F (x) = asin (π x + a) + bcos (π x + β) + 4, and f (1999) = 3, find f (2000)

Because: F (1999) = asin (1999 π + a) + bcos (1999 π + β) + 4 = - Asina bcos β + 4 = 3,
So: Asina + bcos β = 1
So: F (2000) = asin (2000 π + a) + bcos (2000 π + β) + 4
= asina+bcosβ+4
=1+4
=5

Let f (x) = x (x-1) (X-2) (x-3). (x-99), then f '(0) =?
Find the derivative of F (0)

F '(0) = LIM (f (x) - f (0)) / (x-0) x approaches 0
=limf(x)/x
=(x-1) (X-2) (x-3). (x-99) brings 0 into
=-99!
This process is very clear

f(x)=x(x-1)(x-2)… (x-99), how to find f '(0) =?

Let g (x) = (x-1) (X-2) (x-99)
Then f (x) = x · g (x)
f'(x)=g(x)+x·g'(x)
So f '(0) = g (0) = - 99!

Machinery factory to cut the bottom and height are 2 decimeters of right angle triangle iron plate. There is a 21 meter long, 16 meter wide rectangular iron plate, can cut this kind of steel plate at most
How many triangle iron plates?

21÷0.2=105
16÷0.2=80
Total = 105 × 80 × 2 = 16800

How to use science to explain the common phenomena in daily life
It's a problem in daily life. I believe many people have encountered it. After you form the habit of looking at your watch tomorrow, you will find that you will look at your watch at this time every day. The accuracy can reach one minute
For example, someone has developed the habit of looking at his watch. One day, he subconsciously looked at the following table at 13:23. The next day, he happened to look at his watch at 13:23. The third day and the fourth day may all be at this time. But the action of looking at his watch is not continuous. It does not mean that he has been waiting for the arrival of 13:23 since 13:00, but no matter what he is doing, After such a natural look. Found that it was 13:23
Does this kind of imagination belong to the operation of natural reflection? If not, how to explain it?

If x and y are opposite to each other, the value of 2x2 + 2xy-1 is ()
A. -1B. 1C. 2D. 5

∵ X and y are opposite numbers, ∵ x + y = 0, ∵ 2x2 + 2xy-1 = 2x (x + y) - 1 = 0-1 = - 1

Given that a and B are nonzero matrices, and ab = 0, it is proved that (1) the column vectors of a are linearly correlated

If the column vector of a is linearly independent, then AX = 0 only if x = 0, so AB = 0 only if B = 0

The inscribed circle O of right triangle ABC cuts the hypotenuse AB to D, and proves s triangle ABC = ad * BD
The inscribed circle O of right triangle ABC cuts the hypotenuse AB to D, and proves s triangle ABC = ad times BD

Let ad = x, BD = y and the radius of inscribed circle be r
(x+r)^2+(y+r)^2=(x+y)^2
After expansion, we get: 2XR + 2yr + 2R ^ 2 = 2XY, omit 2, and add XY on both sides;
We get: XY + XR + yr + R ^ 2 = 2XY, and decompose on the left
We obtain that: (x + R) (y + R) = 2XY, AC = x + R, BC = y + R combined with tangent length theorem
The rest can be proved

It is known that f (x) = logax, a is greater than 0 and a is not equal to 1. Let f (A1), f (A2), f (an) be the arithmetic sequence of the first 4 tolerance 2
Given that f (x) = logax (a is greater than 0 and not equal to 1), Let f (A1), f (A2),... F (an) (n belongs to n ) be an arithmetic sequence with the first term of 4 and the tolerance of 2. (1) let a be a constant and prove that {an} is an arithmetic sequence
(2) If BN = an f (an), the sum of the first n terms of (BN) is Sn, then SN is obtained when A1 = radical 2

1) F (an) = 4 + 2 (n-1) = 2n + 2log (a, an) = 2n + 2An = a ^ (2n + 2), a is a constant, a (n + 1) / an = A & # 178; therefore, {an} is an equal ratio sequence; 2) A1 = radical 2, a ^ 4 = radical 2, a = 2 ^ (1 / 8) BN = an * f (an) = (2n + 2) * a ^ (2n + 2) = (2n + 2) * 2 ^ [(n + 1) / 4] use dislocation subtraction

Space vector problem
How to do it with space vector
In the cube abcd-a1b1c1d1 with edge length a, find the distance between section bdc1 and plane ab1d1
The answer is (√ 3 / 3) a

It can be solved by the distance from the point to the plane, for example, the distance from the point B (on the plane bdc1) to the plane ab1d1, that is, the length of the normal vector of the plane ab1d1 passing through the point B

F (x) = asin (π x + a) + bcos (π x + β) + 4, and f (1999) = 3, find f (2000)