$\begin{array}{ll}\\ 16.&\textrm{Batas nilai}\: \: m\: \: \textrm{yang memenuhi agar fungsi}\\ &\textrm{kuadrat}\: \: y=mx^{2}+(m+2)x+m\: \: \\ &\textrm{memotong sumbu-X di dua titik yang}\\ &\textrm{berbeda adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\color{red}\displaystyle -\displaystyle \frac{2}{3}<m<2&&&&&\\ \textrm{B}.&\displaystyle -2<m<\displaystyle \frac{2}{3}&&\\ \textrm{C}.&m<-2\: \: \textrm{atau}\: \: m>2\\ \textrm{D}.&m<-2\: \: \textrm{atau}\: \: m>\displaystyle \frac{2}{3}\\ \textrm{E}.&m<-\displaystyle \frac{2}{3}\: \: \textrm{atau}\: \: m>2 \end{array}\\\\ &\textbf{Jawab}:\textbf{A}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: y=mx^{2}+(m+2)x+m\\ &\textrm{dengan}\: \: a=m,\: b=m+2,\: \&\: \: c=m\\ &\textrm{memotong sumbu-X di dua titik berbeda}\\ &\textrm{hal ini artinya nilai Diskriminan }D>0\\ &D=b^{2}-4ac>0\\ &(m+2)^{2}-4m.m>0\\ &\Leftrightarrow m^{2}+4m+4-4m^{2}>0\\ &\Leftrightarrow -3m^{2}+4m+4>0\: \: \: (\textrm{dikali} -1)\\ &\Leftrightarrow 3x^{2}-4m-4<0\\ &\Leftrightarrow (m-2)(3m+2)\color{red}<0\\ &\Leftrightarrow \color{red}-\displaystyle \frac{2}{3}<m<2\\ &\textbf{Anda bisa menggunakan titik uji dulu}\\ &\textrm{untuk memastikannya wilayah yg dimaksud}\\ &\textrm{misalkan pilih}\: \: m=0\\ &\textrm{lalu kita ujikan, yaitu}:\\ &m=0\Rightarrow (0-2)(3.0+2)=-4\color{red}<0\: \: (\textbf{benar})\\\\ &\textrm{Jika kita diletakkan dalam garis bilangan}\\ &\textrm{akan tampak seperti berikut}\\\\ &\begin{aligned}&\underset{\begin{matrix} \color{red}\Downarrow\\ \color{red}\Downarrow \end{matrix} }{\textrm{Daerah Positif}}\\ &\begin{array}{cc|ccc|ccc}\\ &&&&&&\\ +\: +&+\: +&-\: -&-\: -&-\: -&+\: +&+\: +\\\hline &-\displaystyle \frac{2}{3}&&&&2&&\textbf{Sumbu}-m\\ \end{array}\\ &\begin{matrix} \textrm{dengan}\\ \textrm{titik}\: \: \: \: \: \: \\ \textrm{uji}\qquad\\ \color{red}m=0\quad \end{matrix}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah negatif}}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah positif}} \end{aligned} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 17.&\textrm{Batas nilai}\: \: m\: \: \textrm{yang menyebabkan fungsi kuadrat}\\ & y=(m-1)x^{2}-2(m-1)x+(2m+1)\: \: \\ &\textrm{definit positif adalah}\: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle m>-2&&&&&\\ \textrm{B}.&\color{red}\displaystyle m>1&&\\ \textrm{C}.&m<1\\ \textrm{D}.&-2<m<1\\ \textrm{E}.&m<-\displaystyle 2\: \: \textrm{atau}\: \: m>1 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\begin{aligned}&\textrm{Diketahui FK}:\\ &y=(m-1)x^{2}-2(m-1)x+(2m+1)\\ &\textrm{Syarat fungsi kuadrat definit positif}:\\ &\bullet\quad a>0\\ &\bullet \quad D=b^{2}-4ac<0\\ &\textrm{maka}\\ &a=m-1>0\Leftrightarrow m>1,\: \: \textrm{dan}\\ &D=(2(m-1))^{2}-4(m-1)(2m+1)<0\\ &\Leftrightarrow 4(m-1)^{2}-(m-1)(8m+4)<0\\ &\Leftrightarrow (m-1)(4m-4)-(m-1)(8m+4)<0\\ &\Leftrightarrow (m-1)(4m-4-8m-4)<0\\ &\Leftrightarrow (m-1)(-4m-8)<0\: ,\: \textbf{dibagi}\: \: -4\\ &\Leftrightarrow (m-1)(m+2)\color{red}>0 \end{aligned}\\\\ &\textrm{Jika kita diletakkan dalam garis bilangan}\\ &\textrm{akan tampak seperti berikut}\\\\ &\begin{aligned}&\underset{\begin{matrix} \color{red}\Downarrow\\ \color{red}\Downarrow \end{matrix} }{\textrm{Daerah Positif}}\\ &\begin{array}{cc|ccc|ccc}\\ &&&&&&\\ +\: +&+\: +&-\: -&-\: -&-\: -&+\: +&+\: +\\\hline &-\displaystyle 2&&&&1&&\textbf{Sumbu}-m\\ \end{array}\\ &\begin{matrix} \textrm{dengan}\\ \textrm{titik}\: \: \: \: \: \: \\ \textrm{uji}\qquad\\ \color{red}m=0\quad \end{matrix}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah negatif}}\qquad \overset{\begin{matrix} \color{red}\Uparrow\\ \color{red}\Uparrow \end{matrix}}{\textrm{Daerah positif}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 18.