Given that the maximum value of function f (x) = asin ^ 2 (Wx + α) (a > 0) is 2 and the distance between two adjacent symmetrical axes is 2, then f (1) + F (2) +... + F (100)= RT,

Given that the maximum value of function f (x) = asin ^ 2 (Wx + α) (a > 0) is 2 and the distance between two adjacent symmetrical axes is 2, then f (1) + F (2) +... + F (100)= RT,

Y = sin (x) ^ 2, the distance between adjacent symmetry axes = period is pie, so the distance between two adjacent symmetry axes is 2, the period is pie
Original formula = 50 * (f (1) + F (2)) = 100
f(x)=Asin^2(wx+ψ)=A*[1-cos2(wx+ψ)]/2
Maximum 2: i.e. a = 2
If the symmetry axis distance is 2, the period is 4, 2 π / 2W = 4, w = π / 4
The maximum value of function f (x) is 2, and the square of sin (Wx + a) is between 0 and 1, so a = 2, which can be reduced to f (x) = cos (2wx + 2a). Because the distance between two symmetry axes is 2, t = 4, that is, w = 2, f (x) period is 4, in one period: F (1) + F (2) + F (3) + F (4)=
The known function f (x) = asin (x + FAI) (a > 0,0
The maximum value is 1, and a > 0
So a = 1
Data brought into m
sin(π/3 + fai) = 1/2
Zero
Proof of logarithmic property
logb N=loga N/loga B
logb A=1/log a B
Logb n means B is the base
Let x = logb n
Then B ^ x = n
Take the logarithm of a as the base on both sides
loga b^x=loga N
xloga b=loga N
x=(loga N)/(loga b)
That is logb n = (loga n) / (Loga b)
Using the formula proved above, it is easy to get
logb a=lga/lgb
loga b=lgb/lga
Multiply to get
(logb a)×(loga b)=1
So logb a = 1 / loga B
In the following functions (1) C = 2 π R; (2) y = 2x-1; (3) y = 1 / X; (4) y = - 3x; (5) y = x ^ 2 + 1, the number of positive proportional functions is__ One
2 first and second
The second proof process of logarithm operation property
log(a)(M^n)=nlog(a)(M)
Let a ^ n = M
So
n=loga(M)
loga(M^n)=loga(a^n^2)=n^2=nloga(M^n)
Get proof
The image of a positive scale function passes through a point (2, - 3). Its expression is ()
A. y＝−32xB. y＝23xC. y＝32xD. y＝−23x
Let the analytic expression of the function be y = KX. According to the meaning of the problem, we get 2K = - 3. The solution is k = - 32. So the analytic expression of the function is y = - 32x. So we choose a
How to prove the operational property of logarithm?
````Don't you... Just follow the first proof given in the book. Let's talk about it··
logaM/N=logaM-logaN
Prove: let logam = P, Logan = Q
We know the definition of logarithm: P power of a = m, Q power of a = n, PS: the sign of the power will not hit
M / N = P power of a / Q power of a = P-Q power of a
P-Q = loga (M / N) {obtained from the P-Q power of M / N = a ·}
∴p-q=logaM-logaN=loga(M/N)
··I'll give it to me if I hand it_ ∩) O ha! Preview for high school this year... Guess you too···
When k=__ The function y = 2x ^ (k ^ 2-3x-3) is a positive proportional function
(k ^ 2-3x-3) the X here should be K ^ 2!
If so, that's the solution
k^2-3k-3=1
K = - 1 or K = 4
Prove the operational property of logarithm logamn = nlogam
Isn't that the nature?
Property 1: loga m * n = loga m + loga n
So loga Mn = loga m + loga m +... + loga m (n) = nloga M
In the function y = 3x-2, y = 1 + 3 of X, y = - 2x, y = - x squared + 7, there are several positive proportional functions
Give the function relation: y = - 3x-2, y = 2x + 1, y = 2x square + 1, y = 2x + 1 / 3, where is the number of first-order function
The first, the second, the fourth, all