问题 填空题
设f(x)=sin2x+mcos2x,若对一切x∈R,都有f(x)≤f(
π
8
)
,则f(
π
24
)
=______.
答案

由题意知:

f(x)=sin2x+mcos2x=

m2+1
sin(2x+φ),(sinφ=
m
m2+1
,cosφ=
1
m2+1

由题意得:当x=

π
8
时函数f(x)=sin2x+mcos2x取到最值±
m2+1

将x=

π
8
代入可得:sin(2×
π
8
)+mcos(2×
π
8
)=
2
2
(m+1)
m2+1
,即m=1

∴f(x)=sin2x+mcos2x=sin2x+cos2x=

2
sin(2x+
π
4
),

则f(

π
24
)=
2
sin(2×
π
24
+
π
4
)=
2
sin
π
3
=
6
2

故答案为:

6
2

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