Question

Solve the following inequation and answer the questions given below.

\[\frac{1}{2}(2x - 1) \leq 2x + 12 \frac{1}{2} \leq 5 \frac{1}{2} + x\] 

1. Write the maximum and minimum values of x for \(x \in \mathrm{R}\). 
2. What will be the change in maximum and minimum values of x if \(x \in \mathrm{W}\).

Answer 1

Step 1: Solve the given inequation

The given inequation is: \[\frac{1}{2}(2x - 1) \leq 2x + 12\frac{1}{2} \leq 5\frac{1}{2} + x\]

Converting the mixed fractions to improper fractions:

\[12\frac{1}{2} = \frac{25}{2} \quad \text{and} \quad 5\frac{1}{2} = \frac{11}{2}\]

So, the inequation becomes: \[\frac{2x - 1}{2} \leq 2x + \frac{25}{2} \leq \frac{11}{2} + x\]

This can be split into two separate simultaneous linear inequations:

Part I:

\[\frac{2x - 1}{2} \leq 2x + \frac{25}{2}\]

Multiplying both sides by 2:

\[2x - 1 \leq 4x + 25\] 

\[-1 - 25 \leq 4x - 2x\] 

\[-26 \leq 2x\] 

\[-13 \leq x \quad \text{or} \quad x \geq -13\]

Part II:

\[2x + \frac{25}{2} \leq \frac{11}{2} + x\]

Subtracting (x) from both sides:

\[2x - x \leq \frac{11}{2} - \frac{25}{2}\] 

\[x \leq -\frac{14}{2}\]

\[x \leq -7\]

Combining both solutions:

\[-13 \leq x \leq -7\]

a. Maximum and Minimum values of x for \(x \in \mathrm{R}\):

Minimum value of x: –13

Maximum value of x: –7

b. Change in Maximum and Minimum values of x if \(x \in \mathrm{W}\):

The set of whole numbers is \[\mathrm{W} = {0, 1, 2, 3, \dots}\].

Since all elements in the interval \([-13, -7]\) are strictly negative, there are no whole numbers that satisfy the inequation.

Therefore, the solution set for \(x \in \mathrm{W}\) is the empty set: \[\text{Solution Set} = \phi\]

Neither a maximum value nor a minimum value of x exists when \(x \in \mathrm{W}\).