High School Math Emperor, Solution, Math Emperor, High School One Problem

1)f'(x)=3x^2+2ax+b

Since the slope of the tangent at one point p(1,f(1)) on the image of the function f(x) is 12, so f'(1)=3+2a+b=-12

3 is an extreme point of the function f(x).

So f'(3)=27+6a+b=0

The solution yields a=-3 and b=-9

2) From (1) know f(x) = x 3-3x 2-9x + 10 f'(x)=3x 2-6x-9>0.

x<-1 or x>3

So x (-1],[3,+ f'(x) > 0, so the function f(x) increases monotonically.

f at x (-1,3).'(x) > 0, so the function f(x) decreases monotonically.

Knowing that f(x)=x +ax +bx+10, the tangent slope at a point p(1,f(1)) on the image of the function f(x) is 12, and x=3 is the function f(x).

an extreme point; (1) Find the value of the real numbers a, b; (2) Find the monotonic interval of the function f(x).

Solution: (1).f (x) = 3x + 2ax + b, f (1) = 3 + 2a + b = -12, that is, there is 2a + b + 15 = 0....1)

f (3) = 27 + 6a + b = 0, i.e. there are 6a + b + 27 = 0....2)

2)-(1) 4a+12=0, so a=-3, b=-9;

2)f(x)=x³-3x²-9x+10；f′(x)=3x²-6x-9=3(x²-2x-3)=3(x-3)(x+1)

When x -1 or x 3 f (x) 0, so in the interval (-1] [3,+ the inner function f(x) increases monotonically;

When -1 x 3 f (x) 0, the function f(x) decreases monotonically in the interval [-1,3].

What does it mean that 3 is an extreme point of the function f(x)? Does it mean that 3 is an extremum?

1: It is known that f(x) is an odd function defined on r, and when x is greater than 0, f(x) = (x cubic) + x+1

Find the analytic formula for f(x).

Solution: Since f(x) is an odd function defined on r, f(-x) = -f(x).

When x is greater than 0, f(x) = (x cubic) + x+1 Let any x satisfy x<0 so -x is greater than 0

f(-x)=-f(x)=-(x-cubic)-x+1 So when x<0 f(x)=(x's cubic )+x-1 so this is a piecewise function (1)f(x)=(x-cubic)+x+1 x>0

2) f(x) = (x to the third power) + x-1 x<0

2: Let the odd function f(x) be an increasing function over the interval [3,7], and f(3)=5, find the maximum value of the interval [-7,-3].

Solution: The odd function f(x) So f(-x)=-f(x) is an increasing function according to the odd function properties [3,7] and at [-7,-3] is also an increasing function, so the maximum value of [-7,-3] is f(-3)=-f(3)=-5

3: It is known that f(x) is an odd function defined on [-2,2] and an increasing function in the defined domain, and the solution inequality f(square of x-1) + f(3x+1) is greater than or equal to 0

The solution f(x) is an odd function defined on [-2,2] and an increasing function within the defined domain i.e.

f(0)=0 f(x's squared -1) + f(3x+1) is greater than or equal to 0

So f(x's squared -1) is greater than or equal to -f(3x+1) so.

f(x's squared -1) is greater than or equal to f[-(3x+1)] because the multiplication function is defined in [-2,2].

So the square of x-1 is greater than or equal to [-3x+1)] 1).

2 or = x squared - 1 or = 2....2) -2 "or=(3x+1)"or=2...3)

From 123 we get -1" or =x or =0

Problem 1 If x is less than 0, then -x is greater than 0, f(-x)=-f(x)=-(x's cubic power)-x+1, then f(x)=(x's cubic )+x-1

The solution is as follows: since sin 2a+cos 2a=1, the original function can be reduced to:

y=-t^2+at-1/2a-1/2

Because the value range of t is [-1,1], the image of the original function is a segment on the function y=-t 2+at-1 2a-1 2, and it is necessary to discuss the increase and decrease of this function in the interval [-1,1], so that t = 1 and -1 are compared with the size, and the quadratic function about a is obtained, and the function y=-t 2 + at-1 2a-1 2 is obtained by comparing its size, and the increase and decrease of the function y=-t 2+at-1 2a-1 2 in [-1,1] can be obtained, so that the original function y=sin 2x+acosx-1 2a-3 can be known When 2 takes the maximum value, the value of cosx (i.e., t) is brought in, and y=1 is brought in, and the value of a can be solved.

The latter part of the idea is very simple, but it's too troublesome to play on the computer, so I wrote down the idea, ask the landlord to do the math himself, it shouldn't be very difficult.

This question is not very complicated.

First of all, you need to satisfy that x+1-a in 2 is monotonically increasing, because the coefficient of x is greater than zero, so there is no problem of single increase, if you want to satisfy that r is singular increase, so you need to satisfy that x+1-a is greater than or equal to zero (because the critical value of the function in 1 is zero) is true when x=0, that is, a is less than or equal to 1

When x<=0, f(x)=-x 2 is incremented by the function.

f(0)=0

When x>0.

f(x)=x+1-a

This is also an increment function.

