Soal dan Pembahasan Matematika IPA SBMPTN 2014 (Kode 572)

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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 2014 kode 572 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.

Jika a dan $\beta $ adalah akar-akar persamaan kuadrat:$(m - 1){x^2} - (m + 2)x - 1 = 0$,maka ada nilainya untuk...

(A) m > -1
(B) m < 1
(C) -1 < m < 1
(D) m < -1 atau m > 1
(E) $m\,\, < - \frac{2}{3}{\rm{ atau m}}\,\,{\rm{ > }}\,\,\,\frac{2}{3}$

Pembahasan :

Diketahui $\alpha {\rm{ dan }}\beta $ adalah akar-akar dari: PK : $\left( {m - 1} \right){x^2} - \left( {m + 2} \right)x - 10 = 0$ $*\log \left( {1 + \left( {1 - \alpha } \right)\beta + \alpha } \right)$ $ = \log \left( {1 + \beta - a\beta + \alpha } \right)$ $ = \log \left( {1 + \left( {\alpha + \beta } \right) - \alpha \beta } \right)$ $ = \log \left( {1 + \frac{{m + 2}}{{m - 1}} + \frac{1}{{m - 1}}} \right)$ $ = \log \left( {\frac{{2m + 2}}{{m - 1}}} \right)$ * syarat algoritma ada nilainya numerus > 0 $\frac{{2m + 2}}{{m - 1}} > 0$ m < -1 atau m > 1

💥 Kunci Jawaban: D

💦 Soal No 02.

Di antara 20.000 dan 70.000, banyak bilangan genap dan tidak ada digit berulang adalah...

(A) 3.360
(B) 4.032
(C) 7.392
(D) 10.080
(E) 24.998

Pembahasan :

Bilangan genap antara 20.000 dan 70.000 dengan digit tak berulang $20.000{\raise0.7ex\hbox{$s$} \!\mathord{\left/ {\vphantom {s d}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$d$}}29.999 \to = 1x8x7x6x4 = 1344$ $30.000{\raise0.7ex\hbox{$s$} \!\mathord{\left/ {\vphantom {s d}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$d$}}39.999 \to 18765 = 1x8x7x6x5 = 1680$ $40.000{\raise0.7ex\hbox{$s$} \!\mathord{\left/ {\vphantom {s d}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$d$}}49.999 \to 18764 = 1x8x7x6x4 = 1344$ $50.000{\raise0.7ex\hbox{$s$} \!\mathord{\left/ {\vphantom {s d}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$d$}}59.999 \to 18765 = 1x8x7x6x5 = 1680$ $60.000{\raise0.7ex\hbox{$s$} \!\mathord{\left/ {\vphantom {s d}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$d$}}69.999 \to 18764 = 1x8x7x4 = \frac{{1344}}{{7.392}} + $

💥 Kunci Jawaban: C

💦 Soal No 03.

Banyak akar real $f(t) = {t^9} - t$ adalah... buah.

(A) 2
(B) 3
(C) 4
(D) 6
(E) 9

Pembahasan :

$\begin{array}{c}f\left( t \right) = {t^9} - t\\ = t\,.\,\left( {{t^8} + 1} \right)\\ = t\,.\,\left( {{t^4} - 1} \right)\,.\,\left( {{t^4} + 1} \right)\\ = t\,.\,\left( {{t^2} - 1} \right)\,.\,\left( {{t^2} + 1} \right)\,.\,\left( {{t^4} + 1} \right)\end{array}$ $\begin{array}{l} = t.\left( {t - 1} \right){\rm{ }}.{\rm{ }}\left( {t + 1} \right){\rm{ }}.{\rm{ }}\left( {{t^2} + 1} \right).\left( {{t^4} + 1} \right)\\{\rm{ t}}\,{\rm{ = }}\,{\rm{0}}\,\,{\rm{t}}\,{\rm{ = }}\,{\rm{1}}\,\,{\rm{t}}\,{\rm{ = }}\,{\rm{ - 1 }}\begin{array}{*{20}{c}} \downarrow \\{}\end{array}{\rm{ }}\begin{array}{*{20}{c}} \downarrow \\{}\end{array}{\rm{ }}\end{array}$ tidak memiliki akar real banyak akar real = 3

💥 Kunci Jawaban: B

💦 Soal No 04.

Diberikan deret geometri ${u_1}\,\, + \,\,\,{u_2}\,\,\, + \,\,\,{u_3}\,\,\, + \,\,\,...$ jika ${u_5} = 48$ , rasio deret -2, dan $\log {u_1} + \log {u_2} + \log {u_3} + \log {u_4} = 6\log 2 + 4\log 3$ maka nilai $2{u_3} + 3{u_2}$ adalah...

