Given that R (x): SiNx + cosx > m, s (x): x ^ 2 + MX + 1 > 0, if any x belongs to R, R (x) is a false proposition and S (x) is a true proposition, the value range of M is obtained For R (x): ① radical 2sinx (x + &# 188; π) > M ｜ m ＜ - radical 2 Another false proposition 〈 m ≥ - radical 2 ② Root sign 2sinx (x + &# 188; π) > m False proposition The root sign 2sinx (x + &# 188; π) ≤ M ｜ m ≥ root 2 That's right. Why?

Given that R (x): SiNx + cosx > m, s (x): x ^ 2 + MX + 1 > 0, if any x belongs to R, R (x) is a false proposition and S (x) is a true proposition, the value range of M is obtained For R (x): ① radical 2sinx (x + &# 188; π) > M ｜ m ＜ - radical 2 Another false proposition 〈 m ≥ - radical 2 ② Root sign 2sinx (x + &# 188; π) > m False proposition The root sign 2sinx (x + &# 188; π) ≤ M ｜ m ≥ root 2 That's right. Why?

② That's right
The process is not completely reversible, and m ＜ - radical 2 is not a false proposition of the title. 2sinx (x + &# 188; π) > M
It is the false proposition
In addition,
If s (x) is a true proposition, then DALT = m ^ 2-4 > 0, that is, M > 2 or m-2
M > 2
Reminder: the statement above is open to question. The title says: for any x belongs to R, R (x) is a false proposition, not a x makes it false

Two propositions R (x): Sin & nbsp; X + cos & nbsp; X ＞ m, s (x): x2 + MX + 1 ＞ 0 are known. If there is and only one true proposition between R (x) and S (x) for ∀ x ∈ R, the value range of real number m is obtained

∀ SiNx + cosx = 2Sin (x + π 4) ≥ - 2, when R (x) is a true proposition, m ＜ - 2. For ∀ x ∈ R, s (x) is a true proposition, that is, X2 + MX + 1 ＞ 0 is tenable, with △ = M2-4 ＜ 0, ∀ 2 ＜ m ＜ 2

Two propositioners R (x) SiNx + cosx > m s (x): x-square + MX + 1 > 0, if this proposition x belongs to the range where r r (x) and S (x) have and only have one true proposition to find M
Two propositions R (x): SiNx + cosx > m s (x): x square + MX + 1 > 0, if this proposition x belongs to the range where r r (x) and S (x) have and only have one true proposition to find M

It's a difficult question. I can't do it. What's more, there are so few points?
If: x + MX + 1 > 0 is true
So, square M-4

If the function FX = 2m * cos ^ 2 (x / 2) + SiNx, the maximum value of the derivative is equal to the root sign 5
Then the real number m=

If the maximum value of the derivative of the function f (x) = 2m * cos ^ 2 (x / 2) + SiNx is equal to the root sign 5, then M is a real number=
f'(x)=4m*cos(x/2)(-sin(x/2))*1/2＋cosx=-msinx＋cosx=√(m^2＋1)sin(x＋α) (tanα=-1/m)
The solution is m = ± 2