Question

四道数学题我不会做!

1.已知数列｛an}满足an+1/an=n+2/n(n是正整数），且a1=1,.则a5=_______

2.已知sinα+cosα=√3÷3(0＜α＜π），则cos2α=____________

3.函数f(x)=sinx-√3cosx(x属于［-π，0］)的单调递增区间是_________

4.二次不等式kx2-2x+3k<0(k≠0)的解集是全体实数，则k的范围是？

请大家帮我详细地解答每一道题，谢谢!

Answer 1

1.an+1/an=(n+2)/n
an/an-1=(n+1)/(n-1)
..a5/a4=6/4
a4/a3=5/3
a3/a2=4/2
a2/a1=3/1
(an/an-1)(an-1/an-2)..(a4/a3)(a3/a2)(a2/a1)=(n+1)!/[(n-1)!*2]=(n+1)n/2
a1=1,(a5/a4)(a4/a3)(a3/a2)a2/a1)=a5/a1=a5=6*5/2=15

2
sina+cosa=√3/3
√2sin(a+π/4)=√3/3
sin(a+π/4)=√6/6 0<sin(a+π/4)<√(1/2)
π>a+π/4>3π/4
3π/2>a>π/2
π<2a<3π/2,cos2a<0
(sina+cosa)^2=1/3=1+2sin2a,sin2a=-1/3
cos2a=-2√2/3

3
f(x)=2sin(x-π/3)
-π/2<x-π/3<π/2递增
-π/6<=x<=5π/6
x属于[-π,0],那么递增区间[-π/6,0]

4
kx^2-2x+3k<0
k<0
判别式4-12k^2<0
k^2>1/3
K<-√3/3

Answer 2

第一个题是矛盾的，将1带入，会发现2=3,错题吧

Answer 3

鄙视