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Semua nilai x yang memenuhi pertidaksamaan ( 1 − x ) lo g ( x + 5 ) > 2 adalah ...

Semua nilai  yang memenuhi pertidaksamaan  adalah ... 

  1. begin mathsize 14px style negative 1 less than x less than 0 end style 

  2. begin mathsize 14px style negative 1 less than x less than 1 end style 

  3. begin mathsize 14px style 1 less than x less than 2 end style 

  4. begin mathsize 14px style negative 1 less than straight x less than 0 space atau space 2 less than straight x less than 4 end style 

  5. begin mathsize 14px style negative 1 less than straight x less than 0 space atau space 1 less than straight x less than 4 end style 

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B

jawaban yang tepat adalah B

Pembahasan

Pembahasan
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Uji titik untuk setiap daerah interval Untuk , ambil titik Untuk , ambil titik 0 Untuk , ambil titik 5 Sehingga didapatkan seperti berikut : Syarat logaritma Jadi dapat disimpulakan bahwa Jadi jawaban yang tepat adalah B

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript left parenthesis 1 minus x right parenthesis end presuperscript left parenthesis x plus 5 right parenthesis end cell greater than 2 row cell log presuperscript left parenthesis 1 minus x right parenthesis end presuperscript left parenthesis x plus 5 right parenthesis end cell greater than cell log presuperscript left parenthesis 1 minus x right parenthesis end presuperscript left parenthesis 1 minus x right parenthesis squared end cell row cell log presuperscript left parenthesis 1 minus x right parenthesis end presuperscript left parenthesis x plus 5 right parenthesis end cell greater than cell log presuperscript left parenthesis 1 minus x right parenthesis end presuperscript left parenthesis x squared minus 2 x plus 1 right parenthesis end cell row cell x plus 5 end cell greater than cell x squared minus 2 x plus 1 end cell row cell negative x squared plus 3 x plus 4 end cell greater than 0 row cell left parenthesis negative x plus 4 right parenthesis left parenthesis x plus 1 right parenthesis end cell greater than 0 row cell negative x plus 4 end cell equals cell 0 space a t a u space x plus 1 equals 0 end cell row x equals cell 4 space space space space space space space space space space space space space space space x equals negative 1 end cell end table end style 

Uji titik untuk setiap daerah interval

Untuk begin mathsize 14px style x less than negative 1 end style , ambil titik begin mathsize 14px style negative 2 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative x squared plus 3 x plus 4 end cell greater than 0 row cell negative left parenthesis negative 2 right parenthesis squared plus 3 left parenthesis negative 2 right parenthesis plus 4 end cell greater than 0 row cell negative 4 minus 6 plus 4 end cell greater than 0 row cell negative 6 end cell greater than cell 0 space left parenthesis salah right parenthesis end cell end table end style  

Untuk begin mathsize 14px style negative 1 less than x less than 4 end style , ambil titik 0

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative x squared plus 3 x plus 4 end cell greater than 0 row cell 0 plus 0 plus 4 end cell greater than 0 row 4 greater than 0 row 4 greater than cell 0 space left parenthesis benar right parenthesis end cell end table end style 

Untuk begin mathsize 14px style x greater than 4 end style , ambil titik 5

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative x squared plus 3 x plus 4 end cell greater than 0 row cell negative left parenthesis 5 right parenthesis squared plus 3 left parenthesis 5 right parenthesis plus 4 end cell greater than 0 row cell negative 25 plus 15 plus 4 end cell greater than 0 row cell negative 6 end cell greater than cell 0 space left parenthesis salah right parenthesis end cell end table end style 

Sehingga didapatkan seperti berikut :

 

begin mathsize 14px style negative 1 less than x less than 4 end style         

Syarat logaritma 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 1 minus x end cell greater than 0 row cell negative x end cell greater than cell negative 1 space end cell row x less than 1 row blank blank blank row cell x plus 5 end cell greater than 0 row x greater than cell negative 5 end cell end table end style 

Jadi dapat disimpulakan bahwa begin mathsize 14px style HP equals open curly brackets straight x vertical line minus 1 less than straight x less than 1 comma space straight x element of straight R close curly brackets end style 

Jadi jawaban yang tepat adalah B

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