Question

Solve the following inequation and answer the questions given below.

`1/2 (2x - 1) ≤ 2x + 1/2 ≤ 5 1/2 + x`

1. Write the maximum and minimum values of x for x ∈ R.
2. What will be the change in maximum and minimum values of x if x ∈ W?

Answer 1

Given: Solve the inequation `1/2 (2x - 1) ≤ 2x + 1/2 ≤ 5 1/2 + x` and (a) max/min (x) for x ∈ R, (b) change if x ∈ W whole numbers.

Step-wise calculation:

1) Split the double inequality into two parts

`1/2 (2x - 1) ≤ 2x + 1/2` and `2x + 1/2 ≤ 5 1/2 + x`

2) Solve the left inequality

`1/2 (2x - 1) = x - 1/2`

So, `x - 1/2 ≤ 2x + 1/2`.

Subtract (x) from both sides:

`-1/2 ≤ x + 1/2`

Subtract `1/2`: –1 ≤ x 

So, x ≥ –1.

3) Solve the right inequality

Convert `(5 1/2 = 11/2): 2x + 1/2 ≤ 11/2 + x`

Subtract (x): `x + 1/2 ≤ 11/2`

Subtract `1/2`: `x ≤ 10/2 = 5`

4) Combine both results

–1 ≤ x ≤ 5

(a) For x ∈ R

Solution set: (–1, 5)

Minimum value: (–1)

Maximum value: (5)

(b) If x ∈ W whole numbers = {0, 1, 2, ...}

Intersect (–1, 5) with whole numbers: x ∈ {0, 1, 2, 3, 4, 5}

New minimum value: 0 instead of –1

New maximum value: 5 unchanged

Change: minimum increases by 1; maximum change = 0.