一共三道题,求详细解答过程,谢谢!
2个回答
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7
f(x)与g(x)图像完全相同,则周期必须相同,所以
ω=2
f(x)=3sin(2x-π/6)
0≤x≤π/2
-π/6≤x≤5π/6
函数sint在[-π/6,5π/6]上先增后减;
-1/2≤sin(2x-π/6)≤1
-3/2≤3sin(2x-π/6)≤3
f(x)∈[-3/2 ,3]
8
f(x)=cos2x
对称轴为:
2x=π
x=π/2
周期为π
x1+x2=π
x3=x1+π/2
因为 x1,x2,x3成等比数列,所以
x1(x1+π)=(π-x1)²
x1²+x1π=π²+x1²-2x1π
3x1π=π²==>x1=π/3
b=cos2x1=cos(2π/3)= -1/2
9
(1/2)T=(5π/24)-(-π/24)=π/4
T=π/2=2π/ω
ω=4
f(x)=Asin(4x+φ),(-π/24,A)在图像上,所以
A=Asin(-π/6+φ)
sin(-π/6+φ)=1
φ=2π/3
f(x)=Asin(4x+2π/3),(0,√3)在图像上,所以
√3=Asin(2π/3)=A(√3/2)==>A=2
{A=2
{ω=4
{φ=2π/3
p(4,2π/3),k=2
直线L:
y-2π/3=2(x-4)=2x-8
2x-y+(2π/3-8)=0
满意请及时采纳,不明白再追问;
f(x)与g(x)图像完全相同,则周期必须相同,所以
ω=2
f(x)=3sin(2x-π/6)
0≤x≤π/2
-π/6≤x≤5π/6
函数sint在[-π/6,5π/6]上先增后减;
-1/2≤sin(2x-π/6)≤1
-3/2≤3sin(2x-π/6)≤3
f(x)∈[-3/2 ,3]
8
f(x)=cos2x
对称轴为:
2x=π
x=π/2
周期为π
x1+x2=π
x3=x1+π/2
因为 x1,x2,x3成等比数列,所以
x1(x1+π)=(π-x1)²
x1²+x1π=π²+x1²-2x1π
3x1π=π²==>x1=π/3
b=cos2x1=cos(2π/3)= -1/2
9
(1/2)T=(5π/24)-(-π/24)=π/4
T=π/2=2π/ω
ω=4
f(x)=Asin(4x+φ),(-π/24,A)在图像上,所以
A=Asin(-π/6+φ)
sin(-π/6+φ)=1
φ=2π/3
f(x)=Asin(4x+2π/3),(0,√3)在图像上,所以
√3=Asin(2π/3)=A(√3/2)==>A=2
{A=2
{ω=4
{φ=2π/3
p(4,2π/3),k=2
直线L:
y-2π/3=2(x-4)=2x-8
2x-y+(2π/3-8)=0
满意请及时采纳,不明白再追问;
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