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20 17 higher vocational mathematics foundation
1)f(π)=0

f(0)=2^0.5

f(2^0.5)=2^0.5-π

2)f(5)= ff( 1 1)= f( 1 1-2)= F9 = ff(9+6)= ff( 15)= f( 15-2)= f 13 = 13-2 = 1 1

3)a & gt; At 0 o'clock. 1- 1/2*a=a, a=2/3.

A<0, 1/a=a, and a=- 1, or a= 1 (omitted).

So a=- 1 or 2/3.

4)fx=f(x-2), when x >; 0, so there is f 1=f(- 1), and f(- 1)= 1.

f3=f 1……,f 1 = F3 = F5 = F7 =……= f(2n- 1)n & gt; =0, n is an integer

So f3= 1, f(20 17)= 1.

5)fx*f(x+ 1)= 1

F0 * f1=1...1formula

F 1 * F2 = 1...2 formula

F2 * F3 = 1...3 formula

F3 * F4 = 1...4 formula

1 divided by 2 gives f0=f2.

Divide Equation 3 by Equation 4 to get f2=f4.

Divide Equation 2 by Equation 3 to get f 1=f3.

Extended to F0 = F2 = F4 = F6 =...= f (2n)

f 1 = F3 = F5 = F7 =……= f(2n- 1)

ff(5)= ff( 1)= f(-5)= f 1 =-5

f(20 17)=f 1=-5

6) let x=2y.

f(2)= f(2y/y)= f(2y)-f(y)= 1... 1.

1 where y= 1.

F2=f2-f 1= 1, f 1=0.

Let y=2

F4-f2= 1,f4-f 1=2。

You can make y=n

f(2n)-f(n)= 1

Then there must be f (2n)-f1= n.

f(2n)=n

Let n = 2 (n- 1)

Then f (2 n) = 2 (n- 1) n > = 1