Soal dan Pembahasan Matematika IPA SBMPTN 2016 (Kode 220)

Bimbel WIN:

Belajar dari bentuk soal yang sudah pernah ditanyakan membuat persiapan menghadapi ujian yang sebenarnya akan menjadi lebih terarah, lebih fokus dan lebih efektif.

Bentuk soal yang akan diujikan dari tahun ke tahun pada umumnya materinya sama. Pada pelajaran yang menitikberatkan pada hafalan soanya bisa sangat mirip bahkan ada yang persis sama. Sedangkan pada soal hitungan, rumus dan analisanya pada umunya sama.

Oleh karena itu, kami menyarankan bagiadik-adik calon mahasiswa baru (camaba) tahun ini, kuasailah minimal 10 tahun terakhir soal ujian yang sudah pernah keluar.

Pada kesempatan ini, bimbel WIN berbagi soal asli matematika IPA SBMPTN tahun 2016 (Kode 220) lengkap dengan pembahasannya yang mudah untuk dimengerti. Di akhir pembahasan, kami juga mengajak adik-adik camaba untuk tetap berlatih pada soal online yang sudah kami siapkan, Ayouk teruslah berlatih...!!! Semoga tahun ini kalian semuanya yang belajar disini bisa lolos di pilihan pertama kalian, Amiiin... 🙏🙏

- Matematika IPA -

💦 Soal No. 01.

Dua lingkaran mempunyai titik pusat berjarak 25 satuan dan garis singgung persekutuan dalamnya y = 4. Jika lingkaran pertama mempunyai persamaan \({x^2} + {y^2} + 8x - 4y + 16 = 0\), maka persamaan lingkaran kedua yang berpusat di kuadran 1 dengan jari-jari 5 adalah...

(A) \({(\,x - 18\,)^2} + {(\,y - 9\,)^2} = 25\)
(B) \({(\,x - 10\,)^2} + {(\,y - 9\,)^2} = 25\)
(C) \({(\,x - 20\,)^2} + {(\,y - 9\,)^2} = 25\)
(D) \({(\,x - 24\,)^2} + {(\,y - 5\,)^2} = 25\)
(E) \({(\,x - 20\,)^2} + {(\,y - 7\,)^2} = 25\)

\(\begin{array}{l}{{\rm{L}}_1} \equiv {x^2} + {y^2} + 8x - 4y + 16 = 0\\{{\rm{P}}_1}\left( { - 4,2} \right)\,\,\,dan\,\,\,r = \sqrt {16 + 4 - 16} = 2\,\,\\{\rm{Jarak}}\,\,{\rm{antara}}\,\,{\rm{pusat}}\,\,{L_1}\,\,dan\,\,{L_2}\\ = jarak\,\,dua\,\,titik\,\,\, = \,\,25\\\sqrt {{{\left( {a + 4} \right)}^2} + {{\left( {9 - 2} \right)}^2}} = 25\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {a + 4} \right)^2} + 49 = 625\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {a + 4} \right)^2} = 576\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a + 4 = \pm 24\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 20\,\,\left( {{\rm{kuadran}}\,\,{\rm{I}} = a > 0} \right)\\{\rm{Maka}}\,\,{\rm{lingkaran}}\,\,\,{\rm{kedua}}\,\,\,{\rm{dapat}}\,\,\,{\rm{kita tentukan}}\\{\rm{Pusatnya }}{{\rm{P}}_{\rm{2}}}{\rm{ = (20}}{\rm{,0) dan }}\\{\rm{Jari - jarinya }}{{\rm{r}}_{{\rm{ 2}}}}{\rm{ = 5}}\\{\rm{maka persamaan }}{{\rm{L}}_2}\\ \equiv {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\\\,\,\,\,\,\,\,\,\,\,\,{\left( {x - 20} \right)^2} + {\left( {y - 9} \right)^2} = {5^2}\\\,\,\,\,\,\,\,\,\,\,\,{\left( {x - 20} \right)^2} + {\left( {y - 9} \right)^2} = 25\\Kunci\,\,C\end{array}\)

💦 Soal No. 02.

