The point nearest to the origin is () A. （π12，0）B. （-π12，0）C. （-π6，0）D. （π6，0）

The point nearest to the origin is () A. （π12，0）B. （-π12，0）C. （-π6，0）D. （π6，0）

From 4x + 2 π 3 = k π (K ∈ z), we can get x = k π 4 - π 6. When k = 0, we can get point a (- π 6, 0). When k = 1, we can get point (π 12, 0). It is obvious that the nearest point between the image of π 12 ＜ π 6 and the x-axis intersection of the function y = - 52sin (4x + 2 π 3) is (π 12, 0)

If the function f (x) is a decreasing function defined on [- 10,10] and is an odd function, if f (1-m) + F (2m-1) > 0 holds, the value range of real number m is obtained

∵ f (x) is a decreasing function defined on [- 10,10], and is an odd function
∴ -10≤1-m≤10
-10≤2m-1≤10
And f (2m-1) = - f (1-2m)
∴-4.5≤m≤5.5
f(1-m)-f(1-2m)>0
∴f(1-m)>f(1-2m)
∵ f (x) is a decreasing function defined on [- 10,10]
∴1-m

It is known that the odd function f (x) = x ^ 3 + ax ^ 2 + BX + C is an increasing function defined on [- 1,1]
Find the value range of real number B;
2 if B ^ 2-tb + 1 ≥ f (x) is constant for X ∈ [- 1,1], find the value range of real number t

In this interval, the minimum value of F '(x) is f' (0) = b > = 0, so there is: b > = 02. The minimum value of F (x) on [- 1,1] is f (- 1) = - 1-B, the maximum value is f (1) = 1 + B, so there is: B ^ 2-tb + 1 > = 1 + BB

Let f (x) = x power of 4 divided by (x power of 4 + 2), if 0 is less than a and less than 1, try to find the value of F (a) + F (1-A)
Further find the value of F (1 / 1001) + F (2 / 1001) + F (3 / 1001) +. + F (1000 / 1001)

F (a) + F (1-A) = (a power of 4 / a power of 4 + 2) + (1-A power of 4 / 1-A power of 4 + 2)
=2 (a power of 4 + 1-A power of 4 + 4) / (a power of 4 + 2) (1-A power of 4 + 2)
=2 (a power of 4 + 1-A power of 4 + 4) / (a power of 4 + 1-A power of 4 + 4) = 2
five hundred

Is the sum of all elements in the set a = {x | X & # 178; - (a + 2) x + A + 1 = 0, X ∈ r} two solutions or one

The solution is x = 1,
x=a+1
The sum is a + 2

Write out the elements in the set composed of the solutions of the equation x & # 178; - (a + 1) x + a = 0
The answer is clear, but I don't know how to describe this set. I hope the process can be written in detail

The solutions are 1 and a, and the set of solutions is {1, a}

Let a = {x | X & # 178; + (A-1) + B = 0} have only one element a, and find the value of a + B

There is only one element a in the set a = {x | X & # 178; + (A-1) x + B = 0}
therefore
a²=b
2a=1-a
3a=1
a=1/3
b=1/9
therefore
a+b=1/3+1/9=4/9

Let f (x) = x2 + ax + B, a = {x | f (x) = x} = {a}, and the set composed of elements (a, b) be m

∵ a = {x | f (x) = x} = {a}, the two equal roots of the equation x2 + ax + B = x are a. ∵ x2 + (A-1) x + B = 0, both are a. ∵ a + a = - (a − 1) a · a = B, ∵ a = 13b = 19, and the set of ∵ elements (a, b) is m, ∵ M = {(13, 19)}

Let f (x) = x2 + ax + B, a = {x | f (x) = x} = {a}, and the set composed of elements (a, b) be m

∵ a = {x | f (x) = x} = {a}, the two equal roots of the equation x2 + ax + B = x are a. ∵ x2 + (A-1) x + B = 0, both are a. ∵ a + a = - (a − 1) a · a = B, ∵ a = 13b = 19, and the set of ∵ elements (a, b) is m, ∵ M = {(13, 19)}

Let f (x) = x2 + ax + B, a = {x | f (x) = x} = {a}, and the set composed of elements (a, b) be m

∵ a = {x | f (x) = x} = {a}, the two equal roots of the equation x2 + ax + B = x are a. ∵ x2 + (A-1) x + B = 0, both are a. ∵ a + a = - (a − 1) a · a = B, ∵ a = 13b = 19, and the set of ∵ elements (a, b) is m, ∵ M = {(13, 19)}