Math Se 2015 MEXT Monbukagakusho

 

Jawaban: (1) 1 < x < 2; (2) -2; (3) 5; (4) 19, 13; (5) 120; (6) 1/2; (7) 33, 4950; (8) a=2, b=-8; (9) 1, 2; (10) a=5, b=8, c=11.Penjelasan(1) $ax^2 - 3ax + 2a < 0$. Karena $a > 0$, bagi dengan $a$: $x^2 - 3x + 2 < 0 \Rightarrow (x-1)(x-2) < 0$. Maka $1 < x < 2$(2) $4^{3x-1} = 2^{5x-4} \Rightarrow 2^{2(3x-1)} = 2^{5x-4} \Rightarrow 6x-2 = 5x-4 \Rightarrow x = -2$(3) Berdasarkan sifat logaritma $a^{\log_a b} = b$, maka $10^{\log_{10} 5} = 5$(4) Dari $x^2 - 5x + 3 = 0$, $\alpha+\beta=5, \alpha\beta=3$. $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 25 - 6 = 19$. $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta = 25 - 12 = 13$(5) $|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos\theta \Rightarrow 7 = 1 + 4 - 4\cos\theta \Rightarrow 4\cos\theta = -2 \Rightarrow \cos\theta = -1/2$. Maka $\theta = 120^\circ$(6) $\angle A + \angle B + \angle C = 180^\circ \Rightarrow \angle B + \angle C = 150^\circ$. $\sin(150^\circ) = \sin(180^\circ-30^\circ) = \sin 30^\circ = 1/2$(7) Kelipatan 3 dari 100-200: $102, 105, \dots, 198$. $198 = 102 + (n-1)3 \Rightarrow 96 = 3(n-1) \Rightarrow n=33$. Jumlah = $\frac{33}{2}(102+198) = \frac{33}{2}(300) = 4950$(8) Misal $P(x) = x^3+ax^2+bx+5$. $P(1)=0 \Rightarrow 1+a+b+5=0 \Rightarrow a+b=-6$. $P(2)=5 \Rightarrow 8+4a+2b+5=5 \Rightarrow 4a+2b=-8 \Rightarrow 2a+b=-4$. Eliminasi: $a=2, b=-8$(9) $f(0) = |0^2-1| = 1$. $\int_0^2 |x^2-1| dx = \int_0^1 (1-x^2) dx + \int_1^2 (x^2-1) dx = [x - \frac{x^3}{3}]_0^1 + [\frac{x^3}{3} - x]_1^2 = (1-1/3) + ((8/3-2) - (1/3-1)) = 2/3 + 2/3 + 2/3 = 2$(10) $a=b-d, c=b+d$. $a+b+c=3b=24 \Rightarrow b=8$. $(8-d)(8)(8+d)=440 \Rightarrow 64-d^2=55 \Rightarrow d^2=9 \Rightarrow d=3$. Maka $a=5, b=8, c=11$

2.

(1) Garis melalui $A(0,3)$ dan $C(4,0)$: gradien $m = \frac{0-3}{4-0} = -\frac{3}{4}$. Persamaan: $y - 0 = -\frac{3}{4}(x - 4) \Rightarrow 3x + 4y - 12 = 0$$3x + 4y - 12 = 0$. Jawaban: 3, 4, 12.(2) Titik tengah $AB$ adalah $(0,0)$, garis tegak lurus $AB$ adalah $y=0$. Titik tengah $AC$ adalah $(2, 1.5)$, gradien $AC$ adalah $-3/4$, gradien tegak lurus adalah $4/3$. Persamaan: $y - 1.5 = \frac{4}{3}(x - 2) \Rightarrow y = \frac{4}{3}x - \frac{7}{6}$. Titik potong $y=0$ dan $y = \frac{4}{3}x - \frac{7}{6}$ adalah $x = 7/8$

Jawaban: 7/8, 0.(3) $AB=6, BC=\sqrt{4^2+(-3)^2}=5, AC=\sqrt{4^2+3^2}=5$. $OD:DC = AB:AC = 6:5$. Incenter $I = \frac{aA+bB+cC}{a+b+c} = \frac{5(0,3)+5(0,-3)+6(4,0)}{16} = (\frac{24}{16}, \frac{0}{16}) = (1.5, 0)$

Jawaban: 6, 5, 3, 0.3.

(1) $x^2-5x+7 = x+k \Rightarrow x^2-6x+7-k=0$. Diskriminan $D=36-4(7-k)=0 \Rightarrow 8+4k=0 \Rightarrow k=-2$

Jawaban: -2.(2) Untuk $k=-2$: $x^2-6x+9=0 \Rightarrow (x-3)^2=0 \Rightarrow x_P=3$$x^2+3x-1 = x-2 \Rightarrow x^2+2x+1=0 \Rightarrow (x+1)^2=0 \Rightarrow x_Q=-1$$x^2-5x+7 = x^2+3x-1 \Rightarrow 8x=8 \Rightarrow x_R=1$

Jawaban: 3, -1, 1.(3) Luas = $\int_{-1}^{1} ((x^2+3x-1) - (x-2)) dx + \int_{1}^{3} ((x^2-5x+7) - (x-2)) dx$

$= \int_{-1}^{1} (x^2+2x+1) dx + \int_{1}^{3} (x^2-6x+9) dx = [\frac{x^3}{3}+x^2+x]_{-1}^{1} + [\frac{x^3}{3}-3x^2+9x]_{1}^{3}$

$= (\frac{1}{3}+1+1) - (-\frac{1}{3}+1-1) + (9-27+27) - (\frac{1}{3}-3+9) = \frac{7}{3} + \frac{2}{3} = 3$

Jawaban: 3.