Diketahui segitiga ABC dan \(\angle C = {90^{\,o\,}}\). Titik D pada sisi miring AB dan titik E pada AC sehingga AD : BD = AE : EC = 1 : 2 jika p = tan B, maka \(\tan \angle ADC = ...\)

(A) \(\frac{{2p}}{{1 - {p^2}}}\)
(B) \(\frac{{3p}}{{1 - 2{p^2}}}\)
(C) \(\frac{{3p}}{{1 + 2{p^2}}}\)
(D) \(\frac{{2p}}{{1 + {p^2}}}\)
(E) \(\frac{p}{{1 - {p^2}}}\)

\(\begin{array}{*{20}{l}}{dari{\mkern 1mu} {\mkern 1mu} gambar{\mkern 1mu} {\mkern 1mu} diatas:}\\{ED:CB = AE:AC}\\{ED:CB = y:3y}\\{CB = 3ED}\\{ \Rightarrow \tan {\mkern 1mu} {\mkern 1mu} {\rm{B}} = {\rm{P}}}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{{\rm{AC}}}}{{{\rm{CB}}}} = {\rm{P}}}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{3{\rm{y}}}}{{3{\rm{ED}}}} = {\rm{P}}}\\{ \Rightarrow \tan {\mkern 1mu} {\mkern 1mu} \angle {\mkern 1mu} {\rm{ADC}} = \tan \left( {\angle {\mkern 1mu} {\rm{EDC}} + \angle {\mkern 1mu} {\rm{ABC}}} \right)}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \frac{{\tan {\mkern 1mu} {\mkern 1mu} \angle {\mkern 1mu} {\rm{EDC}} + \tan {\mkern 1mu} {\mkern 1mu} {\rm{B}}}}{{1 - \tan {\mkern 1mu} {\mkern 1mu} \angle {\mkern 1mu} {\rm{EDC}} \cdot \tan {\mkern 1mu} {\mkern 1mu} {\rm{B}}}}}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,{\mkern 1mu} = \frac{{\frac{{{\rm{EC}}}}{{{\rm{ED}}}} + P}}{{1 - \frac{{{\rm{EC}}}}{{{\rm{ED}}}} \cdot P}}}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \frac{{\frac{{2y}}{{{\rm{ED}}}} + {\rm{P}}}}{{1 - \frac{{2y}}{{{\rm{ED}}}} \cdot {\rm{P}}}}}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \frac{{2{\rm{P}} + {\rm{P}}}}{{1 - 2{\rm{P}} \cdot {\rm{P}}}}}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \frac{{3{\rm{P}}}}{{1 - 2{\rm{P}}{{\mkern 1mu} ^2}}}}\\{Kunci{\mkern 1mu} {\mkern 1mu} B}\end{array}\)

💦 Soal No 03.

Diketahui \(4{\cos ^2}t + 3\sec t = 3 + 4\cos t\) dengan \(0 \le t \le 2\pi \) . Banyaknya anggota himpunan penelesaian dari persamaan diatas adalah....

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

\(\begin{array}{c}4{\cos ^2}t + 3\sec \,\,t = 3 + 4\cos \,\,t\\4{\cos ^2}t - 4\cos \,\,t = 3 - 3\sec \,\,t\\4\cos \,\,t\left( {\cos \,\,t - 1} \right) = 3\left( {1 - \sec \,\,t} \right)\\ = 3\left( {\frac{{\cos t}}{{\cos t}} - \frac{1}{{\cos t}}} \right)\\\left( {\cos \,\,t - 1} \right)\left( {4\cos \,\,t - \frac{3}{{\cos \,\,t}}} \right) = 0\\*\,\,\cos \,\,t = 1\,\,\,\,\,\\yang\,\,memenuhi\,\,\,t = {0^o}\\*\,4\cos \,\,t - \frac{3}{{\cos \,\,t}} = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\cos ^2}t = \frac{3}{4}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \,\,t = \pm \frac{1}{2}\sqrt 3 \,\\yang\,\,memenuhi\,\,t = \frac{\pi }{6},\frac{{2\pi }}{3},\frac{{4\pi }}{3}\frac{{5\pi }}{3}\\Kunci\,\,E\end{array}\)

💦 Soal No.04

Jika pencerminan titik P (s, t) terhadap garis x = a dan dilanjutkan dengan pencerminan terhadap garis y = b menghasilkan dilatasi sebesar 3 kali, maka ab = ...

