Contoh Soal 4 Fungsi Kuadrat (Kelas X/Fase E Semester Genap) Tahun 2024

$\begin{array}{l}\\ 16.& \text{Batas nilai}m\text{yang memenuhi agar fungsi}\\ & \text{kuadrat}y=m{x}^2+(m+2)x+m\\ & \text{memotong sumbu-X di dua titik yang}\\ & \text{berbeda adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle -{\displaystyle \frac{2}{3}<m<2}}& & & & & \\ \text{B}.& {\displaystyle -2<m<{\displaystyle \frac{2}{3}}}& & \\ \text{C}.& m<-2\text{atau}m>2\\ \text{D}.& m<-2\text{atau}m>{\displaystyle \frac{2}{3}}\\ \text{E}.& m<-{\displaystyle \frac{2}{3}\text{atau}m>2}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{A}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}y=m{x}^2+(m+2)x+m\\ & \text{dengan}a=m,b=m+2,\mathrm{\&}c=m\\ & \text{memotong sumbu-X di dua titik berbeda}\\ & \text{hal ini artinya nilai Diskriminan }D>0\\ & D=b^2-4ac>0\\ & (m+2{)}^2-4m.m>0\\ & \iff m^2+4m+4-4m^2>0\\ & \iff -3m^2+4m+4>0(\text{dikali}-1)\\ & \iff 3x^2-4m-4<0\\ & \iff (m-2)(3m+2)<0\\ & \iff -{\displaystyle \frac{2}{3}<m<2}\\ & \mathbf{\text{Anda bisa menggunakan titik uji dulu}}\\ & \text{untuk memastikannya wilayah yg dimaksud}\\ & \text{misalkan pilih}m=0\\ & \text{lalu kita ujikan, yaitu}:\\ & m=0\Rightarrow (0-2)(3.0+2)=-4<0(\mathbf{\text{benar}})\\ \\ & \text{Jika kita diletakkan dalam garis bilangan}\\ & \text{akan tampak seperti berikut}\\ \\ & \begin{array}{rl}& \underset{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\text{Daerah Positif}}\\ & \begin{array}{c}\\ & & & & & & \\ ++& ++& --& --& --& ++& ++\\ & -{\displaystyle \frac{2}{3}}& & & & 2& & \mathbf{\text{Sumbu}}-m\end{array}\\ & \begin{array}{c}\text{dengan}\\ \text{titik}\\ \text{uji}\\ m=0\end{array}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah negatif}}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah positif}}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Batas nilai}m\text{yang menyebabkan fungsi kuadrat}\\ & y=(m-1)x^2-2(m-1)x+(2m+1)\\ & \text{definit positif adalah}....\\ & \begin{array}{lllllll}\text{A}.& {\displaystyle m>-2}& & & & & \\ \text{B}.& {\displaystyle m>1}& & \\ \text{C}.& m<1\\ \text{D}.& -2<m<1\\ \text{E}.& m<-{\displaystyle 2\text{atau}m>1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \begin{array}{rl}& \text{Diketahui FK}:\\ & y=(m-1)x^2-2(m-1)x+(2m+1)\\ & \text{Syarat fungsi kuadrat definit positif}:\\ & \bullet a>0\\ & \bullet D=b^2-4ac<0\\ & \text{maka}\\ & a=m-1>0\iff m>1,\text{dan}\\ & D=(2(m-1){)}^2-4(m-1)(2m+1)<0\\ & \iff 4(m-1{)}^2-(m-1)(8m+4)<0\\ & \iff (m-1)(4m-4)-(m-1)(8m+4)<0\\ & \iff (m-1)(4m-4-8m-4)<0\\ & \iff (m-1)(-4m-8)<0,\mathbf{\text{dibagi}}-4\\ & \iff (m-1)(m+2)>0\end{array}\\ \\ & \text{Jika kita diletakkan dalam garis bilangan}\\ & \text{akan tampak seperti berikut}\\ \\ & \begin{array}{rl}& \underset{\begin{array}{c}\Downarrow \\ \Downarrow \end{array}}{\text{Daerah Positif}}\\ & \begin{array}{c}\\ & & & & & & \\ ++& ++& --& --& --& ++& ++\\ & -{\displaystyle 2}& & & & 1& & \mathbf{\text{Sumbu}}-m\end{array}\\ & \begin{array}{c}\text{dengan}\\ \text{titik}\\ \text{uji}\\ m=0\end{array}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah negatif}}\stackrel{\begin{array}{c}\Uparrow \\ \Uparrow \end{array}}{\text{Daerah positif}}\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Luas}L\text{suatu segitiga}ABC\text{diketahui}\\ & x(7-x){\text{cm}}^2.\text{Luas maksimum segitiga}\\ & \text{tersebut adalah}...{\text{cm}}^2\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 3{\displaystyle \frac{1}{2}}}& & & & & \text{D}.& {\displaystyle 10{\displaystyle \frac{1}{4}}}\\ \\ \text{B}.& {\displaystyle 5{\displaystyle \frac{1}{2}}}& & \text{C}.& {\displaystyle 7{\displaystyle \frac{1}{4}}}& & \text{E}.& {\displaystyle 12{\displaystyle \frac{1}{4}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{E}}\\ & \text{Diketahui}\left[ABC\right]=x(7-x)=7x-x^2\\ & \text{dengan}a=-1,b=7,\text{dan}c=0\\ & \text{akan}\mathbf{\text{maksimum}},\text{saat}(x_{ss},f(x_{ss}))\\ & \text{yaitu}:\\ & x_{ss}=-{\displaystyle \frac{b}{2a}=-\frac{7}{2.(-1)}=\frac{7}{2},\text{maka}}\\ & f\left({\displaystyle \frac{7}{2}}\right)=7\left({\displaystyle \frac{7}{2}}\right)-{\left({\displaystyle \frac{7}{2}}\right)}^2={\displaystyle \frac{49}{2}-\frac{49}{4}={\displaystyle \frac{49}{4}}}\end{array}$.

