If and only if the function y = AX2 + BX + C (a ≠ 0) passes through the origin______ ．

If and only if the function y = AX2 + BX + C (a ≠ 0) passes through the origin______ ．

If the function y = AX2 + BX + C (a ≠ 0) passes through the origin, that is, O (0, 0) is on the image, and (0, 0) is replaced by the analytic expression, then C = 0. Conversely, if C = 0, then y = AX2 + BX, when x = 0, y = 0, that is, y = AX2 + BX + C (a ≠ 0) passes through the origin, so the answer is: C = 0

We know that the image of quadratic function y = ax ^ 2 + BX + C (a is not equal to 0) has an opening upward and passes through the point (- 1, - 2) (1,0)
(A) When the value of Y increases, the value of x increases
(B) When x > 0, the value of function y decreases with the increase of x value
(C) There is a negative number x 0, so that when x < x 0, the function value y decreases with the increase of x value; when x > x 0, the function value y increases with the increase of x value
(D) There is a positive number x 0, so that when x < x 0, the function value y increases with the increase of x value; when x > x 0, the function value y decreases with the increase of x value

Opening upward, a > 0
y(-1)=a-b+c=-2
y(1)=a+b+c=0
By subtracting the two formulas, we get: 2B = 2, B = 1
So the axis of symmetry is x = - B / (2a) = - 1 / (2a)

The monotonicity of function f (x) = logax + 1 X-1 (a > 0 and a ≠ 1) on (1, + ∞) is discussed and proved

Let u = x + 1 X-1, and let any one of x 2 ＞ x1 ＞ 1, then u 2-u 1 = x 2 + 1x2-1-1-1-1-1-1-x1 + 1x1-1 = (x2 + 1) (x1 + 1) (x + 1-1) (x-1-1-x-1, let u = u = x + 1 + 1-1, let any one of x 2 ＞ x1 ＞ 1, then u 2-u 1 = U 1 = u 2-u 1 = x 2-u 1 = x 2 + 1 + 1-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1 when 0 < a < 1, y= In conclusion, when a > 1, f (x) = logax + 1 X-1 is a decreasing function on (1, + ∞), and when 0 < a < 1, f (x) = logax + 1 X-1 is an increasing function on (1, + ∞)

Let f (x) = a1x + a2x & sup2; + +anxⁿ,a1,a2,a3,… An constitutes the arithmetic sequence, where n is a positive number and f (1) = n & sup2; 1
Let f (x) = a1x + a2x & sup2; + +anxⁿ,a1,a2,a3,… An forms the arithmetic sequence, where n is a positive even number and f (1) = n & sup2;
1) Find an
(2)f(1/3)

1)
f(1)=a1+a2+...+an=n^2
a1+a2+...an-1=(n-1)^2
By subtracting the two formulas, we get an = n ^ 2 - (n-1) ^ 2 = 2N-1
2)
f(x)=x+3x^2+.(2n-1)x^n
f(1/3)=1/3+3(1/3)^2+.(2n-1)(1/3)^n
1/3f(1/3)=(1/3)^2+3(1/3)^3+.(2n-1)(1/3)^(n+1)
The result of subtracting the above two formulas
2/3f(1/3)=1/3+2(1/3)^2+2(1/3)^3+.2(1/3)^n-(2n-1)(1/3)^(n+1)
=1/3+2[(1/3)^2-(1/3)^(n+1)]/(1-1/3)-(2n-1)(1/3)^(n+1)

The function f (x) = a1x + a2x2 + a3x3 + +Anxn (n ∈ n *), and A1, A2 The general formula of the sequence {an} is______ ．

f（1）=a1+a2+a3+… +An = N2, then a1 + A2 + a3 + +An-1 x = (n-1) 2 (n ≥ 2), subtracting the two formulas, an = N2 − (n − 1) 2 = 2N-1 (n ≥ 2), and when n = 1, A1 = 1, so an = 2N-1, so the answer is: an = 2N-1