(A) st
(B) 2st
(C) 3st
(D) 4st
(E) 5st

\(\begin{array}{l} \Rightarrow \left[ \begin{array}{l}2a - s\\2b - t\end{array} \right] = \left[ {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}s\\t\end{array}} \right]\\\,\,\,\,\,\,\left[ \begin{array}{l}2a - s\\2b - t\end{array} \right] = \left[ \begin{array}{l}3s\\3t\end{array} \right]\\ \Rightarrow 2a - s = 3s \to a = 2s\\\,\,\,\,\,\,2b - t = 3t \to b = 2t\\maka\\ \Rightarrow ab = 2s \cdot 2t\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 4st\end{array}\)

💦 Soal No. 05.

Pada kubus ABCD,EFGH, titik P adalah titik potong diagonal AH dan DE, Jika R terletak di tengah rusuk AD, maka nilai \(\sin \angle PBR\) adalah....

(A) \(\frac{{\sqrt 6 }}{6}\)
(B) \(\frac{{\sqrt 6 }}{3}\)
(C) \(\frac{{\sqrt 6 }}{2}\)
(D) \(\frac{{\sqrt 2 }}{2}\)
(E) \(\frac{{\sqrt 2 }}{2}\)

\(\begin{array}{l}Diketahui:\\P = titik\,\,potong\,\,diagonal\,\,AH\,\,dan\,\,DE\\R = Titik\,\,tengah\,\,rusuk\,\,AD\\ \Rightarrow PR = \frac{1}{2}AE = \frac{1}{2}a\\ \Rightarrow BR = \sqrt {A{R^2} + A{B^2}} \\\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{1}{4}{a^2} + {a^2}} \\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}a\sqrt 5 \\ \Rightarrow BP = \sqrt {A{P^2} + A{B^2}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{1}{2}{a^2} + {a^2}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}a\sqrt 6 \\{\rm{Dengan menggunakan aturan kosinus:}}\\\cos \alpha = \frac{{{{\left( {\frac{1}{2}a\sqrt 5 } \right)}^2} + {{\left( {\frac{1}{2}a\sqrt 6 } \right)}^2} - {{\left( {\frac{1}{2}a} \right)}^2}}}{{2 \cdot \frac{1}{2}a\sqrt 5 \cdot \frac{1}{2}a\sqrt 6 }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\frac{5}{4}{a^2} + \frac{6}{4}{a^2} - \frac{1}{4}{a^2}}}{{\frac{1}{2}{a^2}\sqrt {30} }}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\frac{{\frac{5}{2}{a^2}}}{{\frac{1}{2}{a^2}\sqrt {30} }}\,\,\,\,\, = \,\,\,\frac{1}{6}\sqrt {30} \\{\rm{Dimasukkan ke segitiga siku - siku (Gambar }}iii{\rm{)}}\\ \Rightarrow sisi\,\,depan\,\,\, = \sqrt {36 - 30} \,\,\, = \sqrt 6 \\maka\,\,\,\,\,\sin \,\,\alpha = \frac{{\sqrt 6 }}{6}\\Kunci\,\,A\end{array}\)

💦 Soal No. 06.

Diketahui sisa pembagian f(x) oleh \({x^2} + 2x + 4\) adalah \(2x + 3\). Jika sisa pembagian oleh \((x + f{(x)^2}\) adalah \(ax + b\) maka nilai a+b adalah...