(A) 4
(B) 6
(C) 8
(D) 12
(E) 16

Pembahasan :

Diketahui deret geometri${U_5} = 48$ $r = - 2$ $a{r^4} = 48$ $a{\left( { - 2} \right)^4} = 48$ a = 3 $2{u_3} + 3{u_2} = 2a{r^2} + 3ar$ $ = 2.3.4 + 3.3.\left( { - 2} \right)$ =6

💥 Kunci Jawaban: B

💦 Soal No 05.

Diberikan balok ABCD.EFGH dengan AB = AE = 4 dan BC = 3. Titik P dan Q masing-masing titik tengah FG dan GH. Maka tangen sudut bidang diagonal FHDB dan bidang PQDB adalah...

(A) $\frac{1}{{10}}$
(B) $\frac{3}{{10}}$
(C) $\frac{2}{5}$
(D) $\frac{3}{8}$
(E) $\frac{7}{{16}}$

Pembahasan :

HF = $\sqrt {{4^2} + {3^2}} $ HF = 5cm ${L_{\Delta HPF}} = {L_{\Delta HFG}} - {L_{\Delta HPG}}$ $ = \left( {\frac{{HG{\rm{ }}x{\rm{ }}FG}}{2}} \right) - \left( {\frac{{HG{\rm{ x PG}}}}{2}} \right)$ $ = \left( {\frac{{{\rm{4 }}x{\rm{ 3}}}}{2}} \right) - \left( {\frac{{{\rm{4 x 1}}{\rm{,5}}}}{2}} \right)$ ${L_{\Delta HPF = 6 - 3}}$ $\frac{{{\rm{OP x HF}}}}{2} = {L_{\Delta HPF}}$ $\frac{{{\rm{OP x 5}}}}{2} = 3$ OP = $\frac{6}{5}$

$\tan {\rm{ }}\theta {\rm{ = }}\frac{{OP}}{{OM}}$ $ = \frac{{{\raise0.7ex\hbox{$6$} \!\mathord{\left/ {\vphantom {6 5}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$5$}}}}{4}$ $ = \frac{3}{{10}}$

💥 Kunci Jawaban: B

💦 Soal No 06.

Jika A adalah matriks berukuran 2 x 2 dan $\left[ {x{\rm{ 1}}} \right]\, \cdot A\, \cdot \,\left[ \begin{array}{l}x\\1\end{array} \right] = {x^2} - 5x + 8$, maka matriks A yang tepat adalah ...

(A) $\left( {\begin{array}{*{20}{c}}1&{ - 5}\\8&0\end{array}} \right)$
(B) $\left( {\begin{array}{*{20}{c}}1&5\\8&0\end{array}} \right)$
(C) $\left( {\begin{array}{*{20}{c}}1&8\\{ - 5}&0\end{array}} \right)$
(D) $\left( {\begin{array}{*{20}{c}}1&3\\{ - 8}&{ - 8}\end{array}} \right)$
(E) $\left( {\begin{array}{*{20}{c}}1&{ - 3}\\8&8\end{array}} \right)$

Pembahasan :

$\left[ {\begin{array}{*{20}{c}}x&1\end{array}} \right].A.\left[ {\begin{array}{*{20}{c}}x\\1\end{array}} \right] = {x^2} - 5x + 8$ $\left[ {\begin{array}{*{20}{c}}x&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&b\\c&a\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\1\end{array}} \right] = {x^2} - 5x + 8$ $\left[ {\begin{array}{*{20}{c}}{ax + c}&{bx + d}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\1\end{array}} \right] = {x^2} - 5x + 8$ $\left( {ax + c} \right).x + \left( {bx + d} \right) = {x^2} - 5x + 8$ $a{x^2} + \left( {b + c} \right)x + d = {x^2} - 5x + 8$ maka : a = 1 ; b + c = -5 ; d = 8 jawaban yang memenuhi A = $\left[ {\begin{array}{*{20}{c}}1&3\\{ - 8}&8\end{array}} \right]$

💥 Kunci Jawaban: D

💦 Soal No 7.

Jika suku pertama, ke 3, dan ke 6 suatu barisan aritmetika masing-masing adalah ba, a, dan 36 serta jumlah 9 suku pertama barisan tersebut adalah 180, maka beda barisan tersebut adalah...

