Given the function & nbsp; f (x) = x2-2 | x | - 1, try to judge the parity of function f (x) and draw the image of function

Given the function & nbsp; f (x) = x2-2 | x | - 1, try to judge the parity of function f (x) and draw the image of function

The domain of definition is R. for any x ∈ R, f (- x) = (- x) 2-2 | - x | - 1 = x2-2 | - 1 = f (x). Therefore, y = f (x) is an even function. When x > 0, f (x) = x2-2x-1, so the image of the function is as follows:

The graph of the function f (x) = x + ax (a is a constant) passes through the point (2,0), (I) find the value of a and judge the parity of F (x); (II) the function g (x) = LG [f (x) + 2x-m] is meaningful in the interval [2,3], and find the value range of real number m; (III) discuss the number of positive roots of the equation | f (x) | = t + 4x-x2 (t is a constant) about X

(I) according to the meaning of the problem, there is 0 = 2 + A2 {a = - 4, then f (x) = x − 4x, its definition field is x | x ≠ 0, and f (- x) = - f (x), that is, f (x) = x − 4x is an odd function; (II) the function g (x) = LG [f (x) + 2x-m] is meaningful in the interval [2,3], that is, X − 4x + 2x − m > 0 is constant for X ∈ [2,3], and (x − 4x + 2x) min > m is obtained, so that h (x) = x − 4x + 2x, X ∈ [2,3] proves its monotonic increase: let 2 be any 2 If h (x2) − H (x1) = x2-4x2 + 2x2 {4x2} (x1 − 4x1 + 4x1 + 2x1) = (x2 − x1) (x1 (x2 + 4) x (2x2-2x2-2x1) because 2 ≤ x1 ＜ x2 ≤ 3, then H (x2) - H (x1) is more than 0, so h (x) is increasing in X ∈ [2,3] and increasing the minimum value of H (x) = the minimum value of H (x) - 4x + 4x + 2x, H (2) H (2) = 4, H (2) = 4, the minimum value of H (H (x) = the minimum value H (H (H (2) = the minimum value H (H (2) of the minimum value of H (x) - 4x + 4x + 2x + 2x + 2x + 2x) H (H (H (2) = H (2) H (2) H (2) H (2) H (2) H (2) = 4) H (when t = - 4, the number of positive roots is 0 (3) when t ＞ - 4, the number of positive roots is 2

Let f (x) = x + MX and f (1) = 2. (1) judge the parity of F (x) and prove it; (2) judge the monotonicity of F (x) on (1, + ∞) and prove it

(1) ∵ f (x) = x + MX, and f (1) = 2, ∪ 1 + M = 2, the solution is & nbsp; m = 1. The function y = f (x) is odd. It is proved that: F (x) = x + 1X, the domain of definition is (- ∞, 0) ∪ (0, + ∞), symmetric with respect to the origin. And f (− x) = (− x) + 1 − x = − (x + 1x) = − f (x), so y = f (x) is odd

Given the function f (x) = (2x + 3) / (3x) (x > 0), the sequence {an} satisfies A1 = 1, an = f (1 / an-1) (n ∈ n *, and N ＞ 2
1. Find the general formula of an. 2. Find Sn = a1a2-a2a3 + a3a4-a4a5 +. + Anan + 1

1. According to the meaning of the title
an=(2/an-1+3)/(3/an-1)=(3an-1+2)/3
3an=3an-1+2
So an-an-1 = 2 / 3
So the sequence an is an arithmetic sequence with the first term of 1 and the tolerance of 2 / 3. An = 1 + 2 / 3 (n-1) = 2 / 3 N + 1 / 3
2. sn=1*5/3-5/3*7/3 +7/3*9/3-9/3*11/3+…… +anan+1
=1/3(3*5-5*7+7*9-9*11+…… +(2n+1)*（2n+3)

Let f (x) = 2x + 33x (x > 0), and the sequence {an} satisfy A1 = 1, an = f (1, an-1) (n ∈ n *, and N ≥ 2)
Let f (x) = (2x + 3) / (3x) (x > 0), the sequence {an} satisfy A1 = 1, an = f [1 / (a (n-1)] (n ∈ n *, and N ≥ 2)
(1) Find the general term formula of sequence {an};
(2) Let TN = a1a2-a2a3 + a3a4-a4a5 + +(- 1) n-1anan + 1, if TN ≥ tn2 is constant for n ∈ n *, find the value range of real number t;
(3) Is there a sequence {a} with A1 as the first term and Q (0 ＜ Q ＜ 5, Q ∈ n *)_ N {K}, such that n {K}_ Every term in N K} is a different term in the sequence {an}. If it exists, find out the general term formula of all the sequence {NK} that meet the condition; if it does not exist, explain the reason

