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

 $\begin{array}{l}\\ 1.& \text{Jika}a,b\text{bilangan real positif},\\ & \text{tunjukkan bahwa}{a}^a.b^b\ge a^b.b^a\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Diketahui}{a}^a.b^b\ge a^b.b^a\iff {\displaystyle \frac{a^a.b^b}{a^b.b^a}\ge 1}\\ & \iff {\displaystyle \frac{a^{a-b}}{b^{a-b}}\ge 1\iff {\left({\displaystyle \frac{a}{b}}\right)}^{a-b}\ge 1}\\ & \text{Selanjutnya akan ada dua kemungkinan}\\ & \text{yaitu}:a\ge b>0\text{dan}b\ge a>0\end{array}\\ & \begin{array}{|cll|}\hline \text{No}& \text{Kondisi}& \text{Akibat}\\ 1.& a\ge b>0& {\displaystyle \frac{a}{b}\ge 1\text{atau}a-b\ge 0}\\ & & \text{maka}\begin{array}{rl}& {\left({\displaystyle \frac{a}{b}}\right)}^{a-b}\ge 1\end{array}\\ 2.& b\ge a>0& {\displaystyle \frac{a}{b}\le 1\text{atau}a-b\le 0}\\ & & \text{maka}\begin{array}{rl}& {\left({\displaystyle \frac{a}{b}}\right)}^{a-b}\ge 1\end{array}\\ \hline\end{array}\\ & \text{Karena keduanya memiliki hasil yang sama}\\ & \text{maka}{a}^a.b^b\ge a^b.b^a◼\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}a=\sqrt{{\displaystyle \frac{b}{1-b}}}\text{nyatakanlah}b\text{dalam}a\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & a=\sqrt{{\displaystyle \frac{b}{1-b}}}\iff a^2={\displaystyle \frac{b}{1-b}\iff a^2(1-b)=b}\\ & \iff a^2-a^{2}b=b\iff b+a^{2}b=a^2\\ & \iff b(1-a^2)=a^2\iff b={\displaystyle \frac{a^2}{1-a^2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Sederhanakanlah bentuk}{3.4}^4+{3.4}^4+{3.4}^4\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {3.4}^4+{3.4}^4+{3.4}^4=3({3.4}^4)\\ & =9\times 4^4\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui}{a}^2+b^2=1\text{dan}{x}^2+y^2=1\\ & \text{Silahkan lanjutkan proses berikut}\\ & (a^2+b^2)(x^2+y^2)-(ax+by{)}^2=....\\ & \text{a}.\text{Bagaimana hubungan}ax+by\text{dengan}1\\ & \text{b}.\text{Mengapa}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa}\\ & (a^2+b^2)(x^2+y^2)-(ax+by{)}^2\\ & =a^2{x}^2+a^2{y}^2+b^2{x}^2+b^2{y}^2-a^2{x}^2-b^2{y}^2-2abxy\\ & =a^2{y}^2+b^2{x}^2-2abxy\\ & =(ay-bx{)}^2\\ & \text{a. Nilai}ax+by\text{selalu lebih kecil}\\ & \text{atau sama dengan 1}\\ & \text{b. Kita memiliki}(ay-bx{)}^2\ge 0,\text{sehingga}\\ & \iff (a^2+b^2)(x^2+y^2)-(ax+by{)}^2\ge 0\\ & \iff 1\times 1-(ax+by{)}^2\ge 0\\ & \iff 1-(ax+by{)}^2\ge 0\\ & \iff (ax+by{)}^2-1\le 0\\ & \iff (ax+by{)}^2\le 1◼\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Pada segi enam beraturan, berapa banyak}\\ & \text{segitiga yang dapat Anda temukan}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Setiap segitiga dapat terbentuk dari 3 buah}\\ & \text{sebarang, sehingga banyak segitiga bila}\\ & \text{diketahui}n=6,r=3\text{adalah}:\\ & \text{kombinasi 3 titik dari 6 titik yang ada}\\ & \text{Adapun untuk rumus kombinasinya}:\\ & C_r^n=\left(\begin{array}{c}n\\ r\end{array}\right)={\displaystyle \frac{n!}{r!(n-r)!}}\\ & \text{dengan}n!=1\times 2\times 3\times 4\times ...\times n\\ & \text{maka}\\ & C_3^6=\left(\begin{array}{c}6\\ 3\end{array}\right)={\displaystyle \frac{6!}{3!.3!}=\frac{6.5.4.3!}{1.2.3.3!