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Quiz (+50): fungsi, invers, integral, luas daerah ada di gambar soalnya

Berikut ini adalah pertanyaan dari unknown pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Quiz (+50):
fungsi, invers, integral, luas daerah
ada di gambar soalnya
Quiz (+50): fungsi, invers, integral, luas daerah ada di gambar soalnya

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban:

420 SL

Pembahasan

Menentukan invers

\begin{aligned}{\Bigl.}y=f(x)&=\frac{x^2-36x+324}{6}\\{\Biggl.}&=\frac{(x-18)^2}{6}\\{\Biggl.}x-18&=\pm\sqrt{6y}\\{\Bigl.}x&=18\pm\sqrt{6y}\\\\\therefore\ f^{-1}(x)&=\begin{cases}{\Bigl.}\bf g(x)=18+\sqrt{6x}\\{\Bigl.}\bf h(x)=18-\sqrt{6x}\end{cases}\end{aligned}

Menentukan titik potong

\begin{aligned}f^{-1}(x)&=f(x)\\18\pm\sqrt{6x}&=\frac{x^2-36x+324}{6}\\108\pm6\sqrt{6x}&=x^2-36x+324\\{}\pm6\sqrt{6x}&=x^2-36x+216\\36\cdot6x&=\left(x^2-36x+216\right)^2\\216x&=x^4-72x^3+1728x^2-15552x+46656\\0&=x^4-72x^3+1728x^2-15768x+46656\quad....(i)\end{aligned}

\begin{aligned}(i):\ &\bold{(x-6)}\underbrace{\left(x^3-66x^2+1332x-7776\right)}_{\begin{array}{c}(ii)\end{array}}=0\\\\(ii):\ &x^3-66x^2+1332x-7776=0\\&\bold{(x-24)}\underbrace{\left(x^2-42x+324\right)}_{\begin{array}{c}(iii)\end{array}}=0\\\\(iii):\ &x^2-42x=-324\\&x^2-42x+441=-324+441\\&(x-21)^2=117\\&x-21={}\pm\sqrt{117}={}\pm\sqrt{9\cdot13}\\&x=21\pm3\sqrt{13}=\bf3\left(7\pm\sqrt{13}\right)\end{aligned}

Jadi, absis dari titik-titik potongnya adalah:

6, 3(7–√13), 24, dan 3(7+√13)

Menentukan Luas Daerah

  • Absis titik potong terluar adalah x = 6 dan x = 3(7+√13), untuk g(x) dan f(x).
  • Absis titik potong terdalam adalah x = 3(7–√13) dan x = 24, untuk h(x) dan f(x).

\begin{aligned}\Biggl.L&=\overbrace{\int_6^{3b}{|g(x)|\,dx}}^{\begin{array}{c}L_1\end{array}}\\&\quad-\Bigg(\underbrace{\int_{3a}^{24}{|h(x)|\,dx}}_{\begin{array}{c}L_2\end{array}}+\underbrace{\int_6^{3a}{|f(x)|\,dx}}_{\begin{array}{c}L_3\end{array}}+\underbrace{\int_{24}^{3b}{|f(x)|\,dx}}_{\begin{array}{c}L_4\end{array}}\Bigg)\\&\textsf{dengan $a=7-\sqrt{13}\ $ dan $\ b=7+\sqrt{13}$}\end{aligned}

\begin{aligned}&\implies a+b=\bf14\\&\implies a-b=\bf-2\sqrt{13}\\&\implies b-a=\bf2\sqrt{13}\\&\implies ab=49-13=\bf36\\&\implies a^2+b^2=14^2-2(36)=\bf124\\&\implies a^3+b^3=14^3-3(36)(14)=\bf1232\end{aligned}

\begin{aligned}L_1&=\int_6^{3b}{|g(x)|\,dx}\\&=\int_6^{3b}\left|18+\sqrt{6x}\right|\,dx\\&=\int_6^{3b}18\,dx+\int_6^{3b}\sqrt{6}\sqrt{x}\,dx\\&=18\Big[x\Big]_6^{3b}+\frac{2\sqrt{6}}{3}\left[x^{\frac{3}{2}}\right]_6^{3b}\\&=54b-108+\frac{2\sqrt{6}(3b)^{\frac{3}{2}}}{3}-\frac{2\sqrt{6}\left(6^{\frac{3}{2}}\right)}{3}\\&=-108+54b+2\sqrt{6}\sqrt{3}\cdot b^{\frac{3}{2}}-24\\\Biggl.L_1&=\bf-132+54b+6\sqrt{2}\cdot b^{\frac{3}{2}}\end{aligned}

