Contoh Soal 3 Polinom

$\begin{array}{l}\\ 11.& \text{Jika polinom}f(x)\text{dibagi oleh}\\ & (x-a)(x-b)\text{dan}a\ne b,\text{maka}\\ & \text{sisa pembagiannya adalah}....\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle {\displaystyle \frac{x-a}{a-b}f(a)+\frac{x-a}{b-a}f(b)}}\\ \\ & \text{b}.{\displaystyle {\displaystyle \frac{x-a}{a-b}f(b)+\frac{x-a}{b-a}f(a)}}\\ \\ & \text{c}.{\displaystyle {\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}}\\ \\ & \text{d}.{\displaystyle {\displaystyle \frac{x-b}{a-b}f(b)+\frac{x-a}{b-a}f(a)}}\\ \\ & \text{e}.{\displaystyle {\displaystyle \frac{x-a}{b-a}f(b)+\frac{x-a}{b-a}f(a)}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misal sisa pembagiannya}:s(x)=px+q\\ & \text{Saat}f(x)\text{dibagi}(x-a)(x-b)\text{berarti}\\ & \bullet x=a\Rightarrow s(a)=f(a)=ap+q....(1)\\ & \bullet x=b\Rightarrow s(b)=f(b)=bp+q......(2)\\ & \text{Persamaan}(1)\text{dan}(2)\text{dieliminasi}\\ & \begin{array}{l}\\ ap& +& q& =& f(a)\\ bp& +& q& =& f(b)& -\\ ap& -& bp& =& f(a)-f(b)\\ & & p& =& {\displaystyle \frac{f(a)-f(b)}{a-b}}& \end{array}\\ & \text{Dari persamaan}(1),\\ & f(a)=ap+q\\ & f(a)=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+q\\ & q=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+f(a)\\ & q=a\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)+f(a)\left({\displaystyle \frac{a-b}{a-b}}\right)\\ & q={\displaystyle \frac{-bf(a)-af(b)}{a-b}}\\ & \text{Sehingga}\\ & s(x)=px+q\\ & =\left({\displaystyle \frac{f(a)-f(b)}{a-b}}\right)x+\left({\displaystyle \frac{-bf(a)-af(b)}{a-b}}\right)\\ & ={\displaystyle \frac{f(a)x-f(b)x-bf(a)+af(b)}{a-b}}\\ & ={\displaystyle \frac{(x-b)f(a)+(a-x)f(b)}{a-b}}\\ & ={\displaystyle \frac{x-b}{a-b}f(a)+\frac{a-x}{a-b}f(b)}\\ & ={\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Diketahui}f(x)\text{dibagi oleh}x-2\text{bersisa 5},\\ & \text{dan dibagi}x-3\text{bersisa 7. Jia}f(x)\\ & \text{dibagi oleh}{x}^2-5x+6\text{akan memiliki sisa}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x-2}& & \text{d}.{\displaystyle 2x+1}\\ \text{b}.{\displaystyle 2x-4}& \text{c}.{\displaystyle x+2}& \text{e}.2x+3\end{array}\\ \\ & \text{Jawab}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}f(x)& =(x-2).h(x)+5\\ f(x)& =(x-3).h(x)+7\\ f(x)& =(x^2-5x+6).H(x)+s(x)\\ f(x)& =(x-2)(x-3).H(x)+px+q\\ f(2)& =(2-2)(2-3).H(x)+2p+q=5\\ & \Rightarrow 0+2p+q=5.................(1)\\ f(3)& =(3-2)(3-3).H(x)+3p+q=7\\ & \Rightarrow 0+3p+q=7.................(2)\\ \text{Dari}& \text{persamaan}(1)\text{dan}(2)\\ \text{saat}& \text{persamaan (1) dikurangi persamaan (2)}\\ & -p=-2\\ & p=2\\ & \text{maka},q=1\\ & \text{Sehingga},\\ & s(x)=px+q=2x+1\end{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& f(x)\text{dibagi}(x-2)\text{sisa}5\Rightarrow f(2)=5\\ & f(x)\text{dibagi}(x-3)\text{sisa}7\Rightarrow f(3)=7\\ & \text{maka},\\ & s(x)={\displaystyle \frac{x-b}{a-b}f(a)+\frac{x-a}{b-a}f(b)}\\ & ={\displaystyle \frac{x-3}{2-3}(5)+\frac{x-2}{3-2}(7)}\\ & ={\displaystyle \frac{5x-15}{-1}+\frac{7x-14}{1}}\\ & =15-5x+7x-14\\ & =2x+1\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Polinom}f(x)\text{dibagi oleh}(2x-4)\text{bersisa 6},\\ & \text{dibagi oleh}(x+4)\text{bersisa 24}.\\ & \text{Dan polinom}g(x)\text{dibagi oleh}(2x-4)\text{bersisa 5},\\ & \text{dibagi oleh}(x+4)\text{bersisa 2}.\\ & \text{Jika}h(x)=f(x).g(x),\text{maka}h(x)\\ & \text{dibagi}(2x^2+4x-16)\text{akan sisa}....\\ & \begin{array}{l}\\ \text{a}.-3x+24& & \text{d}.-6x+36\\ \text{b}.-3x+36& \text{c}.6x+24& \text{e}.12x+3\end{array}\\ \\ & \text{Jawab}:\\ & \text{Langkah pertama}\\ & \begin{array}{rl}f(x)& =(2x-4).h(x{)}_1+6\\ f(x)& =(x+4).h(x{)}_2+24\\ f(x)& =(2x-4)(x+4).H_1(x)+p_{1}x+q_1\\ & \text{Gunakanlah cara sebagai mana}\\ & \text{contoh soal No. 12 di atas yang}\\ \text{Alte}& \text{natif 2}\\ \text{mak}& \text{a}{p}_{1}x+q_1=-3x+12\end{array}\\ & \text{Langkah kedua}\\ & \begin{array}{rl}g(x)& =(2x-4).h(x{)}_3+5\\ g(x)& =(x+4).h(x{)}_4+2\\ g(x)& =(2x-4)(x+4).H_2(x)+p_{2}x+q_2\\ & \text{Gunakanlah cara sebagai mana}\\ & \text{contoh soal No. 12 di atas yang}\\ \text{Alte}& \text{natif 2}\\ \text{mak}& \text{a}{p}_{2}x+q_2={\displaystyle \frac{1}{2}x+4}\end{array}\\ & \text{Langkah ketiga}\\ & \begin{array}{rl}& h(x)=f(x)\times g(x)\\ & =((2x-4)(x+4)H_1(x)+(-3x+12))\\ & \times ((2x-4)(x+4)H_2(x)+{\displaystyle \frac{1}{2}x+4})\\ & \text{maka}\\ & \bullet h(2)=(0+(-3.2+12))(0+{\displaystyle \frac{1}{2}.2+4})=6.5=30\\ & \bullet h(-4)=(0+(-3.-4+12))(0+{\displaystyle \frac{1}{2}.-4+4})=24.2=48\\ & \text{Dengan pembagi}2x^2+x-16,\text{maka sisanya}:s_3(x)=p_{3}x+q_3\\ & \text{saat}x=2\Rightarrow 2p+q=30\\ & \text{saat}x=-4\Rightarrow -4p+q=48\\ & \text{selanjutnya dengan eliminasi-substitusi diperoleh}p=-3,q=36\\ & \text{sehingga}s(x)=px+q=-3x+36\end{array}\end{array}$

