高一数学题,急!!!!
已知函数f(x)=(x*x-2x-3)的绝对值-3,分别满足下列条件,求实数a的取值范围.①函数有两个零点②函数有三个零点③函数有四个零点已知f(x)=〔√(1-sinx...
已知函数f(x)=(x*x-2x-3)的绝对值-3,分别满足下列条件,求实数a的取值范围.①函数有两个零点 ②函数有三个零点 ③函数有四个零点
已知f(x)=〔√(1-sinx)〕+〔√(1+sinx)〕
①求函数定义域和值域 ②判断奇偶性 ③求最小正周期和单调区间 展开
已知f(x)=〔√(1-sinx)〕+〔√(1+sinx)〕
①求函数定义域和值域 ②判断奇偶性 ③求最小正周期和单调区间 展开
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1)第一题中没有a,如何求范围?
2)
①定义域:
1-sinx≥0
1+sinx≥0
因为sinx≤1,所以x∈R,
f(x)=√[sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2)]+√[sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2)]
=|sin(x/2)-cos(x/2)|+|sin(x/2)+cos(x/2)|
=√2[|sin(x/2-π/4)|+|sin(x/2+π/4)|]
=√2[|sin(x/2-π/4)|+|cos(x/2-π/4)|]
当2kπ+π/4≤x/2≤2kπ+3π/4时,
f(x)=√2[sin(x/2-π/4)+cos(x/2-π/4)]
=2sin(x/2-π/4+π/4)
=2sin(x/2)
f(x)∈[√2,2]
当2kπ+3π/4≤x/2≤2kπ+5π/4时,
f(x)=√2[sin(x/2-π/4)-cos(x/2-π/4)]
=2sin(x/2-π/4-π/4)
=2sin(x/2-π/2)
f(x)∈[√2,2]
当2kπ+5π/4≤x/2≤2kπ+7π/4时,
f(x)=√2[-sin(x/2-π/4)-cos(x/2-π/4)]
=-2sin(x/2-π/4+π/4)
=-2sin(x/2)
f(x)∈[√2,2]
当2kπ+7π/4≤x/2≤2kπ+9π/4时
f(x)=√2[-sin(x/2-π/4)+cos(x/2-π/4)]
=-2sin(x/2-π/4-π/4)
=-2sin(x/2-π/2)
f(x)∈[√2,2]
②奇偶性
f(-x)=√(1-sin(-x))+√(1+sin(-x))=f(x),是偶函数。
③最小正周期=2π/1/2=4π
[kπ+π/2,kπ+π]为增函数,[kπ+π,kπ+3π/2]为减函数
2)
①定义域:
1-sinx≥0
1+sinx≥0
因为sinx≤1,所以x∈R,
f(x)=√[sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2)]+√[sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2)]
=|sin(x/2)-cos(x/2)|+|sin(x/2)+cos(x/2)|
=√2[|sin(x/2-π/4)|+|sin(x/2+π/4)|]
=√2[|sin(x/2-π/4)|+|cos(x/2-π/4)|]
当2kπ+π/4≤x/2≤2kπ+3π/4时,
f(x)=√2[sin(x/2-π/4)+cos(x/2-π/4)]
=2sin(x/2-π/4+π/4)
=2sin(x/2)
f(x)∈[√2,2]
当2kπ+3π/4≤x/2≤2kπ+5π/4时,
f(x)=√2[sin(x/2-π/4)-cos(x/2-π/4)]
=2sin(x/2-π/4-π/4)
=2sin(x/2-π/2)
f(x)∈[√2,2]
当2kπ+5π/4≤x/2≤2kπ+7π/4时,
f(x)=√2[-sin(x/2-π/4)-cos(x/2-π/4)]
=-2sin(x/2-π/4+π/4)
=-2sin(x/2)
f(x)∈[√2,2]
当2kπ+7π/4≤x/2≤2kπ+9π/4时
f(x)=√2[-sin(x/2-π/4)+cos(x/2-π/4)]
=-2sin(x/2-π/4-π/4)
=-2sin(x/2-π/2)
f(x)∈[√2,2]
②奇偶性
f(-x)=√(1-sin(-x))+√(1+sin(-x))=f(x),是偶函数。
③最小正周期=2π/1/2=4π
[kπ+π/2,kπ+π]为增函数,[kπ+π,kπ+3π/2]为减函数
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