Soal dan Pembahasan Matematika IPA SBMPTN 2016 (Kode 220)
- (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) \({(\,x - 20\,)^2} + {(\,y - 7\,)^2} = 25\)
- (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) \(\frac{p}{{1 - {p^2}}}\)
- (A) 1
- (B) 2
- (C) 3
- (D) 4
- (E) 5
\(\begin{array}{c}4{\cos ^2}t + 3\sec \,\,t = 3 + 4\cos \,\,t\\4{\cos ^2}t - 4\cos \,\,t = 3 - 3\sec \,\,t\\4\cos \,\,t\left( {\cos \,\,t - 1} \right) = 3\left( {1 - \sec \,\,t} \right)\\ = 3\left( {\frac{{\cos t}}{{\cos t}} - \frac{1}{{\cos t}}} \right)\\\left( {\cos \,\,t - 1} \right)\left( {4\cos \,\,t - \frac{3}{{\cos \,\,t}}} \right) = 0\\*\,\,\cos \,\,t = 1\,\,\,\,\,\\yang\,\,memenuhi\,\,\,t = {0^o}\\*\,4\cos \,\,t - \frac{3}{{\cos \,\,t}} = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\cos ^2}t = \frac{3}{4}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \,\,t = \pm \frac{1}{2}\sqrt 3 \,\\yang\,\,memenuhi\,\,t = \frac{\pi }{6},\frac{{2\pi }}{3},\frac{{4\pi }}{3}\frac{{5\pi }}{3}\\Kunci\,\,E\end{array}\)
- (A) st
- (B) 2st
- (C) 3st
- (D) 4st
- (E) 5st
- (A) \(\frac{{\sqrt 6 }}{6}\)
- (B) \(\frac{{\sqrt 6 }}{3}\)
- (C) \(\frac{{\sqrt 6 }}{2}\)
- (D) \(\frac{{\sqrt 2 }}{2}\)
- (E) \(\frac{{\sqrt 2 }}{2}\)
- (A) -27
- (B) -15
- (C) 0
- (D) 5
- (E) 9
\(\begin{array}{l}f\left( x \right) = h\left( x \right) \cdot \left( {{x^2} + 2x + 4} \right) + \left( {2x + 3} \right)\\x + f\left( x \right) = h\left( x \right) \cdot \left( {{x^2} + 2x + 4} \right) + \left( {3x + 3} \right)\,\,\,\,\,\,\,\,\,\\{\rm{dikuadratkan }}\,{\rm{lalu bagi dengan }}\,({x^2} + 2x + 4)\\\frac{{{{\left( {x + f\left( x \right)} \right)}^2}}}{{{x^2} + 2x + 4}} = \frac{{{{\left[ {h\left( x \right) \cdot \left( {{x^2} + 2x + 4} \right)} \right]}^2}}}{{{x^2} + 2x + 4}} + \frac{{2 \cdot h\left( x \right) \cdot \left( {{x^2} + {}^2x + 4} \right)\left( {3x + 3} \right)}}{{{x^2} + 2x + 4}} + \frac{{{{\left( {3x + 3} \right)}^2}}}{{{x^2} + 2x + 4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\widehat {tidak\,\,bersisa}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\widehat {\,\,tidak\,\,bersisa}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\widehat {\,\,\,bersisa}\\{x^2} + 2x + 4\mathop{\left){\vphantom{1{\frac{\begin{array}{l}9{x^2} + 18x + 9\\9{x^2} + 18x + 36\end{array}}{{ - 27}}}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{\frac{\begin{array}{l}9{x^2} + 18x + 9\\9{x^2} + 18x + 36\end{array}}{{ - 27}}}}}} \limits^{\displaystyle\,\,\, 9} - \,\,\,\,\,\, \Rightarrow \,\,\,sisanya - 27\\Kunci\,\,A\end{array}\)
- (A) 0 < x < 1
- (B) x > 1
- (C) x > 0
- (D) x > 3
- (E) 1 < x < 3
\(\begin{array}{l}{\rm{Jika grafik}}\,\,\,{y_1}\,\,{\rm{di bawah grafik }}{y_2}{\rm{ maka }}{y_1}{\rm{ < }}{y_2}\\ \Rightarrow {3^{x + 1}} - {\left( {\frac{1}{9}} \right)^x} < {3^x} + 1\\\,\,\,\,\,\,{3^x} \cdot 3 - \frac{1}{{{3^{2x}}}} < {3^x} + 1 \to {\rm{misal}}\,:\,P = 3x\\\,\,\,\,\,\,3P - \frac{1}{{{P^2}}} < P + 1\\\,\,\,\,\,\,\frac{{2P - \frac{1}{{{P^2}}} - 1 < 0}}{{2{P^3} - 1 - {P^2} < 0}}\,\,\,x\,{P^2}\\\,\,\,\,\,\,2{P^3} - {P^2} - 1 < 0\\(p - 1)(2{p^2} + p + 1) < 0\\p < 1\\ \Rightarrow {3^x} < {3^0}\\\,\,\,\,\,\,\,\,x < 0\\Kunci\,\,C\end{array}\)
- (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) \(\frac{1}{{a(a + b)}}\)
