2018-2019学年北师大版选修2-1 1.2.2 充分条件与必要条件习题课 作业
2018-2019学年北师大版选修2-1 1.2.2 充分条件与必要条件习题课 作业第2页

  由q,得B={x|a≤x≤a+1}.

  ∵q是p的必要不充分条件,∴A⊆B,

  ∴{■(a+1≥1"," @a≤1/2 "," )┤∴0≤a≤1/2.

答案:A

★6.已知集合A={x|a-2

A.0≤a≤2 B.-2

C.0

解析:A∩B=⌀⇔{■(a"-" 2≥"-" 2"," @a+2≤4)┤⇔0≤a≤2.

答案:A

7."x=2kπ+π/4(k∈Z)"是"tan x=1"成立的         条件.

解析:当x=2kπ+π/4,k∈Z时,tan x=1;但当tan x=1时,x不一定是2kπ+π/4,如x=5π/4,所以填"充分不必要".

答案:充分不必要

8.在平面直角坐标系xOy中,直线x+(m+1)y=2-m与直线mx+2y=-8互相垂直的充要条件是m=     .

解析:x+(m+1)y=2-m与mx+2y=-8互相垂直⇔1·m+(m+1)·2=0⇔m=-2/3.

答案:-2/3

9.方程x2+2mx-m+12=0的两实根都大于2的充要条件是            .

解析:设函数f(x)=x2+2mx-m+12,其图像如图.

  依条件得{■(Δ≥0"," @f"(" 2")" >0"," @"-" 2m/2>2"," )┤

  解得{■(m≤"-" 4"或" m≥3"," @m>"-" 16/3 "," @m<"-" 2"," )┤即-16/3

  故方程x2+2mx-m+12=0的两根均大于2的充要条件为-16/3

答案:m∈("-" 16/3 ",-" 4]

★10.已知命题p:|x-a|<1,命题q:x2-6x<0,若p是q的既不充分又不必要条件,则实数a的取值范围为            .