High School Math Full Weight Quantifier Questions? 10

1. For the full proposition p:, which contains a quantifier"Arbitrary"The negation of x m,p(x) p is:"exists"x∈m，┐p(x)。

2. For a special proposition p:, which contains a quantifier"There is one"The negation of x m,p(x) p is:"All of them"x∈m，┐p(x)。

Full name proposition Special name proposition.

1.For all x a, p(x) holds 1The presence of x a makes p(x) hold.

2.For everything x a, p(x) holds 2There is at least one x a for p(x) to hold.

3.For each x a, p(x) holds 3For some x a, make p(x) hold.

4.Choose any x a, p(x) to form 4For a certain x a, make p(x) true.

5.Where x a, p(x) is established 5There is an x a, which makes p(x) hold.

Also: The negation of a proposition is a total negation, not a partial negation.

When negating the universal proposition, special attention should be paid to some propositions that omit the universal quantifier, such as the absolute value of the real number is positive. It would be wrong to write "the absolute value of a real number is not positive", and the correct denial would be: "The absolute value of a real number is not a positive number." ”

Commonly used "all" to indicate the full name of the affirmation, its existence negation is "not all", the two are mutual negation, with "neither" to indicate the full name of the negation, its existence affirmation can be expressed by "at least one is"...

In short, it is to remember that the negation of a proposition is a complete negation, not a partial negation. If you grasp this, you will basically not be wrong.

sinx+cosx= 2sin(x+45), because x r, so 2 sinx+cosx 2, then according to the title, m should be the maximum value of sinx+cosx, that is, m 2

1 All solutions: because when a belongs to a, non-p is a true proposition, i.e., when a belongs to a, for any real number x, ax 2 + ax + 1> = 0 is constant.

So a>0

i.e.: a>0

a^2-4a<=0

Get: 0 Another: If a=0, the proposition is also true.

In summary, it can be seen that 0<=a<=4

If P is not true, then P is false, indicating that the result of P is false.

Set A is (0,4).

You can find the square of ax + ax + 1<0 to be true, how to make the square of =a -4a>=0, and the solvable a>=4 or a<=0

To make the value of a a does not belong to the set of a.

A is the set that is (0,4).

That is, find the minimum value of 2x (x2+1), and the minimum value is greater than m. The method of finding the minimum value is to divide the numerator and denominator by x (when x is not equal to zero) and then use the basic inequality to find the minimum value to be -1, so m<-1

Method 2: MX -2x+m<0 is constant, (1) m=0, cannot be constant, (2) m is not equal to 0, mx -2x+m<0 is constant, because m<0, and the discriminant formula < 0

So m -1>0 and m <0 to get m<-1

The question is: m<2x x square + 1 constant formation. Let f(x)=2x x-square+1 to find the minimum value of f(x).

f(x) = 2x x square + 1 = 2 x + x one. It is divided into two cases>0 and x<0, and the basic inequality can be used, and when x=0, just check it.