$\begin{array}{l}\\ 14.& (\text{KSM 2015})\text{Diketahui}f(x)\text{adalah polinom}\\ & (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\\ & \text{dengan}{x}_1,x_2,x_3,x_4,\text{dan}{x}_5\text{adalah}\\ & \text{bilangan bulat berbeda}.\text{Jika}f(104)=2012,\\ & \text{maka nilai}{x}_1+x_2+x_3+x_4+x_5\text{sama dengan}....\\ & \text{a}.13\\ & \text{b}.14\\ & \text{c}.16\\ & \text{d}.17\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketa}& \text{hui bahwa}:\\ f(x)& =(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\\ f(104)& =(104-x_1)(104-x_2)(104-x_3)(104-x_4)(104-x_5)=2012\\ & =2012=1\times 2\times 503\\ & =(-1)\times (1)\times (-2)\times (2)\times (503)\\ \text{maka}& \{\begin{array}{ll}(104-x_1)& =-2\Rightarrow x_1=106\\ (104-x_2)& =-1\Rightarrow x_2=105\\ (104-x_3)& =1\Rightarrow x_3=103\\ (104-x_4)& =2\Rightarrow x_4=102\\ (104-x_5)& =503\Rightarrow x_5=-399\end{array}\\ \text{sehin}& \text{gga},\\ & x_1+x_2+x_3+x_4+x_5=106+105+103+102+(-399)=17\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Tentukanlah suku banyak}f(x)\text{sedemikian}\\ & \text{sehingga}f(x)\text{terbagi oleh}{x}^2+1,\\ & \text{sedangkan}f(x)+1\text{terbagi oleh}{x}^3+x^2+1\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =(x^2+1).h_{1}x\\ f(x)+1& =(x^2+1).h_{1}x+1\\ \text{supaya}& f(x)+1\text{terbagi habis oleh}{x}^3+x^2+1,\\ & \text{maka akan ada bilangan bulat}k,(k\ne 0)\\ k& ={\displaystyle \frac{f(x)+1}{x^3+x^2+1}}\\ & ={\displaystyle \frac{(x^2+1).h_{1}x+1}{x^3+x^2+1}}\\ k=1\Rightarrow & 1={\displaystyle \frac{(x^2+1).h_{1}x+1}{x^3+x^2+1}\text{maka}{h}_{1}x=x}\\ \text{sehingga}& f(x)=x^3+x^2\\ \text{untuk ni}& \text{lai}k\text{yang lain, tak ditemukan}\end{array}\end{array}$

Contoh Soal 4 Polinom

$\begin{array}{l}\\ 16.& \text{Diketahui akar-akar polinom}\\ & x^{2017}+x^{2016}+x^{2015}+...+x^2+x+1=0\\ & \text{adalah}{x}_1,x_2,x_3,...,x_{2017}\\ & \text{Tentukan nilai dari}\\ & {\displaystyle \frac{1}{1-x_1}+\frac{1}{1-x_2}+\frac{1}{1-x_3}+...+{\displaystyle \frac{1}{1-x_{2017}}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \frac{x^{2018}-1}{x-1}}& =x^{2017}+x^{2016}+x^{2015}+...+x^2+x+1=0\\ \text{perlu dii}& \text{ngat bahwa kondisi ini mensyaratkan}x\ne 1,\\ & \text{sehingga}\\ \\ x^{2018}-1& =0\\ x^{2018}& =1\\ x& =\pm 1,\text{pilih}x=-1\\ \text{maka}& \text{nilai dari}\\ {\displaystyle \frac{1}{1-x_1}+}& \frac{1}{1-x_2}+\frac{1}{1-x_3}+...+{\displaystyle \frac{1}{1-x_{2017}}}\\ & =\underset{\text{sebanyak 2017}}{\underbrace{{\displaystyle \frac{1}{1-(-1)}+\frac{1}{1-(-1)}+\frac{1}{1-(-1)}+...+\frac{1}{1-(-1)}}}}\\ & =\underset{\text{sebanyak 2017}}{\underbrace{{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+{\displaystyle \frac{1}{2}}}}}\\ & ={\displaystyle \frac{2017}{2}}\end{array}\end{array}$

Contoh Soal 2 Polinom

$\begin{array}{l}\\ 6.& \text{Diketahui bahwa}\\ & {\displaystyle \frac{f(x)}{x-2}=h(x)+{\displaystyle \frac{3}{x-2}}}\\ & \text{dan}{\displaystyle \frac{f(x)}{x-1}=h(x)+{\displaystyle \frac{2}{x-1},}}\\ & \text{jika}{\displaystyle \frac{f(x)}{(x-2)(x-1)}=h(x)+{\displaystyle \frac{s(x)}{(x-2)(x-1)},}}\\ & \text{maka}s(x)=....\\ & \begin{array}{l}\\ \text{a}.x+1& & \text{d}.2x-1\\ \text{b}.x+2& \text{c}.2x+1& \text{e}.x-2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{f(x)}{x-2}=h(x)+{\displaystyle \frac{3}{x-2}}}\\ & \Rightarrow f(x)=(x-2).h(x)+3\Rightarrow f(2)=3\\ & {\displaystyle \frac{f(x)}{x-1}=h(x)+{\displaystyle \frac{2}{x-1}}}\\ & \Rightarrow f(x)=(x-1).h(x)+2\Rightarrow f(1)=2\\ & {\displaystyle \frac{f(x)}{(x-2)(x-1)}=h(x)+{\displaystyle \frac{s(x)}{(x-2)(x-1)}}}\\ & \text{maka}f(x)=(x-2)(x-1).h(x)+s(x)\\ & f(x)=(x-2)(x-1).h(x)+px+q\\ & f(2)=2p+q=3\\ & f(1)=p+q=2,\\ & \text{sehingga dengan }\text{eliminasi akan diperoleh}\\ p& =1\text{dan}\\ & q=1\\ & \text{Jadi},px+q=x+1\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Jika}{x}^4+2mx-n\text{dibagi}{x}^2-1\\ & \text{bersisa}2x-1,\text{maka nilai}m\\ & \text{dan}n\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.m=-1\text{dan}n=2\\ \text{b}.m=1\text{dan}n=-2\\ \text{c}.m=1\text{dan}n=2\\ \text{d}.m=-1\text{dan}n=-2\\ \text{e}.m=-2\text{dan}n=1\end{array}\\ \\ & \text{Jawab}:\\ & \text{dengan Horner-Kino didapatkan}\end{array}$

$.\{\begin{array}{ll}\text{Suku banyak}:& f(x)=x^4+2mx-n\\ \text{Pembagai}:& p(x)=(x-1)(x+1)=x^2-1\\ & :1\text{dari}-\frac{-1}{1},\text{sedang}0=-\left(\frac{0}{1}\right)\\ \text{Hasil bagi}:& h(x)=x^2+1\\ \text{Sisa bagi}:& s(x)=2mx+(1-n)=2x-1\end{array}$

$.\begin{array}{rl}& \text{Sehingga},\\ & \bullet 2m=2\Rightarrow m=1\\ & \bullet 1-n=-1\Rightarrow n=2\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}f(x)=x^4-k{x}^2+5\text{habis dibagi}\\ & (x-1)\text{maka}f(x)\text{juga habis dibagi oleh}....\\ & \begin{array}{l}\\ \text{a}.x+1& & \text{d}.x+5\\ \text{b}.2x+1& \text{c}.3x+1& \text{e}.2x+5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =x^4-k{x}^2+5\\ f(1)& =(1{)}^4-k(1{)}^2+5\\ 0& =1-k+5\\ k& =6\\ f(x)& =x^4-6x^2+5\\ & =(x^2-1)(x^2-5)\\ & =(x-1)(x+1)(x^2-5)\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika}(m-2)\text{adalah faktor dari}2m^3+3tm+4,\\ & \text{maka nilai}t\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{10}{3}}& & \text{d}.-{\displaystyle \frac{3}{10}}\\ \\ \text{b}.{\displaystyle \frac{1}{3}}& \text{c}.{\displaystyle \frac{3}{10}}& \text{e}.-{\displaystyle \frac{10}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(m)& =2m^3+3tm+4\\ f(2)& =2(2{)}^3+3t(2)+4\\ 0& =16+6t+4\\ -6t& =20\\ t& =-{\displaystyle \frac{10}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{(KSM MA Kab/Kota 2015)Nilai terkecil}n\\ & \text{yang mengkin sehingga}n.(n+1).(n+2)\\ & \text{habis dibagi 24 adalah}....\\ & \begin{array}{c}\\ \text{a}.1\\ \text{b}.2\\ \text{c}.3\\ \text{d}.4\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}k& ={\displaystyle \frac{n.(n+1).(n+2)}{24}}\\ & ={\displaystyle \frac{n.(n+1).(n+2)}{2.(2+1).(2+2)}}\\ & \text{maka}n=2\end{array}\end{array}$

