Continuity of functions in higher numbers Discuss the continuity of the following functions and draw the f (x) graph 1. F (x) = LIM (n approaches infinity) 1 (1 + (cosx) ^ 2n) 2. F (x) = LIM (n approaches infinity) {[1-x ^ (2n)] \ [1 + x ^ (2n)]} * x Graphic words, a little description of what it looks like

Continuity of functions in higher numbers Discuss the continuity of the following functions and draw the f (x) graph 1. F (x) = LIM (n approaches infinity) 1 (1 + (cosx) ^ 2n) 2. F (x) = LIM (n approaches infinity) {[1-x ^ (2n)] \ [1 + x ^ (2n)]} * x Graphic words, a little description of what it looks like

one

Seeking function continuity with higher numbers
Piecewise function f (x) = 3x + 2 (x is less than or equal to 0)
x^2+1 （0

Domain x is not equal to 1
(I just said it on the computer.)
The left limit of X at x = 0: 2; limit: 1. FX is discontinuous at x = 0, which is the first kind of discontinuity (jump discontinuity)
The left limit of X at x = 1: 2; the right limit: 1. FX is also discontinuous at x = 1, which is also the breakpoint of jump

4.75-9.63+（8.25-1.37）

4.75-9.63+（8.25-1.37）=4.75-9.63+8.25-1.37=4.75+8.25-（9.63+1.37）=13-11=2．

Those numbers are rational
098, - π, 22 / 7, - 1245, 36,0, 5, 81, - 6, 500

098,22 / 71245, root 36,0, cubic root 81, - 6

Eight equations (one variable quadratic, one variable higher)
1. It is known that the quadratic equation x ^ 2-4ax + 5A ^ 2-6a = 0 has two real roots, and the absolute value of the difference between the two roots is 6. Find the value of A
2. Given that one root of the equation x ^ 3 + (m-1) x ^ 2 + (2-m) X-2 = 0 is 1 and the square root of the other two roots is 5, find the value of M and the other two roots
3. The ratio of the two elements of the equation x ^ 2 + ax + B = 0 is 3:4, and the discriminant △ = 2 - root sign 3
4. Let X be a real number and prove that the value of (x ^ 2-bc) (2x-b-c) ^ - 1 cannot be between B and C
5. Real numbers a, B, C satisfy a + B = 8, ab-c ^ 2 + (8 times root sign 2) C = 48, and find the root of equation BX ^ 2 + CX-A = 0
6. It is known that α and β are two unequal real roots of the equation x ^ 2 + PX + q = O. if α ^ 2 + α β + β ^ 2 = 3, q is proved
The sum of the squares of the other two roots is 5,
There is a root sign 3 in question 7, which is short of brackets,

1, Δ = 16A ^ 2-4 (5a ^ 2-6a) ≥ 0 | x1-x2 | = √ [(x1 + x2) ^ 2-4x1x2] = 62. Bring in x = 1, is it wrong? The idea is, bring in x = 1, find the value of M, that is, the equation factorization is (x-1) (x ^ 2 + ax + b) = 0, after expansion, get the value of a and B, can find the value of the other two, as for the value of those two

20 simple fifth grade Mathematical Olympiad questions and answers
Hurry!

