Soal dan Pembahasan Matematika IPA SBMPTN 2015 (Kode 509)
- (A) -1
- (B) 0
- (C) 1
- (D) 2
- (E) 3
Kunci Jawaban: D
- (A) \(\frac{{\sqrt 3 }}{2}\left( {2{\alpha ^2} - 1} \right) + \alpha \sqrt {1 - {a^2}} \) \)
- (B) \(\frac{{\sqrt 3 }}{2}\left( {2{\alpha ^2} - 1} \right) - \alpha \sqrt {1 - {a^2}} \)
- (C) \(\frac{{\sqrt 3 }}{2}\left( {{\alpha ^2} - 1} \right) + \alpha \sqrt {1 - {a^2}} \)
- (D) \(\frac{{\sqrt 3 }}{2}\left( {{\alpha ^2} - 1} \right) + \alpha \sqrt {1 + {a^2}} \)
- (E) \(\frac{{\sqrt 3 }}{2}\left( {{\alpha ^2} + 1} \right) + \alpha \sqrt {1 - {a^2}} \)
\(\begin{array}{l}Jika:\cos (x + {15^o}) = \alpha \to {0^o} \le x \le {30^o}\\dari\,\,\,gambar\,\,diperoleh\\\sin (x + {15^o}) = \sqrt {1 - {a^2}} \\Karena:\\\cos 2x = 1 - 2{\sin ^2}x = 2{\cos ^2}x - 1\\\cos {15^o} = \sqrt {\frac{1}{2} + {\rm{ }}\frac{1}{2}{\rm{ }}\cos {\rm{ }}{{30}^0}} \\ = \sqrt {\frac{1}{2}{\rm{ }} + {\rm{ }}\frac{1}{4}{\rm{ }}\sqrt 3 } \\ = \frac{1}{2}\sqrt {2{\rm{ }} + {\rm{ }}\sqrt 3 } \\\sin {15^o} = \sqrt {\frac{1}{2} - {\rm{ }}\frac{1}{2}{\rm{ }}\cos {\rm{ }}{{30}^0}} \\ = \sqrt {\frac{1}{2} - \frac{1}{4}{\rm{ }}\sqrt 3 } \\ = \frac{1}{2}\sqrt {2 - \sqrt 3 } \\maka\\\cos (2x + {60^o}) = {\cos ^2}(x + {30^o})\\ = 2{\cos ^2}(x + {30^o}) - 1\\ = 2{\left[ {\cos ((x + {{15}^o}) + {{15}^o})} \right]^2} - 1\\ = 2.{\left[ {\alpha .\frac{1}{2}\sqrt {2 + \sqrt 3 } - \sqrt {1 - {\alpha ^2}.} \frac{1}{2}\sqrt {2 - \sqrt 3 } } \right]^2} - 1\\ = \frac{2}{4}\left[ {\left( {2 + \sqrt 3 } \right){\alpha ^2} - 2\alpha {{\sqrt {1 - \alpha } }^2} + (1 - {\alpha ^2})(2 - \sqrt 3 } \right] - 1\\ = \frac{1}{2}\left[ {2{\alpha ^2} + \sqrt 3 {\alpha ^2} - 2\alpha \sqrt {1 - {\alpha ^2}} + 2 - \sqrt 3 - 2{\alpha ^2} + \sqrt 3 {\alpha ^2}} \right] - 1\\ = \frac{1}{2}\left[ {2\sqrt 3 {\alpha ^2} - 2\alpha \sqrt {1 - {\alpha ^2}} + 2 - \sqrt 3 } \right] - 1\\ = \sqrt 3 {\alpha ^2} - \alpha \sqrt {1 - {\alpha ^2}} + 1 - \frac{1}{2}\sqrt 3 - 1\\ = \sqrt 3 {\alpha ^2} - \frac{1}{2}\sqrt 3 - \alpha \sqrt {1 - {\alpha ^2}} \\ = \frac{{\sqrt 3 }}{2}(2{\alpha ^2} - 1) - \alpha \sqrt {1 - {\alpha ^2}} \end{array}\)
Kunci Jawaban: B
- (A) \({\rm{ }}{75^o}\)
- (B) \({\rm{ }}{60^o}\)
- (C) \({\rm{ }}{40^o}\)
- (D) \({\rm{ }}{30^o}\)
- (E) \({\rm{ }}{15^o}\)
Kunci Jawaban: B
- (A) y = x + 4
- (B) y = -x + 4
- (C) y = x + 4
- (D) y = x - 2