Contoh Soal 1 Polinom

$\begin{array}{l}\\ 1.& \text{Jika}g(x)=2x^3+x^2-x+1,\\ & \text{maka}g(1)=....\\ & \begin{array}{l}\\ \text{a}.-2& & \text{d}.2\\ \text{b}.-1& \text{c}.1& \text{e}.3\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}g(x)& =2x^3+x^2-x+1\\ g(1)& =2(1{)}^3+(1{)}^2-(1)+1\\ & =2+1-1+1\\ & =3\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika}p(y)=5y^4+2r^2{y}^3+y^2+1\text{dan}\\ & q(y)=4y^5+3r{y}^2-3y-1\\ & \text{serta}p(-1)=q(-1),\text{maka nilai}r\\ & \text{sama dengan}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{3}{2}\text{dan}3}& & \text{d}.-{\displaystyle \frac{3}{2}}\\ \text{b}.{\displaystyle -\frac{3}{2}\text{dan}3}& \text{c}.{\displaystyle \frac{3}{2}\text{dan}-3}& \text{e}.3\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}p(-1)& =q(-1)\\ 5(-1{)}^4+2r^2(-1{)}^3+(-1{)}^2+1& =4(-1{)}^5+3r(-1{)}^2-3(-1)-1\\ 5-2r^2+1+1& =-4+3r+3-1\\ 9-3r-2r^2& =0\\ {\displaystyle \frac{(-6-2r)(-3+2r)}{2}}& =0,\text{ingat pemfaktoran}\\ (-3-r)(-3+2r)& =0\\ r=-3\vee r& ={\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Diketahui}f(x)\text{berderajat}n.\\ & \text{Jika pembaginya berbentuk}(a{x}^2+bx+c),\\ & \text{dengan}a\ne 0,\text{maka hasil baginya berderajat}....\\ & \begin{array}{l}\\ \text{a}.n-1& & \text{d}.3\\ \text{b}.n-2& \text{c}.n-3& \text{e}.2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Suku banyak (polinom)}\\ & =\text{pembagi}\times \text{hasil bagi}+\text{sisa}\\ & x^n+...=(a{x}^2+bx+c)\times (x^{n-2}+...)+(mx+n)\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Hasil bagi dan sisanya jika}(6x^4-3x^2+x-1)\\ & \text{dibagi oleh}(2x-1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.3x^3+{\displaystyle \frac{3}{2}{x}^2-{\displaystyle \frac{3}{4}x+\frac{1}{8}}}& \text{dan}& -{\displaystyle \frac{7}{8}}\\ \text{b}.3x^3+3x^2-{\displaystyle \frac{3}{4}x+1}& \text{dan}& -7\\ \text{c}.x^3+{\displaystyle \frac{3}{2}{x}^2-3x+\frac{1}{8}}& \text{dan}& {\displaystyle \frac{7}{8}}\\ \text{d}.x^3+{\displaystyle \frac{3}{2}{x}^2-{\displaystyle \frac{3}{4}x+1}}& \text{dan}& {\displaystyle \frac{1}{8}}\\ \text{e}.3x^3+{\displaystyle \frac{3}{2}{x}^2-{\displaystyle \frac{3}{4}x-\frac{1}{8}}}& \text{dan}& -{\displaystyle \frac{7}{8}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \begin{array}{rrrrrrr}x=\frac{1}{2}& & 6& 0& -3& 1& -1\\ & & & 3& \frac{3}{2}& -\frac{3}{4}& \frac{1}{8}& +& \\ & & 6& 3& -\frac{3}{2}& \frac{1}{4}& \boxed{{\displaystyle -\frac{7}{8}}}\end{array}\\ & \text{Selanjutnya}\\ & \{\begin{array}{ll}\text{Hasil bagi}:& {\displaystyle \frac{6x^3+3x^2-\frac{3}{2}x+\frac{1}{4}}{2}}\\ & =3x^3+{\displaystyle \frac{3}{2}{x}^2-{\displaystyle \frac{3}{4}x+{\displaystyle \frac{1}{8}}}}\\ & \\ \text{Sisa bagi}:& -{\displaystyle \frac{7}{8}}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Hasil bagi dan sisanya jika}(x^4-x^3-x^2+x-1)\\ & \text{dibagi oleh}(x-2)(x+1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.x^2+1& \text{dan}& 2x+1\\ \text{b}.x^2+1& \text{dan}& 2x-1\\ \text{c}.x^2-1& \text{dan}& 2x+1\\ \text{d}.x^2-1& \text{dan}& 2x-1\\ \text{e}.2x^2-1& \text{dan}& x+1\end{array}\\ \\ & \text{Jawab}:\\ & \text{Dengan cara}\mathbf{\text{Horner-Kino}}\text{diperoleh}\end{array}$

$.\{\begin{array}{ll}\text{Suku banyak}:& f(x)=x^4-x^3-x^2+x-1\\ \text{Pembagai}:& p(x)=(x-2)(x+1)=x^2-x-2\\ & :2\text{dari}-\frac{-2}{1},\text{sedang}1=-\left(\frac{-1}{1}\right)\\ \text{Hasil bagi}:& h(x)=x^2+1\\ \text{Sisa bagi}:& s(x)=2x+1\end{array}$

$\begin{array}{rl}& \text{Sehingga},\\ & x^4-x^3-x^2+x-1\\ & =(x^2-x-2)(x^2+1)+2x+1\end{array}$

Contoh Soal 1 Distribusi Binomial

$\begin{array}{l}\\ 1.& \text{Manakah yang merupakan data diskrit dari pernyataan berikut}\\ & \text{a}.\text{Suhu Badan Anton ketika sakit mencapai}{40}^{\circ }C\\ & \text{b}.\text{Kecepatan mobil yang sedang melaju adalah}100km/jam\\ & \text{c}.\text{Tinggi tiang bendaera di madrasah Budi adalah 4 m}\\ & \text{d}.\text{Jumlah guru yang mengajar di MA Futuhiyah }\\ & \text{sebanyak 30 orang}\\ & \text{e}.\text{Berat bayi yang baru lahir adalah 3.500 gram}\\ \\ & \text{Jawab}:\\ & \text{Alasannya dikarena hasil mencacah}\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika Anda mengumpulkan nilai raport}\\ & \text{teman-teman sekelas Anda untuk pelajaran}\\ & \text{ matematika, maka data yang Anda peroleh }\\ & \text{adalah}....\\ & \text{a}.\text{data diskrit}\\ & \text{b}.\text{data kontinu}\\ & \text{c}.\text{data kualitatif}\\ & \text{d}.\text{Populasi}\\ & \text{e}.\text{Sampel}\\ \\ & \text{Jawab}:\\ & \text{Dengan catatan nilainya cacah}\end{array}$

$\begin{array}{l}\\ 3.& \text{Ukuran yang dihitung dari seluruh data }\\ & \text{dalam populasi adalah}....\\ & \text{a}.\text{data kuantitatif}\\ & \text{b}.\text{data kualitatif}\\ & \text{c}.\text{Statistik}\\ & \text{d}.\text{Statistika}\\ & \text{e}.\text{Parameter}\\ \\ & \text{Jawab}:\\ & \text{Parameter adalah ukuran dari }\\ & \text{seluruh data atau populasi}\end{array}$

