Pertanyaan

Jika x → 0 lim 3 − Ax + B x = 2 , maka ....

Jika $x \to 0 lim \frac{x}{3 - Ax + B} = 2$ , maka ....

A² = –B
3A = B
3A + B = 0
A = 3B
A = B²

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A. Abdul

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Misalkan f(x) = x dan . Perhatikan bahwa f(0) = 0. Jikag(0) ≠ 0, maka . Namun . Sehingga haruslah g(0) = 0. Maka Sehingga Maka didapatkan A = –3 dan B = 9. Perhatikan setiap pilihan yang tersedia. A)A²= –B A² = (–3)² = 9 –B = –9 Sehingga A² ≠ –B (pernyataan A salah) B) 3A = B 3A = 3(–3) = –9 B = 9 Sehingga 3A ≠ B (pernyataan B salah) C) 3A + B = 0 3A + B = 3(–3) + 9 = –9 + 9 = 0 Sehingga 3A + B = 0 (pernyataan C benar) D) A = 3B A = –3 3B = 3(9) = 27 Sehingga A ≠ 3B (pernyataan D salah) E) A = B² A = –3 B² = 9² = 81 Sehingga A ≠ B² (pernyataan E salah) Jadi, jawaban yang tepat adalah C.

Misalkan f(x) = x dan .

Perhatikan bahwa f(0) = 0. Jika g(0) ≠ 0, maka $begin mathsize 14px style limit as straight x rightwards arrow 0 of invisible function application fraction numerator straight x over denominator 3 minus square root of Ax plus straight B end root end fraction equals 0 end style$ .

Namun $begin mathsize 14px style limit as straight x rightwards arrow 0 of invisible function application fraction numerator straight x over denominator 3 minus square root of Ax plus straight B end root end fraction not equal to 0 end style$ . Sehingga haruslah g(0) = 0.

Maka

Sehingga

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 0 of invisible function application fraction numerator straight x over denominator 3 minus square root of Ax plus 9 end root end fraction end cell equals 2 row cell limit as straight x rightwards arrow 0 of invisible function application open parentheses fraction numerator straight x over denominator 3 minus square root of Ax plus 9 end root end fraction times fraction numerator 3 plus square root of Ax plus 9 end root over denominator 3 plus square root of Ax plus 9 end root end fraction close parentheses end cell equals 2 row cell limit as straight x rightwards arrow 0 of invisible function application fraction numerator straight x open parentheses 3 plus square root of Ax plus 9 end root close parentheses over denominator 9 minus open parentheses Ax plus 9 close parentheses end fraction end cell equals 2 row cell limit as straight x rightwards arrow 0 of invisible function application fraction numerator straight x open parentheses 3 plus square root of Ax plus 9 end root close parentheses over denominator 9 minus Ax minus 9 end fraction end cell equals 2 row cell limit as straight x rightwards arrow 0 of invisible function application fraction numerator straight x open parentheses 3 plus square root of Ax plus 9 end root close parentheses over denominator negative Ax end fraction end cell equals 2 row cell limit as straight x rightwards arrow 0 of invisible function application fraction numerator 3 plus square root of Ax plus 9 end root over denominator negative straight A end fraction end cell equals 2 row cell fraction numerator 3 plus square root of straight A open parentheses 0 close parentheses plus 9 end root over denominator negative straight A end fraction end cell equals 2 row cell fraction numerator 3 plus square root of 9 over denominator negative straight A end fraction end cell equals 2 row cell fraction numerator 3 plus 3 over denominator negative straight A end fraction end cell equals 2 row cell fraction numerator 6 over denominator negative straight A end fraction end cell equals 2 row straight A equals cell negative 3 end cell end table end style$

Maka didapatkan A = –3 dan B = 9.

Perhatikan setiap pilihan yang tersedia.

A) A² = –B
A² = (–3)² = 9
–B = –9
Sehingga A² ≠ –B (pernyataan A salah)

B) 3A = B
3A = 3(–3) = –9
B = 9
Sehingga 3A ≠ B (pernyataan B salah)

C) 3A + B = 0
3A + B = 3(–3) + 9 = –9 + 9 = 0
Sehingga 3A + B = 0 (pernyataan C benar)

D) A = 3B
A = –3
3B = 3(9) = 27
Sehingga A ≠ 3B (pernyataan D salah)

E) A = B²
A = –3
B² = 9² = 81
Sehingga A ≠ B² (pernyataan E salah)

Jadi, jawaban yang tepat adalah C.