Jawaban:

Ingat definisi mutlak :

open vertical bar x plus 1 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell x plus 1 greater or equal than 0 rightwards double arrow open vertical bar x plus 1 close vertical bar equals x plus 1... left parenthesis 1 right parenthesis end cell row cell x plus 1 less than 0 rightwards double arrow open vertical bar x plus 1 close vertical bar equals negative open parentheses x plus 1 close parentheses... left parenthesis 2 right parenthesis end cell end table close

Karena yang dicari adalah penyelesaian bulat dan negatif, maka ambil definisi (2). Sehingga,

fraction numerator open vertical bar x plus 1 close vertical bar minus 2 x over denominator x squared plus x minus 12 end fraction less or equal than 0 fraction numerator negative open parentheses x plus 1 close parentheses minus 2 x over denominator x squared plus x minus 12 end fraction less or equal than 0 fraction numerator negative 3 x minus 1 over denominator open parentheses x plus 4 close parentheses open parentheses x minus 3 close parentheses end fraction less or equal than 0 pembuat space nol space colon pembilang space colon space x equals negative 1 third penyebut space colon space x not equal to negative 4 space atau space x not equal to 3

Susun ke pembuat nol ke garis bilangan dan uji interval:

keterangan :

untuk interval x less than negative 4, misal uji x equals negative 5.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative 3 left parenthesis negative 5 right parenthesis minus 1 over denominator open parentheses negative 5 plus 4 close parentheses open parentheses negative 5 minus 3 close parentheses end fraction end cell less or equal than 0 row cell 14 over 8 end cell less or equal than cell 0 space left parenthesis salah right parenthesis end cell end table

untuk interval negative 4 less than x less than negative 1 third, misal uji x equals negative 3.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative 3 left parenthesis negative 3 right parenthesis minus 1 over denominator open parentheses negative 3 plus 4 close parentheses open parentheses negative 3 minus 3 close parentheses end fraction end cell less or equal than 0 row cell fraction numerator 8 over denominator negative 6 end fraction end cell less or equal than cell 0 space open parentheses text benar end text close parentheses end cell end table

untuk interval negative 1 third less than x less than 3, misal uji x equals 0.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative 3 left parenthesis 0 right parenthesis minus 1 over denominator open parentheses 0 plus 4 close parentheses open parentheses 0 minus 3 close parentheses end fraction end cell less or equal than 0 row cell fraction numerator negative 1 over denominator negative 12 end fraction end cell less or equal than 0 row cell 1 over 12 end cell less or equal than cell 0 space open parentheses text salah end text close parentheses end cell end table

untuk interval x greater than 3, misal uji x equals 4

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative 3 left parenthesis 4 right parenthesis minus 1 over denominator open parentheses 4 plus 4 close parentheses open parentheses 4 minus 3 close parentheses end fraction end cell less or equal than 0 row cell fraction numerator negative 13 over denominator 16 end fraction end cell less or equal than cell 0 space open parentheses text benar end text close parentheses end cell end table

Dengan demikian, bilangan bulat negatif pada interval negative 4 less than x less than negative 1 third adalah negative 3 comma space minus 2 space dan space minus 1, yaitu sebanyak 3 buah.

Jadi, jawaban yang tepat adalah B.