$\begin{array}{l}\\ 4.& \text{Diketahui distribusi peluang suatu }\\ & \text{variabel acak diskrit sebagai berikut}\\ & \begin{array}{|ccccc|}\hline x& 0& 1& 2& 3\\ f(x)& m& 0,26& 3m& 0,42\\ \hline\end{array}\\ & \text{Peluang nilai X minimal berharga 2 adalah}\\ & \text{a}.0,24\\ & \text{b}.0,34\\ & \text{c}.0,42\\ & \text{d}.0,58\\ & \text{e}.0,66\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}X\text{adalah variabel }\\ & \text{acak diskrit, maka}\sum f(x)=1\\ & F(c)=P(X\le c)={\displaystyle \sum _{x=0}^{x=c}f(x)}\\ & =f(0)+f(1)+f(2)+f(3)+\cdots +f(c)=1\\ & \text{dalam hal soal}\text{di atas, maka kita tentukan}\\ & \text{nilai}m\text{dulu}\\ & F(3)=P(X\le 3)={\displaystyle \sum _{x=0}^{x=3}f(x)}\\ & =f(0)+f(1)+f(2)+f(3)=1\\ & 1=m+0,26+3m+0,42=4m+0,68\\ & 4m=1-0.68=0,32\\ & m=0.08,\text{sehingga}\\ & P(2\le X\le 3)=f(2)+f(3)=3m+0,42\\ & =3(0,08)+0,42=0,24+0,42=0,66\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Diketahui fungsi peluang suatu }\\ & \text{variabel acak kontinu adalah}\\ & f(y)=\{\begin{array}{c}0,\text{untuk textit{y} yang lain}\\ \\ {\displaystyle \frac{2y+k}{50},\text{untuk}0\le y\le 5}\end{array}\\ & \text{Nilai}P(|Y-1|\le 2)\text{adalah}....\\ & \text{a}.{\displaystyle \frac{7}{25}\text{d}.\frac{14}{25}}\\ \\ & \text{b}.{\displaystyle \frac{9}{25}\text{c}.\frac{12}{25}\text{e}.\frac{18}{25}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\int }_0^5{\displaystyle \frac{2y+k}{50}dy=1}\\ & 1={\int }_0^5{\displaystyle \frac{2y+k}{50}dy}\\ & 50={\int }_0^5(2y+k)dy\\ & 50=y^2+ky{|}_0^5=5^2+5k=25+5k\\ & k=5\\ & P(|Y-1|\le 2)=P(-2\le Y-1\le 2)\\ & =P(-1\le Y\le 3)\\ & =f(-1)+f(0)+f(1)+f(2)+f(3)\\ & ={\int }_0^3\left({\displaystyle \frac{2y+5}{50}}\right)dy\\ & ={\displaystyle \frac{1}{50}(y^2+5y){|}_0^3}\\ & ={\displaystyle \frac{1}{50}(9+15)={\displaystyle \frac{24}{50}=\frac{12}{25}}}\end{array}\end{array}$

Contoh Soal 1 Fungsi Komposisi dan Fungsi Invers

 $\begin{array}{l}\\ 1.& \text{Diketahui fungsi}f(2x)=8x-9\\ & \text{dan}g(3x+1)=6x+3.\\ & \text{Rumus untuk}(f+g)(x)=....\\ & \begin{array}{l}\\ \text{a}.& 6x+8& & & \text{d}.& 14x-6\\ \text{b}.& 6x-8& \text{c}.& 14x+6& \text{e}.& 6x-6\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(2x)=8x-9\\ & \Rightarrow f(x)=f\left(2\left({\displaystyle \frac{x}{2}}\right)\right)\\ & =8\left({\displaystyle \frac{x}{2}}\right)-9=4x-9\\ & g(3x+1)=6x+3\\ & \Rightarrow g(x)=g(3\left({\displaystyle \frac{x-1}{3}}\right)+1)\\ & =6\left({\displaystyle \frac{x-1}{3}}\right)+3\\ & =2x+1\end{array}\\ & (f+g)(x)=(4x-9)+(2x+1)\\ & =6x-8\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui fungsi}f(x)=2x-1\\ & \text{dan}g(x)=x^2.\text{Fungsi}(f+g)(x^2)=....\\ & \begin{array}{l}\\ \text{a}.& x^2+2x-1\\ \text{b}.& x^4+2x^2-1\\ \text{c}.& x^4+2x-1\\ \text{d}.& x^4+(2x-1{)}^2\\ \text{e}.& x^4+2x\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(x)=2x-1\\ & g(x)=x^2\end{array}\\ & (f+g)(x)=(2x-1)+(x^2)\\ & =x^2+2x-1\\ & \text{maka}\\ & (f+g)(x^2)=(x^2{)}^2+2(x^2)-1\\ & =x^4+2x^2-1\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}f(x)=3-x,\text{maka}\\ & f\left(x^2\right)+{(f(x))}^2-2f(x)=....\\ & \begin{array}{l}\\ \text{a}.& 2x^2-6x+4\\ \text{b}.& 2x^2+4x+6\\ \text{c}.& 2x^2-4x-6\\ \text{d}.& 6x+4\\ \text{e}.& -4x+6\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & f(x)=3-x,\text{sehingga}\\ & f\left(x^2\right)+{(f(x))}^2-2f(x)\\ & =(3-x^2)+{(3-x)}^2-2(3-x)\\ & =(3-x^2)+(9-6x+x^2)-(6-2x)\\ & =-x^2+x^2-6x+2x+3+9-6\\ & =-4x+6\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui fungsi}f:\mathbb{R}\to \mathbb{R}\text{dan}\\ & g:\mathbb{R}\to \mathbb{R}\text{dirumuskan dengan}\\ & f(x)=x-1\text{dan}g(x)=x^2+2x-3.\\ & \text{Fungsi komposisi}g\text{atas}f\\ & \text{dinotasikan dengan}\\ & \begin{array}{l}\\ \text{a}.& (g\circ f)(x)=x^2-4\\ \text{b}.& (g\circ f)(x)=x^2-5\\ \text{c}.& (g\circ f)(x)=x^2-6\\ \text{d}.& (g\circ f)(x)=x^2-4x-4\\ \text{e}.& (g\circ f)(x)=x^2-4x-5\end{array}\\ & (\mathbf{\text{UN 2016}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}:\\ & \{\begin{array}{ll}f(x)& =x-1\\ g(x)& =x^2+2x-3\end{array}\\ (g\circ f)& =g(f(x))\\ & ={(f(x))}^2+2(f(x))-3\\ & ={(x-1)}^2+2(x-1)-3\\ & =(x^2-2x+1)+(2x-2)-3\\ & =x^2-4\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Diketahui fungsi}f(x)=6x-3\\ & \text{dan}g(x)=5x+4\text{dan}(f\circ g)(a)=81\\ & \text{Nilai}a\text{adalah}........(\mathbf{\text{Ebtanas 2001}})\\ & \begin{array}{l}\\ \text{a}.& -2& & & \text{d}.& 2\\ \text{b}.& -1& \text{c}.& 1& \text{e}.& 27\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{ll}& f(x)=6x-3\\ & g(x)=5x+4\end{array}\\ & (f\circ g)(a)=81\\ & \text{maka}\\ & (f\circ g)(x)=f(g(x))\\ & =6(g(x))-3\\ & =6(5x+4)-3\\ & =30x+24-3\\ & =30x+21=81\\ & 30x+21=81\\ & 30x=81-21=60\\ & x={\displaystyle \frac{60}{30}=2}\end{array}\end{array}$.

Fungsi Komposisi dan Fungsi Invers

$\text{A. Fungsi Komposisi}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|cc|}\hline \text{Syarat}& \text{Sifat-sifat}\\ \begin{array}{rl}& R_{{}_f}\cap D_{{}_g}\ne \left\{\right\}\end{array}& \begin{array}{rl}1.& \text{Tidak komutatif}\\ & (f\circ g)(x)\ne (g\circ f)(x)\\ 2.& \text{Bersifat asosiatif}\\ & f\circ (g\circ h)(x)=(f\circ g)\circ h(x)\\ 3.& \text{Adanya unsur dentitas}\\ & (f\circ I)(x)=(I\circ f)(x)=f(x)\end{array}\\ \hline\end{array}$.

