Question

已知m,n是实数,且m²+mn+n²=3设t=m²-mn+n²,则t的取值范围

Answer 1

m²+mn+n²=3
(m+ n/2)²+3n²/4=3
令m+ n/2=√3cosa，n=2sina
则戚判m=√3cosa- n/2=√3cosa -sina

t=m²-mn+n²
=(m²+mn+n²)-2mn
=3 -2mn
=3-2(√3cosa-sina)(2sina)
=3-4√3sinacosa+4sin²a
=3-2√3sin(2a)+2[1-cos(2a)]
=3-2√3sin(2a)+2 -2cos(2a)
=5-4[(√3/2)sin(2a)+(1/2)cos(2a)]
=5-4sin(2a+π/6)
sin(2a+π/6)=-1时，t有最大值9；sin(2a+π/灶昌6)=1时，t有最小值高辩改1。
t的取值范围为[1，9]。