1. There are some sweets, which are divided into 5 pieces and 10 pieces more for each person; if the number of existing people increases to 1.5 times of the original number, then 4 pieces for each person will be 2 pieces less?
[analysis and solution] method 1: suppose there are x people at the beginning, and the total number of sugars in the two methods remains the same, there are 5x + 10 = 4 × 1.5x-2, and the solution is x = 12, so there are 12 × 5 + 10 = 70 sugars
Method 2: after the number of people increased 1.5 times, each person was divided into 4 pieces, equivalent to the original number, each person was divided into 1.5 × 4 = 6 pieces
With these sugars, each person is divided into 5 pieces more than 10 pieces, and each person is divided into 6 pieces less than 2 pieces, so the total number of people at the beginning is (10 + 2) / (6-5) = 12, so the total sugar is 12 × 5 + 10 = 70 pieces
2. A and B each have a bag of sugar, less than 20 grains in each bag. If a gives B a certain amount of sugar, a's sugar is twice as many as B's; if B gives a same amount of sugar, a's sugar is three times as many as B's. then, how many grains of sugar do a and B have in total?
The total number of sugar should be a multiple of 3 or 4, that is, a multiple of 12. Because each bag of two bags of sugar does not exceed 20, the total number of sugar is no more than 40. Therefore, the total number of sugar can only be 12, 24 or 36
If the total number of sugars is odd multiple of 12, then "after B gives a the same amount of sugar", the sugar of a is odd multiple of 12 △ (3 + 1) × 3 = 9. Then after a gives B twice the same amount of sugar, the sugar of a is odd multiple of 12 ± (2 + 1) × 2 = 8
That is to say, an odd number plus an even number equals an even number, which is obviously impossible. Therefore, the total number of sugars cannot be an odd multiple of 12
Then a, B two children can only share an even number of times 12 sugar, that is 24
3. There are 42 students in class A and 48 students in class B. It is known that in a certain mathematics examination, the paper is graded according to the percentage system. The results of grading are the same in all classes. The average scores of all classes are integers, and the average scores are higher than 80. How many points is the average score of class a higher than that of class B?
[analysis and solution] method 1: because the average score of each class is an integer, and the total scores of the two classes are equal, the total score is not only a multiple of 42, but also a multiple of 48, so it is a multiple of [42,48] = 336
Because the average score of class B is higher than 80, the total score should be higher than 48 × 80 = 3840
Because of the 100 mark system, the average score of class a will not exceed 100, so the total score should not be higher than 42 × 100 = 4200
The total score of the two classes is 4032
Then the average score of class A is 4032 △ 42 = 96, and that of class B is 4032 △ 48 = 84
So the average score of class A is 96 - 84 = 12 points higher than that of class B
Method 2: the average score of class a × 42 = the average score of class B × 48, that is, the average score of class a × 7 = the average score of class B × 8. Because of the mutual quality of 7 and 8, the average score of class A is 8 times of a certain number, and the average score of class B is 7 times of a certain number. Because the average scores of the two classes are more than 80, not more than 100, the number can only be 12
So the average score of class A is 12 × (8 - 7) = 12 points higher than that of class B
4. A rural hydropower station charges Electricity by households. The specific regulations are as follows: if the monthly electricity consumption is not more than 24 kwh, it will be charged at 9 cents per kwh; if the monthly electricity consumption is more than 24 kwh, the excess part will be charged at 20 cents per kWh. It is known that in a certain month, family a paid 9.6 cents more than family B (the electricity consumption is calculated according to the whole degree). How much electricity did family a and family B pay?
[analysis and solution] if both Party A and Party B use more than 24 degrees of electricity, the difference between their electricity charges should be an integral multiple of 20 cents;
If both a and B use less than 24 kwh, the difference between them should be an integral multiple of 9 cents