$\begin{array}{l}\\ 19.& \text{Perhatikan gambar persegi ABCD dengan}\\ & \text{panjang sisinya 10 cm}\end{array}$.

$.\begin{array}{l}\\ & \text{Jika}BP=DQ=x\text{cm, maka luas }\\ & \text{maksimum segitiga}APQ\text{adalah}...{\text{cm}}^2\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle 35}& & & & & \text{D}.& {\displaystyle 75}\\ \text{B}.& {\displaystyle 50}& & \text{C}.& {\displaystyle 60}& & \text{E}.& 80\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{B}}\\ & \text{Diketahui bahwa}\\ & \begin{array}{rl}\left[APQ\right]& =L_{◻ABCD}-\left[ADQ\right]-\left[QCP\right]-\left[APB\right]\\ & ={10}^2-{\displaystyle \frac{1}{2}x.10-\frac{1}{2}.(10-x{)}^2-\frac{1}{2}x.10}\\ & =100-10x-{\displaystyle \frac{1}{2}(10-x{)}^2}\\ & =10(10-x)-{\displaystyle \frac{1}{2}(10-x{)}^2}\\ & =(10-x)(10-{\displaystyle \frac{1}{2}(10-x)})\\ & =(10-x)(5+{\displaystyle \frac{1}{2}x})\\ & =50-{\displaystyle \frac{1}{2}{x}^2}\end{array}\\ & x_{ss}=-{\displaystyle \frac{b}{2a}=-{\displaystyle \frac{0}{2.{\displaystyle \frac{1}{2}}}=0,\text{maka nilai}}}\\ & f(x_{ss})=f(0)=50-{\displaystyle \frac{1}{2}{.0}^2=50}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Fungsi}f(x)=a(b-c)x^2+b(c-a)x+c(a-b)\\ & \text{akan memotong sumbu-X di titik}(1,0)\text{dan}\\ & \text{memenuhi}....\\ & \begin{array}{llllllll}\text{A}.& {\displaystyle {\displaystyle \frac{b(c-a)}{a(b-c)}}}& & & & & \text{D}.& {\displaystyle {\displaystyle \frac{c(a-b)}{a(b-c)}}}\\ \\ \text{B}.& {\displaystyle {\displaystyle \frac{a(b-c)}{c(a-b)}}}& & \text{C}.& {\displaystyle {\displaystyle \frac{a(b-c)}{b(c-a)}}}& & \text{E}.& {\displaystyle {\displaystyle \frac{c(a-b)}{b(c-a)}}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{D}}\\ & \text{Diketahui bahwa FK}:\\ & f(x)=a(b-c)x^2+b(c-a)x+c(a-b)\\ & \text{maka nilai}\\ & \bullet x_1+x_2=-{\displaystyle \frac{b(c-a)}{a(b-c)}={\displaystyle \frac{b(a-c)}{a(b-c)}}}\\ & \bullet x_1\times x_2={\displaystyle \frac{c(a-b)}{a(b-c)}}\end{array}$