- (E) y = -x - 4
\(\begin{array}{l} \Rightarrow ({x_i},{y_i})\,\,\,\underline {direfleksi\,\,\,y = h} \,\,\,({x_i},2h - {y_i})\\Artinya\\^o)\,\,x\,\,\,nya\,\,tetap\\^o)\,\,y\,\,\,berubah\,\,\,jadi\,\,\,2h - {y_i}\\ \Rightarrow ({x_i},{y_i})\,\,\,\underline {direfleksi\,\,\,x = k} \,\,\,(2h - {x_i},{y_i})\\Artinya\\^o)\,\,y\,\,\,nya\,\,tetap\\^o)\,x\,\,\,berubah\,\,\,jadi\,\,\,2h - {x_i}\\ \Leftrightarrow \\y = - x + 2\,\,\,\,\underline {direfleksi\,\,\,y = 3} \\bayangannya\,\,\,menjadi\,\,\,\left( {2 \times 3 - y} \right) = - x + 2\\y = x + 4\end{array}\)
Kunci Jawaban: A
- (A) 3
- (B) 2
- (C) \(3\frac{1}{3}\)
- (D) \(2\frac{1}{3}\)
- (E) \(2\frac{2}{3}\)
\(\begin{array}{r} \Rightarrow \frac{{RF}}{{RE}} = \frac{{EP}}{{EH}}\\\frac{{RF}}{{RF + 2}} = \frac{1}{2}\\2RF = RF + 2\\RF = 2\end{array}\) \(\begin{array}{l} \Rightarrow {\rm{ }}{V_{EAH,{\rm{ }}FPQ}}{\rm{ }} = {\rm{ }}{V_R} - {V_{R.{\rm{ }}FPQ}}\\ = {\rm{ }}\frac{1}{3}({L_{\Delta {\rm{ }}EAH{\rm{ X ER }}}}) - {\rm{ }}\frac{1}{2}({L_{\Delta {\rm{ FPQ X FR}}}})\\ = {\rm{ }}\frac{1}{3}(\frac{{2{\rm{ }} \times {\rm{ }}2}}{2} \times 4) - \frac{1}{3}(\frac{{1{\rm{ }} \times {\rm{ }}1}}{2} \times {\rm{ }}2)\\ = {\rm{ }}\frac{8}{3} - \frac{1}{3}\\ = \frac{7}{3} = 2\frac{1}{3}\end{array}\)
Kunci Jawaban: D
- (A) \(\frac{{c - ac + {a^2}}}{{c + a - 1}}\)
- (B) \(\frac{{{c^2} + ac - {a^2}}}{{c + 1 - a}}\)
- (C) \(\frac{{c + 2ac - {a^2}}}{{c + 1 - a}}\)
- (D) \(\frac{{2{c^2} + ac - {a^2}}}{{c + 1 - a}}\)
- (E) \(\frac{{c + ac - {a^2}}}{{c + 1 - a}}\)
\(\begin{array}{l}P(x) = {(x - a)^5} + {(x - b)^4} + (x - c)\\habis\,\,\,dibagi\,\\{x^2} - (a + b)x + ab = \left( {x - a} \right)\left( {x - b} \right)\\ \Rightarrow P(a) = 0\\{(a - a)^5} + {(a - b)^4} + a - c = 0\\{(a - b)^4} = c - a\,\,..............1)\\ \Rightarrow \,P(b) = 0\\{(b - a)^5} + {(b - b)^4} + b - c = 0\\ - 1 \cdot {(a - b)^4} \cdot (a - b) + b - c = 0\\sunstitusi\,\,\,........1)\\ - 1 \cdot (c - a) \cdot (a - b) + b - c = 0\\ - 1 \cdot (ac - bc - {a^2}{\rm{ }} + {\rm{ }}ab){\rm{ }} + {\rm{ }}b - c = 0\\b \cdot (c + 1{\rm{ }} - a){\rm{ }} + {\rm{ }}{a^2} - ac - c{\rm{ }} = {\rm{ }}0\\b.(c + 1 - a){\rm{ }} = {\rm{ }}c{\rm{ }} + {\rm{ }}ac - {a^2}\\b{\rm{ }} = {\rm{ }}\frac{{c{\rm{ }} + {\rm{ }}ac - {a^2}}}{{c + 1 - a}}\end{array}\)
Kunci Jawaban: E
- (A) c < - 46
- (B) c < - 33
- (C) c < - 23
- (D) c < 23