(A) 18
(B) 16
(C) 12
(D) 9
(E) 6

Pembahasan :

Barisan aritmatika x = suku pertama y = beda $\begin{array}{l} \Rightarrow {u_1} = x = b - a\\ \Rightarrow {u_3} = x + 2y = a\\ \Rightarrow {u_6} = x + 5y = 36\\ \Rightarrow {s_9} = \frac{9}{2}\left( {2x + 8y} \right) = 180\end{array}$ $\begin{array}{l}2x + 8y = 40\\x + 4y = 20\end{array}$ $\frac{{\begin{array}{*{20}{c}}{x + 5y = 36}\\{x + 4y = 20}\end{array}}}{{y = 16}} - $

💥 Kunci Jawaban: B

💦 Soal No 08.

Jika a dan b merupakan akar-akar persamaan $^{(1 + \left| {\,x\,} \right|)}\log (3x + 7)$ = 2, maka a + b = ...

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

Pembahasan :

$^{\left( {1 + \left| x \right|} \right)}\log \left( {3x + 7} \right) = 2$ $3x + 7 = {\left( {1 + \left| x \right|} \right)^2}$ $3x + 7 = 1 + 2\left| x \right| + {x^2}$ $ - 2\left| x \right| = {x^2} - 3x - 6$ $2\left| x \right| = - {x^2} + 3x + 6$ $\left( {2x + \left( { - {x^2} + 3x + 6} \right)} \right).\left( {2x - ( - {x^2} + 3x + 6} \right) = 0$ $\left( { - {x^2} + 5x + 6} \right).\left( {{x^2} - x - 6} \right) = 0$ $ - 1.\left( {{x^2} - 5x - 6} \right).\left( {{x^2} - x - 6} \right) = 0$ $\left( {x - 6} \right)\left( {x + 1} \right)\left( {x - 3} \right)\left( {x + 2} \right) = 0$ x = 6 ; x = -1 ; x = 3 ; x = -2 TM TM a + b = -1+3 = 2

💥 Kunci Jawaban: C

💦 Soal No 09.

Diketahui suatu parabola simetris terhadap garis x=-2, dan garis singgung parabola tersebut di titik (0, 1) sejajar garis 4x + y = 4. Titik puncak parabola tersebut adalah...

(A) (-2, -3)
(B) (-2, -2)
(C) (-2, 0)
(D) (-2, 1)
(E) (-2, 5)

Pembahasan :

$y = a{x^2} + bx + c$ (i) melalui titik ( 0,1 ) $\begin{array}{l}1 = a{\left( 0 \right)^2} + b\left( 0 \right) + c\\c = 1\end{array}\]\[\begin{array}{l}1 = a{\left( 0 \right)^2} + b\left( 0 \right) + c\\c = 1\end{array}$ maka : 2a = b 2a = -4 a = -2 ( ii ) Sumbu simetri $\begin{array}{l}{x_p} = - 2\\ - \frac{b}{{2a}} = - 2\\b = 2a\\y = - 2{x^2} - 4x + 1\\{y_p} = - 2{\left( { - 2} \right)^2} - 4\left( { - 2} \right) + 1\\{y_p} = - 8 + 8 + 1\\{y_p} = 1\end{array}$ ( iii ) garis singgung parabola sejajar garis $\begin{array}{l}{m_{gs}} = {m_g}\\{y^1} = - 4 \to x = 0\\2ax + b = - 4\\b = - 1\end{array}$ titik puncak = $\left( {{x_p},{y_p}} \right) = \left( { - 2,1} \right)$

💥 Kunci Jawaban: B

💦 Soal No 10.

Diketahui P dan Q suatu polinomial. Jika P(x) berturut-turut memberikan sisa -1 dan 5 apabila dibagi x1 dan dibagi x + 2, dan Q(x) berturut-turut memberikan sisa 1 dan -2 apabila dibagi x + 2 dan dibagi x1, maka P(Q(x)) dibagi ${x^2} + x - 2$ bersisa ...

