, 1, 2, 3, 4, 5,

and

6

, making a total of 7 integers.

Step-by-Step Solution

Solve the quadratic inequality to define the set

S

x^2 - 6x \leq 0 \iff x(x - 6) \leq 0 \iff 0 \leq x \leq 6

. Thus,

S = [0, 6]

The solution set of

x(x-6) \leq 0

lies between the roots

x = 0

and

x = 6

inclusive.

Analyze the inner function

f(x) = |x - 4| - |x + 2|

for

x \in [0, 6]

Critical points of absolute values occur at

x = -2

and

x = 4

. Within

[0, 6]

, we split at

x = 4

The signs of

(x - 4)

and

(x + 2)

determine how the absolute value bars simplify.

Evaluate

g(x) = |f(x)|

on the sub-interval

[0, 4]

For

0 \leq x \leq 4

|x - 4| = 4 - x

and

|x + 2| = x + 2

. Thus,

f(x) = 2 - 2x

and

g(x) = |2 - 2x|

Evaluating

g(x)

at key points gives

g(0) = 2

g(1) = 0

, and

g(4) = 6

. By continuity,

g(x)

covers all values in

[0, 6]

on this interval.

Evaluate

g(x)

on the sub-interval

[4, 6]

For

4 \leq x \leq 6

|x - 4| = x - 4

and

|x + 2| = x + 2

. Thus,

f(x) = -6

and

g(x) = |-6| = 6

g(x)

remains constant at 6 for

x \in [4, 6]

Determine the range of

g(x)

S

and count the integer values of

k

The range of

g(x)

for

x \in [0, 6]

[0, 6]

. The integer values in this interval are

0, 1, 2, 3, 4, 5, 6

The equation

g(x) = k

has a solution in

S

if and only if

k

lies in the range of

g(x)

over

S

. There are

6 - 0 + 1 = 7

such integers.

Key Concept

Absolute value functions case evaluation and finding the range over a restricted domain.

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