It is known that the tolerance D of the arithmetic sequence {an} is greater than 0, and A2 and A5 are two of the equations x2-12x + 27 = 0. The sum of the first n terms of the sequence {BN} is TN, satisfying TN = 2-BN (n ∈ n *) (I) find the general formula of the sequence {an}, {BN}; (II) let the sum of the first n terms of the sequence {an} be Sn, and note CN = (Sn - λ) · BN (λ∈ R, n ∈ n *). If C6 is the largest term in the sequence {CN}, find the value range of the real number λ

It is known that the tolerance D of the arithmetic sequence {an} is greater than 0, and A2 and A5 are two of the equations x2-12x + 27 = 0. The sum of the first n terms of the sequence {BN} is TN, satisfying TN = 2-BN (n ∈ n *) (I) find the general formula of the sequence {an}, {BN}; (II) let the sum of the first n terms of the sequence {an} be Sn, and note CN = (Sn - λ) · BN (λ∈ R, n ∈ n *). If C6 is the largest term in the sequence {CN}, find the value range of the real number λ

(I) the two of the equation x2-12x + 27 = 0 are the two of the equation x2-12x + 27 = 0, A2 + A5 = 12, a2a5 = 12, a2a5 = 27, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\∈ n *) (II) ∵ Sn = n [1 + (2n − 1)] 2 = N2 Then CN = (SN − λ) · BN = (N2 − λ) · (12) n − 1 when n ≥ 2, CN − cn − 1 = (N2 − λ) · (12) n − 1 − [(n − 1) 2 − λ] · (12) n − 2 = − N2 + 4N − 2 + λ 2n − 1 − C6 is the largest term in the sequence {CN}, cn-cn-1 ≤ 0 when n ≥ 7, cn-cn-1 ≥ 0 when n ≤ 23, n ≤ 6, cn-cn-cn-cn-1 ≥ 0 when n ≥ 14 − λ 23

It is known that the isochronous sequence {an}, tolerance is greater than zero, A2 and A5 are two of the equations x ^ 2-12x + 27 = 0, and the first n sum of the other sequence {BN} is Sn, and Sn=
1 - (1 / 2) * BN (1) the general formula for {an}, {BN} respectively (2) CN = an * BN (n = 1,2,3...) Compare CN + 1 with CN

(1) A [2] = 3 = a [1] + D, a [5] = 9 = a [1] + 4D, a [2] = 1 + 2 (n-1) = 2N-1 ∵ the sum of the first n terms of the sequence {B [n]} is s [n], and s [n] = 1-B [n] / 2 ∵ s [n + 1] = 1-B [n + 1] / 2

If the image of the function y = LG (x + a) has intersection with the negative half axis of X axis and the positive half axis of Y axis, then the value range of the real number a is?

The known conditions of the title are changed into:
The definition field of the function is less than 0 and the value field is greater than 0
We can get x0 = LG1
And logarithmic function must have X + a > 0
The solution is: a > 1

The image of the function y = 2x2 and the image of y = one-half of the square of X
A. The opening size is the same, B, opening direction is the same
C. The size and direction of the opening are different. D. the size and direction of the opening are the same

The size of the opening is determined by the absolute value of the quadratic coefficient
The direction is determined by the sign of the quadratic coefficient
The symbols are the same here
But the absolute value is different
So the opening direction is the same
Choose B

Which of the following functions has two intersections between the image and the X axis () a.y = quarter (x-23) square + 155
B. Y = quarter (x + 23) square + 155
C. - quarter (x-23) square - 155
D. - quarter (x + 23) square + 155

D
This is a quadratic function. You bring in the vertex x-coordinate = - 23. First, the opening is downward. And the value of y-coordinate is greater than 0. According to the image, there are two intersections

The coordinate of the intersection of the image of the first-order function y = - one third of X + 1 and the X axis is -- and the coordinate of the intersection of the image and the Y axis is——
The coordinate of the intersection of the image of the first-order function y = - one third of X + 1 and the X axis is -- and the coordinate of the intersection of the image and the Y axis is——

When y = 0, there is (- 1 / 3) x + 1 = 0, x = - 3, so the intersection coordinate of the first-order function and X axis is (- 3,0);
When x = 0, there is y = (- 1 / 3) × 0 + 1, y = 1, so the intersection coordinate of the first-order function and the Y axis is (0,1)

The coordinates of the intersection of the image of the function y = - 4x + 3 and the X axis are___ The coordinate of the intersection point with the y-axis is___ ．

Let y = 0 get: - 4x + 3 = 0, the solution is: x = 34, then the intersection coordinates of function and X axis are (34, 0); let x = 0, the solution is y = 3, then the intersection coordinates of function and Y axis are (0, 3). So the answers are: (34, 0) and (0, 3)

If the first-order function y = 4x-2 is known, then the coordinates of the intersection of the function image and the x-axis are?

The function y = 4x-2 can intersect the x-axis only when y = 0;
Therefore:
When compared with the x-axis, there are:
0=4x-2
Then:
x=（0+2）/4
x=2/4=1/2
So: the intersection coordinates of the first-order function y = 4x-2 and X axis are (1 / 2,0)

The coordinates of the intersection of the image of the function y = - 4x + 3 and the X axis are___ The coordinate of the intersection point with the y-axis is___ ．

Let y = 0 get: - 4x + 3 = 0, the solution is: x = 34, then the intersection coordinates of function and X axis are (34, 0); let x = 0, the solution is y = 3, then the intersection coordinates of function and Y axis are (0, 3). So the answers are: (34, 0) and (0, 3)

Quadratic function y = - x ^ 2 + 4x-2, vertex moves to m point on y = 1 / 2x + 1, image intersects with X axis in a, B, s triangle ABM = 8

Y = - (x-4) (X-8) no matter how you move, the shape of this function is exactly the same as y = x ^ 2. Therefore, the height of the triangle is always the square of (half of the bottom). Let the height be T ^ 2, the half of the bottom be t, and the area of the triangle = height * the half of the bottom = T ^ 3 = 8t = 2. Therefore, the height of the triangle is 4, the ordinate of the function vertex is 4,4 = 0.5x + 1x = 6, and the abscissa of the fixed point is 6