√(-m²+2m+3)∈[0,2];√(-m²+4)∈[0,2]
又:函数f(x)在[0,π]上是递减的,则:
①√(-m²+2m+3)-m²+4
得:m>1/2
②-m²+2m+3≥0
m²-2m-3≤0
得:-1≤m≤3
③-m²+4≥0
m²-4≤0
得:-2≤m≤2
综合①、②、③,得:
1/2
√(-m²+2m+3)∈[0,2];√(-m²+4)∈[0,2]
又:函数f(x)在[0,π]上是递减的,则:
①√(-m²+2m+3)-m²+4
得:m>1/2
②-m²+2m+3≥0
m²-2m-3≤0
得:-1≤m≤3
③-m²+4≥0
m²-4≤0
得:-2≤m≤2
综合①、②、③,得:
1/2