(A) -27
(B) -15
(C) 0
(D) 5
(E) 9

\(\begin{array}{l}f\left( x \right) = h\left( x \right) \cdot \left( {{x^2} + 2x + 4} \right) + \left( {2x + 3} \right)\\x + f\left( x \right) = h\left( x \right) \cdot \left( {{x^2} + 2x + 4} \right) + \left( {3x + 3} \right)\,\,\,\,\,\,\,\,\,\\{\rm{dikuadratkan }}\,{\rm{lalu bagi dengan }}\,({x^2} + 2x + 4)\\\frac{{{{\left( {x + f\left( x \right)} \right)}^2}}}{{{x^2} + 2x + 4}} = \frac{{{{\left[ {h\left( x \right) \cdot \left( {{x^2} + 2x + 4} \right)} \right]}^2}}}{{{x^2} + 2x + 4}} + \frac{{2 \cdot h\left( x \right) \cdot \left( {{x^2} + {}^2x + 4} \right)\left( {3x + 3} \right)}}{{{x^2} + 2x + 4}} + \frac{{{{\left( {3x + 3} \right)}^2}}}{{{x^2} + 2x + 4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\widehat {tidak\,\,bersisa}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\widehat {\,\,tidak\,\,bersisa}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\widehat {\,\,\,bersisa}\\{x^2} + 2x + 4\mathop{\left){\vphantom{1{\frac{\begin{array}{l}9{x^2} + 18x + 9\\9{x^2} + 18x + 36\end{array}}{{ - 27}}}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{\frac{\begin{array}{l}9{x^2} + 18x + 9\\9{x^2} + 18x + 36\end{array}}{{ - 27}}}}}} \limits^{\displaystyle\,\,\, 9} - \,\,\,\,\,\, \Rightarrow \,\,\,sisanya - 27\\Kunci\,\,A\end{array}\)

💦 Soal No. 07.

Grafik \(y = {3^{x + 1}} - {\left( {\frac{1}{9}} \right)^x}\) berada di bawah grafik \(y = {3^x} + 1\) jika ....

(A) 0 < x < 1
(B) x > 1
(C) x > 0
(D) x > 3
(E) 1 < x < 3

\(\begin{array}{l}{\rm{Jika grafik}}\,\,\,{y_1}\,\,{\rm{di bawah grafik }}{y_2}{\rm{ maka }}{y_1}{\rm{ < }}{y_2}\\ \Rightarrow {3^{x + 1}} - {\left( {\frac{1}{9}} \right)^x} < {3^x} + 1\\\,\,\,\,\,\,{3^x} \cdot 3 - \frac{1}{{{3^{2x}}}} < {3^x} + 1 \to {\rm{misal}}\,:\,P = 3x\\\,\,\,\,\,\,3P - \frac{1}{{{P^2}}} < P + 1\\\,\,\,\,\,\,\frac{{2P - \frac{1}{{{P^2}}} - 1 < 0}}{{2{P^3} - 1 - {P^2} < 0}}\,\,\,x\,{P^2}\\\,\,\,\,\,\,2{P^3} - {P^2} - 1 < 0\\(p - 1)(2{p^2} + p + 1) < 0\\p < 1\\ \Rightarrow {3^x} < {3^0}\\\,\,\,\,\,\,\,\,x < 0\\Kunci\,\,C\end{array}\)

💦 Soal No. 08.

\(\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {x + b} - \sqrt {{{\left( {a + b} \right)}^2}} }}{{\left( {{x^2} - {a^2}} \right)\sin (x - a}} = ...\)

