Pertanyaan

Diketahui . Jika ,maka nilai 6 p + 6 q adalah ....

Diketahui $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator p x minus 5 over denominator square root of 2 x plus 1 end root end fraction blank dan blank g open parentheses x close parentheses equals fraction numerator negative 3 x minus 9 over denominator 3 q x minus 1 end fraction end style$ . Jika , maka nilai 6p + 6q adalah ....

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Ingat bahwa . Maka Kemudian, ingat bahwa .Maka Selanjutnya, Subtitusikan persamaan (1) dan (2) ke persamaan ,serta subtitusikan nilai p ke persamaan tersebut sehingga kita peroleh Jadi, didapat Jadi, jawaban yang tepat adalah B.

Ingat bahwa . Maka

$begin mathsize 14px style open parentheses f times g close parentheses open parentheses 4 close parentheses equals f open parentheses 4 close parentheses times g open parentheses 4 close parentheses open parentheses f times g close parentheses open parentheses 4 close parentheses equals open parentheses fraction numerator p open parentheses 4 close parentheses minus 5 over denominator square root of 2 open parentheses 4 close parentheses plus 1 end root end fraction close parentheses times open parentheses fraction numerator negative 3 open parentheses 4 close parentheses minus 9 over denominator 3 q open parentheses 4 close parentheses minus 1 end fraction close parentheses open parentheses f times g close parentheses open parentheses 4 close parentheses equals open parentheses fraction numerator 4 p minus 5 over denominator square root of 8 plus 1 end root end fraction close parentheses times open parentheses fraction numerator negative 12 minus 9 over denominator 12 q minus 1 end fraction close parentheses open parentheses f times g close parentheses open parentheses 4 close parentheses equals open parentheses fraction numerator 4 p minus 5 over denominator 3 end fraction close parentheses times open parentheses fraction numerator negative 21 over denominator 12 q minus 1 end fraction close parentheses open parentheses f times g close parentheses open parentheses 4 close parentheses equals open parentheses fraction numerator negative 28 p plus 35 over denominator 12 q minus 1 end fraction close parentheses horizontal ellipsis horizontal ellipsis horizontal ellipsis open parentheses 1 close parentheses end style$

Kemudian, ingat bahwa $begin mathsize 14px style f over g open parentheses x close parentheses equals fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction end style$ . Maka

$begin mathsize 14px style f over g open parentheses 0 close parentheses equals fraction numerator f open parentheses 0 close parentheses over denominator g open parentheses 0 close parentheses end fraction f over g open parentheses 0 close parentheses equals fraction numerator fraction numerator p open parentheses 0 close parentheses minus 5 over denominator square root of 2 open parentheses 0 close parentheses plus 1 end root end fraction over denominator fraction numerator negative 3 open parentheses 0 close parentheses minus 9 over denominator 3 q open parentheses 0 close parentheses minus 1 end fraction end fraction f over g open parentheses 0 close parentheses equals fraction numerator fraction numerator negative 5 over denominator square root of 1 end fraction over denominator fraction numerator negative 9 over denominator negative 1 end fraction end fraction f over g open parentheses 0 close parentheses equals fraction numerator negative 5 over denominator 9 end fraction horizontal ellipsis horizontal ellipsis horizontal ellipsis open parentheses 2 close parentheses end style$

Selanjutnya,

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses 12 close parentheses plus g open parentheses negative 3 close parentheses end cell equals 0 row cell fraction numerator p open parentheses 12 close parentheses minus 5 over denominator square root of 2 open parentheses 12 close parentheses plus 1 end root end fraction plus fraction numerator negative 3 open parentheses negative 3 close parentheses minus 9 over denominator 3 q open parentheses negative 3 close parentheses minus 1 end fraction end cell equals 0 row cell fraction numerator 12 p minus 5 over denominator square root of 25 end fraction plus fraction numerator 9 minus 9 over denominator negative 9 q minus 1 end fraction end cell equals 0 row cell fraction numerator 12 p minus 5 over denominator 5 end fraction end cell equals 0 row cell 12 p minus 5 end cell equals 0 row cell 12 p end cell equals 5 row p equals cell 5 over 12 horizontal ellipsis horizontal ellipsis horizontal ellipsis left parenthesis 3 right parenthesis end cell row blank blank blank end table end style$

Subtitusikan persamaan (1) dan (2) ke persamaan , serta subtitusikan nilai p ke persamaan tersebut sehingga kita peroleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator negative 28 p plus 35 over denominator 12 q minus 1 end fraction close parentheses end cell equals cell 3 open parentheses fraction numerator negative 5 over denominator 9 end fraction close parentheses end cell row cell open parentheses fraction numerator negative 28 open parentheses 5 over 12 close parentheses plus 35 over denominator 12 q minus 1 end fraction close parentheses end cell equals cell open parentheses fraction numerator negative 5 over denominator 3 end fraction close parentheses end cell row cell open parentheses fraction numerator negative 35 over 3 plus 35 over denominator 12 q minus 1 end fraction close parentheses end cell equals cell open parentheses fraction numerator negative 5 over denominator 3 end fraction close parentheses end cell row cell 3 open parentheses negative 35 over 3 plus 35 close parentheses end cell equals cell negative 5 open parentheses 12 q minus 1 close parentheses end cell row cell negative 35 plus 105 end cell equals cell negative 60 q plus 5 end cell row cell 60 q end cell equals cell 35 minus 105 plus 5 end cell row cell 60 q end cell equals cell negative 65 end cell row q equals cell negative 65 over 60 end cell row q equals cell negative 13 over 12 end cell end table end style$

Jadi, didapat

Jadi, jawaban yang tepat adalah B.