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Himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style 3 less than open vertical bar 2 x minus 1 close vertical bar plus open vertical bar x plus 4 close vertical bar less than 9 end style adalah ...

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

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

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

  5. begin mathsize 14px style open curly brackets x vertical line minus 2 less or equal than x less than 0 blank atau blank x greater than 4 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 C.

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 2 x minus 1 close vertical bar end cell equals cell open curly brackets table row cell open parentheses 2 x minus 1 close parentheses comma blank open parentheses 2 x minus 1 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than 1 half end cell row cell negative open parentheses 2 x minus 1 close parentheses comma blank open parentheses 2 x minus 1 close parentheses less than 0 blank a t a u blank x less than 1 half end cell end table close end cell row cell open vertical bar x plus 4 close vertical bar end cell equals cell open curly brackets table row cell open parentheses x plus 4 close parentheses comma blank open parentheses x plus 4 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than negative 4 end cell row cell negative open parentheses x plus 4 close parentheses comma blank open parentheses x plus 4 close parentheses less than 0 blank a t a u blank x less than negative 4 end cell end table close end cell end table end style   

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

Daerah I
Untuk interval begin mathsize 14px style x less than negative 4 end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 3 less than cell negative open parentheses 2 x minus 1 close parentheses minus open parentheses x plus 4 close parentheses less than 9 end cell row 3 less than cell negative 2 x plus 1 minus x minus 4 less than 9 end cell row 3 less than cell negative 3 x minus 3 less than 9 end cell row cell 3 plus 3 end cell less than cell negative 3 x less than 9 plus 3 end cell row 6 less than cell negative 3 x less than 12 end cell row cell negative 2 end cell greater than cell x greater than negative 4 end cell row cell negative 2 end cell greater than cell x greater than negative 4 end cell end table end style   

Perhatikan bahwa tidak terdapat irisan daerah himpunan penyelesaian begin mathsize 14px style negative 4 less than x less than negative 2 space d a n space x less than negative 4 end style. Maka, himpunan penyelesaian daerah I adalah himpunan kosong ({  }). 

Daerah II
Untuk interval begin mathsize 14px style negative 4 less or equal than x less than 1 half end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 3 less than cell negative open parentheses 2 x minus 1 close parentheses plus open parentheses x plus 4 close parentheses less than 9 end cell row 3 less than cell negative 2 x plus 1 plus x plus 4 less than 9 end cell row 3 less than cell negative x plus 5 less than 9 end cell row cell 3 minus 5 end cell less than cell negative x less than 9 minus 5 end cell row cell negative 2 end cell less than cell negative x less than 4 end cell row 2 greater than cell x greater than negative 4 end cell row cell negative 4 end cell less than cell x less than 2 end cell end table end style   

Irisan daerah himpunan penyelesaian begin mathsize 14px style negative 4 less than x less than 2 space d a n space minus 4 less or equal than x less than 1 half end style. Maka, himpunan penyelesaian daerah II adalah begin mathsize 14px style open curly brackets negative 4 less than x less than 1 half close curly brackets end style

Daerah III
Untuk interval begin mathsize 14px style x greater or equal than 1 half end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 3 less than cell open parentheses 2 x minus 1 close parentheses plus open parentheses x plus 4 close parentheses less than 9 end cell row 3 less than cell 3 x plus 3 less than 9 end cell row cell 3 minus 3 end cell less than cell 3 x plus 3 minus 3 less than 9 minus 3 end cell row 0 less than cell 3 x less than 6 end cell row 0 less than cell x less than 2 end cell end table end style  

Irisan daerah himpunan penyelesaian nilai begin mathsize 14px style x greater or equal than 1 half space d a n space 0 less than x less than 2 end style pada garis bilangan berikut ini. Maka, himpunan penyelesaian daerah III adalah begin mathsize 14px style open curly brackets 1 half less or equal than x less than 2 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 undefined 

Jadi, jawaban yang tepat adalah C.

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