Pembahasan Soal SPMB Matematika IPA tahun 2006 nomor 6 sampai 10

Nomor 6

Jika lingkaran

$x^2 + y^2 + ax + by + c = 0$ berpusat di (1, -1) menyinggung garis

$y = x$ , maka nilai

$a+b+c$ adalah ....

$\spadesuit \,$ Menentukan nilai

$a$ dan

$b$ dengan pusat (1,-1)
pusat : (1,-1) =

$(\frac{-a}{2}, \frac{-b}{2})$

$\frac{-a}{2} = 1 \rightarrow a = -2 \,$ dan

$\, \frac{-b}{2} = -1 \rightarrow b = 2$
sehingga persamaannya :

$x^2 + y^2 -2x + 2y + c = 0$

$\spadesuit \,$ Substitusi

$y = x$ ke persamaan lingkaran

$\begin{align} y=x \rightarrow x^2 + y^2 -2x + 2y + c & = 0 \\ x^2 + x^2 -2x + 2x + c & = 0 \\ 2x^2 + c & = 0 \end{align}$

$\spadesuit \,$ Syarat lingkaran menyinggung garis :

$D = 0$ dari

$2x^2 + c = 0$

$\begin{align} D = 0 \rightarrow b^2-4ac & = 0 \\ 0^2-4.2.c & = 0 \\ c & = 0 \end{align}$
Sehingga nilai

$a + b + c = -2 + 2 + 0 = 0$
Jadi, nilai

$a + b + c = 0 . \heartsuit$

Nomor 7

$15 \int \limits_2^3 x\sqrt{x-2} dx = ....$

$\clubsuit \,$ Konsep dasar :

$\int (ax+b)^n dx = \frac{1}{a(n+1)} (ax+b)^{n+1} + c$

$\clubsuit \,$ Teknik integral tanzalin :

$\int \limits_2^3 x\sqrt{x-2} dx =$

Hasilnya

$\begin{align} & 15 \int \limits_2^3 x\sqrt{x-2} dx \\ & = 15\left[ \frac{2}{3}x(x-2)^\frac{3}{2} - \frac{4}{15}(x-2)^\frac{5}{2} \right]_2^3 \\ & = \left[ 15.\frac{2}{3}x(x-2)^\frac{3}{2} - 15.\frac{4}{15}(x-2)^\frac{5}{2} \right]_2^3 \\ & = \left[ 10x(x-2)^\frac{3}{2} - 4(x-2)^\frac{5}{2} \right]_2^3 \\ & = \left[ 10.3.(3-2)^\frac{3}{2} - 4.(3-2)^\frac{5}{2} \right] - \\ & \left[ 10.2.(2-2)^\frac{3}{2} - 4.(2-2)^\frac{5}{2} \right] \\ & = (10.3.1 - 4.1 )-(0 - 0) \\ & = 30 - 4 = 26 \end{align}$
Jadi, hasil integralnya adalah 26.

$\heartsuit$

Nomor 8

Jika

${}^{81} \log \frac{1}{x} = {}^{x} \log \frac{1}{y} = {}^{y} \log \frac{1}{81} , \,$ maka

$2x - 3y = ....$

$\spadesuit \,$ Konsep logaritma

${}^a \log b = c \rightarrow b = a^c$

${}^a \log b^n = n. {}^a \log b$

${}^a \log b = \frac{1}{{}^b \log a}$

${}^a \log b . {}^b \log c = {}^a \log c$

$\spadesuit \,$ Menyederhanakan soal

$\begin{align} {}^{81} \log \frac{1}{x} & = {}^{x} \log \frac{1}{y} = {}^{y} \log \frac{1}{81} \\ {}^{81} \log x^{-1} & = {}^{x} \log y^{-1} = {}^{y} \log 81^{-1} \\ -1.{}^{81} \log x & = -1. {}^{x} \log y = -1.{}^{y} \log 81 \\ {}^{81} \log x & = {}^{x} \log y = {}^{y} \log 81 \end{align}$

$\spadesuit \,$ Karena ketiga ruas sama, diperoleh

${}^{81} \log x = {}^{y} \log 81 , \,$ ....pers(i)

${}^{x} \log y = {}^{y} \log 81 , \,$ ....pers(ii)

$\spadesuit \,$ Kalikan pers(i) dan pers(ii)

$\begin{align} {}^{81} \log x . {}^{x} \log y & = {}^{y} \log 81 . {}^{y} \log 81 \\ {}^{81} \log y & = ({}^{y} \log 81)^2 \\ \frac{1}{{}^{y} \log 81} & = ({}^{y} \log 81)^2 \\ 1 & = ({}^{y} \log 81)^3 \\ {}^{y} \log 81 & = 1 \rightarrow 81 = y^1 \rightarrow y = 81 \end{align}$

$\begin{align} \text{pers(i) : } \, {}^{81} \log x & = {}^{y} \log 81 \\ {}^{81} \log x & = {}^{81} \log 81 \\ {}^{81} \log x & = 1 \rightarrow x = 81^1 = 81 \end{align}$
Sehingga nilai

$2x-3y = 2.81-3.81 = -81$
Jadi, nilai

$2x-3y = - 81. \heartsuit$

Nomor 9

Diketahui

$x$ dan

$y$ sudut lancip dan

$x - y = \frac{\pi}{6}$ . Jika

$\tan x = 3\tan y$ , maka

$x+y = ...$

$\clubsuit \,$ Konsep dasar :

$\tan (x-y) = \frac{\tan x - \tan y}{1 + \tan x . \tan y}$

$\clubsuit \,$ Menentukan nilai

$y$ dengan

$\tan x = 3\tan y \, \,$ ...pers(i)

$\begin{align} x - y & = \frac{\pi}{6} \, \, \text{...pers(ii)} \\ \tan (x - y) & = \tan ( \frac{\pi}{6} ) \\ \frac{\tan x - \tan y}{1 + \tan x . \tan y} & = \frac{1}{\sqrt{3}} \, \, \text{[substitusi pers(i)]} \\ \frac{3\tan y - \tan y}{1 + 3\tan y . \tan y} & = \frac{1}{\sqrt{3}} \\ \frac{2\tan y}{1 + 3(\tan y)^2} & = \frac{1}{\sqrt{3}} \, \, \text{(misal : } p = \tan y ) \\ \frac{2p}{1 + 3p^2} & = \frac{1}{\sqrt{3}} \\ 2\sqrt{3}p & = 1 + 3p^2 \\ 3p^2 - 2\sqrt{3}p + 1 & = 0 \\ (\sqrt{3}p - 1 )^2 & = 0 \\ p & = \frac{1}{\sqrt{3}} \rightarrow \tan y = \frac{1}{\sqrt{3}} \rightarrow y = 30^\circ \end{align}$
pers(ii):

$x - y = \frac{\pi}{6} \rightarrow x - 30^\circ = 30^\circ \rightarrow x = 60^\circ$
Sehingga nilai

$x + y = 60^\circ + 30^\circ = 90^\circ = \frac{\pi }{2}$
Jadi, nilai

$x + y = \frac{\pi }{2} . \heartsuit$

Nomor 10

Diberikan vektor-vektor

$\vec{a} = x\vec{i} - 3x\vec{j}+6y\vec{k} \,$ dan

$\, \vec{b} = (1-y)\vec{i} +3\vec{j}-(1+x)\vec{k} \,$ dengan

$x > 0$ . Jika

$\vec{a} \,$ dan

$\vec{b}$ sejajar, maka

$\vec{a}+3\vec{b} = ....$

$\spadesuit \,$ Diketahui vektor :

$\vec{a} = (x, -3x, 6y) \,$ dan

$\, \vec{b} = (1-y,3,-(1+x))$

$\spadesuit \,$ Vektor

$\vec{a} \,$ sejajar

$\, \vec{b} \,$ , maka

$\, \vec{b} = n\vec{a}$

$\begin{align} \vec{b} & = n\vec{a} \\ \left( \begin{matrix} 1-y \\ 3 \\ -(1+x) \end{matrix} \right) & = n \left( \begin{matrix} x \\ -3x \\ 6y \end{matrix} \right) \\ \left( \begin{matrix} 1-y \\ 3 \\ -(1+x) \end{matrix} \right) & = \left( \begin{matrix} nx \\ -3nx \\ 6ny \end{matrix} \right) \end{align}$
persamaan yang diperoleh :
*

$3 = -3nx \rightarrow nx = -1 \rightarrow n = \frac{-1}{x}$
*

$1-y = nx \rightarrow 1-y = -1 \rightarrow y = 2$
*

$-(1+x) = 6ny \rightarrow -1-x = 6.\frac{-1}{x}.2$

$\rightarrow x^2+x - 12 = 0 \rightarrow (x+4)(x-3) = 0$

$\rightarrow x = -4 \vee x = 3$
Nilai yang memenuhi :

$x = 3 \,$ (positif) dan

$\, y = 2$
Vektor

$\vec{a}$ dan

$\vec{b}$ menjadi :

$\vec{a} = (3,-9,12) \,$ dan

$\, \vec{b} = (-1,3,-4)$
sehingga :

$\vec{a}+3\vec{b} = \left( \begin{matrix} 3 \\ -9 \\ 12 \end{matrix} \right) + \left( \begin{matrix} -3 \\ 9 \\ -12 \end{matrix} \right) = \vec{0}$
Jadi, nilai

$\vec{a}+3\vec{b} = \vec{0}. \heartsuit$

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