设函数f(x)的定义域为[-1,1],f[cos(α+π30)]=tcos(2α+π15)+sin(α+π5)+cos(α+11π30)(1)若f(0)=-1,求t的值和f(x)的零点;(2)记h(t

题目简介

设函数f(x)的定义域为[-1,1],f[cos(α+π30)]=tcos(2α+π15)+sin(α+π5)+cos(α+11π30)(1)若f(0)=-1,求t的值和f(x)的零点;(2)记h(t

题目详情

设函数f(x)的定义域为[-1,1],f[cos(α+
π
30
)]=tcos(2α+
π
15
)+sin(α+
π
5
)+cos(α+
11π
30
)

(1)若f(0)=-1,求t的值和f(x)的零点;
(2)记h(t),g(t)分别是f(x)的最大值、最小值,求函数F(t)=h(t)-g(t)的解析式.
题型:解答题难度:中档来源:不详

答案

(1)令α=class="stub"7π
15

∴f(cosclass="stub"π
2
)=tcosπ+sin(class="stub"2
3
π
)+cos(class="stub"5
 6
π
)=-t=-1
∴t=1
∴f[cos(α+class="stub"π
30
)]=cos(2a+class="stub"π
15
)+sin(α+class="stub"π
5
)+cos(a+class="stub"11π
30

=cos2(a+class="stub"π
30
)+sin[(a+class="stub"π
30
)+class="stub"π
6
]+cos[(a+class="stub"π
30
)+class="stub"π
3
]
=2cos2(a+class="stub"π
30
)+cos(a+class="stub"π
30
)-1
  令x=cos(a+class="stub"π
30

∴f(x)=2x2+x-1
∵-1≤x≤1
∴x1=-1 x2=class="stub"1
2

(2)f[cos(α+class="stub"π
30
)]=tcos(2a+class="stub"π
15
)+sin(α+class="stub"π
5
)+cos(a+class="stub"11π
30

=tcos2(a+class="stub"π
30
)+sin[(a+class="stub"π
30
)+class="stub"π
6
]+cos[(a+class="stub"π
30
)+class="stub"π
3
]
=2tcos2(a+class="stub"π
30
)+cos(a+class="stub"π
30
)-t 
  令x=cos(a+class="stub"π
30

∴f(x)=2tx2+x-t    x∈[-1,1],
当t>0时,函数f(x)开口向上
-class="stub"1
4t
≤-1时即0<t≤class="stub"1
4
,函数在[-1,1]上为增函数,最大值为h(t)=t+1,最小值为g(t)=t-1
-1<-class="stub"1
4t
<1时即t>class="stub"1
4
,函数在[-1,-class="stub"1
4t
]上为减函数,在[-class="stub"1
4t
,1]上为增函数,最大值为h(t)=t+1,最小值为g(t)=
-8t2-1
8t

当t=0时,函数在[-1,1]上为增函数,最大值为h(t)=1,最小值为g(t)=-1
当t<0时,函数f(x)开口向下
-1<-class="stub"1
4t
<1时即t<-class="stub"1
4
,函数在[-1,-class="stub"1
4t
]上为增函数,在[-class="stub"1
4t
,1]上为减函数,最大值为h(t)=
-8t2-1
8t
,最小值为g(t)=t-1
-class="stub"1
4t
≥1时即0>t≥-class="stub"1
4
,函数在[-1,1]上为减函数,最大值为h(t)=t-1,最小值为g(t)=t+1
∴F(t)=h(t)-g(t)=
2t+class="stub"1
8t
+1  ,t>class="stub"1
4
2               , 0≤t≤class="stub"1
4
-2              ,-class="stub"1
4
≤ t<0
-2t-class="stub"1
8t
-1   ,t<-class="stub"1
4

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