Let f (x) be a periodic function defined on R with a period of 3, and the graph represents the image of the function on the interval [- 2,1], then f (2012) + F (2013) = () A. 3B. 2C. 1D. 0

Let f (x) be a periodic function defined on R with a period of 3, and the graph represents the image of the function on the interval [- 2,1], then f (2012) + F (2013) = () A. 3B. 2C. 1D. 0

Because the period of the function is 3, f (2012) = f (3 × 671-1) = f (- 1), f (2013) = f (3 × 671) = f (0), so according to the image, f (2012) = f (- 1) = 2, f (2013) = f (0) = 0, so f (2012) + F (2013) = 2, so select B

Given the function f (x) = asin3 (π x + a) + bcos3 (π x + β), and f (2012) = - 1, find f (2013)
The problem is wrong and correct: given the function f (x) = asin3 (π x + a) + bcos3 (π x + β) + 1, and f (2012) = - 1, find f (2013)

According to the meaning of the question, we can get f (2012) = asin (6036 π + a) + bcos (6036 π + β) + 1 = asin (a) + bcos (β) + 1 = - 1. Therefore, asin (a) + bcos (β) = 2F (2013) = asin (6039 π + a) + bcos (6039 π + β) + 1 = asin (3 π + a) + bcos (3 π + β) + 1 = - asin (a) -

Given that the function f (x) = follows 3sinx cosx, X belongs to R, ① finding the minimum positive period and maximum value of FX, ② monotone increasing interval of FX, ③ the minimum value of FX on [0, pie]
The process of two three questions

f(x)=2(sinx*√3/2-cosx*1/2)
=2(sinxcosπ/6-cosxsinπ/6)
=2sin(x-π/6)
(1) Minimum positive period T = 2 π
Maximum = 2
（2）-Pai/2+2kPai

Given the function f (x) = cosx2 (3sinx2 + cosx2); (1) find the minimum positive period and monotone increasing interval of function f (x); (2) if f (x) = 1, find the value of COS (2 π 3 − 2x)

(1) The function f (x) = cosx2 (3sinx2 + cosx2) = 32sinx + 12cosx + 12 = sin (x + π 6) + 12, -- (3 points) so the minimum positive period of function f (x) is t = 2 π. - - - (4 points) let 2K π - π 2 ≤ x + π 6 ≤ 2K π + π 2, K ∈ Z, then 2K π - 2 π 3 ≤ x + π 6 ≤ 2K π + π 3, K ∈

Let f (x) = 2cosx (cosx + root 3sinx) - 1. X belong to R

f(x)=2cos²x＋2√3sinxcosx－1=cos2x＋√3sin2x=2sin(2x＋π/6)
The minimum positive period is t = 2 π / 2 = π, the increasing interval is 2 π - π / 2 ≤ 2x + π / 6 ≤ 2K π + π / 2, the increasing interval is obtained: [K π - π / 3, K π + π / 6], where k is an integer

How to calculate the 0.4 power of X

X^0.4=X^4/10=X^4/X^10

(the fifth power of x-1) (the fourth power of X + 1) calculation
There is no mistake in the title. I want to know the process

Original formula = 9th power of X + 5th power of X - 4th power of X - 1

Calculate 2x * (- x) to the fourth power =?

2X * (- x) 4 power = 2x ^ 5

Solving equation 1.3x/2 = 6.5

=6.5 x2
x=(6.5 x2)/1.3
x=10

How to solve the equation X-100 = (3x 100) △ 4

Double four on both sides
4x-400=3x+100
transposition
4x-3x=100+400
x=500