$\text{B. Fungsi Invers}$

$\begin{array}{rl}& \bullet \text{Suatu fungsi}f:A\to B\\ & \text{memiliki fungsi invers}g:B\to A\\ & \mathbf{\text{jika dan hanya jika}}f\\ & \text{merupakan fungsi}\mathbf{\text{bijektif}}\\ & \bullet \text{Jika fungsi}g\text{ada, maka}\\ & g\text{dinyatakan dengan}{f}^{-1}(\text{dibaca}:f\mathbf{\text{invers}})\end{array}$.

$\begin{array}{l}\\ & \mathbf{\text{Catatan}}\\ & \text{Perlu diingat bahwa pada invers }\\ & \text{fungsi komposisi berlaku }\\ & \text{ketentuan sebagai berikut}\\ ⧫& {(g\circ f)}^{-1}(x)=(f^{-1}\circ g^{-1})(x)\\ ⧫& {(f\circ g)}^{-1}(x)=(g^{-1}\circ f^{-1})(x)\\ ⧫& f(x)=({\left(f^{-1}\right)}^{-1}(x))\\ ⧫& x=f^{-1}(f(x))=(f^{-1}\circ f)(x)\\ & =(f\circ f^{-1})(x)=f(f^{-1}(x))\end{array}$.

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah}(f\circ g)(x)\text{dan}(g\circ f)(x)\text{Jika}:\\ & \text{a}.f(x)=2-x\text{dah}g(x)=5x+3\\ & \text{b}.f(x)=2x+1\text{dah}g(x)=x^2-4\\ & \text{c}.f(x)={\displaystyle \frac{5}{x-4}\text{dah}g(x)=3x^2}\\ & \text{d}.f(x)=\sqrt{4-x}\text{dah}g(x)=x^2+x\\ & \text{e}.f(x)=x^3+1\text{dah}g(x)={\displaystyle \frac{x}{x-1}}\\ & \text{f}.f(x)={\displaystyle \frac{3}{x-2}\text{dah}g(x)=\sqrt{x-4}}\end{array}$.

Jawab: 

hanya no. 1 a saja yang dibahas

$\begin{array}{rl}1.\text{a}.(f\circ g)(x)& =f(g(x))\\ & =f(5x+3)\\ & =2-(5x+3)\\ & =-5x-1\text{dan}& \\ (g\circ f)(x)& =g(f(x))\\ & =g(2-x)\\ & =5(2-x)+3\\ & =10-5x+3\\ & =13-10x\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui bahwa}g(x)=3x+2\\ & \text{dan}(g\circ f)(x)=4x-5.\text{Tentukanlah}f(x)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}(g\circ f)(x)& =4x-5\\ g(f(x))& =4x-5\\ 3.f(x)+2& =4x-5\\ 3.f(x)& =4x-7\\ f(x)& ={\displaystyle \frac{4x-7}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui bahwa}g(x)=x+4\text{dan}\\ & (f\circ g)(x)=2x^2+3.\text{Tentukanlah}f(x)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}(f\circ g)(x)& =2x^2+3\\ f(g(x))& =2x^2+3\\ f(x+4)& =2x^2+3,\\ & \text{misalkan}x+4=a\Rightarrow x=a-4,\\ \text{sehingga}& ,\\ f(a)& =2(a-4{)}^2+3\\ f(a)& =2(a^2-8a+16)+3\\ & =2a^2-16a+35\\ f(x)& =2x^2-16x+35\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui}f(x)=3x\text{dan}g(x)=3^x.\\ & \text{Tentukanlah rumus untuk}{}^{{}^{27}}\log (g\circ f)(x)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^{{}^{27}}\log (g\circ f)(x)& ={}^{{}^{27}}\log g(f(x))\\ & ={}^{{}^{27}}\log 3^{3x}\\ & ={}^{{}^{3^3}}\log 3^{3x}\\ & ={}^{{}^{{}^{\left(3^3\right)}}}\log {\left(3^3\right)}^x\\ & =x\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukanlah invers dari}f(x)=6^{2x}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =6^{2x}\\ y& =6^{2x}\\ \log y& =\log 6^{2x}\\ \log y& =2x\log 6\\ {\displaystyle \frac{\log y}{\log 6}}& =2x\\ {\displaystyle \frac{\log y}{2\log 6}}& =x\\ {\displaystyle \frac{\log y}{\log 6^2}}& =x\\ {\displaystyle \frac{\log y}{\log 36}}& =x\\ x& ={\displaystyle \frac{\log y}{\log 36}}& \\ x& ={}^{{}^{36}}\log y\\ f^{-1}(x)& ={}^{{}^{36}}\log x\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukanlah inver dari}\\ & f(x)={\displaystyle \frac{2x+3}{4x-5},x\ne {\displaystyle \frac{5}{4}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{2x+3}{4x-5}}\\ y& ={\displaystyle \frac{2x+3}{4x-5}}\\ (4x-5)y& =2x+3\\ 4xy-5y& =2x+3\\ 4xy-2x& =5y+3\\ x(4y-2)& =5y+3\\ x& ={\displaystyle \frac{5y+3}{4y-2}}\\ f^{-1}(y)& ={\displaystyle \frac{5y+3}{4y-2}}\\ \text{maka},& \\ f^{-1}(x)& ={\displaystyle \frac{5x+3}{4x-2},x\ne {\displaystyle \frac{1}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}f(x)=-2x-4\text{dan}g(x)={\displaystyle \frac{20-3x}{2},}\\ & \text{maka nilai dari}(f\circ g{)}^{-1}(2)=....\\ \\ & \text{Jawab}:\\ & \mathbf{\text{Perhatikan bahwa}}\\ & \begin{array}{rl}& \\ & (f\circ g{)}^{-1}(x)=(g^{-1}\circ f^{-1})(x)\\ & \end{array}\\ & \begin{array}{|cc|}\hline (f\circ g{)}^{-1}(x)& (g^{-1}\circ f^{-1})(x)\\ \begin{array}{rl}(f\circ g)(x)& =f(g(x))\\ y& =-2\left({\displaystyle \frac{20-3x}{2}}\right)-4\\ y& =3x-20-4\\ y& =3x-24\\ y+24& =3x\\ x& ={\displaystyle \frac{y+24}{3}}\\ (f\circ g{)}^{-1}y& ={\displaystyle \frac{y+24}{3}}\\ (f\circ g{)}^{-1}(2)& ={\displaystyle \frac{2+24}{3}}\\ & ={\displaystyle \frac{26}{3}}\end{array}& \begin{array}{rl}(g^{-1}\circ f^{-1})(x)& =g^{-1}(f^{-1}(x))\\ & =......\\ & =....\\ & =....\\ & =....\\ & =....\\ & =....\\ & =....\\ & =....\\ & =....\\ (g^{-1}\circ f^{-1})(2)& =....\\ & =....\end{array}\\ \hline\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Soedyarto, Nugroho, Maryanto. 2008. Matematika 2 untuk SMA dan MA Kelas XI Program IPA. Jakarta: Pusat Perbukuan Departemen Pendidikan Nasional.
2. Sunardi, Waluyo, S., Sutrisno, & Subagya. 2005. Matematika 2untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.

