If the solution set of inequality ax ^ 2-6x + A ^ 2 about X less than 0 is (1, m), then what is m?

If the solution set of inequality ax ^ 2-6x + A ^ 2 about X less than 0 is (1, m), then what is m?

If the solution set of inequality ax ^ 2-6x + A ^ 2 < 0 about X is (1, m)
So obviously a ≠ 0, and know that the parabola is open upward, that is a > 0
So 1 and m are solutions of the equation AX ^ 2-6x + A ^ 2 = 0
So a-6 + A ^ 2 = 0
Then a = 2 or a = - 3 (rounding off)
According to Weida's theorem, 1 * m = a ^ 2 / a = a = 2
So m = 2
If you don't understand, please hi me, I wish you a happy study!

When x tends to 0, to make X-1 and ax of 3 equal infinitesimal, then a=

Because when x → 0, a ^ X-1 ~ x * LNA
So 3 ^ X-1 ~ x * Ln3
So a = Ln3

The sum of the first n terms of the sequence {an} is Sn, A1 = 5, an = Sn + 1. If {BN} satisfies anbn = - 1n, then the general term formula of {an}
If {BN} satisfies an * BN = (- 1) ^ n, then the general term formula of {an}, the first n term and TN of {BN} are given

There is no recurrence relation between an and A1,
And an = Sn + 1 -- a (n + 1) = s (n + 1) + 1 -- a (n + 1) - an = s (n + 1) + 1 - [Sn + 1] = a (n + 1)
——》An = 0, contrary to anbn = (- 1) ^ n