}=20}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukan nilai dari bentuk}\\ & {\displaystyle \frac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...}{1^{-3}+3^{-3}+5^{-3}+7^{-3}+9^{-3}+...}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & x=1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...\text{dan}\\ & y=1^{-3}+3^{-3}+5^{-3}+7^{-3}+9^{-3}+...\\ & \text{Sekarang perhatikan bahwa}\\ & x=1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...\\ & =(1^{-3}+3^{-3}+5^{-3}+...)+(2^{-3}+4^{-3}+6^{-3}+...)\\ & =(1^{-3}+3^{-3}+5^{-3}+...)+2^{-3}(1^{-3}+2^{-3}+3^{-3}+...)\\ & =y+{\displaystyle \frac{1}{8}x}\\ & x-{\displaystyle \frac{1}{8}x=y\iff {\displaystyle \frac{7}{8}x=y\iff {\displaystyle \frac{x}{y}=\frac{8}{7}}}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 8.& \text{Tentukan hasil dari bentuk}\\ & {\sin }^2{1}^{\circ }+{\sin }^2{2}^{\circ }+{\sin }^2{3}^{\circ }+...+{\sin }^2{89}^{\circ }+{\sin }^2{90}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat sudut-sudut yang berelasi pada trigonometri}:\\ & \sin ({90}^0-\alpha )=\cos \alpha \\ & \text{Ingat pula identitas trigonometri}:\\ & {\sin }^2\alpha +{\cos }^2\alpha =1\\ & \underset{―}{\text{maka bentuk}}\\ & {\sin }^2{1}^{\circ }={\sin }^2\left({90}^{\circ }-{89}^{\circ }\right)={\cos }^2{89}^{\circ },\\ & \text{dengan cara yang sama akan diperoleh}\\ & {\sin }^2{2}^{\circ }={\cos }^2{88}^{\circ }\\ & {\sin }^2{3}^{\circ }={\cos }^2{87}^{\circ }\\ & {\sin }^2{4}^{\circ }={\cos }^2{86}^{\circ }\\ & ⋮\\ & {\sin }^2{44}^{\circ }={\cos }^2{46}^{\circ }\\ & \text{Sehingga}\\ & \begin{array}{rl}& {\sin }^2{1}^{\circ }+{\sin }^2{2}^{\circ }+{\sin }^2{3}^{\circ }+...+{\sin }^2{89}^{\circ }+{\sin }^2{90}^{\circ }\\ & ={\sin }^2{1}^{\circ }+...+{\sin }^2{44}^{\circ }+{\sin }^2{46}^{\circ }+...+{\sin }^2{89}^{\circ }+{\sin }^2{45}^{\circ }+{\sin }^2{90}^{\circ }\\ & =\underset{44}{\underbrace{1+1+1+...+1}}+{\sin }^2{45}^{\circ }+{\sin }^2{90}^{\circ }\\ & =44+{\displaystyle \frac{1}{2}+1}\\ & =45{\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Tentukan sisa pembagian}{x}^3-5x^2+3x-4\\ & \text{oleh}2x+1\\ \\ & \mathbf{\text{Jawab}}:\\ & \mathbf{\text{Alternatif 1}}\\ & \text{Misalkan}f(x)=x^3-5x^2+3x-4\\ & \text{Sisa pembagian}f(x)\text{oleh}q(x)=2x+1\text{adalah}:\\ & \text{ambil}q(x)=2x+1=0\Rightarrow x=-{\displaystyle \frac{1}{2},\text{maka}}\\ & \text{penentuan sisa pembagian cukup dengan}\\ & f(-{\displaystyle \frac{1}{2}})={(-{\displaystyle \frac{1}{2}})}^3-5{(-{\displaystyle \frac{1}{2}})}^2+3(-{\displaystyle \frac{1}{2}})-4\\ & =-{\displaystyle \frac{1}{8}-{\displaystyle \frac{5}{4}-\frac{3}{2}-4}}\\ & ={\displaystyle \frac{-1-10-12-32}{8}=-{\displaystyle {\displaystyle \frac{55}{8}}}}\\ & \mathbf{\text{Alternatif 2}}\\ & \text{Dengan metode Horner, yaitu}:\\ & \begin{array}{c}\\ x=-{\displaystyle \frac{1}{2}}& 1& -5& 3& -4& & & \\ & & & & & & & \\ & & -{\displaystyle \frac{1}{2}}& {\displaystyle \frac{11}{4}}& -{\displaystyle \frac{23}{8}}& & & +\\ & 1& -5{\displaystyle \frac{1}{2}}& {\displaystyle \frac{23}{4}}& -{\displaystyle \frac{55}{8}}& & & \end{array}\end{array}$.

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

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