Question

已知函数f(x)=log3mx^2+8x+n/x^2+1的定义域为R，值域为大于等于0小于等于2，求m,n的值

Answer 1

mx^2+8x+n>0的解为x∈R(显然m≠0) m>0 8-4mn>=0(1) m<0那是不可能的 0<=log3（(mx^2+8x+n)/(x^2+1))<=2 即1<=（(mx^2+8x+n)/(x^2+1))<=9 化简后, (m-1)x+8x+(n-1)>=0(x∈R)----->m>1,8-4(m-1)(n-1)>=0(2) (m-9)x+8x+(n-9)<=0(x∈R)----->m<9,8-4(m-9)(n-9)<=0(3) 由(1),(2),(3)得1<m<9, (m-1)(n-1)=16,(m-9)(n-9)=16(判别式等于零才有最值) 所以m=5,n=5