- (E) c < 46
\(\begin{array}{l}{(0,25)^{( - {x^2} + 4x - c)}}{\rm{ < (0}}{\rm{,625}}{{\rm{)}}^{( - x2 - 4x + 5)}}\\{(\frac{1}{4})^{( - {x^2} + 4x - c)}}{\rm{ < }}{\left( {{\rm{ (}}\frac{1}{4}{{\rm{)}}^2}} \right)^{( - {x^2} - 4x + 5)}}\\ - {x^2} + 4x - c{\rm{ }} > {\rm{ - }}2{\rm{ }}{x^2} - 8x + {\rm{ }}10\\{x^2}{\rm{ }} + {\rm{ }}12x - (z + 10) > {\rm{ }}0\,\,\,(definit\,\,positif)\\^o)\,\,\,\,a = 1\,\,\,(a > 0)\\^o)\,\,\,\,D < 0\\{b^{2{\rm{ }}}} - 4ac{\rm{ }} < {\rm{ }}0\\{\rm{144 + 4 (c + 10) < 0}}\\{\rm{144 + 4c + 40 < 0}}\\4{\rm{c + 184 < 0}}\\{\rm{4c < - 184}}\\{\rm{c < - 46}} & & \end{array}\)
Kunci Jawaban: A
- (A) 16
- (B) 18
- (C) 25
- (D) 64
- (E) 100
\(\begin{array}{l}{x_1},\,\,\,{x_2}\\{16^x} - {4^x} - 2 \cdot {4^{x + 2}} - {4^{x + 3}} + a = 0\\misal\,\,\,{4^x} = p\\ \Rightarrow {P^2} - P - (2 \cdot 16 \cdot P) - (64 \cdot P) + A = 0\\{P^2} - 97P + A = 0\\ \Rightarrow {P_1} \cdot {P_2} = \frac{c}{a}\\{4^{{x_1}}} \cdot {4^{{x_2}}} = \frac{a}{1}\,\,\,\,\,\,\,\, \leftarrow {4^{{x_1}}} \cdot {4^{{x_2}}} = {4^{{x_1}}}^{ + {x_2}}\\{4^{{x_1}}}^{ + {x_2}} = a\\{\left( 4 \right)^{^2\log 5 + 1}} = 1\,\,\,\,\,\,\,\, \leftarrow {\,^2}\log 5 + {\,^2}\log 2 = {\,^2}\log 10\\{\left( {{2^2}} \right)^{^2\log 10}} = a\,\,\,\,\,\,\, \leftarrow \,\,{2^{^2\log {{10}^2}}} = {10^2}\\{10^2} = a\\a = 100\end{array}\)
Kunci Jawaban: E
- (A) \( - \frac{1}{2}\)
- (B) \( - \frac{1}{4}\)
- (C) \(\frac{1}{8}\)
- (D) \(\frac{1}{4}\)
- (E) \(\frac{1}{8}\)
\(\begin{array}{l}as\\\frac{{Lim}}{{x \to 1}}{\rm{ }}\frac{{(\sqrt {5 - x} - 2).(\sqrt {2 - x} + 1}}{{1 - x}}\\dalil\,\,\,L'Hospital\\\frac{{Lim}}{{x \to a}}{\rm{ }}\frac{{f(x)}}{{g(x)}}{\rm{ = }}\frac{{f'\,(a)}}{{g'\,(a)}}{\rm{ }}\\ \Rightarrow {\rm{ }}\frac{{\left( {\frac{{ - 1}}{{2\sqrt {5 - x} }}} \right).\left( {\sqrt {2 - x} + 1} \right) + \left( {\sqrt {5 - x - 2} } \right).\left( {\frac{{ - 1}}{{2\sqrt {2 - x} }} + 1} \right)}}{{ - 1}}\\ = {\rm{ }}\frac{{\left( {\frac{{ - 1}}{{2\sqrt {5 - 1} }}} \right).\left( {\sqrt {2 - 1} + 1} \right) + \left( {\sqrt {5 - 1 - 2} } \right).\left( {\frac{{ - 1}}{{2\sqrt {2 - 1} }} + 1} \right)}}{{ - 1}}\\{\rm{ = }}\frac{{( - 1\frac{1}{4}).(2) + (0).(\frac{1}{2})}}{{ - 1}} = {\rm{ }}\frac{1}{2}\end{array}\)
Kunci Jawaban: E
- (A) \(\frac{{{r^3} - {r^2} - r}}{{r - 1}}\)
- (B) \(\frac{{{r^3} - {r^2} + r}}{{r - 1}}\)
