Roboguru

Diketahui  dan pernyataan berikut.

Pertanyaan

Diketahui begin mathsize 14px style f left parenthesis x right parenthesis equals fraction numerator 2 x over denominator x plus 2 end fraction semicolon space x not equal to negative 2 comma space g left parenthesis x right parenthesis equals 3 over x semicolon space x not equal to 0 end style dan pernyataan berikut.

begin mathsize 14px style left parenthesis straight i right parenthesis space space space left parenthesis f minus g right parenthesis left parenthesis x right parenthesis equals fraction numerator x squared minus 3 x minus 6 over denominator x squared plus 2 x end fraction left parenthesis ii right parenthesis space space left parenthesis f cross times g right parenthesis left parenthesis x right parenthesis equals fraction numerator 6 x squared over denominator x squared plus 2 x end fraction left parenthesis iii right parenthesis space open parentheses f over g close parentheses left parenthesis x right parenthesis equals fraction numerator 2 x squared over denominator 3 x plus 6 end fraction left parenthesis iv right parenthesis space open parentheses fraction numerator g over denominator f apostrophe end fraction close parentheses left parenthesis x right parenthesis equals fraction numerator 3 x plus 2 over denominator 2 x squared end fraction end style  

  1. (i) dan (ii)space 

  2. (i) dan (iii)space 

  3. (ii) dan (iii)space 

  4. (ii) dan (iv)space 

  5. (iii) dan (iv)space 

Pembahasan Soal:

Fungsi atau pemetaan adalah suatu relasi (hubungan) dari himpunan A ke himpunan B dimana begin mathsize 14px style x element of A end style dipasangkan (dihubungkan) dengan satu dan hanya satu begin mathsize 14px style y element of B end style

Selisih dua fungsi:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f minus g right parenthesis left parenthesis x right parenthesis end cell equals cell fraction numerator 2 x over denominator x plus 2 end fraction minus 3 over x end cell row blank equals cell fraction numerator 2 x left parenthesis x right parenthesis minus 3 left parenthesis x plus 2 right parenthesis over denominator x squared plus 2 x end fraction end cell row blank equals cell fraction numerator 2 x squared minus 3 x minus 6 over denominator x squared plus 2 x end fraction space space space left parenthesis memenuhi right parenthesis end cell end table end style 

Hasil kali dua fungsi:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f times g right parenthesis left parenthesis x right parenthesis end cell equals cell fraction numerator 2 x over denominator x plus 2 end fraction times 3 over x end cell row blank equals cell fraction numerator 6 x over denominator x squared plus 2 x end fraction space space space left parenthesis tidak space memenuhi right parenthesis end cell end table end style 

Hasil bagi dua fungsi:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f over g close parentheses left parenthesis x right parenthesis end cell equals cell fraction numerator begin display style fraction numerator 2 x over denominator x plus 2 end fraction end style over denominator begin display style 3 over x end style end fraction end cell row blank equals cell fraction numerator 2 x over denominator x plus 2 end fraction times x over 3 end cell row blank equals cell fraction numerator 2 x squared over denominator 3 x plus 6 end fraction space space space space left parenthesis memenuhi right parenthesis end cell end table end style

Dapat dilihat bahwa pernyataan yang benar adalah (i) dan (iii).

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 14 Maret 2021

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