The parity of F (x) = ln (x + the square of 1 + x)

The parity of F (x) = ln (x + the square of 1 + x)

Just judge the value of F (- x) + F (x)
F (- x) + F (x) = ln (- x + root sign (the quadratic power of 1 + x) + ln (x + the square of 1 + x) = ln (1 + x ^ 2-x ^ 2) = ln1 = 0
So f (- x) = - f (x)
That is, the function is odd!

Given the 2x power of y = sec x + A + sin Pie / 3, it is a bit confusing to find dy,

0

0

dy/dx
=(dy/dt)/(dx/dt)
=(3a·cos²t·(-sin t) )/(3a·sin²t·cos t)
=cot t

If a primitive function of F (x) is the cubic power of X, then ∫ f (x) DX =? If dy = 2x power DX of Dy = e is known, then y =?

∫ f (x) DX = x ^ 3. The result of indefinite integral is the original function
dy = e^(2x) dx
dy/dx = e^(2x)
Y = ∫ e ^ (2x) DX = (1 / 2) ∫ e ^ (2x) d (2x) = (1 / 2) e ^ (2x) + C and C are arbitrary constants

Let y = 2E 2x and let x = te, y = t, y = t, dy / DX and D, Y / DX

y=2e^2x
y'=2e^2x*(2x)'=4*e^2x
dy/dt=2t*e^t+t²e^t=(t²+2t)*e^t
d²y/dt²=(2t+2)e^t+(t²+2t)e^t=(t²+4t+2)e^t
dx/dt=(t+1)e^t
d²x/dt²=(t+2)*e^t
So D? Y / DX? 2 = (D? Y / dt? 2) / (D? X / dt? 2) = (t? + 4T + 2) / (T + 2)

Given that a = {y | y = log2x, x > 1}, B = {y | y = (1 / 2) ^ x, x > 1}, then a ∩ B Why not {y | 0 ＜ y ＜ 1} but {y | 0 ＜ y ＜ 1 / 2} Detailed answers

A={y>0},B={y<1/2}
When x > 1, log2x > log2 (1) = 0, that is, Y > 0
Y = (1 / 2) ^ x, when x > 1, (1 / 2) ^ x < (1 / 2) ^ 1 = 1 / 2, that is, y < 1 / 2
So a ∩ B = {y| 0 < y < 1 / 2}

Given the set a = {y | y = log2x, X ＞ 1}, B = {y | y = (1) 2) If x, x > 1}, then a ∩ B=______ .

∵A={y|y=log2x，x＞1}={y|y＞0}，
B={y|y=（1
2）x，x＞1}={y|0＜y＜1
2}，
∴A∩B={y|y＞0}∩{y|0＜y＜1
2}={y|0＜y＜1
2}，
So the answer is: (0, 1
2)

Given that a set is equal to y = log2x, B = y = (1 / 2) * x, then their intersection is

Log2x = (1 / 2) * x, we can use derivative to find the derivative of log2x, 1 / X (LN2) = 1 / 2

If {y} = {x} is {y}, then {x} is {1}, then {y} is equal to {x} 1——————

The range of Y in a set is also [- 1,1]
The value range of Y in B is (- OO, 1]. Pay attention to the translation of the image combined with the inverse scale function
Therefore, the intersection is [- 1,1]

Let a = {1,2,3, K}, B = {4,7, the fourth power of a, the square of a + 3A}, and a belongs to N, K belongs to N, X belongs to a, y belongs to B, map f: A-B, make the element X in B correspond, and calculate the values of a and K

The corresponding rule is y = 3x + 1, so B = {4,7,10,3k + 1} = {4,7,10,3k + 1} = {4,7, a ^ 4, a ^ 2 + 3A} or 10 = a ^ 4, then a is not an integer, so 10 = a ^ 2 + 3a, get, a = 2 or a = - 5, because a belongs to N, so a = - 5 is rounded. So a = 2, then 3K + 1 = a ^ 4 = 16, k = 5A = 2, k = 5, a = 5, a = {1,2,3,5}, B = {4,7,7,7,7,7,7, B = {4,7,7,7,7,7,7,7,7,7,7, 7, 10,16}