Soal dan Pembahasan Seleksi Mandiri Madrasah (KSN-S) 2022

 $\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: a,b\: \:  \textrm{bilangan real positif},\\ &\textrm{tunjukkan bahwa}\: \: a^{a}.b^{b}\geq a^{b}.b^{a}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Diketahui}\: \: a^{a}.b^{b}\geq a^{b}.b^{a}\Leftrightarrow \displaystyle \frac{a^{a}.b^{b}}{a^{b}.b^{a}}\geq 1\\ &\Leftrightarrow \displaystyle \frac{a^{a-b}}{b^{a-b}}\geq 1\Leftrightarrow \left ( \displaystyle \frac{a}{b} \right )^{a-b}\geq 1\\ &\textrm{Selanjutnya akan ada dua kemungkinan}\\ &\textrm{yaitu}:\: a\geq b> 0\: \: \textrm{dan}\: \: b\geq a> 0 \end{aligned}\\ &\begin{array}{|c|l|l|}\hline \textrm{No}&\textrm{Kondisi}&\textrm{Akibat}\\\hline 1.&a\geq b> 0&\displaystyle \frac{a}{b}\geq 1\: \: \textrm{atau}\: \: a-b\geq 0\\ &&\textrm{maka}\: \:  \begin{aligned}&\left (\displaystyle \frac{a}{b}  \right )^{a-b}\geq 1 \end{aligned}\\\hline 2.&b\geq a> 0&\displaystyle \frac{a}{b}\leq  1\: \: \textrm{atau}\: \: a-b\leq  0\\ &&\textrm{maka}\: \:  \begin{aligned}&\left (\displaystyle \frac{a}{b}  \right )^{a-b}\geq 1 \end{aligned}\\\hline \end{array}\\ &\textrm{Karena keduanya memiliki hasil yang sama}\\ &\textrm{maka}\: \: a^{a}.b^{b}\geq a^{b}.b^{a}\qquad \blacksquare  \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: a=\sqrt{\displaystyle \frac{b}{1-b}}\: \:  \textrm{nyatakanlah}\: \: b\: \: \textrm{dalam}\: \: a\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &a=\sqrt{\displaystyle \frac{b}{1-b}}\Leftrightarrow a^{2}=\displaystyle \frac{b}{1-b}\Leftrightarrow a^{2}(1-b)=b\\ &\Leftrightarrow a^{2}-a^{2}b=b\Leftrightarrow b+a^{2}b=a^{2}\\ &\Leftrightarrow b(1-a^{2})=a^{2}\Leftrightarrow b=\color{blue}\displaystyle \frac{a^{2}}{1-a^{2}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Sederhanakanlah bentuk}\: \: 3.4^{4}+3.4^{4}+3.4^{4}\\\\ &\textbf{Jawab}:\\ &\begin{aligned} &3.4^{4}+3.4^{4}+3.4^{4}=3(3.4^{4})\\ &=9\times 4^{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui}\: \: a^{2}+b^{2}=1\: \: \textrm{dan}\: \: x^{2}+y^{2}=1\\ &\textrm{Silahkan lanjutkan proses berikut}\\ &(a^{2}+b^{2})(x^{2}+y^{2})-(ax+by)^{2}=....\\ &\textrm{a}.\quad \textrm{Bagaimana hubungan}\: \: ax+by\: \: \textrm{dengan}\: \: 1\\ &\textrm{b}.\quad \textrm{Mengapa}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\color{blue}\textrm{Perhatikan bahwa}\\ &(a^{2}+b^{2})(x^{2}+y^{2})-(ax+by)^{2}\\ &=a^{2}x^{2}+a^{2}y^{2}+b^{2}x^{2} +b^{2}y^{2}-a^{2}x^{2}-b^{2}y^{2}-2abxy\\ &=a^{2}y^{2}+b^{2}x^{2}-2abxy\\ &=(ay-bx)^{2}\\ &\textrm{a. Nilai}\: \: \color{red}ax+by\: \: \color{black}\textrm{selalu lebih kecil}\\ &\quad\: \textrm{atau sama dengan 1}\\ &\textrm{b. Kita memiliki}\: \: (ay-bx)^{2}\geq 0,\: \: \textrm{sehingga}\\ &\quad\Leftrightarrow \: (a^{2}+b^{2})(x^{2}+y^{2})-(ax+by)^{2}\geq 0\\ &\quad\Leftrightarrow \: \quad 1\times 1-(ax+by)^{2}\geq 0\\ &\quad\Leftrightarrow \: \quad\qquad 1-(ax+by)^{2}\geq 0\\ &\quad\Leftrightarrow \: \quad\qquad (ax+by)^{2}-1\leq  0\\ &\quad\Leftrightarrow \: \: \: \: \: \qquad\qquad (ax+by)^{2}\leq  1\qquad \blacksquare  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Pada