Question

设tanα,tanβ是关于x的一元二次方程mx^2+(2m-3)x+(m-2)=0,当m变化时,求tan(α+β)的最小

Answer 1

方程有两根

Δ=(2m-3)^2-4m(m-2)≥0

-12m+9+8m≥0

m≤9/4

tan(α+β)

=(tana+tanb)/(1-tanatanb)

=-(2m-3)/m / (1-(m-2)/m)

=3-2m / (m-m+2)

=(3-2m)/2

≥(3-2*9/4)/2

=-3/4