(A) \(\frac{1}{{16a(a + b)}}\)
(B) \(\frac{1}{{8a(a + b)}}\)
(C) \(\frac{1}{{4a(a + b)}}\)
(D) \(\frac{1}{{2a(a + b)}}\)
(E) \(\frac{1}{{a(a + b)}}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to a} \frac{{{{\left( {\sqrt {x + b} - \sqrt {a + b} } \right)}^2}}}{{\left( {{x^2} - {a^2}} \right) \cdot \sin \left( {x - a} \right)}}\\ = \mathop {\lim }\limits_{x \to a} \frac{{{{\left( {\sqrt {x + b} - \sqrt {a + b} } \right)}^2}{{\left( {\sqrt {x + b} + \sqrt {a + b} } \right)}^2}}}{{\left( {x - a} \right)\left( {x + a} \right) \cdot \sin \left( {x - a} \right) \cdot {{\left( {\sqrt {x + b} + \sqrt {a + b} } \right)}^2}}}\\ = \mathop {\lim }\limits_{x \to a} \frac{{{{\left( {\left( {x + b} \right) - \left( {a + b} \right)} \right)}^2}}}{{\left( {x - a} \right) \cdot \left( {x + a} \right) \cdot \sin \left( {x - a} \right) \cdot {{\left( {\sqrt {x + b} + \sqrt {a + b} } \right)}^2}}}\\ = \mathop {\lim }\limits_{x \to a} \frac{{\left( {x - a} \right)\left( {x - a} \right)}}{{\left( {x - a} \right) \cdot \left( {x + a} \right) \cdot \sin \left( {x - a} \right) \cdot {{\left( {\sqrt {x + b} + \sqrt {a + b} } \right)}^2}}}\\ = \frac{1}{{\left( {a + a} \right){{\left( {\sqrt {a + b} + \sqrt {a + b} } \right)}^2}}} = \frac{1}{{\left( {a + a} \right){{\left( {\sqrt {a + b} + \sqrt {a + b} } \right)}^2}}}\\ = \frac{1}{{8a\left( {a + b} \right)}}\\Kunci\,\,B\end{array}\)

💦 Soal No. 09.

Misalkan semua suku dari deret geometri adalah positif dan diketahui perbandingan suku ke-6 dan suku ke-4 dari deret tersebut adalah 16. Jika kuadrat suku pertama sama dengan rasionya, maka jumlah empat suku pertama deret geometri adalah.....

(A) 40
(B) 41
(C) 42
(D) 43
(E) 44

\(\begin{array}{l}\\ \Rightarrow \frac{{{U_6}}}{{{U_4}}} = 16\\\,\,\,\,\,\frac{{a{r^5}}}{{a{r^3}}} = 16\\\,\,\,\,\,\,\,\,\,{r^2} = 16\\ \Rightarrow {U_1}^2 = r\\\,\,\,\,\,\,\,\,{a^2} = r\,\,\left( {r > 0} \right)\\\,\,\,\,\,\,\,\,{a^2} = 4\\\,\,\,\,\,\,\,\,\,\,a = 2\\ \Rightarrow {S_4} = \frac{{a \cdot \left( {{r^4} - 1} \right)}}{{r - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \cdot \left( {256 - 1} \right)}}{1}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \cdot \left( {255} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 510\\Kunci...\end{array}\)

💦 Soal No. 10.

Diketahui \(f\left( x \right) = {x^3} - 3{x^2} + a\) memotong sumbu-y di titik (0,10). Maka nilai minimum f(x) untuk \(x \in \left[ {0,1} \right]\) adalah.....

(A) 10
(B) 8
(C) 6
(D) 4
(E) 3

\(\begin{array}{l}f(x) = {x^3} - 3x + a\,\\titik\,potong\,\,sumbu\,\,y(x = 0)\\10 = {0^3} - 3(0) + a \Rightarrow a = 10\\f(x) = {x^3} - 3x + 10\\syarat\,\,maksimum\,\,\\ \Rightarrow f'(x) = 0\\\,\,\,\,\,\,3{x^2} - 3 = 0\\\,\,\,\,\,x = \pm 1\\ \Rightarrow f''(x) = 6x\\\,\,\,\,\,\,f''(1) > 0\\maka\,\,pembuat\,\,\,maksimum\,\,x = 1\\dan\,\,nilai\,\,maksimum = f(1)\\ \Rightarrow {1^3} - 3(1) + 10 = 8\\Kunci\,\,B\end{array}\)

💦 Soal No. 11.