segi enam beraturan, berapa banyak}\\ &\textrm{segitiga yang dapat Anda temukan}\\\\ &\textbf{Jawab}:\\  &\begin{aligned}&\textrm{Setiap segitiga dapat terbentuk dari 3 buah}\\ &\textrm{sebarang, sehingga banyak segitiga bila}\\ &\textrm{diketahui}\: \: n=6,\: \: r=3\: \: \textrm{adalah}:\\ &\color{red}\textrm{kombinasi 3 titik dari 6 titik yang ada}\\ &\textrm{Adapun untuk rumus kombinasinya}:\\ &C_{r}^{n}=\begin{pmatrix} n\\  r \end{pmatrix}=\displaystyle \frac{n!}{r!(n-r)!}\\ &\textrm{dengan}\: \: n!=1\times 2\times 3\times 4\times ...\times n\\ &\textrm{maka}\\ &C_{3}^{6}=\begin{pmatrix} 6\\ 3 \end{pmatrix}=\displaystyle \frac{6!}{3!.3!}=\frac{6.5.4.3!}{1.2.3.3!}=\color{blue}20  \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tentukan nilai dari bentuk}\\ &\displaystyle \frac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...}{1^{-3}+3^{-3}+5^{-3}+7^{-3}+9^{-3}+...}\\\\ &\textbf{Jawab}:\\  &\begin{aligned}&\textrm{Misalkan}\\ &x=1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...\: \: \textrm{dan}\\ &y=1^{-3}+3^{-3}+5^{-3}+7^{-3}+9^{-3}+...\\ &\textrm{Sekarang perhatikan bahwa}\\ &x=1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...\\ &\: \: \: =(1^{-3}+3^{-3}+5^{-3}+...)\: +\: (2^{-3}+4^{-3}+6^{-3}+...)\\ &\: \: \: =(1^{-3}+3^{-3}+5^{-3}+...)\: +\: 2^{-3}(1^{-3}+2^{-3}+3^{-3}+...)\\ &\: \: \: =y+\displaystyle \frac{1}{8}x\\ &x-\displaystyle \frac{1}{8}x=y\Leftrightarrow \displaystyle \frac{7}{8}x=y\Leftrightarrow \displaystyle \frac{x}{y}=\color{blue}\frac{8}{7}\\  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\\ &\textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textbf{Jawab}:\\ &\color{purple}\textrm{Perhatikanlah bahwa}\\ &\color{blue}\begin{aligned}&\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases} \end{aligned} \\ &\color{purple}\textrm{Selanjutnya}\\ &\color{purple}\textrm{Untuk}\: \: (1-a-b),\: \textrm{maka}\\ &\color{purple}\begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned} \\ &\color{purple}\textrm{Untuk}\: \: 2(1-b),\: \textrm{maka}\\ &\color{purple}\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned} \\ &\color{blue}\textrm{Untuk}\: \: \left ( \displaystyle \frac{1-x-y}{2(1-b)} \right ),\: \textrm{maka}\\ &\color{purple}\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\ &\color{purple}\textrm{Jadi},\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}=12^{^{\: ^{12}\log 2}}=2 \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Tentukan hasil dari bentuk}\\ &\sin ^{2}1^{\circ}+\sin ^{2}2^{\circ}+\sin ^{2}3^{\circ}+...+\sin ^{2}89^{\circ}+\sin ^{2}90^{\circ}\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat sudut-sudut yang berelasi pada trigonometri}:\\ &\color{red}\sin \left ( 90^{0}-\alpha  \right )=\cos \alpha \\ &\textrm{Ingat pula identitas trigonometri}:\\ & \color{red}\sin ^{2}\alpha +\cos ^{2}\alpha =1\\ &\underline{\textrm{maka bentuk}}\\ &\sin ^{2}\color{blue}1^{\circ}\color{black}=\sin^{2} \left (\color{blue}90^{\circ}-89^{\circ}\color{black}  \right )  =\cos ^{2}89^{\circ},\\ & \textrm{dengan cara yang sama akan diperoleh}\\ &\sin ^{2}\color{blue}2^{\circ}\color{black}=\cos ^{2}88^{\circ}\\ &\sin ^{2}\color{blue}3^{\circ}\color{black}=\cos ^{2}87^{\circ}\\ &\sin ^{2}\color{blue}4^{\circ}\color{black}=\cos ^{2}86^{\circ}\\ &\qquad\qquad \vdots \\ &\sin ^{2}\color{blue}44^{\circ}\color{black}=\cos ^{2}46^{\circ}\\ &\textrm{Sehingga}\\ &\begin{aligned}&\sin ^{2}1^{\circ}+\sin ^{2}2^{\circ}+\sin ^{2}3^{\circ}+...