题目内容
有两个函数f(x)=asin(kx+
),g(x)=btan(kx-
)(k>0),它们的周期之和为
π且f(
)=g(
),f(
)=-
g(
)+1求这两个函数,并求g(x)的单调递增区间.
| π |
| 3 |
| π |
| 3 |
| 3 |
| 2 |
| π |
| 2 |
| π |
| 2 |
| π |
| 4 |
| 3 |
| π |
| 4 |
由条件得
+
=
π,∴k=2.
由f(
)=g(
),得a=2b①
由f(
)=-
g(
)+1,得a=2-2b②
∴由①②解得a=1,b=
.
∴f(x)=sin(2x+
),g(x)=
tan(2x-
).
∴当-
+kπ<2x-
<
+kπ,k∈Z时,g(x)单调递增.
∴g(x)的单调递增区间为:(
-
,
+
π)k∈Z.
| 2π |
| k |
| π |
| k |
| 3 |
| 2 |
由f(
| π |
| 2 |
| π |
| 2 |
由f(
| π |
| 4 |
| 3 |
| π |
| 4 |
∴由①②解得a=1,b=
| 1 |
| 2 |
∴f(x)=sin(2x+
| π |
| 3 |
| 1 |
| 2 |
| π |
| 3 |
∴当-
| π |
| 2 |
| π |
| 3 |
| π |
| 2 |
∴g(x)的单调递增区间为:(
| kπ |
| 2 |
| π |
| 12 |
| kπ |
| 2 |
| 5 |
| 12 |
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