Now 9.6 cents is neither an integral multiple of 2 cents nor an integral multiple of 9 cents, so the electricity consumption of a family exceeds 24 degrees, and that of B family does not exceed 24 degrees
Suppose that a uses 24 + X kWh and B uses 24-y kwh, with 20x + 9y = 96, x = 3 and y = 4
That is to say, if a uses 27 kilowatt hours of electricity and B uses 20 kilowatt hours of electricity, then B should pay 20 × 9 = 180 points = 1 yuan and 8 Jiao, then a pays 180 + 96 = 276 points = 2 yuan and 7 Jiao and 6 jiao
That is, Party A and Party B each pay 2 yuan 7 Jiao 6 Fen and 1 yuan 8 Jiao
5. The number of spring outings in primary one and secondary two schools is an integral multiple of 10. When traveling, the staff of the two schools can't take one car, and each car is as full as possible. Now we know that if both schools rent 14 seat coaches, they need to rent 72 such coaches; if both schools rent 19 seat coaches, they need to rent 14 seat coaches, Then the second primary school will rent seven of these cars more than the first?
[analysis and solution] let m be the number of tourists in the second spring festival and n be the number of tourists in the first spring
It is also known that the two schools need to rent a total of 72 14 seat vans, so 70 × 14 + 2 ≤ m + n ≤ 72 × 14, that is 982 ≤ m + n ≤ 1008
At the same time, it is known that m and N are both multiples of 10, so we have
The other four groups are solved because m and N are not multiples of 10
After testing, only meet the requirements
Therefore, 430 people participated in spring outing in primary school and 570 people participated in spring outing in secondary school
6. A tourist rowed a boat from the dock at 10:15, and he wanted to return to the dock no later than 13:00. The flow rate of the river was 1.4 kilometers per hour, and the speed of the boat in still water was 3 kilometers per hour. He took a 15 minute rest every 30 minutes, didn't change direction, and rowed back after a rest. How far can he paddle from the dock at most?
[analysis and solution] starting from 10:15, you must return no later than 13:00, so you can row for 2 hours and 45 minutes at most, that is 165 minutes. 165 = 4 × 30 + 3 × 15, you can row for 4 30 minutes at most, and rest for 3 15 minutes
The downstream speed is 3 + 1.4 = 4.4km/4, so the downstream half-hour rowing distance is 4.4 × 0.5 = 2.2km;
The speed of countercurrent is 3-1.4 = 1.6 km / 4, so the half-hour rowing distance of countercurrent is 1.6 × 0.5 = 0.8 km
After 15 minutes' rest, the distance of the boat is 1.4 × 0.25 = 0.35 km
In the first case, when the boat starts downstream, it will paddle for at least half an hour and drive 2.2km. During the three hours of rest, the boat will drift downstream for a distance of 0.35 × 3 = 1.05km. Therefore, it is necessary to paddle 2.2 + 1.05 = 3.25km when the boat returns upstream
It takes at least 30 + 15 × 3 + 121.875 = 196.875 minutes > 165 minutes, so it is too late to return the ship on time
The second situation: when the reverse current starts, every half an hour, the boat will travel 0.8 km. After three times of reverse current, the boat will travel 0.8 × 3 = 2.4 km. When the tourists rest, the boat will float 1.05 km downstream. So when rowing back, it will only take 2.4-1.05 = 1.35 km, 1.35 △ 4.4 ≈ 0.3068 hours ≈ 18.41 minutes. The total time is 3 × 30 + 3 × 15 + 18.41 = 153.41 minutes

How to find LIM (n →∞) cos (x / 2) cos (x / 4)... Cos (x / 2n)

A 2008 digit composed of 2 in 2008. After dividing this 2008 digit by 7, what is the remainder?
A 2008 digit composed of 2 in 2008. After dividing this 2008 digit by 7, what is the remainder?

A: the remainder is 3
031746 2008 divided by 6, the remaining 4 bits 2222 correspond to 0317, so the remainder is 3
two hundred and twenty-two thousand two hundred and twenty-two

How to calculate 24 with 3, - 5,7 and - 13

((-5)*(-13)+7)/3=24

Mathematica's solution of differential equation "Y '' + 4Y '+ 2 = 0" is written as dsolve [y' '[x] + 4 y' [x] + 2 y = = 0, y [x], x]. Why is it wrong to say "no variable"?

All punctuation in the code should be half angle punctuation in English state. Obviously, the two commas in your program are punctuation in Chinese. In addition, your two writing methods are not consistent. In the previous differential equation, there is no zero order derivative of Y. how can it be generated out of thin air later? Dsolve [y '' [x] + 4 y '[x] + 2 = = 0