已知函数f(x)=2sin(1/2x+π/3)+1,x∈R的最大值和最小值分别为3和-1

答：
f(x)=2sin(x/2+π/3)+1
最大值为2+1=3,最小值为-2+1=-1
F(x)=f(x)+lnk=2sin(x/2+π/3)+1+lnk=0在[-π/6,π]上有且仅有2个零点
所以：-(1+lnk)=2sin(x/2+π/3)
因为：-π/6<=x<=π
所以：-π/12<=x/2<=π/2,-π/12+π/3<=x/2+π/3<=π/2+π/3
所以：π/4<=x/2+π/3<=5π/6
所以：sin(5π/6)<=sin(x/2+π/3)<=sin(π/2)
所以：1/2<=sin(x/2+π/3)<=1
所以：2*sin(π/4)<=-(1+lnk)<2*1
所以：√2<=-(1+lnk)<2
所以：-2<1+lnk<=-√2
所以：-3解得：1/e^3本题需要绘制简图结合理解