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Himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style open vertical bar open vertical bar 3 x minus 2 close vertical bar minus open vertical bar x plus 3 close vertical bar close vertical bar greater than x minus 1 end style adalah ...

  1. begin mathsize 14px style open curly brackets x vertical line blank x less than 4 comma blank x element of R close curly brackets end style 

  2. begin mathsize 14px style open curly brackets x vertical line 2 less than x less than 4 comma blank x element of R close curly brackets end style 

  3. begin mathsize 14px style open curly brackets x vertical line minus 3 less than x less than 2 comma blank x element of R close curly brackets end style 

  4. begin mathsize 14px style open curly brackets x vertical line x less than 2 blank atau blank x greater than 4 comma blank x element of R close curly brackets end style 

  5. begin mathsize 14px style open curly brackets x vertical line x less than negative 3 blank atau blank x greater than negative 2 comma blank x element of R close curly brackets end style 

K. Putri

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Ganesha

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

Pembahasan

Dengan definisi nilai mutlak diperoleh,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 3 x minus 2 close vertical bar end cell equals cell open curly brackets table row cell open parentheses 3 x minus 2 close parentheses comma blank open parentheses 3 x minus 2 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than 2 over 3 end cell row cell negative open parentheses 3 x minus 2 close parentheses comma blank open parentheses 3 x minus 2 close parentheses less than 0 blank a t a u blank x less than 2 over 3 end cell end table close end cell row cell open vertical bar x plus 3 close vertical bar end cell equals cell open curly brackets table row cell open parentheses x plus 3 close parentheses comma blank open parentheses x plus 3 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than negative 3 end cell row cell negative open parentheses x plus 3 close parentheses comma blank open parentheses x plus 3 close parentheses less than 0 blank a t a u blank x less than negative 3 end cell end table close end cell end table end style  

Gabungan daerah untuk interval nilai x di atas menghasilkan 3 daerah sebagai berikut ini.

Daerah I
Untuk interval begin mathsize 14px style x less than negative 3 end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar negative open parentheses 3 x minus 2 close parentheses minus open square brackets negative open parentheses x plus 3 close parentheses close square brackets close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar negative 3 x plus 2 plus x plus 3 close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar negative 2 x plus 5 close vertical bar end cell greater than cell x minus 1 end cell row cell open parentheses negative 2 x plus 5 close parentheses squared end cell greater than cell open parentheses x minus 1 close parentheses squared end cell row cell 4 x squared minus 20 x plus 25 end cell greater than cell x squared minus 2 x plus 1 end cell row cell 4 x squared minus x squared minus 20 x plus 2 x plus 25 minus 1 end cell greater than 0 row cell 3 x squared minus 18 x plus 24 end cell greater than 0 row cell x squared minus 6 x plus 8 end cell greater than 0 row cell open parentheses x minus 2 close parentheses open parentheses x minus 4 close parentheses end cell greater than 0 end table end style   

Perhatikan daerah himpunan penyelesaian dari interval nilai begin mathsize 14px style x less than negative 3 space d a n space open parentheses x minus 2 close parentheses open parentheses x minus 4 close parentheses greater than 0 end style pada garis bilangan berikut ini.

Maka, daerah himpunan penyelesian untuk daerah I adalah undefined 

Daerah II
Untuk interval begin mathsize 14px style negative 3 less or equal than x less than 2 over 3 end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar negative open parentheses 3 x minus 2 close parentheses minus open parentheses x plus 3 close parentheses close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar negative 3 x plus 2 minus x minus 3 close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar negative 4 x minus 1 close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar negative open parentheses 4 x plus 1 close parentheses close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar 4 x plus 1 close vertical bar end cell greater than cell x minus 1 end cell end table end style   

Perhatikan bahwa untuk berapapun nilai x, pertidaksamaan begin mathsize 14px style open vertical bar 4 x plus 1 close vertical bar greater than x minus 1 end style bernilai benar. Maka, daerah himpunan penyelesaian pertidaksamaan untuk daerah II irisan antara undefined dan x∈R yaitu begin mathsize 14px style open curly brackets negative 3 less or equal than x less than 2 over 3 close curly brackets end style.

Daerah III
Untuk interval begin mathsize 14px style x greater or equal than 2 over 3 end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar open parentheses 3 x minus 2 close parentheses minus open parentheses x plus 3 close parentheses close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar 2 x minus 5 close vertical bar end cell greater than cell x minus 1 end cell row cell open parentheses 2 x minus 5 close parentheses squared end cell greater than cell open parentheses x minus 1 close parentheses squared end cell row cell 4 x squared minus 20 x plus 25 end cell greater than cell x squared minus 2 x plus 1 end cell row cell 4 x squared minus x squared minus 20 x plus 2 x plus 25 minus 1 end cell greater than 0 row cell 3 x squared minus 18 x plus 24 end cell greater than 0 row cell x squared minus 6 x plus 8 end cell greater than 0 row cell open parentheses x minus 2 close parentheses open parentheses x minus 4 close parentheses end cell greater than 0 end table end style   

Perhatikan daerah himpunan penyelesaian dari interval nilai begin mathsize 14px style x less or equal than 2 over 3 end style dan begin mathsize 14px style open parentheses x minus 2 close parentheses open parentheses x minus 4 close parentheses greater than 0 end style pada garis bilangan berikut ini.

Maka, daerah himpunan penyelesian untuk daerah III adalah begin mathsize 14px style open curly brackets 2 over 3 less or equal than x less than 2 blank atau blank x greater than 4 close curly brackets end style 

Jika kita gabungkan daerah himpunan penyelesaian dari ketiga daerah di atas dalam sebuah garis bilangan diperoleh gambar berikut ini.

Maka, himpunan penyelesaian pertidaksamaan undefined adalah begin mathsize 14px style open curly brackets x vertical line x less than 2 blank atau blank x greater than 4 blank x element of R close curly brackets end style 

Jadi, jawaban yang tepat adalah D.

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