Quick high Chinese mathematics, the process is detailed

Hello! Because a=0, b=-1, so.

f（x）=e^x-x

The derivative of the composite function f(x)=f(x)+g(x) is f'(x)=f'(x)+g'(x).

So f'(x)=e x-1

Let f'(x) 0, and solve x to belong to (0,+

Let f'(x) 0, and solve x to belong to (0,+

So the increasing interval of f(x) is that x belongs to (0,+

The minus interval is that x belongs to (- 0).

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by a=0, b=-1, f(x)=e x-x

The derivative yields f'(x)=e x-1

f'(x)=0, i.e., when x=0, the function can obtain an extremum.

f'(x)>0 is an increasing function, that is, when x belongs to (0, + is the increasing function f'(x) 0, the solution is that x belongs to (- 0) is a decreasing function.

f（x）=e^x+ax^2+bx

Finding the derivative, f '(x)=e x+2ax+bWhen a=0, b=-1, we get.

f ‘（x）=e^x-1

Let f '(x)=e x-1=0, x=0, and when x>0, f '(x)>0, and when x<0, f '(x)<0, and f(x) have a monotonic increase interval of [0,+

The monotonic reduction interval is (- 0).

In fact, the above person's approach is also possible, but it is estimated that this is the first sub-problem, if you substitute a and b values too early, it is not very convenient to solve the two sub-problems of 2 and 3.

Derivative, f'x=e x-1 is less than 0 from negative infinity to 0, and greater than 0 from 0 to positive infinity, fx decreases monotonically from negative infinity to 0 and increases monotonically from 0 to positive infinity.

p q is a true proposition, then only one of p and q is true, first look at the proposition p: subtract the function on r, then 2a-5<0, get a <; Looking at the proposition q, let the function y=x -ax+2 so that y<0 at xx (1,2), that is, x=1 and x=2 bring in the similar y<0, so that 1-a+2<0 and 4-2a+2<0 get a>3; So the range of the real number a is a (negative infinity, positive infinity).

If p is false, then 2a-5>=0, a>=5 2; If q is a true proposition, then 1*1-1*a+2<=0; 2*2-2*a+2<=0;The solution is a>=3; If q is false, a<3Then when p is false and q is false, <=a<3, then when "p q" is a true proposition, a《or a>=3

p q is a true proposition, so only one of p and q is true.

<> look at the process, there should be no problem, dry shouting finch (o).

13 typical area-type probability problems. Compare the shadow area to the circle area.

Let the radius of the circle be r, the side length of the inscribed square is a, and the side length of the regular octagon is b. Then a= 2 r, i.e. r=a2. Meet in a regular octagon:

ae+ee'+e'b=ab, i.e., b+ 2 2 b+ 2 2 b=a. Get b = a (1 + 2).

The area of the regular octagon s1=(2+2 2)*b

The area of the circle s2= *r

Substitute to find it.