∵2k+×2=2550,∴k=50, 查看更多

 

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已知函數(shù)f(x) =2x+1,x∈R.規(guī)定:給定一個實(shí)數(shù)x0,賦值x1= f(x0),若x1≤255,則繼續(xù)賦值x2= f(x1) …,以此類推,若x n-1≤255,則xn= f(xn-1),否則停止賦值,如果得到xn后停止,則稱賦值了n次(n∈N *).已知賦值k次后該過程停止,則x0的取值范圍是 

(A)(2k-9 ,2 k-8]  (B)(2 k-8 -1, 2k-9-1](C)(28-k -1, 29-k-1] (D)(27-k -1, 28-k-1]

 

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已知函數(shù)f(x) =2x+1,x∈R.規(guī)定:給定一個實(shí)數(shù)x0,賦值x1= f(x0),若x1≤255,則繼續(xù)賦值x2=" f(x1)" …,以此類推,若x n-1≤255,則xn= f(xn-1),否則停止賦值,如果得到xn后停止,則稱賦值了n次(n∈N *).已知賦值k次后該過程停止,則x0的取值范圍是 

A.(2k-9 ,2 k-8]B.(2 k-8 -1, 2k-9-1]C.(28-k -1, 29-k-1]D.(27-k -1, 28-k-1]

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已知函數(shù)f(x) =2x+1,x∈R.規(guī)定:給定一個實(shí)數(shù)x0,賦值x1= f(x0),若x1≤255,則繼續(xù)賦值x2= f(x1) …,以此類推,若x n-1≤255,則xn= f(xn-1),否則停止賦值,如果得到xn后停止,則稱賦值了n次(n∈N *).已知賦值k次后該過程停止,則x0的取值范圍是  (       )

(A)(2k-9 ,2 k-8]       (B)(2 k-8 -1, 2k-9-1]      (C)(28-k -1, 29-k-1]      (D)(27-k -1, 28-k-1]

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已知函數(shù)f(x) =2x+1,x∈R.規(guī)定:給定一個實(shí)數(shù)x0,賦值x1= f(x0),若x1≤255,則繼續(xù)賦值x2= f(x1) …,以此類推,若x n-1≤255,則xn= f(xn-1),否則停止賦值,如果得到xn后停止,則稱賦值了n次(n∈N *).已知賦值k次后該過程停止,則x0的取值范圍是  (       )

(A)(2k-9 ,2 k-8]       (B)(2 k-8 -1, 2k-9-1]      (C)(28-k -1, 29-k-1]      (D)(27-k -1, 28-k-1]

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已知函數(shù)f(x) =2x+1,x∈R.規(guī)定:給定一個實(shí)數(shù)x0,賦值x1= f(x0),若x1≤255,則繼續(xù)賦值x2=" f(x1)" …,以此類推,若x n-1≤255,則xn= f(xn-1),否則停止賦值,如果得到xn后停止,則稱賦值了n次(n∈N *).已知賦值k次后該過程停止,則x0的取值范圍是


  1. A.
    (2k-9 ,2 k-8]
  2. B.
    (2 k-8 -1, 2k-9-1]
  3. C.
    (28-k -1, 29-k-1]
  4. D.
    (27-k -1, 28-k-1]

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