UN 2010 Matematika SMA/MA IPA - B

Thread starter suhu
Start date Jan 30, 2011

suhu

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Jan 30, 2011

#1

suhu

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Jan 30, 2011

#2

suhu

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Jan 31, 2011

#3

Nomor 2

Nomor 2
$\\f(x)=3x-1\:;\:g(x)=2x^2+3\\(g_{\circ} f)(1)=....\\Jawab:\\ \begin{array}{rllccccc} (g_{\circ} f)(x)&=g(f(x))\\ &=2(f(x))^2+3\\ &=2(3x-1)^2+3\\ &=2(9x^2-6x+1)+3\\ &=18x^2-12x+2+3\\ &=18x^2-12x+5\\ (g_{\circ} f)(1)&=18(1)^2-12(1)+5\\ &=18-12+5\\ &=11 \end{array}$
$\therefore$ Jawab C

Last edited: Feb 4, 2011

suhu

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Feb 4, 2011

#4

Nomor 3

Nomor 3

$\\\begin{matrix}f(x)=\frac{9x+4}{6x-5}\:;\:x\neq \frac{5}{6}\:\:\:\:\:\:\:\:\:\:\\f^{-1}(-2)=....\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\\\mbox{Solusi:}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\\ \begin{array}{crlcccccc} &y&=\frac{9x+4}{6x-5}\\\Leftrightarrow& 6xy-5y&=9x+4\\\Leftrightarrow& 6xy-9x&=5y+4\\ \Leftrightarrow& x(6y-9)&=5y+4\\\Leftrightarrow& x&=\frac{5y+4}{6y-9}\\\Rightarrow& f^{-1}(x)&=\frac{5x+4}{6x-9}\\ \Leftrightarrow& f^{-1}(-2)&=\frac{5(-2)+4}{6(-2)-9}\\\Leftrightarrow& f^{-1}(-2)&=\frac{-6}{-21}\\\Leftrightarrow& f^{-1}(-2)&=\frac{6}{21}=\frac{2}{7}\\ \end{array}\end{matrix}$
$\therefore$ Jawab C

suhu

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Feb 4, 2011

#5

Nomor 1

Nomor 1

$\\PK\equiv 2x^2+mx+16=0\\ akar^2nya\:\alpha \:\beta\:positif ; \:\alpha=2\beta\\Dit:m\\Jawab:\\ \begin{array}{crllccccc} &\alpha\times\beta&=\frac{c}{a}&=\frac{16}{2}\\ \Leftrightarrow &2\beta\times\beta&=2\beta^2&=8\\ \Leftrightarrow &\beta=2&\Leftrightarrow \alpha&=4\\ \Rightarrow &\alpha+\beta&=-\frac{m}{2}\\ \Leftrightarrow &4+2&=-\frac{m}{2}\\ \Leftrightarrow &m&=-12 \end{array}$

$\therefore$ Jawab A

suhu

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Feb 6, 2011

#6

Nomor 4

Nomor 4

$\\\begin{array}{c} \end{array}\\ \begin{array}{rcll} Adi&\equiv &4b+3p&=19500\\ Budi&\equiv &2b+p&=\:\:\,8500\\ Candra&\equiv &b+p-10000&=....\:\:\:\:\:\:\\ \end{array}\\\mbox{Penyelesaian:}\\ \begin{array}{rcllcccccc} Adi&\equiv &4b+3p&=19500\\ Budi\times3&\equiv &6b+3p&=25500_-\\\hline &\Rightarrow &2b&=\:\:\,6000\\ &\Leftrightarrow &b&=\:\:\,3000\\ Budi&\equiv &2(3000)+p&=\:\:\,8500\\ &\Leftrightarrow &\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\,p&=\:\:\,2500\\ Candra&\equiv &b+p-10000&=-4500 \end{array}\\\\ \therefore \mbox{kembalian Candra Rp 4500\\\\ ; Jawaban C}$

suhu

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Feb 8, 2011

#7

Nomor 5

Nomor 5

$\\\begin{array}{ccl} \left.\begin{array}{r}f(x)=x^2+bx+4\\\mbox{menyinggung}\\y=3x+4\end{array}\right \}&\Leftrightarrow & x^2+bx+4=3x+4\\ &\Leftrightarrow & x^2+bx-3x+4-4=0\\ &\Leftrightarrow & x^2+(b-3)x=0\\\\ &\Rightarrow &D=0\\ &\Leftrightarrow &(b-3)^2-4(1)(0)=0\\ &\Leftrightarrow &(b-3)^2=0\\ &\Leftrightarrow &b-3=0\\ &\Leftrightarrow &b=3 \end{array}\\ \therefore \mbox{Jawab D}$

