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Himpunan penyelesaian dari pertidaksamaan  adalah ...

Pertanyaan

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

  1. begin mathsize 14px style open curly brackets x vertical line minus 3 less than x less than 1 comma blank x element of R close curly brackets end style  

  2. begin mathsize 14px style open curly brackets x vertical line minus 1 less than x less than 3 comma blank x element of R close curly brackets end style  

  3. begin mathsize 14px style open curly brackets x vertical line minus 1 less than x less than 0 blank atau blank x greater than 2 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 negative 1 blank atau blank x greater than 2 blank x element of R close curly brackets end style  

  5. begin mathsize 14px style open curly brackets x vertical line minus 2 less than x less than negative 1 blank atau blank x greater than 0 blank x element of R close curly brackets end style 

K. Putri

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Ganesha

Jawaban terverifikasi

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 plus 3 close vertical bar end cell equals cell open curly brackets table row cell open parentheses 2 x plus 3 close parentheses comma blank open parentheses 2 x plus 3 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than negative 3 over 2 end cell row cell negative open parentheses 2 x plus 3 close parentheses comma blank open parentheses 2 x plus 3 close parentheses less than 0 blank a t a u blank x less than negative 3 over 2 end cell end table close end cell row cell open vertical bar x minus 1 close vertical bar end cell equals cell open curly brackets table row cell open parentheses x minus 1 close parentheses comma blank open parentheses x minus 1 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than 1 end cell row cell negative open parentheses x minus 1 close parentheses comma blank open parentheses x minus 1 close parentheses less than 0 blank a t a u blank x less than 1 end cell end table close end cell row cell open vertical bar x minus 2 close vertical bar end cell equals cell open curly brackets table row cell open parentheses x minus 2 close parentheses comma blank open parentheses x minus 2 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than 2 end cell row cell negative open parentheses x minus 2 close parentheses comma blank open parentheses x minus 2 close parentheses less than 0 blank a t a u blank x less than 2 end cell end table close end cell end table end style   

Gabungan daerah untuk interval nilai x di atas menghasilkan empat kemungkinan interval nilai x.

Kemungkinan I
Untuk interval begin mathsize 14px style x less than negative 3 over 2 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 2 x plus 3 close parentheses minus open square brackets negative open parentheses x minus 2 close parentheses close square brackets close vertical bar end cell greater than cell negative open parentheses x minus 1 close parentheses end cell row cell open vertical bar negative 2 x minus 3 plus x minus 2 close vertical bar end cell greater than cell negative x plus 1 end cell row cell open vertical bar negative x minus 5 close vertical bar end cell greater than cell negative x plus 1 end cell row cell open parentheses negative x minus 5 close parentheses squared end cell greater than cell open parentheses negative x plus 1 close parentheses squared end cell row cell x squared plus 10 x plus 25 end cell greater than cell x squared minus 2 x plus 1 end cell row cell 10 x plus 25 end cell greater than cell negative 2 x plus 1 end cell row cell 10 x plus 2 x end cell greater than cell 1 minus 25 end cell row cell 12 x end cell greater than cell negative 24 end cell row x greater than cell negative 2 end cell end table end style   

Perhatikan daerah irisan begin mathsize 14px style x less than negative 3 over 2 space d a n space x greater than negative 2 end style pada garis bilangan berikut ini.

 

Maka, daerah himpunan penyelesian untuk kemungkinan I adalah begin mathsize 14px style open curly brackets negative 2 less than x less than negative 3 over 2 close curly brackets end style 

Kemungkinan II
Untuk interval begin mathsize 14px style negative 3 over 2 less or equal than x less than 1 end style maka diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 3 minus open square brackets negative open parentheses x minus 2 close parentheses close square brackets close vertical bar end cell greater than cell negative open parentheses x minus 1 close parentheses end cell row cell open vertical bar 2 x plus 3 plus x minus 2 close vertical bar end cell greater than cell negative x plus 1 end cell row cell open vertical bar 3 x plus 1 close vertical bar end cell greater than cell negative x plus 1 end cell row cell open parentheses 3 x plus 1 close parentheses squared end cell greater than cell open parentheses negative x plus 1 close parentheses squared end cell row cell 9 x squared plus 6 x plus 1 end cell greater than cell x squared minus 2 x plus 1 end cell row cell 9 x squared minus x squared plus 6 x plus 2 x end cell greater than 0 row cell 8 x squared plus 8 x end cell greater than 0 row cell 8 x open parentheses x plus 1 close parentheses end cell greater than 0 row blank blank blank end table end style   

Perhatikan daerah irisan undefined dan pertidaksamaan begin mathsize 14px style 8 x open parentheses x plus 1 close parentheses greater than 0 end style pada garis bilangan berikut ini.

Maka, daerah himpunan penyelesian untuk kemungkinan II adalah begin mathsize 14px style open curly brackets negative 1 less than x less than 0 close curly brackets end style 

Kemungkinan III
Untuk interval begin mathsize 14px style 1 greater or equal than x greater than 2 end style maka diperoleh 

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

Perhatikan bahwa ruas kiri pertidaksamaan tersebut akan selalu positif dan untuk berapapun nilai x, pertidaksamaan tersebut akan bernilai benar. Sehingga penyelesaian pertidaksamaan tersebut adalah x∈R. Kemudian kita iriskan 1 ≥ x > 2  dengan x∈R pada garis bilangan. 

 

Maka, daerah himpunan penyelesian untuk kemungkinan III adalah begin mathsize 14px style open curly brackets 1 greater or equal than x greater than 2 blank close curly brackets end style 

Kemungkinan IV
Untuk interval begin mathsize 14px style x greater or equal than 2 end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 3 minus open parentheses x minus 2 close parentheses close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar 2 x plus 3 plus x plus 2 close vertical bar end cell greater than cell x minus 1 end cell row cell open vertical bar 3 x plus 5 close vertical bar end cell greater than cell x minus 1 end cell row cell open parentheses 3 x plus 5 close parentheses squared end cell greater than cell open parentheses x minus 1 close parentheses squared end cell row cell 9 x squared plus 30 x plus 25 end cell greater than cell x squared minus 2 x plus 1 end cell row cell 9 x squared minus x squared plus 30 x plus 2 x plus 25 minus 1 end cell greater than 0 row cell 8 x squared plus 32 x plus 24 end cell greater than 0 row cell x squared plus 4 x plus 3 end cell greater than 0 row cell open parentheses x plus 1 close parentheses open parentheses x plus 3 close parentheses end cell greater than 0 end table end style  

Perhatikan daerah irisan x ≥ 2 dan pertidaksamaan begin mathsize 14px style open parentheses x plus 1 close parentheses open parentheses x plus 3 close parentheses greater than 0 end style pada garis bilangan berikut ini.

 

Maka, daerah himpunan penyelesian untuk kemungkinan IV adalah begin mathsize 14px style open curly brackets x greater or equal than 2 close curly brackets end style 

Jika kita gabungkan daerah himpunan penyelesaian dari keempat kemungkinan atas dalam sebuah garis bilangan hingga diperoleh gambar berikut ini.

Maka,himpunan penyelesaian pertidaksamaan undefined adalah undefined  Jadi, jawaban yang tepat adalah E.

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