&\textrm{Luas}\: \: L\: \: \textrm{suatu segitiga}\: \: ABC\: \: \textrm{diketahui}\\ & x(7-x)\: \: \textrm{cm}^{2}.\: \textrm{Luas maksimum segitiga} \\ &\textrm{tersebut adalah}\: \: ...\: \: \textrm{cm}^{2}\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 3\displaystyle \frac{1}{2}&&&&&\textrm{D}.&\displaystyle 10\displaystyle \frac{1}{4}\\\\ \textrm{B}.&\displaystyle 5\displaystyle \frac{1}{2}&&\textrm{C}.&\displaystyle 7\displaystyle \frac{1}{4}&&\textrm{E}.&\color{red}\displaystyle 12\displaystyle \frac{1}{4} \end{array}\\\\ &\textbf{Jawab}:\textbf{E}\\ &\textrm{Diketahui}\: \: \left [ ABC \right ]=x(7-x)=7x-x^{2}\\ &\textrm{dengan}\: \: a=-1,\: \: b=7,\: \textrm{dan}\: \: c=0\\ &\textrm{akan}\: \textbf{maksimum},\: \textrm{saat}\: \: \left (x_{ss},f(x_{ss})\right )\\ &\textrm{yaitu}:\\ &x_{ss}=-\displaystyle \frac{b}{2a}=-\frac{7}{2.(-1)}=\frac{7}{2},\: \: \textrm{maka}\\ &f\left ( \displaystyle \frac{7}{2} \right )=7\left ( \displaystyle \frac{7}{2} \right )-\left ( \displaystyle \frac{7}{2} \right )^{2}=\displaystyle \frac{49}{2}-\frac{49}{4}=\color{red}\displaystyle \frac{49}{4} \end{array}$.
$\begin{array}{ll}\\ 19.&\textrm{Perhatikan gambar persegi ABCD dengan}\\ &\textrm{panjang sisinya 10 cm} \end{array}$.
$.\quad\begin{array}{ll}\\ &\textrm{Jika}\: \: BP=DQ=x\: \: \textrm{cm, maka luas }\\ &\textrm{maksimum segitiga}\: \: APQ\: \: \textrm{adalah}\: ...\: \: \textrm{cm}^{2}\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle 35&&&&&\textrm{D}.&\displaystyle 75\\ \textrm{B}.&\color{red}\displaystyle 50&&\textrm{C}.&\displaystyle 60&&\textrm{E}.&80 \end{array}\\\\ &\textbf{Jawab}:\textbf{B}\\ &\textrm{Diketahui bahwa}\\ &\begin{aligned}\left [ APQ \right ]&=L_{\square ABCD}-\left [ ADQ \right ]-\left [ QCP \right ]-\left [ APB \right ]\\ &=10^{2}-\displaystyle \frac{1}{2}x.10-\frac{1}{2}.(10-x)^{2}-\frac{1}{2}x.10\\ &=100-10x-\displaystyle \frac{1}{2}(10-x)^{2}\\ &=10(10-x)-\displaystyle \frac{1}{2}(10-x)^{2}\\ &=(10-x)\left ( 10-\displaystyle \frac{1}{2}(10-x) \right )\\ &=(10-x)\left ( 5+\displaystyle \frac{1}{2}x \right )\\ &=50-\displaystyle \frac{1}{2}x^{2} \end{aligned}\\ &x_{ss}=-\displaystyle \frac{b}{2a}=-\displaystyle \frac{0}{2.\displaystyle \frac{1}{2}}=0,\: \: \textrm{maka nilai}\\ &f(x_{ss})=f(0)=50-\displaystyle \frac{1}{2}.0^{2}=\color{red}50 \end{array}$.
$\begin{array}{ll}\\ 20.&\textrm{Fungsi}\: \: f(x)=a(b-c)x^{2}+b(c-a)x+c(a-b)\\ &\textrm{akan memotong sumbu-X di titik}\: \: (1,0)\: \: \textrm{dan}\\ &\textrm{memenuhi}\: \: ....\\ &\begin{array}{lllllllll}\textrm{A}.&\displaystyle \displaystyle \frac{b(c-a)}{a(b-c)}&&&&&\textrm{D}.&\color{red}\displaystyle \displaystyle \frac{c(a-b)}{a(b-c)}\\\\ \textrm{B}.&\displaystyle \displaystyle \frac{a(b-c)}{c(a-b)}&&\textrm{C}.&\displaystyle \displaystyle \frac{a(b-c)}{b(c-a)}&&\textrm{E}.&\displaystyle \displaystyle \frac{c(a-b)}{b(c-a)} \end{array}\\\\ &\textbf{Jawab}:\textbf{D}\\ &\textrm{Diketahui bahwa FK}:\\ &f(x)=a(b-c)x^{2}+b(c-a)x+c(a-b)\\ &\textrm{maka nilai}\\ &\bullet \quad x_{1}+x_{2}=-\displaystyle \frac{b(c-a)}{a(b-c)}=\displaystyle \frac{b(a-c)}{a(b-c)}\\ &\bullet \quad x_{1}\times x_{2}=\color{red}\displaystyle \frac{c(a-b)}{a(b-c)} \end{array}$
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