Roboguru

Pertanyaan

Diketahui size 14px f begin mathsize 14px style left parenthesis x right parenthesis end style size 14px equals fraction numerator size 14px b size 14px x over denominator square root of size 14px 3 size 14px minus size 14px x end root end fraction dan begin mathsize 14px style g left parenthesis x right parenthesis equals fraction numerator x plus 4 over denominator a end fraction end style. Jika begin mathsize 14px style open parentheses f minus g close parentheses open parentheses negative 1 close parentheses equals negative 2 over 3 times g over f open parentheses 2 close parentheses end style dan size 14px f begin mathsize 14px style left parenthesis 2 right parenthesis end style size 14px plus size 14px g begin mathsize 14px style left parenthesis negative 2 right parenthesis end style size 14px equals size 14px 0, maka nilai begin mathsize 14px style a squared minus a b plus 3 b end style adalah ....

  1. begin mathsize 14px style negative 13 over 80 end style 

  2. size 14px 13 over size 14px 80 

  3. size 14px 147 over size 14px 80 

  4. size 14px 237 over size 14px 80 

  5. size 14px 397 over size 14px 80 

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Ingat bahwa begin mathsize 14px style left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals f left parenthesis x right parenthesis minus g left parenthesis x right parenthesis end style, maka

begin mathsize 14px style left parenthesis f minus g right parenthesis left parenthesis negative 1 right parenthesis equals f left parenthesis negative 1 right parenthesis minus g left parenthesis negative 1 right parenthesis left parenthesis f minus g right parenthesis left parenthesis negative 1 right parenthesis equals fraction numerator b left parenthesis negative 1 right parenthesis over denominator square root of 3 minus left parenthesis negative 1 right parenthesis space end root end fraction minus left parenthesis fraction numerator left parenthesis negative 1 right parenthesis plus 4 over denominator a end fraction left parenthesis f minus g right parenthesis left parenthesis negative 1 right parenthesis equals negative fraction numerator b over denominator square root of 4 end fraction minus 3 over a left parenthesis f minus g right parenthesis left parenthesis negative 1 right parenthesis equals negative b over 2 minus 3 over a space space space horizontal ellipsis space space space left parenthesis 1 right parenthesis end style  

Kemudian, ingat bahwa begin mathsize 14px style g over f left parenthesis x right parenthesis equals fraction numerator g left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction end style, maka

begin mathsize 14px style g over f open parentheses 2 close parentheses equals fraction numerator g open parentheses 2 close parentheses over denominator f open parentheses 2 close parentheses end fraction g over f open parentheses 2 close parentheses equals fraction numerator fraction numerator 2 plus 4 over denominator a end fraction over denominator fraction numerator b open parentheses 2 close parentheses over denominator square root of 3 minus 2 end root end fraction end fraction g over f open parentheses 2 close parentheses equals fraction numerator 6 over a over denominator fraction numerator 2 b over denominator square root of 1 end fraction end fraction g over f open parentheses 2 close parentheses equals fraction numerator 6 over a over denominator fraction numerator 2 b over denominator 1 end fraction end fraction g over f open parentheses 2 close parentheses equals fraction numerator 6 over denominator 2 a b end fraction equals fraction numerator 3 over denominator a b end fraction space space space horizontal ellipsis space space 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 2 close parentheses plus g open parentheses negative 2 close parentheses end cell equals 0 row cell fraction numerator b open parentheses 2 close parentheses over denominator square root of 3 minus 2 end root end fraction plus fraction numerator open parentheses negative 2 close parentheses plus 4 over denominator a end fraction end cell equals 0 row cell fraction numerator 2 b over denominator square root of 1 end fraction plus 2 over a end cell equals 0 row cell 2 b plus 2 over a end cell equals 0 row cell 2 b end cell equals cell negative 2 over a end cell row cell 2 a b end cell equals cell negative 2 end cell row cell a b end cell equals cell negative 1 space space horizontal ellipsis space space left parenthesis 3 right parenthesis end cell end table end style  

Subtitusikan persamaan (3) ke persamaan (2) sehingga kita peroleh

begin mathsize 14px style g over f open parentheses 2 close parentheses equals fraction numerator 3 over denominator negative 1 end fraction equals negative 3 end style  

Selanjutnya, subtitusikan persamaan (1) ke begin mathsize 14px style open parentheses f minus g close parentheses open parentheses negative 1 close parentheses equals negative 2 over 3 g over f open parentheses 2 close parentheses end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative b over 2 minus 3 over a end cell equals cell negative 2 over 3 open parentheses negative 3 close parentheses end cell row cell fraction numerator negative a b minus 6 over denominator 2 a end fraction end cell equals 2 row cell negative a b minus 6 end cell equals cell 4 a space space space horizontal ellipsis space space open parentheses 4 close parentheses end cell end table end style  

Lalu persamaan (3) ke persamaan (4) sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses negative 1 close parentheses minus 6 end cell equals cell 4 a end cell row cell 1 minus 6 end cell equals cell 4 a end cell row cell negative 5 end cell equals cell 4 a end cell row a equals cell negative 5 over 4 end cell end table end style  

Untuk mendapatkan nilai b, kita subtitusikan nilai a ke persamaan (3) maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 5 over 4 b end cell equals cell negative 1 end cell row b equals cell 4 over 5 end cell end table end style  

Maka diperoleh,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a squared minus a b plus 3 b end cell equals cell open parentheses negative 5 over 4 close parentheses squared minus open parentheses negative 1 close parentheses plus 3 open parentheses 4 over 5 close parentheses end cell row blank equals cell 25 over 16 plus 1 plus 12 over 5 end cell row blank equals cell fraction numerator 125 plus 80 plus 192 over denominator 80 end fraction end cell row blank equals cell 397 over 80 end cell end table end style  

Jadi, jawaban yang tepat adalah E.

18

0.0 (0 rating)

Pertanyaan serupa

Diketahui f(x)=2x−4dan(x)=2x2−x+5. Nilai dari gf​(−3) adalah ....

18

0.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia