Jumat, 20 Februari 2015

Pembahasan Soal SPMB Matematika Dasar tahun 2005


Nomor 1
$\displaystyle \lim_{x \to 3} \frac{9-x^2}{2\sqrt{x^2+3}-4\sqrt{3}} = .... $
$\clubsuit \, $ Merasionalkan penyebut
$\begin{align} \displaystyle \lim_{x \to 3} \frac{9-x^2}{2\sqrt{x^2+3}-4\sqrt{3}} & = \displaystyle \lim_{x \to 3} \frac{9-x^2}{2\sqrt{x^2+3}-4\sqrt{3}} . \frac{2\sqrt{x^2+3}+4\sqrt{3}}{2\sqrt{x^2+3}+4\sqrt{3}} \\ & = \displaystyle \lim_{x \to 3} \frac{(9-x^2).(2\sqrt{x^2+3}+4\sqrt{3})}{4(x^2+3) - 48} \\ & = \displaystyle \lim_{x \to 3} \frac{-(x^2-9).(2\sqrt{x^2+3}+4\sqrt{3})}{4(x^2+3-12)} \\ & = \displaystyle \lim_{x \to 3} \frac{-(x^2-9).(2\sqrt{x^2+3}+4\sqrt{3})}{4(x^2-9)} \\ & = \displaystyle \lim_{x \to 3} \frac{-(2\sqrt{x^2+3}+4\sqrt{3})}{4} \\ & = \frac{-(2\sqrt{3^2+3}+4\sqrt{3})}{4} \\ & = \frac{-8\sqrt{3}}{4} = -2\sqrt{3} \end{align}$
Jadi, nilai limitnya adalah $ -2\sqrt{3}. \heartsuit $

Cara II
$\clubsuit \, $ Menggunakan turunan
$\begin{align} \displaystyle \lim_{x \to 3} \frac{9-x^2}{2\sqrt{x^2+3}-4\sqrt{3}} & = \displaystyle \lim_{x \to 3} \frac{-2x}{2.\frac{2x}{2\sqrt{x^2+3}}} \\ & = \displaystyle \lim_{x \to 3} - \sqrt{x^2+3} = - \sqrt{3^2+3} \\ & = - \sqrt{12} = -2\sqrt{3} \end{align}$
Jadi, nilai limitnya adalah $ -2\sqrt{3}. \heartsuit $
Nomor 2
Jumlah dua bilangan $p \, $ dan $q \, $ adalah 6. Nilai minimum dari $p^2+q^2 = ..... $
$p+q = 6 \rightarrow p = 6-q \, \, $
$\spadesuit \, $ Nilai maks/min : $f^\prime (q) = 0 \, \, $ (turunan = 0)
$\begin{align} p^2 + q^2 & = (6-q)^2 + q^2 \\ & = 36 -12q + q^2 + q^2 \\ f(q) & = 2q^2 - 12q + 36 \rightarrow f^\prime (q) = 4q -12 \\ f^\prime (q) & = 0 \\ 4q -12 & = 0 \rightarrow q = 3 \\ p & = 6-q \rightarrow p = 6-3 = 3 \end{align}$
Sehingga : $ p^2+q^2 = 3^2 + 3^2 = 9 + 9 = 18 $
Jadi, nilai minimum $p^2+q^2 \, $ adalah 18 . $ \heartsuit $
Nomor 3
Garis singgung pada kurva $y=\frac{2x+1}{2-3x} \, \, $ di titik (1, -3) adalah ....
$\clubsuit \, $ Menentukan turunan
$\begin{align*} y & = \frac{U}{V} \rightarrow y^\prime = \frac{U^\prime . V - U . V^\prime }{V^2} \\ y & =\frac{2x+1}{2-3x} \\ y^\prime & = \frac{2(2-3x) - (2x+1).(-3)}{(2-3x)^2} \end{align*}$
$\clubsuit \, $ Menentukan gradien : $m = f^\prime (1) $
$\begin{align*} m & = f^\prime (1) = \frac{2(2-3.1) - (2.1+1).(-3)}{(2-3.1)^2} \\ & = \frac{-2+9}{1} = 7 \end{align*}$
$\clubsuit \, $ Persamaan garis singgung di (1, -3)
$\begin{align*} y-y_1 & = m(x-x_1) \\ y-(-3) & = 7 (x-1) \\ y-7x+10 & = 0 \end{align*}$
Jadi, PGS nya adalah $ y-7x+10 = 0 . \heartsuit $
Nomor 4
Jika fungsi $f(x)=\sin ax + \cos bx \, $ memenuhi $f^\prime (0) = b \, \, $ dan $ f^\prime \left( \frac{\pi}{2a} \right) = -1 \, \, $ , maka $a+b = ....$
$\spadesuit \, $ Turunan fungsinya
$ f(x)=\sin ax + \cos bx \rightarrow f^\prime (x) = a\cos ax - b\sin bx $
$\spadesuit \, $ Menentukan hubungan $a \, $ dan $b $
$\begin{align} f^\prime (0) & = b \\ a\cos a.0 - b\sin b.0 & = b \\ a-0 & = b \rightarrow a = b \end{align}$
$\spadesuit \, $ Substitusi $ a = b \, $ dan gunakan $ f^\prime \left( \frac{\pi}{2a} \right) = -1 $
$\begin{align} f^\prime \left( \frac{\pi}{2a} \right) & = -1 \\ a\cos a.\left( \frac{\pi}{2a} \right) - b\sin b.\left( \frac{\pi}{2a} \right) & = -1 \, \, \text{(substitusi } a=b ) \\ b\cos b.\left( \frac{\pi}{2b} \right) - b\sin b.\left( \frac{\pi}{2b} \right) & = -1 \, \, \\ b\cos \left( \frac{\pi}{2} \right) - b\sin \left( \frac{\pi}{2} \right) & = -1 \, \, \\ b.0 - b. 1 & = -1 \\ -b & = -1 \rightarrow b=1 \end{align}$
Sehingga : $ a = b = 1 $
Jadi, nilai $ a+b = 1+ 1 = 2 . \heartsuit $
Nomor 5
Nilai $x \, \, $ yang memenuhi pertidaksamaan $ {}^{\frac{1}{6}} \log (x^2-x) > -1 \, \, $ adalah ....
$\clubsuit \, $ Syarat logaritma
$x^2-x > 0 \rightarrow x(x-1) > 0 \rightarrow x=0 \vee x = 1 $
spmb_matdas_1_2005.png
$HP_1 = \{ x < 0 \vee x > 1 \} $
$\clubsuit \, $ Menyelesaikan logaritmanya
$\begin{align*} {}^{\frac{1}{6}} \log (x^2-x) & > -1 \\ {}^{\frac{1}{6}} \log (x^2-x) & > {}^{\frac{1}{6}} \log \left( \frac{1}{6} \right)^{-1} \, \, \text{(coret } \, {}^{\frac{1}{6}} \log ) \\ x^2-x & < 6 \, \, \text{(ketaksamaan dibalik)} \\ x^2-x-6 & < 6 \\ (x+2)(x-3) & < 6 \\ x=-2 & \vee x=3 \end{align*}$
spmb_matdas_1a_2005.png
$HP_2 = \{ -2 < x < 3 \} $
Sehingga : $ HP = HP_1 \cap HP_2 = \{ -2 < x < 0 \vee 1 < x < 3 \} $
Jadi, solusinya adalah $ \{ -2 < x < 0 \vee 1 < x < 3 \} . \heartsuit$
Nomor Soal Lainnya : 1-5 6-10 11-15 16-20 21-25

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