- (C) \(\frac{{{r^3} + {r^2} + r}}{{r + 1}}\)
- (D) \(\frac{{{r^3} + {r^2} - r}}{{r - 1}}\)
- (E) \(\frac{{{r^3} - {r^2} + r}}{{r + 1}}\)
\(\begin{array}{l}Diketahui:\,\,\,\,{U_1},\,\,\,{U_2},\,\,\,{U_3},{\rm{ }}....... \to {\rm{ barisan geometri}}\\ \Rightarrow {\rm{ }}{{\rm{U}}_3} - {U_6} = x\\ \Rightarrow {\rm{ }}{{\rm{U}}_2} - {U_4} = y\\\frac{x}{y} = \frac{{a{r^2} - a{r^5}}}{{ar - a{r^3}}}\\ = {\rm{ }}\frac{{a{r^2}.(1 - {r^3})}}{{ar.(1 - {r^2})}}\\ = {\rm{ }}\frac{{R.(1 - R).(1 + R + {R^2})}}{{(1 - R).(1 + R)}}\\ = \frac{{{r^3} + {r^2} + r}}{{1 + r}}\end{array}\)
Kunci Jawaban: E
- (A) \\(0 < x < \frac{{5\pi }}{{12}}\)
- (B) \(0 < x < \frac{\pi }{{12}}\)
- (C) \(\frac{\pi }{6} < x < \frac{\pi }{3}\)
- (D) \(\frac{{5\pi }}{{12}} < x < \frac{{7\pi }}{{12}}\)
- (E) \( - \frac{{7\pi }}{{12}} < x < \frac{\pi }{{12}}\)
\(\begin{array}{l}f(x) = - \sqrt {{\rm{co}}{{\rm{s}}^2}x + \frac{x}{2} + \pi } ,\,\,\,\, - \pi < x < 2\pi \\syarat\,\,\,fungsi\,\,\,turun:f'(x) < 0\\f'(x) = - \frac{1}{2}{\left( {{\rm{co}}{{\rm{s}}^2}x + \frac{x}{2} + \pi } \right)^{ - 1/2}}\left( {2\cos x \cdot \sin x + \frac{1}{2}} \right)\\ - \left[ {\frac{{ - 2.\cos {\rm{ }}x{\rm{ }}.\sin {\rm{ x + }}\frac{1}{2}}}{{2 \cdot \sqrt {{{\cos }^2}x + \frac{x}{2} + \pi } }}} \right] < 0\\{\rm{2 cos x }}{\rm{. sin x }}\left( {{\rm{ - }}\frac{1}{2}} \right) < 0\\ & {\rm{sin 2x < }}\frac{1}{2}\\Pembuat\,\,\,nol\,\,\,untuk:\\\sin 2x = \frac{1}{2}\\ \Rightarrow \,\,\sin 2x = \sin \frac{\pi }{6}.......1)\\2x = \frac{\pi }{6} + 2{\rm{ k }}\pi \\x = \frac{\pi }{{12}}{\rm{ + k }}\pi \\untuk\,\,\,nilai\,\,\,k:\,\,\\k = - 1 \to x = 1\frac{{11}}{{12}}\pi \\k = 0 \to x = \frac{1}{{12}}\pi \\k = 1 \to x = \frac{{13}}{{12}}\pi \\ \Rightarrow \,\,\sin 2x = \sin \frac{{5\pi }}{6}.......2)\\2x = \frac{{5\pi }}{6} + 2k\pi \\x = \frac{{5\pi }}{{12}} + k{\rm{ }}\pi \\k = - 1 \to x = - \frac{7}{{12}}\pi \\k = 0 \to x = \frac{5}{6}\pi \\k = 1 \to x = \frac{{17}}{{12}}\pi \\dari\,\,\,garis\,\,\,bilangan\,\,\,dibawah\\HP = \left\{ { - \frac{{7\pi }}{{12}} < x < \frac{\pi }{{12}}} \right\}\end{array}\)
Kunci Jawaban: E
- (A) \(7\frac{1}{3}\)
- (B) \(6\frac{2}{3}\)
- (C) 6
- (D) \(5\frac{2}{3}\)
- (E) 5
- (A) 9
- (B) 12
- (C) 15
- (D) 18
- (E) 21
