Pertanyaan

Diketahui: sin A + sin B = 2 sin 2 1 ( A + B ) ⋅ cos 2 1 ( A − B ) dan cos A − cos B = − 2 sin 2 1 ( A + B ) ⋅ sin 2 1 ( A − B ) maka nilai dari x → 0 lim cos 4 x − cos 2 x x ( sin 6 x + sin 4 x ) = …

Diketahui: $sin A + sin B = 2 sin \frac{1}{2} (A + B) \cdot cos \frac{1}{2} (A - B)$ dan $cos A - cos B = - 2 sin \frac{1}{2} (A + B) \cdot sin \frac{1}{2} (A - B)$ maka nilai dari $x \to 0 lim \frac{x ( sin 6 x + sin 4 x )}{cos 4 x - cos 2 x} = \dots$

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Habis dalam

Klaim

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

Diketahui: Ditanya: nilai dari ? Penyelesaian: Ingat kembali bahwa: Misalkan: , maka nilai dari adalah Jadi, nilai dari adalah Oleh karena itu, jawaban yang benar adalah A.

Diketahui:

Ditanya:

nilai dari $limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction$ ?

Penyelesaian:

Ingat kembali bahwa:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator sin space x over denominator x end fraction end cell equals 1 row cell limit as x rightwards arrow 0 of space fraction numerator sin space a x over denominator b x end fraction end cell equals cell a over b end cell end table$

Misalkan: $M equals limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction$ , maka nilai dari adalah

$table attributes columnalign right center left columnspacing 0px end attributes row M equals cell limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator x open parentheses 2 space sin space 1 half open parentheses 6 x plus 4 x close parentheses times cos space 1 half open parentheses 6 x minus 4 x close parentheses close parentheses over denominator negative 2 space sin space 1 half open parentheses 4 x plus 2 x close parentheses times sin space 1 half open parentheses 4 x minus 2 x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator x open parentheses 2 space sin space 1 half open parentheses 10 x close parentheses times cos space 1 half open parentheses 2 x close parentheses close parentheses over denominator negative 2 space sin space 1 half open parentheses 6 x close parentheses times sin space 1 half open parentheses 2 x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator x open parentheses 2 space sin space 5 x times cos space x close parentheses over denominator negative 2 space sin space 3 x times sin space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator x times sin space 5 x times cos space x over denominator negative sin space 3 x times sin space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator begin display style x over x end style times begin display style fraction numerator sin space 5 x over denominator x end fraction end style over denominator negative begin display style fraction numerator sin space 3 x over denominator x end fraction end style times begin display style fraction numerator sin space x over denominator x end fraction end style end fraction times limit as x rightwards arrow 0 of cos space x end cell row blank equals cell fraction numerator 1 times 5 over denominator negative 3 times 1 end fraction times 1 end cell row M equals cell negative 5 over 3 end cell end table$

Jadi, nilai dari $limit as x rightwards arrow 0 of space fraction numerator x open parentheses sin space 6 x plus sin space 4 x close parentheses over denominator cos space 4 x minus cos space 2 x end fraction$ adalah negative 5 over 3.

Oleh karena itu, jawaban yang benar adalah A.