高分问几道高二的数学题

来源:百度知道 编辑:UC知道 时间:2024/05/13 08:50:59
1.求满足cos(sinx)=1/2 (-π/2<x<π/2)的角X的集合

2.f(x)=-acos2x-√3*asin2x+2a+b, a>0, 0《x《π/2 f(x)∈[-5,1],
求当t∈[-1,0]时函数y=a*t^2+bt-3的最小值

3.f(x)=a[2cos^2(x/2)+sinx]+b ,
A.若a=1求函数(x)的单调增区间
B.若a<0且当x∈[0,π]时,函数f(x)的值域是[3,4],求a和b的值

4.f(x)=sin(2x+φ) (-π<φ<0) ,y=f(x)图像的一个对称中心坐标为(-π/8,0)
A.求φ
B.求x∈R时函数f(x)的最小值,并求相应的x的取值集合

已知向量OA=e1 向量OB=e2 且O,A,B不共线,若向量AC等于3倍的向量AB 求向量OC

6.非零向量e1和e2不共线,要使ke1+e2和e1+ke2共线,确定实数K的值

1.求满足cos(sinx)=1/2 (-π/2<x<π/2)的角X的集合
-π/2<x<π/2
-1<sinx<1
cos(sinx)=1/2
sinx=π/3或sinx=-π/3
无解
角X的集合是空集

2.f(x)=-acos2x-√3*asin2x+2a+b, a>0, 0<=x<=π/2 f(x)∈[-5,1],
求当t∈[-1,0]时函数y=a*t^2+bt-3的最小值
f(x)=-acos2x-√3*asin2x+2a+b
=-2a[cos(π/3)cos2x+sin2xsin(π/3)]+2a+b
=-2acos(2x-π/3)+2a+b
0<=x<=π/2
-π/3<=2x-π/3<=2π/3
-1/2<=cos(2x-π/3)<=1
-5=b<=f(x)<=3a+b=1
a=2
b=-5

t∈[-1,0]
y=a*t^2+bt-3
=2tt-5t-3>=-3

3.f(x)=a[2cos^2(x/2)+sinx]+b
A.若a=1求函数(x)的单调增区间
f(x)=[2cos^2(x/2)+sinx-1+1]+b
=(cosx+sinx)+1+b
=√2sin(x+π/4)+(1+b)
单调增区间
(2kπ-3π/4,2kπ+π/4)

B.若a<0且当x∈[0,π]时,函数f(x)的值域是[3,4],求a和b的值
f(x)=a[2cos^2(x/2)+sinx-1+1]+b
=a(cosx+sinx)+(a+b)
=asin(x+π/4)+(a+b)
x∈[0,π]
sin(x+π/4)∈[-1,1]
3=b+2a<=f(x)<=b=4
b=4
a=-1/2

4.f(x)=sin(2x+