If the volume of the cone and the sphere is equal, and the radius of the cone bottom is equal to the diameter of the sphere, the ratio of the cone side area to the sphere area is () A. 2：2B. 2：1C. 5：2D. 3：2

Let the radius of the ball be: R, so the volume of the ball be: 4 π 3r3. Let the height of the cone be: H, because the volume of the cone and the ball is equal, ∧ 4 π 3r3 = 13 π (2R) 2h, ∧ H = R, the generatrix of the cone is: R2 + (2R) 2 = 5R, the surface area of the ball is: 4 π R2, the side area of the cone is: 12 × 4 π R · 5R = 25 π R2, and the ratio of the side area of the cone and the spherical area is 5:2

A hemisphere has a common bottom. If the volume of the cone is equal to that of the hemisphere, find the ratio of the height to the radius of the cone

Let the common radius be r, the common surface be s, the height of the cone be h, the volume of the hemisphere be 1 / 2 * 4 / 3 π ^ 3 ------ the volume of the cubic cone of one-half times four-thirds π R is 1 / 3SH, because s = π R ^ 2, the volume of the cone is converted to 1 / 3 * π R ^ 2 * h ------ the square of one-third π r times h, the hemisphere is known

It is known that the radius of a ball is r, the height of a cone is equal to the diameter of the ball, the surface area is equal to the surface area of the ball, and the volume of the cone

Cone surface area = Πr √ (R ^ 2 + 4R ^ 2) + Πr ^ 2
Sphere area = 4 Π R ^ 2
Then Π R √ (R ^ 2 + 4R ^ 2) + Πr ^ 2 = 4 Π R ^ 2
r^2=R^2/2
Volume of cone
=∏r^2*2R/3=∏R^2/2 *2R/3=∏R^3/3

Given that the inverse function of the function f (x) = 3 ^ (x + 1) + 9 ^ X-12 is f ^ - 1 (x), then the value of f ^ - 1 (6) is

Let f (x) = 6
Then 3 ^ (x + 1) + 9 ^ X-12 = 6
That is, (3 ^ x) ^ 2 + 3 (3 ^ x) - 18 = 0
The solution is (3 ^ x) = 3 or - 6
So (3 ^ x) = 3
x=1
F ^ - 1 (6) = 1

If f (x) = x ^ 2 + 2aX + 1 (x > B-1) has inverse function, find the minimum value of a ^ 2 + B ^ 2

The axis of symmetry of F (x) is x = - A
According to the question, B-1 > = - A
a+b>=1
2(a^2+b^2）>=(a+b)^2>=1
a^2+b^2>=1/2
When a = b = 1 / 2, the equal sign holds
So the minimum value is 1 / 2

If f ^ - 1 (x) is the inverse function of the function f (x) = 1 / 2 (a ^ x-a ^ - x), (a > 1), then the value range of X that makes f ^ - 1 (x) > 1 hold

Find the inverse function of y = 1 / 2 (a ^ x-a ^ - x), that is, 2Y = a ^ x-a ^ - x, and take logarithm on both sides to get
Ln (2Y) = xlna + xlna = 2xlna, x = ln2y ln (a ^ 2)
So f ^ - 1 (x) = ln2x - ln (a ^ 2)
So if f ^ - 1 (x) > 1, that is ln2x - ln (a ^ 2) > 1, both sides are deformed by E
2X > e + A ^ 2, i.e. x > 1 / 2 (E + A ^ 2)

Let F-1 (x) be the inverse function of the function f (x) = 1 / 2 (a ^ x-a ^ - x) (a > 1), then the value range of X with F-1 (x) > 1 is

F-1 (x) is the inverse of F (x), so
Make the value range of F-1 (x) larger than the X definition field of 1
namely
The range of F (x) over the domain of definition greater than 1
Therefore, when x > 1, f (x) = 1 / [2 (a ^ X - A ^ - x)] range
(here, I would like to ask you whether a ^ X - A ^ - x is in denominator or numerator. You don't have a clear expression. I understand it in terms of denominator. If it is numerator, it will be simpler.)
Let a ^ x = t
Because a > 1, x > 1
t > 1
g(t) = t - 1/t
When t approaches 1, G (T) approaches 0; when t approaches positive infinity, G (T) also approaches positive infinity
therefore
g(t) ∈(0, +∞）
That is, a ^ X - A ^ - x ∈ (0, + ∞)
1/[2(a^x - a^-x) ∈ (0, +∞)
That is to say, the range of F (x) in the domain x > 1 is (0, + ∞)
Then the range of its inverse function on (0, + ∞) is x > 1
The range of X is x > 0

If the inverse function of function f (x) = 2 ^ - X-2 ^ x is F-1 (x), what is the value range of X that makes F-1 (x) greater than 2

The domain of inverse function is the domain of original function
The domain of inverse function is the domain of original function
The range of inverse function is known here, and the domain of definition is obtained
It is known that the original function definition field is x > 2, and the value range of F (x) is calculated
Y = - x is a decreasing function, base 2 is greater than 1, and 2 ^ x is an increasing function
So y = 2 ^ - x is a decreasing function
And - 2 ^ x is also a decreasing function
So f (x) = 2 ^ - X - 2 ^ x is a decreasing function
x>2
So f (x) 2 is X

If the inverse function image of function f (x) = a ^ x passes (2, - 1)

a=1/2

Given the inverse function g (x) = (A-1) x + 1-1 of function f (x), if f (3) = - 2, find the value of F (9)
Is g (x) = (A-1) ^ (x + 1) - 1

f(3)=-2
The results show that G (- 2) = 3 = (A-1) ^ (- 1) - 1 = 1 / (A-1) - 1,
The solution is a = 5 / 4
So g (x) = 1 / 4 ^ (x + 1) - 1 = 9,
The solution is: x = - 1-log4 (10)
That is, f (9) = - 1-log4 (10)

If the volume of the cone and the sphere is equal, and the radius of the cone bottom is equal to the diameter of the sphere, the ratio of the cone side area to the sphere area is () A. 2：2B. 2：1C. 5：2D. 3：2