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
,且f(
)=g(
),f(
)=-
g(
)+1
(1)求这两个函数的解析式;
(2)求f(x)的单调区间.
| π |
| 3 |
| π |
| 3 |
| 3π |
| 2 |
| π |
| 2 |
| π |
| 2 |
| π |
| 4 |
| 3 |
| π |
| 4 |
(1)求这两个函数的解析式;
(2)求f(x)的单调区间.
(1)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-
).
(2)当-
+2kπ<2x+
<
+2kπ,k∈Z时,f(x)单调递增.
单调增区间为〔kπ-
,kπ+
〕(k∈Z);单调减区间为〔kπ+
,kπ+
〕(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 |
(2)当-
| π |
| 2 |
| π |
| 3 |
| π |
| 2 |
单调增区间为〔kπ-
| 5π |
| 12 |
| π |
| 12 |
| π |
| 12 |
| 7π |
| 12 |
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