Then as long as 0+1-a>=f(0)=0, f(x) is guaranteed to increment a<=1 on r

a<=1

f(x)=①.-x squared (x less than or equal to 0). x+1-a(x is greater than 0).

x squared (x less than or equal to 0), this quadratic function opens downward, the axis of symmetry is x=0, and increases monotonically in the part less than 0, the maximum limit value is at x=0, f(x)=0 (note: the limit maximum), so in this interval f(x) < 0

x+1-a (x is greater than 0), this primary function is a straight line, the slope is 1, and it increases monotonically in the part greater than 0, so its minimum limit value is at x=0, f(x)=1-a (note: the limit minimum), so it is in this range.

f(x)>1-a

Ensuring that the limit minimum of the segmentation function is greater than or equal to the maximum value of the interval limit ensures that the function is an incrementing function over the entire defined domain.

So a<=1

The graph solution makes the line always convenient on a quadratic function, and the solution should be less than 1

(1) Only one person from A and B: C21 C83; A and B are both selected: C22 C82 2 All three of them stand alone:

a73；There are two people standing together: C32 A72 3 AB is not selected: C54; AB chooses one of them:

C21 C53 4 The three regions are divided into 1, 1 and 2 college students: C41 C31 C22 A22 A33; 1,1 is evenly distributed

5 Negative considerations: C93 C53

The first question, choose any 4 out of 10 people, for C10 4, here you need to subtract the situation that A and B are not selected, that is, these four people are selected among the remaining 8 people, and 4 out of 8 people are selected arbitrarily, for C8 4, then the answer is C10 4 - C8 4

The second question, considering it this way, if 3 people stand on the steps, there must be one person standing, for C3 1, and if one person stands, there are 7 ways to stand, that is, C7 1, and there are six steps left, and the remaining two people, standing casually, are C6 1*C6 1, so the answer is C3 1*C7 1*(C6 1*C6 1).

The third question is to use the method of elimination, using the possibility of all course selection minus the possibility of ab choosing both subjects, all the course selection methods are, c7 4, the combination of ab two subjects, is to choose two of the remaining 5 subjects, that is, c5 2, therefore, the answer is c7 4 -c5 2 (ps, there is no order of course selection).

The fourth question, this question is reversed, 4 people, divided into three places, in terms of places, there is a region divided into two college students, this place is C3 1, the college students are divided into C4 2, and there are two college students left, after the whereabouts of one of them are determined, the other is determined, that is, C2 1, so the answer is C3 1 * C4 3 * C2 1

The fifth question, this question is also in no order, as long as the red ball is taken out, then all the ways to take it are c9 3, but here you need to subtract the ball that is taken out, it is the possibility of white balls, if they are all white balls, it is c5 3, so the answer is c9 3-c5 3

Brother, there is something I want to say to you, you must not have listened to the class! It's not so simple, but it's certain that you don't need others to help you, you can figure it out, but the college entrance examination won't test any simple questions, for this kind of question, you should look at the list of questions, read a few more, figure it out, it's OK, remember to rely on yourself.

I don't have time, I'll answer the first question for you first: The opposite of at least one person from A and B is that at most 0 people from A and B will not participate, so that only 8 students can participate, and 4 people from 8 people will be selected, C8 4Then calculate the unrestricted selection method and select 4 people out of 10 people, C104

Then the calculation is c10 4-c8 4

Step-by-step counting principle high school elective textbook Do more questions and you will understand how to use permutations and combinations of formulas.

y= (so?)

Then the root number in the number should be greater than or equal to 0, and the true number should be greater than 0, i.e

4x^2-3x＞0 ②

Solution -1 4 x 1

Solution x 3 4 or x 0

So define the domain as -1 4 x 0 or 3 4 x1

The root number should be non-negative, so it should be greater than zero, so 0<4x 2-3x<=1, we get 3 4

Vector ac=(cosx-2,sinx).

Vector bc = (cosx, sinx-2).

AC vertical BC, then (cosx-2)*cosx+sinx*(sinx-2)=0

Then sinx+cosx=1 2, combined with sin x+cos x=1 to solve sinx and cosx, and then get tanx mobile phone questions can only enter 100 words, and the result is calculated by yourself.

Vector ac = (cosx -2, sinx) vector bc = (cosx, sinx -2) both perpendicular, (cosx) 2 - 2cosx + sinx) 2 -2sinx = 0

sinx + cosx = 1/2

2)sin(x +y) = 1/2, tgy = 1, y = π/4

x = arcsin( 2 4) - 4 - 4 The rest is calculated by yourself.

ac perpendicular to bc has: (cosx-2) x cosx + sinx x(sinx-2)=0;

So: 1-2(cosx+sinx)=0, again(cosx) 2+(sinx) 2=1;Combine the two formulas, and 0