(A) 2x - 3
(B) 2x + 3
(C) 3x + 2
(D) -3x + 2
(E) 3x + -2

Pembahasan :

$P\left( x \right):\left( {x - 1} \right) \to s = - 1 \to P\left( 1 \right) = - 1$ $\begin{array}{l}P\left( x \right):\left( {x + 2} \right) \to s = 5 \to P\left( { - 1} \right) = 5\\Q\left( x \right):\left( {x - 1} \right) \to s = - 2 \to Q\left( { - 1} \right) = - 2\\Q\left( x \right):\left( {x + 2} \right) \to s = 1 \to Q\left( { - 2} \right) = 1\\P\left( {Q\left( x \right)} \right):{x^2} + x - 2 \to s = ax + b\\p\left( {Q\left( x \right)} \right) = H\left( x \right)\,.\,\left( {{x^2} + x - 2} \right) + \left( {ax + b} \right)\\x = - 2 \to P\left( {Q\left( 1 \right)} \right) = H\left( 1 \right).\left( {1 + 1 - 2} \right) + a + b\\P\left( { - 2} \right) = a + b\\5 = a + b\end{array}$ $x = - 2 \to P\left( {Q\left( { - 2} \right)} \right) = H\left( { - 2} \right)\left( {4 - 2 - 2} \right) - 2a + b$ P (1) =-2a+b -1 = -2 + b $\frac{{\begin{array}{*{20}{c}}{a + b = 5}\\{ - 2a + b = - 1}\end{array}}}{{\begin{array}{*{20}{c}}{3a = 6}\\{a = 2}\end{array}}}$ a + b = 5 2 + b = 5 b = 3 maka : s (x) = 2x + 3

💥 Kunci Jawaban: B

💦 Soal No 11.

Persamaan garis lurus yang melalui titik potong lingkaran-lingkaran yang melalui titik (-2, -1) dan menyinggung sumbu -X dan sumbu -Y adalah...

(A) x + 2y + 4 = 0
(B) x + y + 3 = 0
(C) 3x + y + 7 = 0
(D) x + 3y + 5 = 0
(E) 2x + y + 5 = 0

Pembahasan :

pers, lingkaran melalui titik (-2,-1)dengan pusat (-p,-p)dan r = p $\begin{array}{l}{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\\{\left( { - 2 + p} \right)^2} + {\left( { - 1 + p} \right)^2} = {p^2}\\4 - 4p + {p^2} + 1 - 2p + {p^2} = {p^2}\\{p^2} - 6p + = 0\\\left( {p - 5} \right)\left( {p - 1} \right) = 0\\p = 5\,atau\,p = 1\\\end{array}$ $P = 5\,atau\,P = 1$ $p = 5 \to pusat\left( { - 5, - 5} \right)dan{\rm{ r = 5}}$ ${L_1} = {\left( {x + 5} \right)^2} + {\left( {y + 5} \right)^2} = {5^2}$ ${x^2} + 10x + 25 + {y^2} + 10y + 25 = 25$ ${x^2} + {y^2} + 10x + 10y + 25 = 0$ $p = 1 \to pusat\left( { - 1, - 1} \right)dan{\rm{ r = 1}}$ $\begin{array}{l}{L_2} = {\left( {x + 1} \right)^2} + {\left( {y + 1} \right)^2} = {1^2}\\{x^2} + 2x + 1 + {y^2} + 2y + 1 = 1\\{x^2} + {y^2} + 2x + 2y + 1 = 0\end{array}$ pers ling melalui titik potong dua lingkaran, $\frac{{\begin{array}{*{20}{c}}{{x^2} + {y^2} + 10x + 10y + 25 = 0}\\{{\rm{ }}{x^2} + {y^2} + 2x + 2y + 1 = 0}\end{array}}}{{{\rm{ }}\begin{array}{*{20}{c}}{8x + 8y + 24 = 0}\\{{\rm{ }}x + y + 3 = 0}\end{array}}}$

💥 Kunci Jawaban: A

💦 Soal No 12.

Bila $\tan x = \frac{{ - 3}}{4},\frac{{3\pi }}{2} < x < 2\pi $, maka$\sin (\frac{\pi }{3} - x) = $ ...

(A) $\left( {2\sqrt 3 + 3} \right)/10$
(B) $\left( {3\sqrt 3 + 3} \right)/10$
(C) $\left( {4\sqrt 3 + 3} \right)/10$
(D) $\left( {3\sqrt 3 - 3} \right)/10$
(E) $\left( {4\sqrt 3 - 3} \right)/10$

Pembahasan :

Diketahui : tan x $ - \frac{3}{4}\dim ana{\rm{ }}\frac{{3\pi }}{2} < x < 2\pi $ sin $\left( {{\raise0.7ex\hbox{$\pi $} \!\mathord{\left/ {\vphantom {\pi 3}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$3$}} - x} \right) = \sin \frac{\pi }{3}.\cos x - \cos \frac{\pi }{3}.\sin x$ $ = \left( {\frac{1}{2}\sqrt 3 .\frac{4}{5}} \right) - \left( {\frac{1}{2}.\left( {\frac{{ - 3}}{5}} \right)} \right)$ $ = \frac{{4\sqrt 3 }}{{10}} + \frac{3}{{10}}$ $ = \frac{1}{{10}}\left( {4\sqrt 3 + 3} \right)$

💥 Kunci Jawaban: C

💦 Soal No 13.

