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Pertanyaan

Nilai x yang memenuhi ∣ x + 3 ∣ < 9 − x 2 ​ adalah ...

Nilai  yang memenuhi  adalah ...

  1. negative 3 less than x less than 3 

  2. 0 less than x less than 3 

  3. negative 3 less than x less than 0 space atau space x greater than 3 

  4. x less than negative 3 space atau space straight x greater than 9 

  5. negative 3 less than x less than 0 

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S. Eka

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

jawaban yang tepat adalah E.

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Pembahasan

Ingat bahwa: Nilai mutlak memiliki sifat jika maka . Selain itu bentuk dan bentuk akar memiliki sifat . Sehingga Perhatikan syarat bentuk akar dimana , sehingga bentuk akar pada soal memiliki syarat Lakukan uji titik untuk menentukan daerah penyelesaiannya. Misal Misal Berdasarkan pemisalan dan syarat daerah penyelesaian pertidaksamaan di atas diperoleh bahwa pertidaksamaan adalah . Jadi, jawaban yang tepat adalah E.

Ingat bahwa:

Nilai mutlak memiliki sifat jika open vertical bar f left parenthesis x right parenthesis close vertical bar less than a maka negative a less than f left parenthesis x right parenthesis less than a. Selain itu bentuk open parentheses a plus b close parentheses squared equals a squared plus 2 a b plus b squared dan bentuk akar memiliki sifat open parentheses square root of a close parentheses squared equals a.

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 3 close vertical bar end cell less than cell square root of 9 minus x squared end root end cell row cell negative open parentheses square root of 9 minus x squared end root close parentheses end cell less than cell x plus 3 less than square root of 9 minus x squared end root end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative square root of 9 minus x squared end root close parentheses squared end cell less than cell open parentheses x plus 3 close parentheses squared end cell row cell 9 minus x squared end cell less than cell x squared plus 6 x plus 9 end cell row cell negative x squared minus x squared minus 6 x plus 9 minus 9 end cell less than 0 row cell negative 2 x squared minus 6 x end cell less than 0 row cell 2 x squared plus 6 x end cell greater than 0 row cell 2 x open parentheses x plus 3 close parentheses end cell greater than 0 row cell 2 x end cell greater than 0 row x greater than 0 row blank blank atau row cell straight x plus 3 end cell greater than 0 row straight x greater than cell negative 3 end cell end table

   table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x plus 3 close parentheses squared end cell less than cell open parentheses square root of 9 minus x squared end root close parentheses squared end cell row cell x squared plus 6 x plus 9 end cell less than cell 9 minus x squared end cell row cell x squared plus x squared plus 6 x plus 9 minus 9 end cell less than 0 row cell 2 x squared plus 6 x end cell less than 0 row cell 2 x left parenthesis x plus 3 right parenthesis end cell less than 0 row cell 2 x end cell less than 0 row x less than 0 row blank blank atau row cell x plus 3 end cell less than 0 row x less than cell negative 3 end cell end table

Perhatikan syarat bentuk akar dimana square root of a comma space a greater or equal than 0, sehingga bentuk akar pada soal memiliki syarat

table attributes columnalign right center left columnspacing 0px end attributes row cell 9 minus x squared end cell greater or equal than 0 row cell negative x squared end cell greater or equal than cell negative 9 end cell row cell x squared end cell greater or equal than 9 row cell negative 3 end cell less or equal than cell x less or equal than 3 end cell end table

Lakukan uji titik untuk menentukan daerah penyelesaiannya.

Misal x equals negative 2

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar negative 2 plus 3 close vertical bar end cell less than cell square root of 9 minus left parenthesis negative 2 right parenthesis squared end root end cell row cell open vertical bar 1 close vertical bar end cell less than cell square root of 5 end cell row 1 less than cell square root of 5 horizontal ellipsis open parentheses benar close parentheses end cell end table

Misal x equals 1

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 1 plus 3 close vertical bar end cell less than cell square root of 9 minus left parenthesis 1 right parenthesis squared end root end cell row cell open vertical bar 4 close vertical bar end cell less than cell square root of 8 end cell row 4 less than cell square root of 8 horizontal ellipsis open parentheses salah close parentheses end cell end table

Berdasarkan pemisalan dan syarat daerah penyelesaian pertidaksamaan di atas diperoleh bahwa pertidaksamaan open vertical bar x plus 3 close vertical bar less than square root of 9 minus x squared end root adalah negative 3 less than x less than 0.

Jadi, jawaban yang tepat adalah E.

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