Contoh Soal 3 Fungsi

$\begin{array}{l}\\ 11.& \text{Fungsi berikut yang tidak mempunyai }\\ & \text{asimtot vertikal adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)={\displaystyle \frac{x+2}{x^2-3}}& \\ \text{b}.& f(x)={\displaystyle \frac{x}{(x-2{)}^2}}\\ \text{c}.& f(x)={\displaystyle \frac{x^2-9}{x+3}}\\ \text{d}.& f(x)={\displaystyle \frac{-3}{x}}\\ \text{e}.& \text{semuanya mempunyai asimtot vertikal}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Perh}& \text{atikanlah opsi jawaban}c,\text{yaitu}:\\ f(x)& ={\displaystyle \frac{x^2-9}{x+3}}\\ \text{Jika}& \text{disederhanakan akan menjadi }\\ & \mathbf{\text{fungsi linear}}\text{yaitu}:\\ f(x)& ={\displaystyle \frac{x^2-9}{x+3}={\displaystyle \frac{(x+3)(x-3)}{x+3}=x-3}}\\ \text{sehi}& \text{ngga fungsi pada opsi}c\\ & \text{adalah berupa persamaan linear}\\ & \text{yang secara otomatis }\\ & \mathbf{\text{tidak akan memiliki asimtot}}\end{array}\end{array}$

Contoh Soal 2 Fungsi

$\begin{array}{l}\\ 6.& \text{Diketahui bahwa}\\ & f(x)=\{\begin{array}{ll}0,& \text{untuk}x<0\\ x^2,& \text{untuk}0\le x<1\\ 2x-1,& \text{untuk}x\ge 1\end{array}\\ & \text{Nilai dari}f(-1)+f\left({\displaystyle \frac{1}{2}}\right)-f(3)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -5{\displaystyle \frac{1}{4}}& & & \text{d}.& 4{\displaystyle \frac{3}{4}}\\ \\ \text{b}.& -4{\displaystyle \frac{3}{4}}& \text{c}.& 4& \text{e}.& 5{\displaystyle \frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \\ f(x)& =\{\begin{array}{ll}0,& \text{untuk}x<0\\ x^2,& \text{untuk}0\le x<1\\ 2x-1,& \text{untuk}x\ge 1\end{array}\\ \text{maka nilai}& f(-1)+f\left({\displaystyle \frac{1}{2}}\right)-f(3)\\ & =0+{\left({\displaystyle \frac{1}{2}}\right)}^2-(2(3)-1)\\ & ={\displaystyle \frac{1}{4}-5}\\ & =-4{\displaystyle \frac{3}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika diketahui}f(x+{\displaystyle \frac{1}{x}})=x^3+{\displaystyle \frac{1}{x^3},}\\ & \text{maka nilai dari}f\left({\displaystyle \frac{5}{2}}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2{\displaystyle \frac{1}{8}}& & & \text{d}.& 8{\displaystyle \frac{1}{8}}\\ \\ \text{b}.& 2{\displaystyle \frac{1}{2}}& \text{c}.& 4{\displaystyle \frac{1}{8}}& \text{e}.& 12{\displaystyle \frac{1}{8}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Perhatikan}& \text{bahwa}:\\ (p+q{)}^3& =p^3+3p^{2}q+3p{q}^2+q^3\\ \text{Jika kita su}& \text{bstitusikan}p=x^3\text{dan}q={\displaystyle \frac{1}{x^3}}\\ {(x+{\displaystyle \frac{1}{x}})}^3& =x^3+3x^2\left({\displaystyle \frac{1}{x}}\right)+3x{\left({\displaystyle \frac{1}{x}}\right)}^2+{\left({\displaystyle \frac{1}{x}}\right)}^3\\ & =x^3+{\left({\displaystyle \frac{1}{x}}\right)}^3+3x+{\displaystyle \frac{3}{x}}\\ & =(x^3+{\displaystyle \frac{1}{x^3}})+3(x+{\displaystyle \frac{1}{x}})\\ \text{sehingga}& \\ f(x+{\displaystyle \frac{1}{x}})& =x^3+{\displaystyle \frac{1}{x^3}}\\ & ={(x+{\displaystyle \frac{1}{x}})}^3-3(x+{\displaystyle \frac{1}{x}})\\ f(u)& =u^3-3u,\text{maka}\\ f\left({\displaystyle \frac{5}{2}}\right)& ={\left({\displaystyle \frac{5}{2}}\right)}^3-3\left({\displaystyle \frac{5}{2}}\right)\\ & ={\displaystyle \frac{125}{8}-\frac{15}{2}}\\ & ={\displaystyle \frac{65}{8}}\\ & =8{\displaystyle \frac{1}{8}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Misal fungsi}f\text{terdefinisi untuk }\\ & \text{seluruh bilangan real}x.\\ & \text{Jika}f(p+q)=f(pq)\text{untuk semua}p,q\\ & \text{bilangan bulat positif dan}f(1)=2,\\ & \text{maka nilai}f(2021)=....\\ & \begin{array}{l}\\ \text{a}.& 0& & & \text{d}.& 3\\ \\ \text{b}.& 1& \text{c}.& 2& \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}\\ f(1)& =2\text{dan}\\ f(p+q)& =f(pq)\\ \text{maka}& \\ f(2)& =f(1+1)=f(1.1)=f(1)=2\\ f(3)& =f(1+2)=f(1.2)=f(2)=f(1)=2\\ f(4)& =f(1+3)=f(1.3)=f(3)=f(2)=f(1)=2\\ f(5)& =f(1+4)=f(1.4)=f(4)=f(3)=f(2)=f(1)=2\\ ⋮& \\ f(2021)& =\cdots =\cdots =\cdots =f(2)=f(1)=2\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}{a}_{{}_0}={\displaystyle \frac{2}{5}\text{dan}{a}_{n+1}=2\left|a_n\right|-1,\text{maka nilai}{a}_{{}_{2022}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -0,6& & & \text{d}.& 0,4\\ \\ \text{b}.& -0,2& \text{c}.& 0,2& \text{e}.& 0,6\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}\\ a_{{}_0}& ={\displaystyle \frac{2}{5}=0,4\text{dan}{a}_{n+1}=2\left|a_n\right|-1,\text{maka}}\\ a_{{}_1}& =2\left|a_{{}_0}\right|-1=2|0,4|-1=0,8-1=-0,2\\ a_{{}_2}& =2\left|a_{{}_1}\right|-1=2|-0,2|-1=2(0,2)-1\\ & =0,4-1=-0,6\\ a_{{}_3}& =2\left|a_{{}_2}\right|-1=2|-0,6|-1=1,2-1=0,2\\ a_{{}_4}& =2\left|a_{{}_3}\right|-1=2|0,2|-1=0,4-1=-0,6={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}\\ a_{{}_5}& =2\left|a_{{}_4}\right|-1=2|-0,6|-1=1,2-1=0,2={\mathbf{\text{a}}}_{{}_{\mathbf{\text{3}}}}\\ a_{{}_6}& =2\left|a_{{}_5}\right|-1=2|0,2|-1=0,4-1=-0,6={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}\\ a_{{}_7}& =2\left|a_{{}_6}\right|-1=2|-0,6|-1=1,2-1=0,2={\mathbf{\text{a}}}_{{}_{\mathbf{\text{3}}}}\\ ⋮& \\ a_{{}_{2022}}& =\cdots =\cdots ={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}=-0,6\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Kurva}f(x)={\displaystyle \frac{10}{x^2-10x+25}}\\ & \text{mempunyai asimtot vertikal pada}....\\ & \begin{array}{l}\\ \text{a}.& x=0\text{saja}& & & \\ \text{b}.& x=5\text{saja}& \\ \text{c}.& x=10\text{saja}\\ \text{d}.& x=0\text{dan}x=5\text{saja}\\ \text{e}.& x=0,x=5,\text{dan}x=10\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\mathbf{\text{Asimtot vertikal}}& (\mathbf{\text{tegak}})\text{diperoleh saat}\\ x^2-10x+25& =0\\ (x-5{)}^2& =0\\ x-5& =0\\ x& =5\\ \\ \text{Ilustrasinya gamba}& \text{rnya adalah sebagai berikut}:\end{array}\end{array}.$

Limit Fungsi Aljabar (XI Matematika Wajib)

