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# Pembahasan soal snmptn 2012 matematika ipa kode 634

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• 1. Pembahasan Soal SNMPTN 2012 SELEKSI NASIONAL MASUK PERGURUAN TINGGI NEGERI Disertai TRIK SUPERKILAT dan LOGIKA PRAKTIS Matematika IPA Disusun Oleh : Pak Anang
• 2. Bimbel SBMPTN 2013 Matematika IPA by Pak Anang (http://pak-anang.blogspot.com) Halaman 1 Kumpulan SMART SOLUTION dan TRIK SUPERKILAT Pembahasan Soal SNMPTN 2012 Matematika IPA Kode Soal 634 By Pak Anang (http://pak-anang.blogspot.com) lim 𝑥→0 1 − cos2 𝑥 𝑥2 tan (𝑥 + 𝜋 3 ) A. −√3 B. 0 C. √3 3 D. √3 2 E. √3 Penyelesaian: Ingat: lim 𝑥→0 𝑥 sin 𝑥 = lim 𝑥→0 sin 𝑥 𝑥 = lim 𝑥→0 𝑥 tan 𝑥 = lim 𝑥→0 tan 𝑥 𝑥 = 1 1 = sin2 𝑥 + cos2 𝑥 Substitusi 𝑥 = 0 pada limit: lim 𝑥→0 1 − cos2 𝑥 𝑥2 tan (𝑥 + 𝜋 3 ) = 1 − 1 02√3 = 0 0 (bentuk tak tentu) Jadi limit tersebut diselesaikan menggunakan identitas trigonometri: lim 𝑥→0 1 − cos2 𝑥 𝑥2 tan (𝑥 + 𝜋 3 ) = lim 𝑥→0 (sin2 𝑥 + cos2 𝑥) − cos2 𝑥 𝑥2 tan (𝑥 + 𝜋 3 ) = lim 𝑥→0 sin2 𝑥 𝑥2 tan (𝑥 + 𝜋 3 ) = lim 𝑥→0 sin 𝑥 𝑥 ∙ sin 𝑥 𝑥 ∙ 1 tan (𝑥 + 𝜋 3 ) = lim 𝑥→0 sin 𝑥 𝑥 ∙ lim 𝑥→0 sin 𝑥 𝑥 ∙ lim 𝑥→0 1 tan (𝑥 + 𝜋 3 ) (Ingat lim 𝑥→0 sin 𝑥 𝑥 = 1) = 1 ∙ 1 ∙ lim 𝑥→0 1 tan (𝑥 + 𝜋 3 ) = 1 tan (0 + 𝜋 3 ) = 1 tan 60° = 1 √3 (Ingat rasionalisasi bentuk akar) = 1 √3 × √3 √3 = √3 3 1. TRIK SUPERKILAT: lim 𝑥→0 1 − cos2 𝑥 𝑥2 tan (𝑥 + 𝜋 3 ) = 𝑥2 𝑥2 tan 𝜋 3 = 1 √3 = √3 3