suhu

New member

Feb 9, 2011

#8

Nomor 6

Nomor 6

$\begin{array}{rl} 6).\:Pers\bigodot \equiv &x^2+y^2+Ax+By+C=0\Rightarrow x^2+y^2+6x-4y-7=0\\ Pusat_{\bigodot}\equiv &(-\frac{1}{2}A,-\frac{1}{2}B)=(-\frac{1}{2}(6),-\frac{1}{2}(-4))=(-3,2)\\ Jari^2\,_{\bigodot}\equiv &\sqrt{\frac{1}{4}A^2+\frac{1}{4}B^2-C}=\sqrt{\frac{1}{4}(6)^2+\frac{1}{4}(-4)^2-(-7)}=\sqrt{20} \end{array}\\$
$\begin{array}{crcccc} {\color{green} \rightarrow }&\mbox{grs singgung}&\perp &\mbox{grs }y=7-2x&\Leftrightarrow& m_{g.singgung}=-\frac{1}{(-2)}=\frac{1}{2}\\ &\mbox{grs singgung}&\equiv &y=mx+n&\Leftrightarrow&y=\frac{1}{2}x+n\\ &\mbox{grs singgung}&\equiv &ax+by+c&\Leftrightarrow&\frac{1}{2}x-y-n=0\\ &Jari^2\,_{\bigodot}&\equiv&d&=&\left | \frac{ax_P+by_P+c}{\sqrt{a^2b^2}} \right |\\ &&\Leftrightarrow&\sqrt{20}&=&\left | \frac{(\frac{1}{2})(-3)+(-1)(2)+(n)}{\sqrt{(\frac{1}{2})^2+(1)^2}} \right |\\ &&\Leftrightarrow&\sqrt{20}&=&\left | \frac{-1\frac{1}{2}-2+n}{\sqrt{(\frac{1}{4})+1}} \right |\\ &&\Leftrightarrow&\sqrt{20}&=&\left | \frac{-3\frac{1}{2}+n}{\sqrt{\frac{5}{4}}} \right |\\ &&\Leftrightarrow&(\sqrt{20})(\sqrt{\frac{5}{4}})&=&\left | -3\frac{1}{2}+n \right |\\ &&\Leftrightarrow&n&=&5\mp 3\frac{1}{2}\\ &&\Leftrightarrow&n=1\frac{1}{2}&\vee &n=8\frac{1}{2} \end{array}$
$\begin{array}{lcccccc} n=1\frac{1}{2}&\Rightarrow&\mbox{grs singgung}&\equiv &\frac{1}{2}x-y-1\frac{1}{2}=0&\equiv &x-2y-3=0\\ \vee\\ n=8\frac{1}{2}&\Rightarrow&\mbox{grs singgung}&\equiv &\frac{1}{2}x-y-8\frac{1}{2}=0&\equiv &x-2y-17=0 \end{array}\\\\ \therefore \mbox{ Pilihan C}$

suhu

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Feb 10, 2011

#9

Nomor 7

Nomor 7

$\\Akar\:x^2-5x-1=0\rightarrow p\:\vee \:q\\ PK\:baru\:2p+1\vee2q+1\\ PK\:baru?\\ p+q=-\frac{(-5)}{1}=5\\ p\times q=\frac{-1}{1}=-1\\ \begin{array}{lcrcccccccccc} PK\:baru\Rightarrow &\bigoplus & -\frac{b}{a}=\\ &&2p+1+2q+1=\\ &&2(p+q)+2=\\ &&2(5)+2=\\&&12\\ &\bigotimes&\frac{c}{a}=\\ &&2p+1\times\,2q+1=\\ &&4pq+2p+2q+1=\\ &&4pq+2(p+q)+1=\\ &&4(-1)+2(5)+1=\\&&7\\ PK\:baru\:a=1&\Rightarrow &x^2-12x+7=0 \end{array}\\ \therefore\,\mbox{Jawab D}$

suhu

New member

Feb 11, 2011

#10

Nomor 8

Nomor 8

$\\\left ( \begin{matrix} p &1 \\ 2 &-1 \end{matrix} \right )\times \left ( \begin{matrix} 2 &3 \\ 3 &q \end{matrix} \right )= \left ( \begin{matrix} 11 &10 \\ 1 &r \end{matrix} \right )\\\\ \mbox{Nilai }2p+3q-r=....\\ \mbox{Jawab:}\\ \begin{array}{crll} (1)&2p+3&=11&\\(2)&3p+q&=10&\\(3)&6-q&=r&\\ (1)&2p&=11-3&\rightarrow p=4\\ (2)&q&=10-2(4)&\rightarrow q=2\\ (3)&r&=6-2&\rightarrow r=4\\ \end{array}\\ \begin{array}{rr} \mbox{Nilai }&2p+3q-r=\\ &2(4)+3(2)-4=\\&10 \end{array}$

$\therefore$ Jawaban E

agen_pale

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Apr 10, 2011

#11

mantap nih....thread semoga berguna buat adik-2 untuk menghadapi UN nanti....

Adamsuhada

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Aug 15, 2012

#12

kalo soal ttg integral aljabar ada ga?

tiaseptiani

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Dec 21, 2014

#13

nyimak ah

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