Question

已知函数f(x)=a^x-1/a^x (a>1) 且a^t*f(2t)+m*f(t)>=0 在t属于[1,2]上恒成立

Answer 1

a^t*f(2t)+m*f(t)

= (a^t)*[a^(2t)-1/(a^(2t))] + m*[a^t-1/(a^t)]

=a^(3t) - 1/(a^t) + m*[a^t - 1/(a^t)],let y = a^t,so y at [a,a^2]

上式 = y^3 - 1/y + m*(y - 1/y)

= y^3 +m*y -(m+1)/y = f(y) >=0,

m*(y-1/y) >= 1/y - y^3,

m >= [1/y - y^3]/(y-1/y) = (1-y^4)/(y^2-1) = -(y^2 + 1) >= -(a^2 + 1) ,because of y at [a,a^2]

so m>= -(a^2 + 1),