+\sin ^{2}89^{\circ}+\sin ^{2}90^{\circ}\\ &=\sin ^{2}1^{\circ}+...+\sin ^{2}44^{\circ}+\sin ^{2}46^{\circ}+...+\sin ^{2}89^{\circ}\color{red}+\sin ^{2}45^{\circ}+\sin ^{2}90^{\circ}\\ &=\underset{44}{\underbrace{1+1+1+...+1}}\color{red}+\sin ^{2}45^{\circ}+\sin ^{2}90^{\circ}\\ &=44+\displaystyle \frac{1}{2}+1\\ &=\color{blue}45\displaystyle \frac{1}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Tentukan sisa pembagian}\: \: x^{3}-5x^{2}+3x-4\\ &\textrm{oleh}\: \: 2x+1\\\\ &\textbf{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Misalkan}\: \: f(x)=x^{3}-5x^{2}+3x-4\\ &\textrm{Sisa pembagian}\: \: f(x)\: \: \textrm{oleh}\: \: q(x)=2x+1\: \: \textrm{adalah}:\\ &\textrm{ambil}\: \: q(x)=2x+1=0\Rightarrow x=-\displaystyle \frac{1}{2},\: \: \textrm{maka}\\ &\textrm{penentuan sisa pembagian cukup dengan}\\ &f\left ( -\displaystyle \frac{1}{2} \right )=\left ( -\displaystyle \frac{1}{2} \right )^{3}-5\left ( -\displaystyle \frac{1}{2} \right )^{2}+3\left ( -\displaystyle \frac{1}{2} \right )-4\\ &\qquad\qquad =-\displaystyle \frac{1}{8}-\displaystyle \frac{5}{4}-\frac{3}{2}-4\\ &\qquad\qquad =\displaystyle \frac{-1-10-12-32}{8}=\color{red}-\displaystyle \displaystyle \frac{55}{8}\\ &\color{blue}\textbf{Alternatif 2}\\ &\textrm{Dengan metode Horner, yaitu}:\\ &\begin{array}{c|cccccccccc}\\ x=-\displaystyle \frac{1}{2}&1&-5&3&-4&&&\\ &&&&&&&\\ &&-\displaystyle \frac{1}{2}&\displaystyle \frac{11}{4}&-\displaystyle \frac{23}{8}&&&+\\\hline &1&-5\displaystyle \frac{1}{2}&\displaystyle \frac{23}{4}&\color{red}-\displaystyle \frac{55}{8}&&& \end{array} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Fungsi}\: \: f\: \: \textrm{memiliki sifat untuk tiap bilangan}\\ &\textrm{bilangan real}\: \: x\: \:  \textrm{berlaku}\: \: f(x)+f(x-1)=x^{2}\\ &\textrm{Jika}\: \: f(19)=94,\: \textrm{tentukan sisa pembagian}\\ & f(94)\: \: \textrm{oleh}\: \: 100\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\: \: f(x)=x^{2}-f(x-1),\\ &\textrm{maka}\\ &\begin{aligned}f(94)&=94^{2}-f(93)=94^{2}-\left ( 93^{2}-f(92) \right )\\ &=94^{2}-93^{2}+f(92)=94^{2}-93^{2}+(92^{2}-f(91))\\ &=94^{2}-93^{2}+92^{2}-91^{2}+f(90)\\ &=94^{2}-93^{2}+92^{2}-91^{2}+90^{2}-\cdots +20^{2}-f(19)\\ &\quad \textrm{ingat bentuk}\: \: \color{red}a^{2}-b^{2}=(a+b)(a-b),\: \color{black}\textrm{maka}\\ &=(94+93)+(92+91)+\cdots +400-94\\ &=\underset{4255}{\underbrace{94+93+...+22+21}}+400-94\\ &=4255+306\\ &=\color{blue}4561 \end{aligned}\\ &\textrm{Sehingga sisa pembagian}\: \: f(94)\: \: \textrm{oleh 100 adalah}\: \: \color{blue}61 \end{array}$.

DAFTAR PUSTAKA

1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
2. Susyanto, N. 2012. Tutor Senior Olimpiade Matematika Lima Benua Tingkat SMP. Yogyakarta: KENDI MAS MEDIA
3. ..........Η Στήλη των Μαθηματικών, έτος 2007, τεύχη 46-94