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Diketahui lim x → 0 ​ x 2 ax 2 + b ​ − 3 ​ = 4 1 ​ , maka nilai dari 3 a + b = ....

Diketahui  , maka nilai dari ....

  1. begin mathsize 14px style 27 over 2 end style 

  2. begin mathsize 14px style 21 over 2 end style 

  3. begin mathsize 14px style 18 over 2 end style 

  4. begin mathsize 14px style 15 over 2 end style 

  5. begin mathsize 14px style 9 over 2 end style 

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

jawaban yang tepat adalah A.

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Pembahasan

Perhatikan bahwa jika dilakukan substitusi x = 0 , maka didapatkan hasil Didapat bentuk limit yang jika dilakukan substitusi x = 0 , maka penyebutnya bernilai 0. Tetapi, nilai limitnya ada. Maka haruslah pembilangnya juga bernilai 0 jika disubstitusi x = 0 . Sehingga Maka didapat bahwa Sehingga Jadi, jawaban yang tepat adalah A.

Perhatikan bahwa jika dilakukan substitusi = 0, maka didapatkan hasil

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator square root of ax squared plus straight b end root minus 3 over denominator straight x squared end fraction end cell equals cell fraction numerator square root of straight a open parentheses 0 close parentheses squared plus straight b end root minus 3 over denominator 0 squared end fraction end cell row blank equals cell fraction numerator square root of 0 plus straight b end root minus 3 over denominator 0 end fraction end cell row blank equals cell fraction numerator square root of straight b minus 3 over denominator 0 end fraction end cell end table end style 

Didapat bentuk limit yang jika dilakukan substitusi = 0, maka penyebutnya bernilai 0. Tetapi, nilai limitnya ada.
Maka haruslah pembilangnya juga bernilai 0 jika disubstitusi = 0. Sehingga
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell square root of straight b minus 3 end cell equals 0 row cell square root of straight b end cell equals 3 row straight b equals 9 end table end style 

Maka didapat bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 0 of fraction numerator square root of ax squared plus 9 end root minus 3 over denominator straight x squared end fraction end cell equals cell 1 fourth end cell row cell limit as straight x rightwards arrow 0 of open parentheses fraction numerator square root of ax squared plus 9 end root minus 3 over denominator straight x squared end fraction times fraction numerator square root of ax squared plus 9 end root plus 3 over denominator square root of ax squared plus 9 end root plus 3 end fraction close parentheses end cell equals cell 1 fourth end cell row cell limit as straight x rightwards arrow 0 of fraction numerator open parentheses ax squared plus 9 close parentheses minus 9 over denominator straight x squared open parentheses square root of ax squared plus 9 end root plus 3 close parentheses end fraction end cell equals cell 1 fourth end cell row cell limit as straight x rightwards arrow 0 of fraction numerator ax squared over denominator straight x squared open parentheses square root of ax squared plus 9 end root plus 3 close parentheses end fraction end cell equals cell 1 fourth end cell row cell limit as straight x rightwards arrow 0 of fraction numerator straight a over denominator square root of ax squared plus 9 end root plus 3 end fraction end cell equals cell 1 fourth end cell row cell fraction numerator straight a over denominator square root of straight a open parentheses 0 close parentheses squared plus 9 end root plus 3 end fraction end cell equals cell 1 fourth end cell row cell fraction numerator straight a over denominator square root of 0 plus 9 end root plus 3 end fraction end cell equals cell 1 fourth end cell row cell fraction numerator straight a over denominator square root of 9 plus 3 end fraction end cell equals cell 1 fourth end cell row cell fraction numerator straight a over denominator 3 plus 3 end fraction end cell equals cell 1 fourth end cell row cell straight a over 6 end cell equals cell 1 fourth end cell row straight a equals cell 6 over 4 end cell row straight a equals cell 3 over 2 end cell end table end style 

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 straight a plus straight b end cell equals cell 3 open parentheses 3 over 2 close parentheses plus 9 end cell row blank equals cell 9 over 2 plus 9 end cell row blank equals cell 9 over 2 plus 18 over 2 end cell row blank equals cell 27 over 2 end cell end table end style 

Jadi, jawaban yang tepat adalah A.

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