How to prove that xsinx is unbounded It is proved that xsinx is bounded on (0, + infinity)

How to prove that xsinx is unbounded It is proved that xsinx is bounded on (0, + infinity)

Let x = k * PI + pi / 2, then SiNx = 1, let n be any one; if k > n / PI + 1 / 2, then xsinx > n

How to prove f (x) = xsinx unbounded with more rigorous method?

If we want to use a rigorous method, we can only use the definition. So, according to the definition, for any given m greater than 0, I can find x0 = 2 * ([M] + 1) * PI + pi / 2, so that the absolute value of F (x0) is greater than m, so it is unbounded

In RT △ ABC, ∠ ACB = 90 °, CD ⊥ AB in D, AE bisection ⊥ BAC = BC in E, through C, e, D to make a circular intersection, AE in G, CD and AE in F, proving: Ag = FG

It is proved that: as shown in the figure, if GD is connected, then ∠ DCE = ∠ DGE = ∠ DAG + ∠ ADG, ∧ ACB = 90 °, CD ⊥ AB is in D, ∧ BDC = ∠ ADC, ∧ ACB = ∠ BCD + ∠ ACD = 90 °, ∧ DAC + ∠ ACD = 90 °

Let the total distance be an unknown number
A group of students from a to B walk at a speed of 4 km / h. After walking for 1 km, a student returns to a at a speed of 5 km / h, takes the schoolbag he forgot in a, and then rushes back to the team at the same speed. As a result, he catches up with the team at a distance of 1.5 km from B to find the distance between a and B

Let the total distance be s
After 1 km, the team and the ordered students use the same time
Team walk s-1-1.5
Students ordered to walk 1 + s-1.5
Then (s-2.5) / 4 = (s-0.5) / 5
S = 10.5km
A: the distance from place a to place B is 10.5km
Hope to help you!

Given a (2,3), B (5,4), C (7,10), if AP = AB + λ AC, point P is in the fourth quadrant, then the value range of λ is___ ．

∵ a (2,3), B (5,4) ∵ AB = (5-2,4-3) = (3,1). Similarly, we can get AC = (5,7) let P (x, y), then AP = (X-2, Y-3) ∵ AP = AB + λ AC ∵ X-2 = 3 + 5 λ Y-3 = 1 + 7 λ, that is, x = 5 + 5 λ y = 4 + 7 λ, we can get P (5 + 5 λ, 4 + 7 λ) ∵ point P is in the fourth quadrant, ∵ 5 + 5 λ {% >%

In the equilateral triangle ABC, e is a point in the triangle, AE = 3, be = 4, CE = 5
In the equilateral triangle ABC, e is a point in the triangle, AE = 3, be = 4, CE = 5. Find the degree of angle AEB

Rotate the triangle AEC around point a to triangle AFB and connect Fe
Then the triangle AFB is equal to the triangle AEC
So AF = AE, FB = CE, angle Fab = angle EAC
So angle Fab + angle BAE = angle EAC + angle BAE
So angle FAE = angle BAC
Because the middle angle BAC of equilateral triangle ABC is 60 degrees
So the angle FAE is 60 degrees
Because AF = AE
So the triangle AFE is an equilateral triangle
So the angle FEA = 60 degrees, Fe = AE = 3
Because FB = CE = 5, be = 4
So FB ^ 2 = be ^ 2 + Fe ^ 2
So the angle Feb = 90 degrees
Because the angle FEA = 60 degrees
So angle AEB = angle FEA + angle Feb = 60 + 90 = 150 degrees

The sixth power of x plus 6 times the fifth power of x plus 15 times the fourth power of x plus 20 times the cube of x plus 15 times the square of x plus 6 times x plus 1
How to merge

x^6+6x^5+15x^4+20x^3+15x^2+6x+1
=x^4(x^2+2x+1)+4x^3(x^2+2x+1)+6x^2(x^2+2x+1)+4x(x^2+2x+1)+(x^2+2x+1)
=x^4(x+1)^2+4x^3(x+1)^2+6x^2(x+1)^2+4x(x+1)^2+(x+1)^2
=(x+1)^2(x^4+4x^3+6x^2+4x+1)
=(x+1)^2[x^2(x^2+2x+1)+2x(x^2+2x+1)+(x^2+2x+1)]
=(x+1)^2[x^2(x+1)^2+2x(x+1)^2+(x+1)^2]
=(x+1)^4(x^2+2x+1)
=(x+1)^6

Find a point on the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1 so that its distance to the right focus is equal to twice the distance to the left focus
Process and result to be calculated

Let (m, n) (M-3) ^ 2 + n ^ 2 = 4 [(M + 3) ^ 2 + n ^ 2] (M-3) ^ 2-4 (M + 3) ^ 2 = 3N ^ 2 (M-3 + 2m + 6) (m-3-2m-6) = 3N ^ 2 (M + 1) (- M-9) = n ^ 2 substitute ellipse m ^ 2 / 25 + (M + 1) (- M-9) / 16 = 19m ^ 2 + 250m + 625 = 0m = - 25, M = - 25 / 9m = - 25, n have no solution, M = - 2

In the triangle ABC, ab = 15, BC = 14, CA = 13, find the height ad on the side of BC

The ∵ ad is high, and ∵ abd and ∵ ACD are right triangles
∴AD²=AB²-BD²=AC²-DC²
And DC = bc-bd
∴AB²-BD²=AC²-(BC-BD)²
∵AB=15 BC=14 CA=13
∴15²-BD²=13²-(14-BD)²
15²-BD²=13²-14²+28BD-BD²
∴28BD=15²+14²-13²=252
∴BD=9
∵AD²=AB²-BD²
∴AD²=15²-9²=144
That is, ad = 12
The height on BC side ad = 12

If we know that the imaginary number (X-2) + Yi is the root 3, we can find the maximum value of Y / X
I haven't learned anti sine, please use other methods

(X-2) ^ 2 + y ^ 2 = 3 center coordinate (2,0) radius √ 3
k=y/x
K (max) is the slope of the tangent of the line y = KX and the circle (X-2) ^ 2 + y ^ 2 = 3
Just draw a picture
tanα=k=√3/1=√3
k(min)=-√3