题目
题型:解答题难度:一般来源:不详
3 |
3 |
(1)求函数f(x)的对称轴方程;
(2)当x∈(0,
π |
2 |
(3)若f(x0) =
2 |
5 |
π |
4 |
π |
2 |
答案
3 |
π |
3 |
令2x+
π |
3 |
π |
2 |
π |
12 |
kπ |
2 |
∴对称轴方程为:x=
π |
12 |
kπ |
2 |
(2)∵x∈(0,
π |
2 |
π |
3 |
π |
3 |
4π |
3 |
∴sin(2x+
π |
3 |
| ||
2 |
∴2sin(2x+
π |
3 |
3 |
∵函数g(x)=f(x)+m有零点,即f(x)=-m有解.(8分)
即-m∈(-
3 |
3 |
(3)f(x0)=
2 |
5 |
π |
3 |
2 |
5 |
2 |
5 |
π |
3 |
4 |
5 |
4 |
5 |
∵x0∈(
π |
4 |
π |
2 |
∴2x0+
π |
3 |
5π |
6 |
4π |
3 |
又∵sin(2x0+
π |
3 |
4 |
5 |
∴2x0+
π |
3 |
4π |
3 |
∴cos(2x0+
π |
3 |
3 |
5 |
∴sin2x0=sin[(2x0+
π |
3 |
π |
3 |
=sin(2x0+
π |
3 |
π |
3 |
π |
3 |
π |
3 |
=(-
4 |
5 |
1 |
2 |
3 |
5 |
| ||
2 |
=
3
| ||
10 |
核心考点
试题【已知函数f(x)=2sinxcosx+23cos2 x-3+2(1)求函数f(x)的对称轴方程;(2)当x∈(0,π2)时,若函数g(x)=f(x)+m有零点,】;主要考察你对函数的零点存在定理等知识点的理解。[详细]
举一反三
A.(0,1) | B.(1,2) | C.(2,3) | D.(3,4) |
1 |
x |
A.(0,1] | B.(1,2] | C.(2,3] | D.(3,4] |