The axis area of a cone is an equilateral triangle whose side length is twice the root 3. The volume of the cone is equal to

The axis area of a cone is an equilateral triangle whose side length is twice the root 3. The volume of the cone is equal to

π 3 times the root

It is known that the cross section of a cone passing through its axis is an equilateral triangle, and the cross section area is 3cm & # 178;, so the bottom area of the cone can be obtained

The section of the axis is an equilateral triangle,
Area of equilateral triangle = 1 / 2 * a * (√ 3) / 2A = (√ 3) / 4A & # 178; = 3 (√ 3), where a is the side length
a²=12
The bottom area of the cone = π * (A / 2) & # 178; = π A & # 178 / 4 = 3, π = 9.42cm
The mathematics tutoring group will answer for you,

If the ratio of the side area to the bottom area of the cone is radical 2, then the apex angle of the cone shaft section is

Let R be the bottom radius of the cone and l be the generatrix length
So π RL = √ 2 π R & # 178;
L=√2R
The section of the cone axis is an isosceles triangle,
The waist is l and the bottom is 2R
L²+L²=（2R）²
Therefore, the apex angle of conical shaft section is 90 degrees

If the generatrix of a cone is 2 cm long and the cross-sectional area passing through the apex and the center of the bottom is 2 square cm, then the side area of the cone is? The answer is 2 (radical 2) π

The top solution of the section is a, the section area = 2 = 1 / 2 * 2 * Sina, Sina = 1, then a = 90 degrees
Bottom diameter 2R = inclined side length of section = 2 √ 2cm
Bus length L = 2
Cone side area s = R / L * π * L ^ 2 = √ 2 / 2 * π * 4 = 2 √ 2 π cm ^ 2

It is known that the image vertex coordinates of quadratic function f (x) = ax ^ 2 + BX + C are (1, - 2), and the analytic expression of function f (x) is obtained through point P (2,1) ①
② If x ∈ [- 4,2], find the maximum and minimum of the function

Let this function be
y=a(x-1)^2-2
P (2,1) generation
A = 3
So y = 3 (x-1) ^ 2-2
f(x)=3x^2-6x+1

Let f (x) = AX2 + BX + C (a, B, C ∈ R). If x = - 1 is an extreme point of the function y = f (x) ex, then the following image cannot be ()
A. B. C. D.

From y = f (x) ex = ex (AX2 + BX + C) {y ′ = f ′ (x) ex + ExF (x) = ex [AX2 + (B + 2a) x + B + C], from x = - 1 as an extreme point of function f (x) ex, - 1 is a root of equation AX2 + (B + 2a) x + B + C = 0, so there is a - (B + 2a) + B + C = 0 {C = a

Given the function f (x) = x ^ 3-ax-1, please prove that the image of F (x) = x ^ 3-ax-1 cannot always be above the line y = a

Certification:
If f (x) = x ^ 3-ax-1, the image cannot always be above the line y = a, that is, when x takes a certain value, f (x) may be less than y, f (x)

In the following figures, it is impossible to be the image of a function y = f (x)______

In the function y = f (x), for every x value, there can only be a unique y corresponding to it. The image of the function y = f (x) and the line parallel to the Y axis can only have at most one intersection point

If we want to know the function f (x) = LNX ax & # 178; - x, a ∈ R, if the function y = f (x) is a monotone increasing function in its domain of definition, we can find the value range of A
We try to use derivatives

Y = LNX ax & # 178; - x, domain x > 0,
y'=1/x-2ax-1=-(2ax^2+x-1)/x
A is not equal to 0
Let y '> = 0
Let 2aX ^ 2 + X-10 be constant
If a > 0, then x = - 1 / (4a) 0 is an increasing function
If A0,
Y = 2aX ^ 2 + X-1 is an increasing function on (0, - 1 / (4a)), and a decreasing function on (- 1 / (4a), positive infinity),
So y (max) = y (- 1 / (4a)) = - (8a + 1) / (8a)

Known 1

1
Let y = f (x) = log (a) (x + √ (x ^ 2-1))
Then:
x + √(x^2-1) = a^y
√(x^2-1) = a^y - x
x^2 - 1 = (a^y - x)^2 = x^2 - 2x*a^y + a^2y
2x*a^y = a^2y + 1
x = (a^2y + 1) / 2a^y = [ a^y + a^(-y) ] / 2
So the inverse function of F (x) is:
f^(-1)(x) = [ a^x + a^(-x) ] / 2
The domain of the inverse function of F (x) is the domain of F (x)
Obviously, x + √ (x ^ 2-1) is an increasing function of X, and x > 1
x+√(x^2-1) > 1+√(1^2-1) = 1
But a > 1
f(x) = log(a)( x + √(x^2-1) ) > log(a)(1) = 0
The range of F (x) is: (0, + ∞)
The domain of f ^ (- 1) (x) is also: (0, + ∞)
2、 The second question needs to take advantage of the first
be aware:
g(x) = [ 2^x + 2^(-x) ] / 2 …… （1）
f^(-1)(x) = [ a^x + a^(-x) ] / 2 …… （2）
The two expressions are very similar except that the base of the former is 2 and the latter is a
For (2), the inverse function of F (x) is itself
x = log(a)( f^(-1)(x) + √( [f^(-1)(x)]^2 - 1) ) …… （3）
Similarly, for (1), there is:
x = log(2)( g(x) + √(g^2(x)-1) ) …… （4）
The topic is to compare the size of G (x) and f ^ (- 1) (x) when x is equal
Therefore, we compare g (x) with f ^ (- 1) (x) under the condition that (3) (4) is equal
log(a)( f^(-1)(x) + √( [f^(-1)(x)]^2 - 1) ) = log(2)( g(x) + √(g^2(x)-1) )
Due to a