问题 解答题
已知向量
a
=(cos(x-
π
4
),sin(x-
π
4
))
b
=(cos(x+
π
4
),-sin(x+
π
4
))
f(x)=
a
b
-k|
a
+
b
|
,x∈[0,π].
(1)若x=
12
,求
a
b
|
a
+
b
|

(2)若k=1,当x为何值时,f(x)有最小值,最小值是多少?
(3)若f(x)的最大值为3,求k的值.
答案

(1)由题意可知

a
b
=(cos(x-
π
4
),sin(x-
π
4
))• (cos(x+
π
4
),-sin(x+
π
4
))

=cos(x-

π
4
)•cos(x+
π
4
)- sin(x-
π
4
)•sin(x+
π
4
)

=cos2x,∵x=

12
,∴
a
b
=cos2x=-
3
2

|

a
+
b
|=|(cos(x-
π
4
)+cos(x+
π
4
),- sin(x-
π
4
)+sin(x+
π
4
))|

=

(cos(x-
π
4
)+cos(x+
π
4
)
2
+(-sin(x-
π
4
)+sin(x+
π
4
))
2

=

2+2cos2x
=
2-
3
=
6
-
2
2

(2)k=1,f(x)=

a
b
-k|
a
+
b
|=
a
b
-|
a
+
b
|

=2cos2x-2|cosx|-1

当x=

π
3
或x=
3
时,函数f(x)有最小值f(x)min=-
3
2

(3)由(2)可知f(x)=2cos2x-2k|cosx|-1

设|cosx|=t,由x∈[0,π]

则:f(x)=g(t)=2t2-2kt-1,t∈[0,1]

当:

k
2
1
2
⇒k≤1时,f(x)max=g(1)=2-2k-1=3⇒k=-1

k≤1
k=-1
⇒k=-1,

当:

k
2
1
2
⇒k>1时,f(x)max=g(0)=-1≠3,

综上之:k=-1.

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