If k is a real constant such that the inequality |2x - 3| + |x + 4| \le k has no real solu… — GRE General Test

If $k$ is a real constant such that the inequality $|2x - 3| + |x + 4| \le k$ has no real solutions for $x$ , which of the following inequality statements expresses all possible values of $k$ ?

$k < \frac{11}{2}$ Answer
B
$k \le \frac{11}{2}$
C
$k < 7$
D
$k < 11$
E
$k > \frac{11}{2}$

Answer

The statement expressing all possible values of

k

k < \frac{11}{2}

The function

f(x) = |2x - 3| + |x + 4|

represents a continuous piecewise linear curve. Evaluating

f(x)

at its critical points

x = -4

and

x = \frac{3}{2}

yields

f(-4) = 11

and

f\left(\frac{3}{2}\right) = 5.5 = \frac{11}{2}

. Since the slope is

-3

for

x < -4

-1

for

-4 < x < \frac{3}{2}

, and

+3

for

x > \frac{3}{2}

, the global minimum value of

f(x)

across all real numbers is

\frac{11}{2}

. Consequently, the inequality

f(x) \le k

has no real solutions if and only if

k

is strictly less than this minimum value, leading to

k < \frac{11}{2}

Step-by-Step Solution

Identify the critical points of the absolute value terms.

The terms

|2x - 3|

and

|x + 4|

change behavior at

x = \frac{3}{2}

and

x = -4

, respectively.

Absolute value functions

f(x) = |ax + b|

reach zero and change slope at their roots.

Evaluate

f(x) = |2x - 3| + |x + 4|

at the critical points and analyze its piecewise behavior.

x = -4

f(-4) = |-11| + |0| = 11

. At

x = \frac{3}{2}

f\left(\frac{3}{2}\right) = |0| + |\frac{11}{2}| = \frac{11}{2}

. For

x < -4

f(x) = (3 - 2x) - (x + 4) = -3x - 1

. For

-4 \le x \le \frac{3}{2}

f(x) = (3 - 2x) + (x + 4) = -x + 7

. For

x > \frac{3}{2}

f(x) = (2x - 3) + (x + 4) = 3x + 1

Because

f(x)

is a convex piecewise linear function that grows to

\infty

x \to \pm\infty

, its global minimum must occur at one of its critical points.

Determine the global minimum value of

f(x)

Comparing values,

f\left(\frac{3}{2}\right) = \frac{11}{2}

is smaller than

f(-4) = 11

, so the minimum value of

|2x - 3| + |x + 4|

for all real

x

\frac{11}{2}

The function output is always greater than or equal to

\frac{11}{2}

for any real number

x

Apply the condition for no real solutions.

For

|2x - 3| + |x + 4| \le k

to have no solutions,

k

must be strictly less than the absolute minimum value of the expression, so

k < \frac{11}{2}

k \ge \frac{11}{2}

, there is at least one

x

value (such as

x = \frac{3}{2}

) satisfying the inequality.

Key Concept

Minimizing Sums of Absolute Values and Boundary Conditions of Inequalities

Question

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