Diketahui fungsi \(f(x) = f(x + 2)\) untuk setiap x elemen bilangan real. Jika \(\smallint _0^2f(x)dx = B\) maka \(\smallint _3^7f(x + 8)dx = ....\)

(A) B
(B) 2B
(C) 3B
(D) 4B
(E) 5B

\(\begin{array}{l}x \in \left[ {0,1} \right]\\f\left( x \right) = f\left( {x + 2} \right){\rm{merupakan}}\,\,{\rm{fungsi}}\,\,{\rm{konstan}}\\\int\limits_0^2 {f\left( x \right)dx = B} \\{\rm{misal}}\,\,f\left( x \right) = c\\\int\limits_0^2 {c\,\,dx = \left. {cx} \right|_0^2} \Rightarrow B = 2c\\maka\\\int\limits_3^7 {f\left( {x + 8} \right)dx = \int\limits_3^7 {c\,\,dx} } \\ = \left. {cx} \right|\begin{array}{*{20}{c}}7\\3\end{array} = 4c\\ = 2\left( {2c} \right) = 2B\\Kunci\,\,B\end{array}\)

💦 Soal No. 12.

Luas daerah diantara kurva \(y = 2x + 1\) dan kurva \(y = {x^2} + 2a\) selalu bernilai konstan, yaitu k. Nilai dari k adalah...

(A) \(\frac{1}{3}\)
(B) \(\frac{2}{3}\)
(C) \(\frac{4}{3}\)
(D) \(\frac{5}{3}\)
(E) \(\frac{7}{3}\)

\(\begin{array}{l}{x^2} + 2a = 2a + 1\\{x^2} = 1\,\langle \begin{array}{*{20}{c}}{\,\,{x_1} = - 1}\\{{x_2} = 1}\end{array}\\L = \int\limits_{ - 1}^1 {2a + 1 - \left( {{x^2} + 2a} \right)dx} \\\,\,\, = \int\limits_{ - 1}^1 {1 - {x^2}\,dx} \\K = x - \frac{1}{3}\left. {{x^3}} \right]_{ - 1}^1\\\,\,\,\,\, = \left( {1 - \frac{1}{3}} \right) - \left( { - 1 + \frac{1}{3}} \right)\\\,\,\,\,\, = 1 - \frac{1}{3} + 1 - \frac{1}{3}\\\,\,\,\,\, = \frac{4}{3}\\Kunci\,\,C\end{array}\)

💦 Soal No. 13.

Banyaknya bilangan genap n = abc dengan 3 digit sehingga 3< b < c adalah...

(A) 48
(B) 54
(C) 60
(D) 64
(E) 72

\(\begin{array}{l}{\rm{Angka}}\,\,{\rm{di}}\,\,{\rm{C}}\,\,hanya\,\,{\rm{terisi}}\,\,\,{\rm{dengan}}\,\,\,{\rm{angka}}\,\,\,{\rm{6}}\,\,{\rm{atau}}\,\,{\rm{8}}\\{\rm{jika}}\,\,\,C = 6\,\,\,\, \Rightarrow \left[ 9 \right] \times \left[ 2 \right] \times \left[ 1 \right]\,\,\,\, = \,\,\,18\\{\rm{jika}}\,\,\,C = 8\,\,\,\, \Rightarrow \left[ 9 \right] \times \underline {\left[ 4 \right] \times \left[ 1 \right]\,\,\,\, = \,\,\,36} \,{\,_{\, + }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,54\\Kunci\,\,\,B\end{array}\)

💦 Soal No. 14.

Garis singgung kurva \(y = 3 - {x^2}\) di titik P(-a, b) dan Q(a, b) memotong sumbu -y membuat segitia PQR sama sisi adalah....

