Pembahasan Soal UMPTN Matematika Dasar tahun 2001 nomor 6 sampai 10

Nomor 6

Supaya sistem persamaan linear

$\left\{ \begin{array}{c} 2x + 3y = 6 \\ (1+a)x-6y = 7 \end{array} \right.$
merupakan persamaan dua garis yang saling tegak lurus, maka

$\, a = ....$

$\spadesuit \,$ Menentukan gradien

$2x+3y=6 \rightarrow m_1 = \frac{-a}{b} = \frac{-2}{3}$

$(1+a)x-6y = 7 \rightarrow m_2 = \frac{-a}{b} = \frac{-(1+a)}{-6} = \frac{1+a}{6}$

$\spadesuit \,$ Kedua garis tegak lurus

$\begin{align} m_1.m_2 & = - 1 \\ \frac{-2}{3} . \frac{1+a}{6} & = -1 \\ 1+a & = 9 \\ a & = 8 \end{align}$
Jadi, nilai

$a = 8 . \heartsuit$

Nomor 7

Pada

$\Delta$ ABC diketahui

$a+b = 10, \,$ sudut A =

$30^\circ \,$ dan sudut B =

$45^\circ \,$ , maka panjang sisi

$\, b = ....$

$\clubsuit \,$ Gambar

$a+b = 10 \, \,$ ....pers(i)

$\clubsuit \,$ Aturan sinus pada segitiga ABC

$\begin{align} \frac{a}{\sin A} & = \frac{b}{\sin B} \\ \frac{a}{\sin 30 ^\circ} & = \frac{b}{\sin 45 ^\circ} \\ \frac{a}{\frac{1}{2}} & = \frac{b}{\frac{1}{2} \sqrt{2}} \\ a & = \frac{1}{2} b \sqrt{2} \, \, \, \text{...pers(ii)} \end{align}$

$\clubsuit \,$ Substitusi pers(ii) ke pers(i)

$\begin{align} a+b & = 10 \\ \frac{1}{2} b \sqrt{2} + b & = 10 \, \, \text{(kali 2)} \\ b \sqrt{2} + 2b & = 20 \\ b (2 + \sqrt{2}) & = 20 \\ b & = \frac{20}{ 2 + \sqrt{2} } \\ b & = \frac{20}{ 2 + \sqrt{2} } . \frac{2 - \sqrt{2}}{ 2 - \sqrt{2} } = 10 (2 - \sqrt{2}) \end{align}$
Jadi, nilai

$b = 10 (2 - \sqrt{2}) . \heartsuit$

Nomor 8

Jika

$\tan ^2 x + 1 = a^2 \, ,$ maka

$\, \sin ^2 x = ....$

$\spadesuit \,$ Menentukan nilai tan

$x$

$\begin{align} \tan ^2 x + 1 & = a^2 \\ \tan ^2 x & = a^2 - 1 \\ \tan x & = \frac{\sqrt{a^2 - 1}}{1} \end{align}$

$\spadesuit \,$ Menentukan nilai

$\sin ^2 x$

$\begin{align} \sin x & = \frac{de}{mi} \\ \sin x & = \frac{\sqrt{a^2 - 1}}{a} \\ \sin ^2 x & = \frac{a^2 - 1}{a^2} \end{align}$
Jadi, nilai

$\sin ^2 x = \frac{a^2 - 1}{a^2} . \heartsuit$

Nomor 9

Penyelesaian

$\, \frac{x^2-3x-18}{(x-6)^2(x-2)} < 0 \,$ adalah ....

$\clubsuit \,$ Menentukan akar-akar (pembuat nol)

$\begin{align} \frac{x^2-3x-18}{(x-6)^2(x-2)} & < 0 \\ \frac{(x+3)(x-6)}{(x-6)^2.(x-2)} & < 0 \\ x=-3, \, x & = 6 , \, x = 2 \end{align}$

Jadi, penyelesaiannya adalah

$HP = \{ x < -3 \vee 2 < x < 6 \} . \heartsuit$

Nomor 10

Pertidaksamaan

$\, \left| \frac{2x-1}{x+5} \right| \leq 3 \,$ mempunyai penyelesaian ....

$\spadesuit \,$ Konsep Dasar :

$|x|^2 = x^2 \,$ dan

$\, p^2-q^2 = (p-q)(p+q)$

$\spadesuit \,$ Kuadratkan kedua ruas

$\begin{align} \left| \frac{2x-1}{x+5} \right| & \leq 3 \\ \left| \frac{2x-1}{x+5} \right|^2 & \leq 3^2 \\ \left( \frac{2x-1}{x+5} \right)^2 & \leq 3^2 \\ \left( \frac{2x-1}{x+5} \right)^2 - 3^2 \leq 0 \\ \left( \frac{2x-1}{x+5} - 3 \right) \left( \frac{2x-1}{x+5} + 3 \right) \leq 0 \\ \left( \frac{2x-1-3(x+5)}{x+5} \right) . \left( \frac{2x-1+3(x+5)}{x+5} \right) \leq 0 \\ \left( \frac{-x-16}{x+5} \right) . \left( \frac{5x+14}{x+5} \right) \leq 0 \\ x=-16, \, x= -\frac{14}{5} , \, x & = -5 \end{align}$