设函数y=sin(paix/2+pai/3),若对任意x∈R,存在X1,X2使f(x1)

设函数y=sin(paix/2+pai/3),若对任意x∈R,存在X1,X2使f(x1)
设函数y=sin(paix/2+pai/3),若对任意x∈R,存在X1,X2使f(x1)

设函数y=sin(paix/2+pai/3),若对任意x∈R,存在X1,X2使f(x1)
由题意可知f(x1)=f(x)min=-1
=>sin(π/2x1+π/3)=-1
=>π/2x1+π/3=2k1π-π/2
=>x1=1/(4k1-5/3)
同理f(x2)=f(x)max=1
=>sin(π/2x2+π/3)=1
=>π/2x2+π/3=2k2π+π/2
=>x2=1/(4k2+1/3)
|x1-x2|=|1/(4k1-5/3)-1/(4k2+1/3)|
当k1,k2趋近于无穷大时|x1-x2|趋近于0无最小值
应该是
y=sin(πx/2+π/3)吧,可得
x1=4k1-5/3
x2=4k2+1/3
则|x1-x2|=|4(k1-k2)-2|(因k1,k2为整数）
k1-k2=0,or,1时取最小值
此时|x1-x2|=2