f(x)=2/3+1/x
an=2/3+a(n-1)
So an-a (n-1) = 2 / 3
So {an} is an arithmetic sequence
First term A1 = 1, d = 2 / 3
So an = 1 + 2 (n-1) / 3 = (2n + 1) / 3
(1) N is even
Sn=(a1a2-a2a3)+(a3a4-a4a5)+.+[a(n-1)an-ana(n+1)]
= -a2(a3-a1)-a4(a3+a5)+.-a(n)[a(n+1)-a(n-1)]
=-4/3*(a2+a4+.+an)
=-(4/3) *[5/3+(2n+1)/3]*n/4
=(-4/3)*n(n+3)/6
=-2n(n+3)/9
(2) N is an odd number
Sn=S(n-1)+an*a(n+1)
=-2(n-1)(n+2)/9+(2n+1)(2n+3)/9

Let f (x) = 2x + 3 3x (x > 0), the sequence {an} satisfy A1 = 1, an = f (1, an-1)
Let f (x)=
2x+3
3x
(x > 0), the sequence {an} satisfies A1 = 1, an = F（
one
an-1
）(n ∈ n *, and N ≥ 2)
(1) Find the general term formula of sequence {an};
(2) Let TN = a1a2-a2a3 + a3a4-a4a5 + +(- 1) n-1anan + 1, if TN ≥ tn2 is constant for n ∈ n *, find the value range of real number t;
(3) Is there a sequence {a} with A1 as the first term and Q (0 ＜ Q ＜ 5, Q ∈ n *)_ N, K}, K ∈ n *, such that the sequence {a}_ Every term in N K} is a different term in the sequence {an}. If it exists, find out the general term formula of all the sequence {NK} that meet the condition; if it does not exist, explain the reason

Solution: the function f (x) = 2 / 3 + 1 / the one = 2 / + (n-1) such a (n-1) = 2 / 3 {} is an arithmetic sequence, the first A1 = 1, d = 2 / 3, so a = 1 2 (n-1) / 3 = (2n + 1) / 3 (1) n is an even number, Sn = (a1a2-a2a3) + (a3a4-a4a5) +. +

If the tolerance D of the arithmetic sequence an is not equal to 0 and A1 A2 is two of the X equation x ^ 2-a3x + A4 = 0, then the general term formula an of the arithmetic sequence an=

a1+a2=a3
a1a2=a4
That is a1 + A1 + D = a1 + 2D
a1(a1+d)=a1+3d
The solution is A1 = D = 2
an=2n

Given the sequence an, A1 = 1, an = & A (n-1) + & - 2 (n is greater than or equal to 2), an can form an arithmetic sequence with tolerance not 0 when & what is the value
(2) If & = 3 let BN = an + 1 / 2 find the sum of the first n terms of BN and the formula SN

(1) A (n) - A (n-1) = (& - 1) a (n-1) + & - 2 because it is an arithmetic sequence, the right side of all equal signs should be a constant term. If a (n-1) = 0, then the tolerance of sequence a (n) is 0, which does not hold, so it can only be & = 1 (2) a (n) = 2A (n-1) + 1a (n) + 1 = 2 [a (n-1) + 1] = 2 ^ 2 [a (n-2) + 1] =... = 2 ^ (n-1) [a (1) + 1] = 2 ^ Na (n) = 2 ^

Given the function f (x) = loga (x), (a > 0 and not equal to 1), if the sequence 2, f (A1), f (A2) F (an), 2n + 4 is equal difference, and the tolerance D of equal difference sequence is calculated

Arithmetic sequence 2, f (A1), f (A2) F (an), 2n + 4 has n + 2 items, tolerance d = (2n + 4-2) / (n + 1) = 2

1. In the arithmetic sequence {an}, given Sn = m, SM = n, (n is not equal to m), find s (M + n)?

Let the first term be A1, the tolerance be D, Sn = Na1 + n * (n-1) d / 2 = M,
Sm=ma1+m*(m-1)d/2=n,
By subtracting the two formulas, we get (n-m) a1 + [(n-m) (n + m) - (n-m)] d / 2 = - (n-m)
a1+[n+m-1]d/2=-1
S(m+n)=(n+m)a1+[(n+m)(n+m-1)]d/2=(n+m)[a1+[n+m-1]d/2]=-(n+m)