\(Ax + By - 4C = 0\) nilai A, B, dan C dipilih dari bilangan (0,1, 4, ...16) dan tidak boleh sama Suatu garis dianggap beda bila perbandingan A, B dan 4C tidak berulang. \(\begin{array}{l}A = 0 \Rightarrow y = \frac{{4c}}{B} \to \frac{{4(1)}}{4} \ne \frac{{4(1)}}{{16}} \ne \frac{{4(4)}}{1} \ne \frac{{4(16)}}{1} \to 4{\rm{ }}garis\\B = 0 \Rightarrow x = \frac{{4c}}{A} \to \frac{{4(1)}}{4} \ne \frac{{4(1)}}{{16}} \ne \frac{{4(4)}}{1} \ne \frac{{4(16)}}{1} \to 4{\rm{ }}garis\\C = 0 \Rightarrow {\rm{ }}\frac{{Ax}}{y} \to \frac{{ - B}}{A} \to \frac{1}{4} \ne \frac{{ - 1}}{{16}} \ne \frac{{ - 4}}{1} \ne \frac{{ - 16}}{1} \to 4{\rm{ }}garis\\A \ne B \ne C \ne 0 \to A:B:4C \to {3^p}3 = 6{\rm{ garis}}\\Total = 4 + 4 + 4 + 6 = 16\,\,\,garis\end{array}\)
Kunci Jawaban: D
- (A) \(\frac{{161}}{{180}}\)
- (B) \\(\frac{{155}}{{180}}\)
- (C) \(\frac{{25}}{{180}}\)
- (D) \(\frac{{19}}{{180}}\)
- (E) \(\frac{{11}}{{180}}\)
Kelas A (30 orang) \( < _{30 - x = banyak{\rm{ wanita kelas A}}}^{x = banyak{\rm{ pria kelas A}}}\)
Kelas B (30 orang) \( < _{30 - Y = banyak{\rm{ wanita kelas B}}}^{y = banyak{\rm{ pria kelas B}}}\) \(\begin{array}{r}P{\rm{ }}(1{\rm{ }}pria{\rm{ }}A\,\,\,dan\,\,\,1\,\,\,pria{\rm{ }}B{\rm{ }}){\rm{ }} = \frac{{11}}{{36}}\\\frac{x}{{30}} \cdot \frac{y}{{30}}{\rm{ = }}\frac{{11}}{{36}}{\rm{ }}\\\frac{{xy}}{{900}}{\rm{ = }}\frac{{11}}{{36}}{\rm{ }}\\\frac{{xy}}{{25}}{\rm{ = }}11\\xy = 25 \times 11 < _{y = 11}^{x = 25}{\rm{ }}\\P{\rm{ (minimal}}\,\,\,1\,\,\,pria{\rm{ }}) = \\1 - P(1\,\,\,wanita\,\,\,A\,\,\,dan\,\,\,1\,\,\,wanita\,\,\,B)\\ = 1 - \left[ {\frac{{(30 - x)}}{{30}} \times \frac{{(30 - y)}}{{30}}} \right]\\{\rm{ = 1 - }}\frac{{5 \times 19}}{{30 \times 30}}{\rm{ }}\\ = \frac{{161}}{{180}}\end{array}\)
Kunci Jawaban: A
- (A) -2
- (B) -1
- (C) 1
- (D) 2
- (E) 3
\(\begin{array}{r}f(x) = - {x^3} + 3x - c \to - 1 \le x \le 2\\ \Rightarrow f'\,(x) = 0\\ - 3{x^2} + 3 = 0\\{x^2} - 1 = 0\\x = 1\,\,\,dan\,\,\,x = - 1\\ \Rightarrow f''\,(x) = - 6x\\f''\,( - 1) = - 6( - 1) = 6\\jadi\,\,\min imum\,\,\,saat\,\,\,x = - 1\\ \Rightarrow f( - 1) = - 1{( - 1)^3} + 3( - 1) - c\\f( - 1) = - 2 - c\\ \Rightarrow r = 1 - \sqrt 3 \\ \Rightarrow {U_2} - {U_1} = f'(0)\\ar - a = - 3{(0)^2} + 3\\a(r - 1) = 3\\a(1 - \sqrt 3 - 1) = 3\\a = \frac{3}{{ - \sqrt 3 }} = - \sqrt 3 \\ \Rightarrow f( - 1) = \frac{a}{{1 - r}}\,\,\,\\ - 2 - c = \frac{{{\rm{ - }}\sqrt 3 }}{{1 - \left( {1 - \sqrt 3 } \right)}}\\ - 2 - c = - 1\\c = - 1\end{array}\)
Kunci Jawaban: A