A∩B= empty set

x？ -3x+2≤0

1≤x≤2

∴y<； 1 or y>2

y=x？ -2x+a=(x- 1)？ +a- 1

B={yly≥a- 1}

∴y>； 2

a- 1 >2

A> III

∴a>； three

P∧q is true.

P and G are telling the truth.

A∩B≠ empty set

a- 1≤2

a≤3

x？ -ax-4≤0

(a-√a？ + 16)/2≤x≤(a+√a？ + 16)/2

A is contained in C.

∴(a-√a？ + 16)/2≤ 1

a-2≤√a？ + 16

When a≤2, it is obviously true.

When a>2

Answer? -4a+4≤a？ + 16

a≥-3

∴a∈R

(a+√a？ + 16)/2≥2

√a？ + 16≥4-a

When a≥4, it is obviously true.

When a<4,

Answer? + 16≥a？ -8a+ 16

a≥0

∴a≥0

finally

0≤a≤3