问题 解答题
已知向量
a
=(cos
3
2
x,sin
3
2
x)
b
=(cos
1
2
x,sin
1
2
x)
,x∈[0,π].
(1)当x=
π
4
时,求
a
b
|
a
+
b
|
的值;
(2)求f(x)=m|
a
+
b
|-
a
b
(m∈R)的最大值.
答案

(1)∵

a
=(cos
3
2
x,sin
3
2
x)
b
=(cos
1
2
x,sin
1
2
x)

a
b
=cos
3
2
xcos
1
2
x+sin
3
2
xsin
1
2
x=cos(
3
2
x-
1
2
x)
=cosx

x=

π
4
时,
a
b
=
2
2

|

a
+
b
|2=
a
2
+
b
2
+2
a
b
=2+2cosx

x=

π
4
时,|
a
+
b
|
=
2+
2

(2)∵x∈[0,π],∴0≤cos

x
2
≤1

f(x)=m|

a
+
b
|-
a
b
=2m|cos
x
2
|-cosx
=-2cos2
x
2
+2mcos
x
2
-1

令t=cos

x
2
(0≤t≤1)则f(x)=-2t2+2mt-1=-2(t-
m
2
)2+
m2
2
-1

∴当

m
2
>1即m>2时,此时t=1,f(x)max=2m-3

当0≤

m
2
≤1即0≤m≤2时,此时t=
m
2
f(x)max=
m2
2
-1

m
2
<0即m<0时,此时t=0,f(x)max=-1

f(x)max=

2m-3(m>2)
m2
2
-1(0≤m≤2)
-1(m<0)

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