\begin{aligned}\Biggl.L_2&=\int_{3a}^{24}{|h(x)|\,dx}\\&=\int_{3a}^{24}\left|18-\sqrt{6x}\right|\,dx\\&=\int_{3a}^{24}18\,dx-\int_{3a}^{24}\sqrt{6}\sqrt{x}\,dx\\&=18\Big[x\Big]_{3a}^{24}-\frac{2\sqrt{6}}{3}\left[x^{\frac{3}{2}}\right]_{3a}^{24}\\&=432-54a-\frac{2\sqrt{6}}{3}\left(24^{\frac{3}{2}}\right)+\frac{2\sqrt{6}}{3}(3a)^{\frac{3}{2}}\\&=432-54a-\frac{2(36)(8)}{3}+6\sqrt{2}\cdot a^{\frac{3}{2}}\\\Biggl.L_2&=\bf240-54a+6\sqrt{2}\cdot a^{\frac{3}{2}}\end{aligned}

\begin{aligned}\Biggl.L_3&=\int_6^{3a}{|f(x)|\,dx}\\&=\int_6^{3a}{\left|\frac{x^2-36x+324}{6}\right|\,dx}\\&=\frac{1}{6}\cdot\int_6^{3a}\left(x^2-36x+324\right)dx\\&=\frac{1}{6}\left[\frac{x^3}{3}-18x^2+324x\right ]_6^{3a}\\&=\frac{1}{6}\left(9a^3-72-162a^2+648+972a-1944\right)\\&=\frac{1}{6}\left(-1368+9a^3-162a^2+972a\right)\\\Biggl.L_3&=\bf\frac{3a^3}{2}-27a^2+162a-228\end{aligned}

\begin{aligned}\Biggl.L_4&=\int_{24}^{3b}{|f(x)|\,dx}\\&=\frac{1}{6}\cdot\int_{24}^{3b}\left(x^2-36x+324\right)dx\\&=\frac{1}{6}\left[\frac{x^3}{3}-18x^2+324x\right ]_{24}^{3b}\\&=\frac{1}{6}\left(9b^3-4608-162b^2+10368+972b-7776\right)\\&=\frac{1}{6}\left ( -2016+9b^3-162b^2+972b \right )\\\Biggl. L_4&=\bf\frac{3b^3}{2}-27b^2+162b-336\end{aligned}

LUAS DAERAH:

\begin{aligned}L&=L_1-(L_2+L_3+L_4)\\&=L_1-L_2-(L_3+L_4)\\L&=-132+54b+6\sqrt{2}\cdot b^{\frac{3}{2}}-\left(240-54a+6\sqrt{2}\cdot a^{\frac{3}{2}}\right)\\&\quad-\left(\frac{3a^3}{2}-27a^2+162a-228 + \left( \frac{3b^3}{2}-27b^2+162b-336 \right)\right)\\L&=-372+54(b+a)+6\sqrt{2}\left(b^{\frac{3}{2}}-a^{\frac{3}{2}}\right)\\&\quad-\left( \frac{3\left(a^3+b^3\right)}{2} - 27\left(a^2+b^2\right) + 162(a+b) - 564 \right)\end{aligned}

\begin{aligned}L&=-372+54(14)+6\sqrt{2}\left(b^{\frac{3}{2}}-a^{\frac{3}{2}}\right)\\&\quad-\left( \frac{3(1232)}{2} - 27(124) + 162(14) - 564 \right)\\L&=-372+756+6\sqrt{2}\left(b^{\frac{3}{2}}-a^{\frac{3}{2}}\right)\\&\quad-(1848-3348+2268-564)\end{aligned}

\begin{aligned}L&=384+6\sqrt{2}\left(b^{\frac{3}{2}}-a^{\frac{3}{2}}\right)-204\\&=180+6\sqrt{2}\left(b^{\frac{3}{2}}-a^{\frac{3}{2}}\right)\\&=180+6\sqrt{2}\sqrt{\left(b^{\frac{3}{2}}-a^{\frac{3}{2}}\right)^2}\\&=180+6\sqrt{2}\sqrt{b^3+a^3-2(ab)^{\frac{3}{2}}}\\&=180+6\sqrt{2}\sqrt{1232-2\left(36^{\frac{3}{2}}\right)}\\&=180+6\sqrt{2}\sqrt{1232-2\cdot6^3}\\&=180+6\sqrt{2}\sqrt{1232-432}\\&=180+6\sqrt{2}\sqrt{800}\\&=180+6\sqrt{2}\cdot20\sqrt{2}\\&=180+240\\&=\bf420\ SL\end{aligned}

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Last Update: Fri, 03 Jun 22
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