Known function f x Acos 2 x 2 2 help me, thanks

Because f(x) max=3, a=1

f(0)=cos(2 )+2=2, so the spacing of the adjacent axes of symmetry = 4 is half the period, so =4, = 4;

So, f(x)=cos(x 2 + 2)+2;

Because m=log 4( 2)=log 4(4) 1 4=1 4, so f(m)+f(m+1)=cos( 8+ 2)+2+cos(5 8+ 2)+2

sin(π/8)-sin(5π/8)+4-sin(3π/8-π/4)-sin(3π/8+π/4)+4-2sin(3π/8)cos(π/4)+4=4-√(1+√2)/2

f(x)max=3, so a=1

f(0)=cos(2)+2=2, so the spacing of the adjacent symmetry axes of = 4 is half of the minimum positive period, so 2 = 2 In summary, f(x)=cos(x 2 + 2)+2 can't see what is the bottom.

If 2 is the base, there is m=4 m+1=5

f(m)+f(m+1)=cos(5 2)+cos3 +4=3 If 4 is the bottom, there is m=1 4 m+1=5 4f(m)+f(m+1)=cos(5 8)+cos(9 8)+4, the result is more troublesome, don't fight.

From the figure: a=2, from the coordinates of Nazhen (0,1) =pi 3 from the coordinates (11pi hole rolling 12,0) w=14 11

Solve a and then substitute it to calculate it.

The function f(x)=acos( x+ )a>0, >0 , 0)1)f(x) is an odd function.

then f(0)=0

i.e. cos =0

kπ+π2,k∈z

Not a pure age must be = 2

2) If the pants are lost = 2

then f(x)=acos(x+2)=-asin(x)f(-x)=-f(x).

i.e. f(x) is an odd function.

f(x) is a necessary but not sufficient condition for the odd function to = two scum gods.

(3+A Hand Macro Judgment) [ 3 (3+A )*sinx+a (3+A )*cosx)]

3+a²)sin(x+φ)

where cos = 3 (a +3) and sin = a (a +3) f(x) the maximum value is (a +3).

From (a +3)=2, a +3=4, a =1 and a>0, a=1

f(x)=2sin(x+π/6)

Obtained from - 2+2k x+6 2+2k, k z.

2 3+2k x 3+2k , k z function completion modification number f(x) on r monotonically increasing interval is.

2π/3+2kπ, 3+2kπ],k∈z

First question: f(x) = sin(x+ )acos(x+2 )

a= 2, =4:

f(x) =sin(x+π/4)+√2cos(x+π/2)

2/2sinx+√2/2cosx-√2cosx

2/2sinx -√2/2cosx

sin(x-π/4)

x∈【0，π】

x-π/4∈【-4，3π/4】

When x-4 = 2, f(x) takes the maximum value of the bucket, and f(x)max = sin(2)=1

When x-4=-4, f(x) takes the minimum value, and f(x)min = sin(- 4)=-2 2

The second question: f( 2)=0, i.e.:

sin(π/2+θ)acos(π/2+2θ) cosθ-asin(2θ) cosθ-2asinθcosθ =cosθ(1-2asinθ)=0

Pin roll- 2, 2), cos 0

1-2asinθ=0

sinθ=1/(2a)

f( )1, i.e.: burning bucket mill.

sin(π+acos(π+2θ) sinθ-acos2θ =sinθ -a(1-2sin²θ)1

1/(2a) -a [1-2/(4a²)]1

1/(2a) -a + 1/(2a)] a = 1

a = 1sinθ =1/(2a) =1/2

The value of the function of f(x)=acos(x+) at x=1 is 0, i.e., cos(w+)0

w+φ=kπ+π2

kπ+π2-w，k∈z

With reedin f(x) = acos( x+k + 2-w) asin(wx-w).

asin[w(x-1)]

Note: When k is an even number, k=2n

cos(ωx+2nπ+π2-w)

cos(π/2+wx-w)

sin(wx-w)

When k is an odd number, k=2n+1

cos(ωx+2nπ+3π/2-w)

cos(3π/2+wx-w)

sin(wx-w) 】

f(x+1)= asinwx is an odd function.

I choose Cf(x+1) must be an odd function.

is an odd function.

then f(0)=0

acosφ=0

kπ+π/2

Not necessarily = 2

Not sufficiently. And = 2

then f(x)=-asin x

is an odd function. Therefore, it is a necessary but not sufficient condition.

Originally, there should have been 2 situations in this question:

Solution: According to the meaning of the question, a cannot be equal to 0, otherwise f(3) =x =1, 3=1 is obviously not true, so a is not equal to 0;

f(x) =x + acosx

f(3) =1 Substituting 3 into the equation:

1 = 3 + acos3

acos3 = 2

Since cosx is an even function, then acosx is also an even function, symmetrical:

Then: acos(-3) =2

f(-3) =3)+(2) =5

First of all, determine that this function is an odd function... x is an odd function, and acosx is also an odd function, so this function is an odd function. According to the properties of the odd function, f(-3)=-1 is obtained