(A) \(\sqrt[2]{3}\)
(B) \(\sqrt 3 \)
(C) \(\frac{1}{2}\sqrt 3 \)
(D) \(\frac{1}{3}\sqrt 3 \)
(E) \(\frac{1}{4}\sqrt 3 \)

\(\begin{array}{l}{\rm{Gradien garis singgung kurva}}\,\\y = 3 - {x^2}\,\,adalah\,\,\,m = y' = 2x\\{\rm{I}}{\rm{.}}\,\,{\rm{Persamaan garis singgung}}\,\,{\rm{di }}\left( { - a,b} \right)\\*\,m = - 2a\\\, \Rightarrow y - b = - 2a\left( {x + a} \right)\\\,\,\,\,\,\,\,y = - 2a\,\,x - 2{a^2} + b\\{\rm{II}}{\rm{.}}\,\,{\rm{Persamaan garis singgung}}\,\,{\rm{di }}\left( {a,b} \right)\\*\,m = 2a\\\, \Rightarrow y - b = 2a\left( {x + a} \right)\\\,\,\,\,\,\,\,y = 2a\,\,x + 2{a^2} + b\\{\rm{Kita eliminasi kedua persamaan tersebut}}\\{\rm{untuk mendapatkan titik R yaitu R}}\,\,{\rm{ = }}\,\,{\rm{(0}}{\rm{,2}}{{\rm{a}}^{\rm{2}}}{\rm{ + b)}}\\{\rm{Jadi}}\,\,RS = (2{a^2} + b - b) = 2{a^2}\,\,dan\,\,\,SQ = a\\\tan \,\,Q = \frac{{RS}}{{SQ}} = \tan \,\,60\,(sama\,\,sisi)\\\,\frac{{2{a^2}}}{a} = \sqrt 3 \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,a = \frac{1}{2}\sqrt 3 \\Kunci jawaban\,\,C\end{array}\)

💦 Soal No. 15.

Misalkan g adalah garis singgung lingkaran \({x^2} + {y^2} = 25\) di titik A(3, 4). Jika garis singgung tersebut ditransformasikan dengan matriks rotasi \(\left( {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{4}{5}}\\{ - \frac{4}{5}}&{\frac{3}{5}}\end{array}} \right)\) , maka absis dari titik potong antara garis singggung ingkaran dengan hasil transformasi adalah...

(A) \({\frac{7}{2}}\)
(B) \({\frac{18}{5}}\)
(C) 4
(D) \({\frac{24}{5}}\)
(E) 5

Persamaan garis singgung:

*) \({x^2} + {y^2} = {r^2}\) di titik \(\left( {{x_1},{y_1}} \right)\)

adalah: \({x_1}x + {y_1}y = {r^2}\)

Jadi persamaan garis singgung \({x^2} + {y^2} = 25\,\,{\rm{di}}\,\,A\left( {3,4} \right)\)

\(3x + 4y = 25\)

*) Garis \(3x + 4y = 25\) ditransformasi oleh \(\left( {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{4}{5}}\\{\frac{{ - \,4}}{5}}&{\frac{3}{5}}\end{array}} \right)\,\)

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{x'}\\{y'}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{4}{5}}\\{ - \frac{4}{5}}&{\frac{3}{5}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right) = \frac{1}{{\frac{9}{{25}} + \frac{{16}}{{25}}}}\left( {\begin{array}{*{20}{c}}{\frac{3}{5}}&{\frac{{ - 4}}{5}}\\{\frac{4}{5}}&{\frac{3}{5}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{x'}\\{y'}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\frac{3}{5}x' - \frac{4}{5}y'}\\{\frac{4}{5}x' + \frac{3}{5}y'}\end{array}} \right)\end{array}\)

*) Untuk mencari bayangannya kita substitusi ke benda (garis) \(3x + 4y = 25\) menjadi:

\(\begin{array}{c}3\left( {\frac{3}{5}x' - \frac{4}{5}y'} \right) + \left( {\frac{4}{5}x' + \frac{3}{5}y'} \right) = 25\\\,\frac{9}{5}x' + \frac{{16}}{5}y' + \frac{{12}}{5}y' - \frac{{12}}{5}y' = 0\\\,x = 5\end{array}\)

*) Absis itik potong garis singgung lingkaran \({x_1}x + {y_1}y = {r^2}\) dan hasil transformasinya x = 5 adalah x = 5

Kunci Jawaban : E