题目

y=(6cosx+sinx-5)/(2cosx-3sinx-5) 求值域

提问时间：2021-05-04

答案

设y=(6 Cos[x] + Sin[x] - 5)/(2 Cos[x] - 3 Sin[x] - 5)
则y'=(20 (-Cos[x] + Cos[x]^2 + Sin[x] + Sin[x]^2))/(-5 + 2 Cos[x] - 3 Sin[x])^2
令y'=0得
-Cos[x] + Cos[x]^2 + Sin[x] + Sin[x]^2=0,
即
1 + Sin[x] - Cos[x] = 0,
Cos[x] - Sin[x] = 1,
Sqrt[2] Sin[\[Pi]/4 - x] = 1,
Sin[π/4 - x] = 1/Sqrt[2]
π/4 - x = π/4+2kπ,或3π/4+2kπ,
- x = 2kπ,或π/2+2kπ,
x = 2kπ,或-π/2+2kπ,
代x = 2kπ,得Cos[x]=1,Sin[x]=0;
y=(6 - 5)/(2 - 5)=-1/3;
代x = -π/2+2kπ,得Cos[x]=0,Sin[x]=-1;
y=(-1 - 5)/(3 - 5)=3,
所以一个是最大值3,一个是最小值-1/3.