Question

3道高一下数学的解答题（大家帮忙啊！！）

Answer 1

1)y=f(x)的周期为2π,只要算出[0,2π]上的解,再加上周期就可以了
x∈[0,2π]时,x/2∈[0,π],f(x/2)=1/2,
∴x/2=π/6或5π/6,
x=π/3或5π/3
解集为{x|x=2kπ+π/3或2kπ+5π/3,k∈Z},就是{x|x=2kπ±π/3,k∈Z}
2)T=2π/m,
最大值对应的x是2kπ/m,最小值对应的x是(2kπ+π)/m(k是整数)
①2kπ/m<=2π<3π<=(2kπ+π)/m,解得k<=m<=(2k+1)/3
要使有解,k<=(2k+1)/3,(2k+1)/3>=0,-1/2<
k<=1,
∴k=0或k=1
k=0,
0<m<=1/3;
k=1,1<=m<=1
∴0<m<=1/3,或m=1,即(0,1/3]或m=1
②(2k-1)π/m<=2π<3π<=2kπ/m,解得(2k-1)/2<=m<=2k/3
要使有解,(2k-1)/2<=2k/3,2k/3>0,0<
k<=3/2,
∴k=1
k=1,1/2<=m<=2/3
∴1/2<=m<=2/3,即[1/2,2/3]
综上m∈(0,1/3]或[1/2,2/3]或m=1
3)方程应该是9^x+2a(3^x)+2=0吧
令t=3^x,
t^2+2at+2=0(x>0,t>1)
f(t)=t^2+2at+2,f(t)=0在(1,无穷)上有两个根
f(1)=3+2a>0,
a>-3/2①
对称轴x=-a>1,
a<-1②
△=4a^2-8>=0,
a>=√2或a<=-√2③
由①②③得
-3/2<a<=-√2,即(-3/2,-√2]

Answer 2

1.该题就是在问sin(x/2)=1/2，在x∈[0,2π]区间内，取值就可以了。
2.因为：f(x)=cosmx(m>0)没有取到1，也没有取到-1，
所以：mx不等于π的倍数，且x在区间(2π，3π)内
则：可以求得m
的值
3.方程两边同加a^2,得：9x^2+2a·(3^x)+2+a^2=a^2
（3x+a）^2=a^2-2
3x1=(a^2-2)^0.5-a
,
3x2=-(a^2-2)^0.5-a
(且a^2-2大于等于0)

Answer 3

1)y=f(x)的周期为2π,只要算出[0,2π]上的解,再加上周期就可以了
x∈[0,2π]时,x/2∈[0,π],f(x/2)=1/2,
∴x/2=π/6或5π/6,
x=π/3或5π/3
解集为{x|x=2kπ+π/3或2kπ+5π/3,k∈Z},就是{x|x=2kπ±π/3,k∈Z}
2)T=2π/m,
最大值对应的x是2kπ/m,最小值对应的x是(2kπ+π)/m(k是整数)
①2kπ/m<=2π<3π<=(2kπ+π)/m,解得k<=m<=(2k+1)/3
要使有解,k<=(2k+1)/3,(2k+1)/3>=0,-1/2<
k<=1,
∴k=0或k=1
k=0,
0<m<=1/3;
k=1,1<=m<=1
∴0<m<=1/3,或m=1,即(0,1/3]或m=1
②(2k-1)π/m<=2π<3π<=2kπ/m,解得(2k-1)/2<=m<=2k/3
要使有解,(2k-1)/2<=2k/3,2k/3>0,0<
k<=3/2,
∴k=1
k=1,1/2<=m<=2/3
∴1/2<=m<=2/3,即[1/2,2/3]
综上m∈(0,1/3]或[1/2,2/3]或m=1
3)方程应该是9^x+2a(3^x)+2=0吧
令t=3^x,
t^2+2at+2=0(x>0,t>1)
f(t)=t^2+2at+2,f(t)=0在(1,无穷)上有两个根
f(1)=3+2a>0,
a>-3/2①
对称轴x=-a>1,
a<-1②
△=4a^2-8>=0,
a>=√2或a<=-√2③
由①②③得
-3/2<a<=-√2,即(-3/2,-√2]

Answer 4

1)y=f(x)的周期为2π,只要算出[0,2π]上的解,再加上周期就可以了
x∈[0,2π]时,x/2∈[0,π],f(x/2)=1/2,
∴x/2=π/6或5π/6,
x=π/3或5π/3
解集为{x|x=2kπ+π/3或2kπ+5π/3,k∈Z},就是{x|x=2kπ±π/3,k∈Z}
2)T=2π/m,
最大值对应的x是2kπ/m,最小值对应的x是(2kπ+π)/m(k是整数)
①2kπ/m<=2π<3π<=(2kπ+π)/m,解得k<=m<=(2k+1)/3
要使有解,k<=(2k+1)/3,(2k+1)/3>=0,-1/2<
k<=1,
∴k=0或k=1
k=0,
0<m<=1/3;
k=1,1<=m<=1
∴0<m<=1/3,或m=1,即(0,1/3]或m=1
②(2k-1)π/m<=2π<3π<=2kπ/m,解得(2k-1)/2<=m<=2k/3
要使有解,(2k-1)/2<=2k/3,2k/3>0,0<
k<=3/2,
∴k=1
k=1,1/2<=m<=2/3
∴1/2<=m<=2/3,即[1/2,2/3]
综上m∈(0,1/3]或[1/2,2/3]或m=1
3)方程应该是9^x+2a(3^x)+2=0吧
令t=3^x,
t^2+2at+2=0(x>0,t>1)
f(t)=t^2+2at+2,f(t)=0在(1,无穷)上有两个根
f(1)=3+2a>0,
a>-3/2①
对称轴x=-a>1,
a<-1②
△=4a^2-8>=0,
a>=√2或a<=-√2③
由①②③得
-3/2<a<=-√2,即(-3/2,-√2]