Contoh Soal 3 Peluang Kejadian Majmuk

$\begin{array}{ll}11.& \text{Jika kejadian}A\text{dan}B\text{adalah dua kejadian}\\ & \text{dengan}P(A)={\displaystyle \frac{8}{15},P(B)={\displaystyle \frac{7}{12},\text{dan}}}\\ & P(A|B)={\displaystyle \frac{4}{7},\text{maka nilai}P(B|A)=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{8}{45}}& & & \text{d}.& {\displaystyle \frac{5}{8}}\\ \\ \text{b}.& {\displaystyle \frac{1}{3}}& \text{c}.& {\displaystyle \frac{3}{8}}& \text{e}.& {\displaystyle \frac{7}{15}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah contoh}\\ & \text{kejadian}\mathbf{\text{tidak saling bebas (bersyarat)}}.\\ & \text{Diketahui bahwa}\\ & P(A)={\displaystyle \frac{8}{15},P(B)={\displaystyle \frac{7}{12},P(A|B)={\displaystyle \frac{4}{7}}}}\\ & \text{Ditanyakan nilai}P(B|A)=...?\\ & \text{maka}\\ & \begin{array}{rl}P(A\cap B)& =P(A\cap B)\\ P(A)\times P(B|A)& =P(B)\times P(A|B)\\ P(B|A)& ={\displaystyle \frac{P(B)\times P(A|B)}{P(A)}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{7}{12}}\right)\times \left({\displaystyle \frac{4}{7}}\right)}{{\displaystyle \frac{8}{15}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{8}{15}}}}\\ & ={\displaystyle \frac{1}{3}\times \frac{15}{8}}\\ & ={\displaystyle \frac{5}{8}}\end{array}\end{array}\end{array}$

Contoh Soal 2 Peluang Kejadian Majmuk

$\begin{array}{ll}6.& \text{Sebuah kantong berisi 7 kelereng merah,}\\ & \text{5 kelereng hijau, dan 4 kelereng biru}\\ & \text{Diambil sebuah kelereng secara acak.}\\ & \text{Peluang yang terambil merah atau hijau}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{16}}& & & \text{d}.& {\displaystyle \frac{3}{4}}\\ \\ \text{b}.& {\displaystyle \frac{7}{16}}& \text{c}.& {\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle \frac{2}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah contoh}\\ & \text{kejadian}\mathbf{\text{saling lepas}}.\text{Misalkan}\\ & A=\text{kejadian terambil 1 kelereng merah}\\ & n(A)=C(7,1)=\left(\begin{array}{c}7\\ 1\end{array}\right)=7\\ & B=\text{kejadian terambil 1 kelereng hijau}\\ & n(B)=C(5,1)=\left(\begin{array}{c}5\\ 1\end{array}\right)=5\\ & S=\text{semua dianggap identik}\\ & n(S)=C(16,1)=\left(\begin{array}{c}16\\ 1\end{array}\right)=16\\ & \text{maka}\\ & P(A\cup B)=P(A)+P(B)\\ & ={\displaystyle \frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}}\\ & ={\displaystyle \frac{7}{16}+\frac{5}{16}=\frac{12}{16}={\displaystyle \frac{3}{4}}}\end{array}\end{array}$

$\begin{array}{ll}7.& \text{Dari 100 orang yang mengikuti kegiatan}\\ & \text{jalan santai terdapat 60 orang memakai}\\ & \text{topi dan 45 orang yang berkacamata.}\\ & \text{Peluang bahwa seorang yang dipilih dari}\\ & \text{kelompok orang itu memakai topi dan}\\ & \text{kacamata adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{20}}& & & \text{d}.& {\displaystyle \frac{11}{20}}\\ \\ \text{b}.& {\displaystyle \frac{2}{5}}& \text{c}.& {\displaystyle \frac{9}{20}}& \text{e}.& {\displaystyle \frac{3}{5}}\end{array}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah ilustrasi}\mathbf{\text{Diagram Venn}}\\ & \text{berikut ini}\\ & \begin{array}{|ccl|}\hline \begin{array}{|l|}\hline \text{S}=100\\ \hline\end{array}& & \\ & \text{A}\text{B}& \\ & \begin{array}{|lcr|}\hline 60-n& n& 45-n\\ \hline\end{array}& \\ & & \\ \hline\end{array}\\ & \begin{array}{rl}& \text{Kejadian di atas adalah contoh}\\ & \text{kejadian}\mathbf{\text{tidak saling lepas}}.\\ & A=\text{kejadian terpilih seorang bertopi}\\ & n(A)=C(60,1)=\left(\begin{array}{c}60\\ 1\end{array}\right)=60\\ & B=\text{kejadian terpilih seorang berkacamata}\\ & n(B)=C(45,1)=\left(\begin{array}{c}45\\ 1\end{array}\right)=45\\ & A\cap B=\text{terpilih seorang bertopi dan}\\ & \text{berkacamata}\\ & n(A\cap B)=x\\ & S=\text{semua dianggap identik}\\ & n(S)=C(100,1)=\left(\begin{array}{c}100\\ 1\end{array}\right)=100\\ & \text{maka}\\ & P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & {\displaystyle \frac{n(A\cup B)}{n(S)}={\displaystyle \frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-{\displaystyle \frac{n(A\cap B)}{n(S)}}}}\\ & {\displaystyle \frac{100}{100}=\frac{60}{100}+\frac{45}{100}-\frac{x}{100}}\\ & {\displaystyle \frac{x}{100}={\displaystyle \frac{105}{100}-\frac{100}{100}}}\\ & ={\displaystyle \frac{5}{100}=\frac{1}{20}}\end{array}\end{array}$

$\begin{array}{ll}8.& \text{Diketahui dua buah kotak A dan B}\\ & \text{berisi 5 bola putih dan 3 bola merah.}\\ & \text{Kotak B berisi 4 bola putih dan 2 bola}\\ & \text{merah. Jika diambil secara acak 1 kotak,}\\ & \text{kemudian diambil secara acak 1 bola dari}\\ & \text{kotak tersebut, maka peluang terambilnya}\\ & \text{bola putih adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{16}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle \frac{1}{3}}& \text{c}.& {\displaystyle \frac{7}{16}}& \text{e}.& {\displaystyle \frac{31}{48}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah kejadian}\\ & \mathbf{\text{saling lepas}}\text{dari dua kejadian Q}\\ & \text{dan R. Misalkan}:\\ & \text{Pada kotak A}\\ & Q=\text{Terambil 1 bola putih di kotak A}\\ & n(Q)=C(5,1)=\left(\begin{array}{c}5\\ 1\end{array}\right)=5\\ & S_Q=\text{Terambil 1 bola saja di kotak A}\\ & n(S_Q)=C(8,1)=\left(\begin{array}{c}8\\ 1\end{array}\right)=8\\ & \text{Pada kotak B}\\ & R=\text{Terambil 1 bola putih di kotak B}\\ & n(R)=C(4,1)=\left(\begin{array}{c}4\\ 1\end{array}\right)=4\\ & S_R=\text{Terambil 1 bola saja di kotak B}\\ & n(S_R)=C(6,1)=\left(\begin{array}{c}6\\ 1\end{array}\right)=6\\ & \text{Karena kejadian pengambilan sebuah}\\ & \text{bola putih di atas adalah dari pilihan}\\ & \text{dua buah kotak yang ada, maka peluang}\\ & \text{pengambilannya adalah 1 dari 2 kotak}\\ & \text{peluang kejadian ini adalah}={\displaystyle \frac{1}{2}.}\\ & \text{Sehingga peluang kasus di atas adalah}:\\ & {\displaystyle \frac{1}{2}P(Q\cup R)={\displaystyle \frac{1}{2}(P(Q)+P(R))}}\\ & ={\displaystyle \frac{1}{2}\left({\displaystyle \frac{n(Q)}{n(S_Q)}+\frac{n(R)}{n(S_R)}}\right)}\\ & ={\displaystyle \frac{1}{2}\left({\displaystyle \frac{5}{8}+\frac{4}{6}}\right)}\\ & ={\displaystyle \frac{1}{2}\left({\displaystyle \frac{31}{24}}\right)={\displaystyle \frac{31}{48}}}\end{array}\end{array}$

