Math Se 2016 MEXT Monbukagakusho

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Jawaban dari : Aplikasi gauth pro

Jawaban: (1) 4; (2) 3; (3) 18, 76; (4) 5; (5) 10; (6) 96, 36; (7) 55, 764; (8) -2; (9) (i) 1, -2, (ii) -1, -2, (iii) 2x-1, (iv) 4,5PenjelasanUntuk $x^2-5x+1<0$, akar-akar dari $x^2-5x+1=0$ adalah $x = \frac{5 \pm \sqrt{25-4}}{2} = \frac{5 \pm \sqrt{21}}{2}$

Karena $\sqrt{21} \approx 4,58$, maka $x \approx 0,21$ dan $x \approx 4,79$. Bilangan bulat yang memenuhi adalah $\{1, 2, 3, 4\}$, total 4 bilangan.(2) $\sqrt{(a+1)^2} + \sqrt{(a-2)^2} = |a+1| + |a-2|$. Karena $-1 < a < 2$, maka $|a+1| = a+1$ dan $|a-2| = -(a-2) = 2-a$. Hasilnya $(a+1) + (2-a) = 3$(3) Diketahui $2^x - 2^{-x} = 4$. Kuadratkan: $(2^x - 2^{-x})^2 = 4^2 \Rightarrow 2^{2x} - 2 + 2^{-2x} = 16 \Rightarrow 2^{2x} + 2^{-2x} = 18$. Untuk $2^{3x} - 2^{-3x}$, gunakan $(2^x - 2^{-x})^3 = 2^{3x} - 3(2^x)(2^{-x})(2^x - 2^{-x}) - 2^{-3x} \Rightarrow 4^3 = (2^{3x} - 2^{-3x}) - 3(4) \Rightarrow 64 = (2^{3x} - 2^{-3x}) - 12 \Rightarrow 76$(4) $\log_3(x-3) = \log_9(x-1) \Rightarrow \log_3(x-3) = \frac{\log_3(x-1)}{\log_3 9} = \frac{1}{2}\log_3(x-1)$. Maka $\log_3(x-3)^2 = \log_3(x-1) \Rightarrow x^2 - 6x + 9 = x - 1 \Rightarrow x^2 - 7x + 10 = 0 \Rightarrow (x-5)(x-2) = 0$. Karena syarat logaritma $x-3 > 0$, maka $x=5$(5) Aturan kosinus: $(x+2)^2 = x^2 + (x-2)^2 - 2(x)(x-2)\cos(120^\circ)$. $x^2+4x+4 = x^2 + x^2-4x+4 + x(x-2) \Rightarrow 4x = x^2-4x + x^2-2x \Rightarrow 2x^2 - 10x = 0 \Rightarrow 2x(x-5)=0$. Karena $x-2 > 0$, maka $x=5$(6) (i) Posisi ribuan 4 pilihan (1-4), ratusan 4, puluhan 3, satuan 2. $4 \times 4 \times 3 \times 2 = 96$. (ii) Satuan harus 1 atau 3 (2 pilihan). Ribuan 3 pilihan (sisa dari $\{1,2,3,4\}$ kecuali satuan), ratusan 3, puluhan 2. $2 \times 3 \times 3 \times 2 = 36$(7) (i) $1+4+9+16+25 = 55$. (ii) $\sum_{n=1}^{13} n^2 - 55 = \frac{13(14)(27)}{6} - 55 = 819 - 55 = 764$(8) $2\vec{a}+3\vec{b} = (-2+3, 4+3x) = (1, 4+3x)$. $\vec{a}-2\vec{b} = (-1-2, 2-2x) = (-3, 2-2x)$. Karena paralel, $\frac{1}{-3} = \frac{4+3x}{2-2x} \Rightarrow 2-2x = -12-9x \Rightarrow 7x = -14 \Rightarrow x = -2$(9) (i) $x^2+2x-1 = x+1 \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=-2, 1$. (ii) $y = (x+1)^2 - 2$, vertex $(-1, -2)$. (iii) $f'(x) = 2x+2$. Di $x=0$, $f'(0)=2$. Titik $(0, -1)$, persamaan $y - (-1) = 2(x-0) \Rightarrow y=2x-1$. (iv) $\int_{-2}^{1} (x+1 - (x^2+2x-1)) dx = \int_{-2}^{1} (-x^2-x+2) dx = [-\frac{1}{3}x^3 - \frac{1}{2}x^2 + 2x]_{-2}^{1} = (-\frac{1}{3}-\frac{1}{2}+2) - (\frac{8}{3}-2-4) = \frac{7}{6} - (-\frac{10}{3}) = \frac{27}{6} = 4,5$

Berikut adalah penyelesaian untuk pertanyaan tersebut:Soal 2:

Diketahui $AB=AC=13$ dan $BC=10$. Titik $P$ membagi $BC$ menjadi dua, $BP=PC=5$$AR = AB - BR = 13 - 5 = 8$Tinggi segitiga $h = \sqrt{13^2 - 5^2} = 12$. Luas $L = \frac{1}{2} \times 10 \times 12 = 60$. Jari-jari $r = L/s = 60/18 = 10/3$. Pada $\triangle AOR$ (siku-siku di $R$),

$AR=8, OR=10/3, AO=\sqrt{8^2+(10/3)^2} = \sqrt{724}/3 = 2\sqrt{181}/3$

$\sin \angle AOR = AR/AO = 8 / (2\sqrt{181}/3) = 12/\sqrt{181}$

$\tan \angle AOR = AR/OR = 8 / (10/3) = 24/10 = 2.4$

Jari-jari $r = 10/3$$\vec{AB} \cdot \vec{AO} = AB \cdot AO \cos(\angle BAR) = 13 \cdot (8/\cos(\angle A/2)) \cdot \cos(\angle A/2) = 13 \cdot 8 = 104$$\vec{AB} \cdot \vec{BC} = |\vec{AB}||\vec{BC}| \cos(180^\circ - \angle B) = 13 \cdot 10 \cdot (-5/13) = -50$Jawaban: (1) 8, (2) $12/\sqrt{181}$, (3) 2.4, (4) 10/3, (5) 104 dan -50.Soal 3:

(1) Titik puncak $(-2, 1)$, melalui $(0, 5)$. $y = a(x+2)^2 + 1 \Rightarrow 5 = a(4) + 1 \Rightarrow a=1$. $y = x^2+4x+5$. $a=1, b=4, c=5$

(2) Melalui $(-3, 0)$ dan $(0, 9)$. $y = ax^2+bx+c$. Karena garis lurus (asumsi grafik linear), $y = 3x+9$. Jika kuadrat, data kurang, namun jika linear $a=0, b=3, c=9$

(3) Akar $-1$ dan $3$, melalui $(1, 6)$. $y = a(x+1)(x-3) \Rightarrow 6 = a(2)(-2) \Rightarrow a = -1.5$. $y = -1.5(x^2-2x-3) = -1.5x^2+3x+4.5$. $a=-1.5, b=3, c=4.5$Jawaban: (1) $a=1, b=4, c=5$; (2) $a=0, b=3, c=9$; (3) $a=-1.5, b=3, c=4.5$