题目
题型:解答题难度:一般来源:不详
1 |
3 |
a |
(1+x1)(1+x2) |
2 |
(1-4a-x1)(1-4a-x2) |
(1)求集合B;
(2)若x∈B,且x∈Z,求证:tan
1 |
x |
1 |
x |
(3)比较sin
1 | ||
|
1 | ||
|
答案
1 |
3 |
∴f′(x)=x2+4ax+a,
∵x1,x2∈A,∴f′(x)=0有两个实根,
∴x1+x2=-4a,x1x2=a,△=16a2-4a>0,
∴a>
1 |
4 |
∵(1+x1)(1+x2)=1+(x1+x2)+x1x2=1-4a+a=1-3a,
(1-4a-x1)(1-4a-x2)=1-8a+16a2+(4a-1)(x1+x2)+x1x2
=1-3a.
∵B={a|
a |
(1+x1)(1+x2) |
2 |
(1-4a-x1)(1-4a-x2) |
∴
a |
1-3a |
2 |
1-3a |
a-2 |
1-3a |
∴
(a-2)(1-1+3a) |
1-3a |
3a(a-2) |
3a-1 |
解得0<a<
1 |
3 |
综上所述,B={a|
1 |
4 |
1 |
3 |
(2)∵x∈Z,且x∈B,∴x≥2,∴
1 |
x |
1 |
2 |
令t=
1 |
x |
π |
2 |
则R′(t)=
cos2t+sin2t |
cos2t |
∴R(t)在(0,
π |
2 |
∴R(t)>R(0)=0,∴tant-a>0,
∴tan
1 |
x |
1 |
x |
(3)由(2)得x≥2时,tan
1 |
x |
1 |
x |
∵
2012 |
∴tan
1 | ||
|
1 | ||
|
1 | ||
|
1 |
2012 |
∴
sin2
| ||||
cos2
|
1 |
2012 |
1 | ||
|
1 | ||
|
∴2012•sin′(
1 | ||
|
1 | ||
|
∴2013sin′(
1 | ||
|
∵sin′(
1 | ||
|
1 |
2013 |
∵
1 | ||
|
π |
2 |
∴sin
1 | ||
|
1 | ||
|
核心考点
试题【已知函数f(x)=13x3+2ax2+ax+b(a≠0),A={x∈R|f′(x)=0},B={a|a(1+x1)(1+x2)-2(1-4a-x1)(1-4a-】;主要考察你对集合间的关系问题等知识点的理解。[详细]
举一反三
A.2 | B.2或4 | C.2或3或4 | D.无穷多个 |
(1)若M⊆N,求实数a的取值范围;(2)若M⊇N,求实数a的取值范围.
x |
x-1 |
A.M=N | B.M⊇N | C.N⊊M | D.M∩N=φ |