$\begin{array}{ll}9.& \text{Kotak I berisi 3 bola merah dan 2 bola }\\ & \text{putih. Kotak II berisi 3 bola hijau dan 5}\\ & \text{biru. Dari tiap-tiap kotak diambil 2 bola}\\ & \text{sekaligus secara acak. Peluang terambil 2}\\ & \text{bola merah pada kotak I dan 2 bola biru}\\ & \text{dari kotak II adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{10}}& & & \text{d}.& {\displaystyle \frac{3}{8}}\\ \\ \text{b}.& {\displaystyle \frac{3}{28}}& \text{c}.& {\displaystyle \frac{4}{15}}& \text{e}.& {\displaystyle \frac{57}{140}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah kejadian}\\ & \mathbf{\text{saling bebas}}\text{dari dua kejadian A}\\ & \text{dan B. Misalkan}:\\ & \text{Pada kotak I}\\ & A=\text{Terambil 2 bola merah di kotak I}\\ & n(A)=C(3,2)=\left(\begin{array}{c}3\\ 2\end{array}\right)=3\\ & S=\text{Terambil 2 bola saja di kotak I}\\ & n(S)=C(5,2)=\left(\begin{array}{c}5\\ 2\end{array}\right)=10\\ & \text{Pada kotak II}\\ & B=\text{Terambil 2 bola biru di kotak II}\\ & n(B)=C(5,2)=\left(\begin{array}{c}5\\ 2\end{array}\right)=10\\ & S=\text{Terambil 2 bola saja di kota II}\\ & n(S)=C(8,2)=\left(\begin{array}{c}8\\ 2\end{array}\right)=28\\ & \text{Sehingga peluang kasus di atas adalah}:\\ & {\displaystyle P(A\cap B)=P(A)\times P(B)}\\ & ={\displaystyle \frac{n(A)}{n(S)}\times \frac{n(B)}{n(S)}}\\ & ={\displaystyle \frac{3}{10}\times \frac{10}{28}}\\ & ={\displaystyle \frac{3}{28}}\end{array}\end{array}$

$\begin{array}{ll}10.& \text{Jika kejadian}A\text{dan}B\text{dapat terjadi secara}\\ & \text{bersamaan. Jika}P(A)=0,6,P(B)=0.75,\\ & \text{dan}P(A\cap B)=0,43,\text{maka}P(A\cup B)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,98}& & & \text{d}.& {\displaystyle 0,92}\\ \\ \text{b}.& {\displaystyle 0,96}& \text{c}.& {\displaystyle 0,94}& \text{e}.& {\displaystyle 0,91}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah contoh}\\ & \text{kejadian}\mathbf{\text{tidak saling lepas}}.\\ & \text{Diketahui bahwa}\\ & P(A)=0,6,P(B)=0,75,P(A\cap B)=0,43\\ & \text{Ditanyakan nilai}P(A\cup B)=...?\\ & \text{maka}\\ & P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & =0,6+0,75-0,43\\ & =0,92\end{array}\end{array}$

Contoh Soal 1 Peluang Kejadian Majmuk

$\begin{array}{ll}1.& \text{Banyak anggota ruang sampel dari}\\ & \text{pelemparan sebuah dadu dan dua }\\ & \text{keping mata uang secara bersamaan}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}& & & \text{d}.& {\displaystyle 24}\\ \\ \text{b}.& {\displaystyle 6}& \text{c}.& {\displaystyle 12}& \text{e}.& {\displaystyle 36}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}1& \text{mata dadu}=P(6,1)=6\\ 2& \text{keping mata uang}=P(2,1)\times P(2,1)=4\\ \text{R}& \text{uang sampelnya adalah}:6\times 4=24\end{array}\end{array}$

$\begin{array}{ll}2.& \text{Setumpuk kartu remi diambil sebuah}\\ & \text{kartu secara acak. Peluang agar kartu}\\ & \text{yang terambil bukan kartu king}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0}& & & \text{d}.& {\displaystyle {\displaystyle \frac{12}{13}}}\\ \\ \text{b}.& {\displaystyle {\displaystyle \frac{1}{13}}}& \text{c}.& {\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle 1}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misalkan}A=\text{Kejadian muncul kartu king}\\ & n(A)=\text{banyak kartu king ada}=4\\ & n(S)=\text{total kartu}=4\times 13\\ & A^{\prime }=\text{kejadian muncul bukan kartu king}\\ & \text{maka peluangnya bukan kartu king}:\\ & P(A^{\prime })=1-P(A)=1-{\displaystyle \frac{4}{4\times 13}=\frac{12}{13}}\end{array}\end{array}$

$\begin{array}{ll}3.& \text{Sebuah dadu dilempar sekali. Peluang}\\ & \text{muncul mata dadu 3 atau lebih adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{6}}}& & & \text{d}.& {\displaystyle {\displaystyle \frac{3}{5}}}\\ \\ \text{b}.& {\displaystyle {\displaystyle \frac{1}{3}}}& \text{c}.& {\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle \frac{2}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misal}A=\text{muncul mata dadu 3 atau lebih}\\ & A=\{3,4,5,6\}\\ & S=\{1,2,3,4,5,6\}\\ & \text{maka}n(A)=4\text{dengan}(S)=6\\ & P(A)={\displaystyle \frac{n(A)}{n(S)}=\frac{4}{6}=\frac{2}{3}}\end{array}\end{array}$

$\begin{array}{ll}4.& \text{Sebuah dadu dan sebuah mata uang logam}\\ & \text{dilempar bersama-sama. Peluang muncul}\\ & \text{gambar pada mata uang dan mata 1 pada}\\ & \text{dadu adalah}....\\ & \begin{array}{llllll}\text{a}.& {\displaystyle \frac{1}{12}}& & & \text{d}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{1}{6}}& \text{c}.& {\displaystyle \frac{1}{4}}& \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \mathbf{\text{Cara pertama}}\\ & \text{Perhatikan tabel berikut}\\ & \begin{array}{|ccccccc|}\hline ◻& 1& 2& 3& 4& 5& 6\\ A& (A,1)& (A,2)& (A,3)& (A,4)& (A,5)& (A,6)\\ G& (G,1)& (G,2)& (G,3)& (G,4)& (G,5)& (G,6)\\ \hline\end{array}\\ & \text{dari tabel di atas didapatkan bahwa}:\\ & A=\text{kejadian muncul mata 1 pada dadu}\\ & n(A)=2\\ & B=\text{kejadian muncul gambar pada uang}\\ & n(B)=6\\ & A\cap B=\text{kejadian muncul 1 pada dadu}\\ & \text{gambar pada koin}\\ & n(A\cap B)=1\\ & \text{dengan}n(S)=12,\\ & \text{maka peluang muncul mata 1 dan gambar}\\ & P(A\cap B)={\displaystyle \frac{n(A\cap B)}{n(S)}=\frac{1}{12}}\\ & \mathbf{\text{Cara kedua}}\\ & \text{Karena ini dua kejadian}\mathit{\text{saling bebas}}\\ & \text{maka}\\ & P(A\cap B)=P(A)\times P(B)\\ & ={\displaystyle \frac{n(A)}{n(S)}\times \frac{n(B)}{n(S)}}\\ & ={\displaystyle \frac{2}{12}\times \frac{6}{12}=\frac{12}{144}=\frac{1}{12}}\end{array}\end{array}$

$\begin{array}{ll}5.& \text{Peluang Dika lulus ujian adalah}0,75\text{dan}\\ & \text{peluang Tutik lulus ujian adalah}0,80.\\ & \text{Peluang keduanya lulus ujian adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,4}& & & \text{d}.& {\displaystyle 0,7}\\ \\ \text{b}.& {\displaystyle 0,5}& \text{c}.& {\displaystyle 0,6}& \text{e}.& {\displaystyle 0,8}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Dua kejadian ini adalah}\mathit{\text{saling bebas}}\\ & \text{Misal}A=\text{Kejadian Dika lulus}\\ & n(A)=\cdots \text{tidak diketahui, tetapi}\\ & P(A)=0,75={\displaystyle \frac{3}{4},\mathbf{\text{diketahui}}}\\ & \text{dan}B=\text{Tutik lulus}\\ & n(B)=\cdots \text{juga tidak diketahui}\\ & P(B)=0,8={\displaystyle \frac{4}{5}}\\ & P(A\cap B)=P(A)\times P(B)\\ & ={\displaystyle \frac{n(A)}{n(S)}\times \frac{n(B)}{n(S)}}\\ & ={\displaystyle \frac{3}{4}\times \frac{4}{5}=\frac{3}{5}=0,6}\end{array}\end{array}$