If f (x) is an odd function defined on R with period 3 and f (1) = 2, then f (5) =?

If f (x) is an odd function defined on R with period 3 and f (1) = 2, then f (5) =?

Odd function
f(-1)=-f(1)=-2
The cycle is three
So f (5) = f (5-2 × 3) = f (- 1) = - 2
Because f (x) is an odd function on R, then f (- x) = - f (x). If f (x) is a periodic function with period 3, then f (x + 3) = f (x). Because we also know that f (1) = 2. And. F (5) = f (2 + 3) = f (2) = f (- 1 + 3) = f (- 1) = - f (1) = - 2.
It is known that a = (5, - 2), B = (- 4, - 3) C = (x, y). If a-2b + 3C = 0, then c=
Is it a vector? The addition and subtraction of vector 0 (0,0) should be learned
5-2（-4）+3x=0
-2-2(-3)+3y=0
c=(-13/3,-4/3)
I'm sorry I made a mistake at the beginning
If f (x) is an odd function with period 5, f (- 3) = 4 and cos α = 12, then f (4cos2 α)=______ .
∵ cos α = 12, ∵ 4cos2 α = 8cos2 α - 4 = - 2 ∵ f (4cos2 α) = f (- 2) ∵ f (x) is an odd function with a period of 5, f (- 3) = 4 ∵ f (- 2) = - f (2) = - f (- 3 + 5) = - f (- 3) = - 4, so the answer is: - 4
A * b = a + 2B b * C = 2B + 3C c * d = 3C + 4D, then (3 * 2) * (4 * 5) =?
Amount Why are you again?
3 * 2 = 3 times 3 + 2 times 2 = 9 + 4 = 13
4 * 5 = 4 times 4 + 5 times 5 = 16 + 25 = 41
13 * 41 = 13 times 13 + 41 times 41 = 169 + 1681 = 1850
The answer is 1850
If f (x) is an odd function with a period of 4, f (1 / 2) = 1 and Sina = 1 / 4, then what is f (4cos2a) equal to
4cos2a=4(1-2sin²a)=7/2
T=4
So the original formula = f (7 / 2)
=f(7/2-4)
=f(-1/2)
It's an odd function
So = - f (1 / 2)
=-1
The definition field of y = FX is (0.1),
Then 0 in F (X & # 178;)
A = (5, - 2), B = (- 4, - 3), C = x, y), if a + 2B + 3C = O, then C is equal to
A、(13/3,8/3)
B、(1,8/3)
C、(13/3,4/3)
D、(-13/3,-4/3)
There should be a solution or process
Let me tell you that a + 2B + 3C = 0 is equal to X (a) + 2x (b) + 3x (c) = 0, y (a) + 2Y (b) + 3Y (c) = 0
According to the X-Series equation 5 + 2 * (- 4) + 3 * x = 0, we get - 3 + 3 * x = 0 = > x = 1
According to the Y Series equation - 2 + 2 * (- 3) + 3 * y = 0, we get - 8 + 3 * y = 0 = > y = 8 / 3
So point C is (1,8 / 3)
B is the answer
B
B
5+2*（－4）+3x=0
X=1
（－2）+2*（－3）+3y=0
y=8/3
If f (x) is an odd function with a period of four, f (1 / 2) = 1 and Sina = 1 / 4, then what is f (4cos2a) equal to
4cos2a=4(1-2sin²a)=7/2
T=4
Original formula = f (7 / 2)
=f(7/2-4)
=f(-1/2)
Odd function
=-f(1/2)
=-1
Given that a + 2B divided by 3B = 3 / 2, find the value of a in B,
a+2b/3b=3/2
Namely: 1 / 3 * A / B + 2 / 3 = 3 / 2
a/b=5/6*3
=5/2
It's five out of two, right!
Let f (x) be a periodic function with period 2 and an odd function, and f (- 2 / 5) = 3, Sina = √ 5 / 5, then the value of F (4cos2a) is
How to do, please analyze in detail
cos2a = 1 - 2*(sina)^2 = 1 - 2 * (√5/5)^2
= 1 - 2 * 1/5
= 3/5
4*cos2a = 4*3/5 = 12/5
F (x) is a periodic function with period 2
f(12/5) = f(2 + 2/5) = f(2/5)
F (x) is an odd function, so
f(-x) = - f(x)
f(2/5) = - f(-2/5) = -3
The value of F (4cos2a) is - 3
From Sina = √ 5 / 5, we know cosa = (2 √ 5) / 5 or - (2 √ 5) / 5
cos2a=(cosa)^2-(sina)^2=3/5
So f (4cos2a) = f (12 / 5) = f (2 + (2 / 5))
Because f is an odd function, f (- 2 / 5) = - f (2 / 5) = 3
Because f is a periodic function and the minimum period is 2,
So f (2 + (2 / 5)) = f (2 / 5) = - 3
Given a ^ 2 + 10A + 25 + (B-3) ^ 2 = 0, find the algebraic formula B ^ 4 / (a-b) ^ 2 * a ^ 3 + AB ^ 2-2a ^ 2B / b ^ 3 divided by B ^ 2-A ^ 2 / AB + B ^ 2
{b^4/(a-b)^2}
*{a^3+ab^2-2a^2b/b^3}
Divide by {B ^ 2-A ^ 2 / AB + B ^ 2}
Value of
The two nonnegative numbers are 0 and only each can be 0 respectively. So a = -5, B = 3 (a + 25 + 25 (B-3) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\a (a-b) &
It's better to put in brackets. It's hard for you to read the question
If a ^ 2 + 10A + 25 + (B-3) ^ 2 = (a + 5) ^ 2 + (B-3) ^ 2 = 0, then a = - 5, B = 3. Your writing is a bit messy. For example, B ^ 4 / (a-b) ^ 2 * a ^ 3 may have ambiguity, so you can bring it in by yourself
a^2+10a+25+(b-3)^2=0
(a+5)²+(b-3)²=0
Two nonnegative numbers added to 0 can only be 0 each
So a = - 5, B = 3
Just substitute it into the following formula
(since the following formula is not very clear, I will not replace it.)
a^2+10a+25+(b-3)^2=0
(a+5)²+(b-3)²=0
Two nonnegative numbers added to 0 can only be 0 each
So a = - 5, B = 3
Just substitute it into the following formula