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

$x^2-x > 0 \rightarrow x(x-1) > 0 \rightarrow x=0 \vee x = 1$

$HP_1 = \{ x < 0 \vee x > 1 \}$

$\clubsuit \,$ Menyelesaikan logaritmanya

$\begin{align*} {}^{\frac{1}{6}} \log (x^2-x) & > -1 \\ {}^{\frac{1}{6}} \log (x^2-x) & > {}^{\frac{1}{6}} \log \left( \frac{1}{6} \right)^{-1} \, \, \text{(coret } \, {}^{\frac{1}{6}} \log ) \\ x^2-x & < 6 \, \, \text{(ketaksamaan dibalik)} \\ x^2-x-6 & < 6 \\ (x+2)(x-3) & < 6 \\ x=-2 & \vee x=3 \end{align*}$