题目内容
已知函数f(x)=sin2x+2
sinxcosx-
cos2x,x∈R.
(I)求f(x)的最小正周期和值域;
(II)若x0(0≤x0≤
)为f(x)的一个零点,求sin2x0的值.
| 3 |
| 1 |
| 2 |
(I)求f(x)的最小正周期和值域;
(II)若x0(0≤x0≤
| π |
| 2 |
(I)由题意得,f(x)=
+
sin2x-
cos2x
=
sin2x-cos2x+
=2sin(2x-
)+
,
∴f(x)的最小正周期为π,且最大值为2+
=
,最小值为-2+
=-
,
,则f(x)的值域为[-
,
],
(II)由f(x0)=2sin(2x0-
)+
=0得,
sin(2x0-
)=-
<0,
又由0≤x0≤
得,-
≤2x0-
≤
,
∴-
≤2x0-
≤0,
∴cos(2x0-
)=
=
,
sin2x0=sin[(2x0-
)+
]=sin(2x0-
)cos
+cos(2x0-
)sin
=-
×
+
×
=
.
| 1-cos2x |
| 2 |
| 3 |
| 1 |
| 2 |
=
| 3 |
| 1 |
| 2 |
| π |
| 6 |
| 1 |
| 2 |
∴f(x)的最小正周期为π,且最大值为2+
| 1 |
| 2 |
| 5 |
| 2 |
| 1 |
| 2 |
| 3 |
| 2 |
,则f(x)的值域为[-
| 3 |
| 2 |
| 5 |
| 2 |
(II)由f(x0)=2sin(2x0-
| π |
| 6 |
| 1 |
| 2 |
sin(2x0-
| π |
| 6 |
| 1 |
| 4 |
又由0≤x0≤
| π |
| 2 |
| π |
| 6 |
| π |
| 6 |
| 5π |
| 6 |
∴-
| π |
| 6 |
| π |
| 6 |
∴cos(2x0-
| π |
| 6 |
1-sin2(2x0-
|
| ||
| 4 |
sin2x0=sin[(2x0-
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
=-
| 1 |
| 4 |
| ||
| 2 |
| ||
| 4 |
| 1 |
| 2 |
| ||||
| 8 |
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