Question

帮我解决几道高一数学题 跪求 3

帮我解决几道高一数学题 跪求

Answer 1

13.(1)定义域R
（2）偶函数 f(-x)=2^[(-x)^2-1]=2^(x^2-1)=f(x)
(3)考虑到y=2^x是单调增函数 f(x)>=4即2^(x^2-1)>=2^2
故x^2-1>=2 所以 x<=-根下3或x>=根下3
8.sin2(x-π/4)=sin（2x-π/2）=-sin（π/2-2x）=-cos2x=2（sinx）^2-1=2-根下5
9. (tan10°-√3)sin40°
=(tan10°-tan60°)sin40°
=(sin10°/cos10°-sin60°/cos60°)sin40°
=sin40°*(sin10°*cos60-cos10°*sin60°)/(cos10°*cos60°)
=sin40°*sin(10°-60°)/(cos10°*1/2)
=-2sin40°*sin50°/cos10°
=[cos(40°+50°)-cos(40°-50°)]/cos10°
=(cos90°-cos10°)/cos10°
=(0-cos10°)/cos10°
=-1
10.
(1)定义域(x+5)/(x-5)>0且x-5不等于0
故定义域{x|x<-5或x>5}
(2)奇函数
f(-x)=lg[(-x+5)/(-x-5)]=lg[(x-5)/(x+5)]=-lg[(x+5)/(x-5)]=-f(x)....这里用到lg(1/x)=-lgx
（3）因为是奇函数 f(-a)=-f(a)=-2

Answer 2

13.(1)定义域R
（2）偶函数 f(-x)=2^[(-x)^2-1]=2^(x^2-1)=f(x)
(3)考虑到y=2^x是单调增函数 f(x)>=4即2^(x^2-1)>=2^2
故x^2-1>=2 所以 x<=-根下3或x>=根下3
8.sin2(x-π/4)=sin（2x-π/2）=-sin（π/2-2x）=-cos2x=2（sinx）^2-1=2-根下5
9. (tan10°-√3)sin40°
=(tan10°-tan60°)sin40°
=(sin10°/cos10°-sin60°/cos60°)sin40°
=sin40°*(sin10°*cos60-cos10°*sin60°)/(cos10°*cos60°)
=sin40°*sin(10°-60°)/(cos10°*1/2)
=-2sin40°*sin50°/cos10°
=[cos(40°+50°)-cos(40°-50°)]/cos10°
=(cos90°-cos10°)/cos10°
=(0-cos10°)/cos10°
=-1

Answer 3

看不亲