Let f (x) = a1x + a2x & sup2; + +anxⁿ,a1,a2,a3,… An forms the arithmetic sequence, where n is a positive even number and f (1) = n & sup2;
1)an
(2)f(½)

The k-th coefficient AK = 2k-1, that is, an = 2N-1;
F (1 / 2) can refer to the formula of "dislocation subtraction" to find the sum of sequence and prove the problem

It is known that f (x) = a1x + a2x + a3x + +Anx, and A1, A2, A3 And then f (1) = the square of N, f (-)
It is known that f (x) = a1x + a2x + a3x + +Anx, and A1, A2, A3 Then f (1) = the square of N, f (- 1) = n, find the general term an of the sequence

f(1)=a1+a2+a3+.+an=n^2=S
a=S-S=n^2-(n-1)^2=2n-1
a=S=1
So a = 2N-1

Let f (x) = 2x cosx, {an} be an arithmetic sequence with tolerance tt / 8, f (A1) + F (A2) + If f (A5) = 5tt, then f [(A3)] ^ 2-a1a3=

f(a1)+f(a2)+f(a3)+f(a4)+f(a5)=2(a1+a2+a3+a4+a5)-(cosa1+cosa2+cosa3+cosa4+cosa5)
=10a3-(cosa1+cosa2+cosa3+cosa4+cosa5)
=10a3-[cos(a3-2π/8)+cos(a3-π/8)+cosa3+cos(a3+π/8)+cos(a3+2π/8)]
=5π
10a3-5π=[cos(a3-2π/8)+cos(a3-π/8)+cosa3+cos(a3+π/8)+cos(a3+2π/8)]
=[cos(a3-2π/8)+cos(a3+2π/8)]+cosa3+[cos(a3-π/8)+cos(a3+π/8)]
=2cosa3cos(π/4)+cosa3+2cosa3cos(π/8)
=[1+2cos(π/4)+2cos(π/8)]cosa3
=[1+√2+√(2+√2)]cosa3
Let g (x) = - [1 + √ 2 + √ (2 + √ 2)] cosx + 10x-5 π
g'(x)=[1+√2+√(2+√2)]sinx+10＞0
G (x) has no inflection point, monotonically increasing and at most one solution
g‘’(x)=-[1+√2+√(2+√2)]cosx
G '(x) has an inflection point at x = k π + π / 2,
f[(a3)]^2-a1a3=(2a3-cosa3)^2-a1a3
=[2(a1+π/4)-cos(a1+π/4)]^2-a1(a1+π/4)
=4(a1+π/4)^2+[cos(a1+π/4)]^2-4(a1+π/4)cos(a1+π/4)-a1(a1+π/4)

Let f (x) = 2x cosx, {an} be an arithmetic sequence with tolerance π, f (A1) + F (A2) + If f (A5) = 5 π, then [f (a)] ^ 2-a1 * A3=

Because cos (- 3 π / 2) + cos (- π / 2) + cos π / 2 + cos3 π / 2 + cos5 π / 2 = 0
2(-3π/2-π/2+π/2+3π/2+5π/2)=5π
So A1 = - 3 π / 2
a2=-π/2
a3=π/2
a4=3π/2
a5=5π/2

Let f (x) = 2x cosx, {an} be an arithmetic sequence with tolerance of π 8, f (A1) + F (A2) + +If f (A5) = 5 π, then [f (A3)] 2-a1a5=______ ．

∵ f (x) = 2x cosx, ∵ let g (x) = 2x + SiNx, ∵ {an} be an arithmetic sequence with tolerance of π 8, f (A1) + F (A2) + +f（a5）=5π∴g（a1-π2）+g（a2-π2）+… +G (A5 - π 2) = 0, then A3 = π 2, A1 = π 4, A5 = 3 π 4 ℅ [f (A3)] 2-a1a5 = π 2 - π 4 · 3 π 4 = 13 π 216, so the answer is: 13 π 216