The solution set of inequality (m2-2m-3) X2 - (M-3) X-1 ＜ 0 of X is r, and the value range of real number m is obtained

The solution set of inequality (m2-2m-3) X2 - (M-3) X-1 ＜ 0 of X is r, and the value range of real number m is obtained

(1) When m2-2m-3 = 0, i.e. M = - 1 or M = 3, if the solution set of the original inequality is r, then M = 3 (2 points) (2) when m2-2m-3 ≠ 0, to make the solution set of the original inequality R, there are: M2 − 2m − 3 ＜ 0 (m − 3) 2 + 4 (M2 − 2m − 3) ＜ 0 {− 1 ＜ m ＜ 3} 15 ＜ m ＜ 3 {− 15 ＜ m ＜ 3 (10 points) the range of m in synthesis (1) (2) is − 15 ＜ m ≤ 3 (12 points)

If the inequality x ^ 2-2mx-3 > 0 holds for all m (M is between 0 and 1, including 0 and 1), find the value range of X

Consider the left as a function of degree M
Because it is once, so both ends can be satisfied
That is, x ^ 2-3 > 0 and x ^ 2-2x-3 > 0
We can find the range of X we need
Classical solution

The inequality x ^ 2-2mx-1 > 0 holds for all 1 ≤ x ≤ 3, and the range of M is obtained

The answer is complex or simple. Let me show you which one

Given 0 ＜ a ＜ 1, the inequality a & # 178; X-1 ≤ a (solution set of x-1)

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