1.|x| is always non-negative, so the set A is actually R+ (positive real number set) plus 0.

2-x 2 is a constant ≤2, so we draw its image on the number axis and get it.

A∩B={y|0≤y≤2}

2. The first group S can easily be regarded as an upward-opening polyline passing through the origin.

Then calculate | x | = x+b.

Conduct classified discussions.

When x≥0, x=x+b,b=0,∴b 1=0.

When x < 0, there are 2x=-b, x=-b/2 and -b/2 < 0, that is, b > 0.

Next, the result of polymerization is enough.

3. According to A∩B={-2} branch, one generation enters -2, and both equations hold.

Equation 1: 4+2p-2 = 0, p =- 1.

Equation 2: 4-2q+r = 0

r=2q-4

When p=- 1, the set brought in is

x^2+x-2=0

(x- 1)(x+2)=0

Yes, x 1= 1, x2=-2.

There is no 5 in A∪B,

Therefore, it must be true that 5 is substituted into x 2+qx+2q-4 = 0 (R has been replaced by 2q-4).

The substitution is 25+5q+2q-4 = 0, 7q =-2 1, and q =-3.

So P =- 1, Q =-3, R = 2 *-3-4 =- 10.