帮忙解题,谢谢。
2个回答
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A= { x|x^2-3x-10≤0}
={ x| (x-5)(x+2)≤0}
={ x| -2≤x≤5}
B= { x|m+1≤x≤ 2m-1}
B is subset of A
case 1: B=Φ
2m-1 < m+1
m< 2
case 2: B≠Φ , ie m≥2
B is subset of A
-2≤m+1 and 2m-1≤5
m≥-1 and m≤3
-1≤m≤3
solution for case 2: 2≤m≤3
B is subset of A
=>
case 1 or case 2
(m< 2) or 2≤m≤3
m≤3
={ x| (x-5)(x+2)≤0}
={ x| -2≤x≤5}
B= { x|m+1≤x≤ 2m-1}
B is subset of A
case 1: B=Φ
2m-1 < m+1
m< 2
case 2: B≠Φ , ie m≥2
B is subset of A
-2≤m+1 and 2m-1≤5
m≥-1 and m≤3
-1≤m≤3
solution for case 2: 2≤m≤3
B is subset of A
=>
case 1 or case 2
(m< 2) or 2≤m≤3
m≤3
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