题目内容
设函数f(x)=
sin2x+cos2x+1
(Ⅰ)求函数f(x)的最小正周期;
(Ⅱ)求函数f(x)的增区间
(Ⅲ)当x∈[-
,
]时,求函数f(x)的最大最小值并求出相应的x的值.
| 3 |
(Ⅰ)求函数f(x)的最小正周期;
(Ⅱ)求函数f(x)的增区间
(Ⅲ)当x∈[-
| π |
| 6 |
| π |
| 3 |
(I)函数f(x)=
sin2x+cos2x+1
=2(
sin2x+
cos2x)+1
=2sin(2x+
)+1
∴T=
=π,
∴函数f(x)的最小正周期为π;
(II)由-
+2kπ≤2x+
≤
+2kπ,
解得-
+kπ≤x≤
+kπ,
∴函数f(x)的增区间为[-
+kπ,
+kπ](k∈Z).
(III)由x∈[-
,
],可得(2x+
)∈[-
,
],
∴-
≤sin(2x+
)≤1.
当且仅当2x+
=-
,即x=-
,ymin=2×(-
)+1=0;
当且仅当2x+
=
,即x=
,ymax=2×1+1=3.
| 3 |
=2(
| ||
| 2 |
| 1 |
| 2 |
=2sin(2x+
| π |
| 6 |
∴T=
| 2π |
| 2 |
∴函数f(x)的最小正周期为π;
(II)由-
| π |
| 2 |
| π |
| 6 |
| π |
| 2 |
解得-
| π |
| 3 |
| π |
| 6 |
∴函数f(x)的增区间为[-
| π |
| 3 |
| π |
| 6 |
(III)由x∈[-
| π |
| 6 |
| π |
| 3 |
| π |
| 6 |
| π |
| 6 |
| 5π |
| 6 |
∴-
| 1 |
| 2 |
| π |
| 6 |
当且仅当2x+
| π |
| 6 |
| π |
| 6 |
| π |
| 6 |
| 1 |
| 2 |
当且仅当2x+
| π |
| 6 |
| π |
| 2 |
| π |
| 6 |
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