问题 解答题
已知函数f(x)=sinx,x∈R
(1)函数g(x)=2sinx•(sinx+cosx)-1的图象可由f(x)的图象经过怎
样的平移和伸缩变换得到;
(2)设h(x)=f(
π
2
-2x)+4λf(x-
π
2
)
,是否存在实数λ,使得函数h(x)
在R上的最小值是-
3
2
?若存在,求出对应的λ值;若不存在,说明理由.
答案

(1)g(x)=2sin2x+sin2x-1=sin2x-cos2x=

2
sin(2x-
π
4
)

先将f(x)的图象向右平移

π
4
个单位长度得到y=sin(x-
π
4
)
的图象;

再将y=sin(x-

π
4
)图象上各点的横坐标变为原来的
1
2
倍,得到

函数y=sin(2x-

π
4
)的图象;最后将曲线上各点的纵坐标变为

原来的

2
倍得到函数g(x)的图象.

(2)h(x)=cos2x-4λcosx=2cos2x-4λcosx-1=2(cosx-λ)2-2λ2-1

λ<-1
1+4λ=-
3
2
-1≤λ≤1
-2λ2-1=-
3
2
λ>1
1-4λ=-
3
2

λ=±

1
2

单项选择题
完形填空
完形填空。
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