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Agar titik (3, 2) merupakan anggota himpunan penyelesaian sistem pertidaksamaan y≤2x2+mx−1 dan y≥x2+2mx+5 maka haruslah...

Pertanyaan

Agar titik open parentheses 3 comma space 2 close parentheses merupakan anggota himpunan penyelesaian sistem pertidaksamaan y less or equal than 2 x squared plus m x minus 1 dan y greater or equal than x squared plus 2 m x plus 5 maka haruslah...

  1. m less or equal than negative 2 space atau space m greater or equal than 5

  2. m less or equal than negative 5 space atau space m greater or equal than negative 2

  3. m less or equal than 2 space atau space m greater or equal than 5

  4. negative 2 less or equal than m less or equal than 5

  5. negative 5 less or equal than m less or equal than negative 2

O. Rahmawati

Master Teacher

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah E.

Pembahasan

Diketahui: titik left parenthesis 3 comma space 2 right parenthesis

y less or equal than 2 x squared plus m x minus 1... left parenthesis 1 right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row y greater or equal than cell x squared plus 2 m x plus 5 end cell row cell negative y end cell less or equal than cell negative x squared minus 2 m x minus 5... left parenthesis 2 right parenthesis end cell end table

Ditanya: nilai m?

Jawab:

  • Substitusi ke pertidaksamaan (1)

table attributes columnalign right center left columnspacing 0px end attributes row y less or equal than cell 2 x squared plus m x minus 1 end cell row 2 less or equal than cell 2 open parentheses 3 close parentheses squared plus m left parenthesis 3 right parenthesis minus 1 end cell row 2 less or equal than cell 18 plus 3 m minus 1 end cell row 2 less or equal than cell 17 plus 3 m end cell row cell 2 minus 17 end cell less or equal than cell 3 m end cell row cell negative 15 end cell less or equal than cell 3 m end cell row cell 3 m end cell greater or equal than cell negative 15 end cell row m greater or equal than cell negative 5 end cell end table

  • Substitusi ke pertidaksamaan (2):

table attributes columnalign right center left columnspacing 0px end attributes row cell negative y end cell less or equal than cell negative x squared minus 2 m x minus 5 end cell row cell negative 2 end cell less or equal than cell negative open parentheses 3 close parentheses squared minus 2 m open parentheses 3 close parentheses minus 5 end cell row cell negative 2 end cell less or equal than cell negative 9 minus 6 m minus 5 end cell row cell negative 2 end cell less or equal than cell negative 6 m minus 14 end cell row cell negative 2 plus 14 end cell less or equal than cell negative 6 m end cell row 12 less or equal than cell negative 6 m end cell row cell negative 6 m end cell greater or equal than 12 row m less or equal than cell negative 2 end cell end table

Jadi, nilai m haruslah negative 5 less or equal than m less or equal than negative 2.

Oleh karena itu, jawaban yang benar adalah E.

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