solusi 17 simak ui mat das kode 1 2015

(1)

|jejakseribupena.com, Soal dan Solusi Simak UI Matematika Dasar, 2014

17. SIMAK UI Matematika Dasar Kode 1, 2015

Misalkan

 

2 , 0 1 2

f x  x  x

_dan

 

2 2 ,1 1 2

f x   x  x

.

2

 



 



f x f f x

dan

f n1

 

x  fn



f x

 



, maka pernyataan berikut yang BENAR ....

(1)

f n

 

0 0

(2)

f n

 

1 0,n1

(3)

  1 0, 2 2

n

f   _{ } n

 

(4)

  1

0, 3

4

n

f   _{ } n

 

Solusi: [E]

(1)

f

 

0   2 0 0

 



 



 

2

0 0 0 0

f  f f  f 

 



 





 





 



 

3 ₀ 2 ₀ ₀ ₀ ₀ ₀

f  f f f f f  f f  f 

...

 n

_{ }

₀ ₀ f

 

_(Benar)

(2)

f

 

1    2 2 1 0

 



 



 

2

1 1 0 2 0 0

f  f f  f   

 



 





 





 



 

3 ₁ 2 ₁ ₁ ₀ ₀ ₀

f  f f  f f f  f f  f 

...

 n

_{ }

₁ _0, ₁

f n

  

_(Benar)

(3)

  1 0, 2 2

n

f   _{ } n

 

1

2 0 0

2

f    _{ }

 

 

2 1 1 ₀ _{2 0} ₀

2 2

f  _{ } f_f _{ }_ f   

    

 





 

3 1 2 1 1

0 0 0

2 2 2

f _{ }  f _f _{ }_ f_f_f_{ } __ f f  f 

        

...

  1

0, 2

2

n

f   n

 _{ } 

 

(Benar)

(4)

  1 0, 3 4

n

f   _{ } n

  1

2 0 0

4

f    _{ }

 

 

2 1 1

0 0

4 4

f  _{ } f _f _{ }_ f 

    

 





 

3 1 2 1 1 ₀ ₀ ₀

4 4 4

f _{ }  f _f _{ }_ f__f_f_{ } ___ f f  f 

        

 











 



 

4 1 3 1 1

0 0 0 0

4 4 4

f _{ }  f _f_{ } _ f _f_f_f _{ }___ f f f  f f  f 

         

...

  1

0, 3

4

n

f   n

 _{ } 