Pertanyaan

Diketahui f ( x ) = 3 − x b x dan g ( x ) = a x + 4 . Jika dan f ( 2 ) + g ( − 2 ) = 0 ,maka nilai a 2 − ab + 3 b adalah ....

Diketahui $f (x) = \frac{b x}{3 - x}$ dan $g (x) = \frac{x + 4}{a}$ . Jika dan $f (2) + g (- 2) = 0$ , maka nilai $a^{2} - ab + 3 b$ adalah ....

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Ingat bahwa ,maka Kemudian, ingat bahwa , maka Selanjutnya, Subtitusikan persamaan (3) ke persamaan (2) sehingga kita peroleh Selanjutnya, subtitusikan persamaan (1) ke , maka Lalu persamaan (3) ke persamaan (4) sehingga Untuk mendapatkan nilai b, kita subtitusikan nilai a ke persamaan (3) maka Maka diperoleh, Jadi, jawaban yang tepat adalah E.

Ingat bahwa , 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 , 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.