Pertanyaan

.... (SBMPTN 2018)

$begin mathsize 14px style limit as straight x rightwards arrow 3 of invisible function application fraction numerator sin invisible function application open parentheses 2 straight x minus 6 close parentheses over denominator square root of 4 minus straight x end root minus 1 end fraction equals end style$ ....

(SBMPTN 2018)

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Klaim

R. Diah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Jawaban

jawabannya adalah E.

Pembahasan

Perhatikan bahwa Kemudian, ingat kembali bahwa Jika , maka mengakibatkan atau . Oleh karena itu, diperoleh Dengan demikian, didapat Jadi, jawabannya adalah E.

Perhatikan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 3 of space fraction numerator sin left parenthesis 2 x minus 6 right parenthesis over denominator square root of 4 minus x end root minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space open parentheses fraction numerator sin left parenthesis 2 x minus 6 right parenthesis over denominator square root of 4 minus x end root minus 1 end fraction times fraction numerator square root of 4 minus x end root plus 1 over denominator square root of 4 minus x end root plus 1 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator sin left parenthesis 2 x minus 6 right parenthesis open parentheses square root of 4 minus x end root plus 1 close parentheses over denominator 4 minus x minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator sin left parenthesis 2 x minus 6 right parenthesis open parentheses square root of 4 minus x end root plus 1 close parentheses over denominator 3 minus x end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space open parentheses fraction numerator sin left parenthesis 2 x minus 6 right parenthesis open parentheses square root of 4 minus x end root plus 1 close parentheses over denominator 3 minus x end fraction times 2 over 2 close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator 2 sin left parenthesis 2 x minus 6 right parenthesis open parentheses square root of 4 minus x end root plus 1 close parentheses over denominator 6 minus 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator 2 sin left parenthesis 2 x minus 6 right parenthesis open parentheses square root of 4 minus x end root plus 1 close parentheses over denominator negative open parentheses 2 x minus 6 close parentheses end fraction end cell row blank equals cell open parentheses limit as x rightwards arrow 3 of space 2 close parentheses times open parentheses negative limit as x rightwards arrow 3 of fraction numerator sin left parenthesis 2 x minus 6 right parenthesis over denominator 2 x minus 6 end fraction close parentheses times open parentheses limit as x rightwards arrow 3 of space square root of 4 minus x end root plus 1 close parentheses end cell end table end style$

Kemudian, ingat kembali bahwa

$begin mathsize 14px style limit as a rightwards arrow 0 of space fraction numerator sin space a over denominator a end fraction equals 1 end style$

Jika , maka mengakibatkan atau .

Oleh karena itu, diperoleh

$begin mathsize 14px style limit as x rightwards arrow 3 of fraction numerator sin left parenthesis 2 x minus 6 right parenthesis over denominator 2 x minus 6 end fraction equals 1 end style$

Dengan demikian, didapat

$begin mathsize 14px style open parentheses limit as x rightwards arrow 3 of space 2 close parentheses times open parentheses negative limit as x rightwards arrow 3 of space fraction numerator sin left parenthesis 2 x minus 6 right parenthesis over denominator 2 x minus 6 end fraction close parentheses times open parentheses limit as x rightwards arrow 3 of square root of 4 minus x end root plus 1 close parentheses equals 2 times open parentheses negative 1 close parentheses times left parenthesis square root of 4 minus 3 end root plus 1 right parenthesis equals negative 2 times 2 equals negative 4 end style$

Jadi, jawabannya adalah E.