已知A={x|x²-2x-3≤0,x∈R},B={x|x²-2mx+m²-9≤,X∈R,m∈R}.若A∪B=[1,3],求实数m的值
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![已知A={x|x²-2x-3≤0,x∈R},B={x|x²-2mx+m²-9≤,X∈R,m∈R}.若A∪B=[1,3],求实数m的值](/uploads/image/z/2700245-29-5.jpg?t=%E5%B7%B2%E7%9F%A5A%3D%7Bx%7Cx%26%23178%3B-2x-3%E2%89%A40%2Cx%E2%88%88R%7D%2CB%3D%7Bx%7Cx%26%23178%3B-2mx%2Bm%26%23178%3B-9%E2%89%A4%2CX%E2%88%88R%2Cm%E2%88%88R%7D.%E8%8B%A5A%E2%88%AAB%3D%5B1%2C3%5D%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%80%BC)
已知A={x|x²-2x-3≤0,x∈R},B={x|x²-2mx+m²-9≤,X∈R,m∈R}.若A∪B=[1,3],求实数m的值
已知A={x|x²-2x-3≤0,x∈R},B={x|x²-2mx+m²-9≤,X∈R,m∈R}.若A∪B=[1,3],求实数m的值
已知A={x|x²-2x-3≤0,x∈R},B={x|x²-2mx+m²-9≤,X∈R,m∈R}.若A∪B=[1,3],求实数m的值
A:-1<=x<=3
B:m-3<=x<=m+3
因为A∪B=[1,3],
m-3=1,m+3>3
所以m=4
A={x|x²-2x-3≤0,x∈R} -> A={x|-1<=x<=3}
B={x|x²-2mx+m²-9≤0,X∈R,m∈R} -> B={x|-3+m<=x<=3+m}
若A∩B=[1,3],则-3+m=1 -> m=4
A∪B≠[1,3],因为A本身已经是[-1,3]了