2211.11 – Absolutely Unequal


Tags:Problem Set 9Algebra/TrigonometryGeometry-Analytical

Which of the following gives the set of real solutions to the inequality: x1+x+2<5? \vert x-1\vert + \vert x+2\vert < 5?

a. 3<x<2-3 < x < 2,

b. 1<x<2-1 < x < 2,

c. 2<x<1-2 < x < 1,

d. 1.5<x<1-1.5 < x < 1,

e. NONO real solution.


Solution

There are three cases.

Case 1: x1x ≥ 1. Then:

x1+x+2<5x1+x+2<5,2x+1<5,2x<4,x<2. \begin{aligned} \vert x-1\vert + \vert x+2\vert &<& 5 \\ x-1+x+2 &<& 5, \\ 2x+1 &<& 5, \\ 2x<4,\\ x<2. \end{aligned}

In summary: 1x<21≤x<2.

Case 2: 2<x<1-2 < x < 1. Then:

x1+x+2<51x+x+2<5,3<5. \begin{aligned} \vert x-1\vert + \vert x+2\vert &<& 5 \\ 1-x+x+2 &<& 5, \\ 3 &<& 5. \end{aligned}

Since this is alwaysalways true, we simply have 2<x<1-2 < x < 1, that is, we include the whole case.

Case 3: x2.x ≤ -2. Then:

x1+x+2<51xx2<5,2x1<5,2x<6,x>3. \begin{aligned} \vert x-1\vert + \vert x+2\vert &<& 5 \\ 1-x-x-2 &<& 5, \\ -2x-1 &<& 5, \\ -2x &<& 6, \\ x &>& -3. \end{aligned}

In summary, 3<x2-3 < x ≤ -2.

Putting the three cases together: 3<x<2-3 < x < 2. The answer is a.