Diberikan limas T.ABC Misalkan $u = \overline {TA} $, $v = \overline {TB} $,$w = \overline {TC} $ . Jika P adalah titik berat $\Delta \,ABC$, maka $\overline {TP} $ ...

(A) $\frac{1}{3}\left( {u + v + w} \right)$
(B) $\frac{1}{2}\left( {u + v + w} \right)$
(C) $\frac{2}{3}\left( {u + v + w} \right)$
(D) $\frac{3}{4}\left( {u + v + w} \right)$
(E) $u + v + w$

Pembahasan :

$\begin{array}{l}\overline {OP} = \frac{1}{3}\overline {OA} + \frac{1}{3}\overline {OB} + \frac{1}{3}\overline {OC} \\\overline {TP} = \overline {TO} + \overline {OP} \\ = \frac{1}{3}\overline {TO} + \frac{1}{3}\overline {TO} + \frac{1}{3}\overline {TO} + \frac{1}{3}\overline {OA} + \frac{1}{3}\overline {OB} + \frac{1}{3}\overline {OC} \\TP = \,\frac{1}{3}\left( {\overline {TO} + \overline {OA} } \right) + \frac{1}{3}\left( {\overline {TO} + \overline {OB} } \right) + \frac{1}{3}\left( {\overline {TO} + \overline {OC} } \right)\\ = \frac{1}{3}\overline {TA} + \frac{1}{3}\overline {TB} + \frac{1}{3}\overline {TC} \\ = \frac{1}{3}\left( {\overline U + \overline V + \overline W } \right)\end{array}$

💥 Kunci Jawaban: A

💦 Soal No 14.

Misalkan A(t) meytakan luas daerah di bawah kurva$y = b{x^2}$, $0 \le x \le t$ .Jika titik $P({x_0},\,\,0)$ sehingga $A({x_0}):A(1) = 1\,\,:\,\,8$, maka perbandingan luas trapesium ABPQ : DCPQ = ...

(A) 2 : 1
(B) 3 : 1
(C) 6 : 1
(D) 8 : 1
(E) 9 : 1

Pembahasan :

Diketahui : $\frac{{A(xo)}}{{A(1)}} = \frac{1}{8}$ $\frac{{\int\limits_0 {^{xo}b{x^2}{\rm{ dx}}} }}{{\int\limits_0 {^1b{x^2}{\rm{ dx}}} }} = \frac{1}{8}$ $\frac{{\frac{1}{3}b{\rm{ }}{{\rm{x}}_o}^3}}{{\frac{1}{3}b{{.1}^3}}} = \frac{1}{8}$ ${x_o}^3 = \frac{1}{8}$ ${x_o} = \frac{1}{2}$ $\frac{{{L_{ABPQ}}}}{{{L_{DCPQ}}}} = \frac{{\frac{1}{2}\left( {\frac{1}{4}b + b} \right).{\raise0.7ex\hbox{$3$} \!\mathord{\left/ {\vphantom {3 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}}}{{\frac{1}{2}.\left( {\frac{1}{4}b + b} \right).{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 3}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$3$}}}}$ $ = \frac{3}{1}$

💥 Kunci Jawaban: B

💦 Soal No 15.

Jika$\frac{{f({x^3}) - f({a^3})}}{{x - a}} = - 1$ $\mathop {\lim }\limits_{x \to a} $ , maka${f^/}(1\,) = ...$

(A) - 1
(B) $ - \frac{1}{3}$
(C) $\frac{1}{3}$
(D) 1
(E) 2

Pembahasan :

$\begin{array}{l}\mathop {Lim}\limits_{x \to a} \frac{{f\left( {{x^3}} \right) - f\left( {{a^3}} \right)}}{1} = - 1\\\frac{{3{x^2}.{f^1}\left( {{x^3}} \right) - 0}}{1} = - 1\\x = a \to 3{a^2}.{f^1}\left( {{a^3}} \right) = - 1\\{\rm{ }}{{\rm{f}}^1}\left( {{a^3}} \right) = \frac{{ - 1}}{{3{a^2}}}\\a = 1 \to {f^1}\left( {{1^3}} \right) = - \frac{1}{{3{{\left( 1 \right)}^2}}}\\f'(1) = - \frac{1}{3}\end{array}$

💥 Kunci Jawaban: B