\(\begin{array}{l}\mathop {\lim }\limits_{x \to a} \frac{{{{\left( {\sqrt {x + b} - \sqrt {a + b} } \right)}^2}}}{{\left( {{x^2} - {a^2}} \right) \cdot \sin \left( {x - a} \right)}}\\ = \mathop {\lim }\limits_{x \to a} \frac{{{{\left( {\sqrt {x + b} - \sqrt {a + b} } \right)}^2}{{\left( {\sqrt {x + b} + \sqrt {a + b} } \right)}^2}}}{{\left( {x - a} \right)\left( {x + a} \right) \cdot \sin \left( {x - a} \right) \cdot {{\left( {\sqrt {x + b} + \sqrt {a + b} } \right)}^2}}}\\ = \mathop {\lim }\limits_{x \to a} \frac{{{{\left( {\left( {x + b} \right) - \left( {a + b} \right)} \right)}^2}}}{{\left( {x - a} \right) \cdot \left( {x + a} \right) \cdot \sin \left( {x - a} \right) \cdot {{\left( {\sqrt {x + b} + \sqrt {a + b} } \right)}^2}}}\\ = \mathop {\lim }\limits_{x \to a} \frac{{\left( {x - a} \right)\left( {x - a} \right)}}{{\left( {x - a} \right) \cdot \left( {x + a} \right) \cdot \sin \left( {x - a} \right) \cdot {{\left( {\sqrt {x + b} + \sqrt {a + b} } \right)}^2}}}\\ = \frac{1}{{\left( {a + a} \right){{\left( {\sqrt {a + b} + \sqrt {a + b} } \right)}^2}}} = \frac{1}{{\left( {a + a} \right){{\left( {\sqrt {a + b} + \sqrt {a + b} } \right)}^2}}}\\ = \frac{1}{{8a\left( {a + b} \right)}}\\Kunci\,\,B\end{array}\)
- (A) 40
- (B) 41
- (C) 42
- (D) 43
- (E) 44
\(\begin{array}{l}\\ \Rightarrow \frac{{{U_6}}}{{{U_4}}} = 16\\\,\,\,\,\,\frac{{a{r^5}}}{{a{r^3}}} = 16\\\,\,\,\,\,\,\,\,\,{r^2} = 16\\ \Rightarrow {U_1}^2 = r\\\,\,\,\,\,\,\,\,{a^2} = r\,\,\left( {r > 0} \right)\\\,\,\,\,\,\,\,\,{a^2} = 4\\\,\,\,\,\,\,\,\,\,\,a = 2\\ \Rightarrow {S_4} = \frac{{a \cdot \left( {{r^4} - 1} \right)}}{{r - 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \cdot \left( {256 - 1} \right)}}{1}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \cdot \left( {255} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 510\\Kunci...\end{array}\)
- (A) 10
- (B) 8
- (C) 6
- (D) 4
- (E) 3
\(\begin{array}{l}f(x) = {x^3} - 3x + a\,\\titik\,potong\,\,sumbu\,\,y(x = 0)\\10 = {0^3} - 3(0) + a \Rightarrow a = 10\\f(x) = {x^3} - 3x + 10\\syarat\,\,maksimum\,\,\\ \Rightarrow f'(x) = 0\\\,\,\,\,\,\,3{x^2} - 3 = 0\\\,\,\,\,\,x = \pm 1\\ \Rightarrow f''(x) = 6x\\\,\,\,\,\,\,f''(1) > 0\\maka\,\,pembuat\,\,\,maksimum\,\,x = 1\\dan\,\,nilai\,\,maksimum = f(1)\\ \Rightarrow {1^3} - 3(1) + 10 = 8\\Kunci\,\,B\end{array}\)
- (A) B
- (B) 2B
- (C) 3B
- (D) 4B
- (E) 5B
\(\begin{array}{l}x \in \left[ {0,1} \right]\\f\left( x \right) = f\left( {x + 2} \right){\rm{merupakan}}\,\,{\rm{fungsi}}\,\,{\rm{konstan}}\\\int\limits_0^2 {f\left( x \right)dx = B} \\{\rm{misal}}\,\,f\left( x \right) = c\\\int\limits_0^2 {c\,\,dx = \left. {cx} \right|_0^2} \Rightarrow B = 2c\\maka\\\int\limits_3^7 {f\left( {x + 8} \right)dx = \int\limits_3^7 {c\,\,dx} } \\ = \left. {cx} \right|\begin{array}{*{20}{c}}7\\3\end{array} = 4c\\ = 2\left( {2c} \right) = 2B\\Kunci\,\,B\end{array}\)
- (A) \(\frac{1}{3}\)
- (B) \(\frac{2}{3}\)
- (C) \(\frac{4}{3}\)
- (D) \(\frac{5}{3}\)
- (E) \(\frac{7}{3}\)
\(\begin{array}{l}{x^2} + 2a = 2a + 1\\{x^2} = 1\,\langle \begin{array}{*{20}{c}}{\,\,{x_1} = - 1}\\{{x_2} = 1}\end{array}\\L = \int\limits_{ - 1}^1 {2a + 1 - \left( {{x^2} + 2a} \right)dx} \\\,\,\, = \int\limits_{ - 1}^1 {1 - {x^2}\,dx} \\K = x - \frac{1}{3}\left. {{x^3}} \right]_{ - 1}^1\\\,\,\,\,\, = \left( {1 - \frac{1}{3}} \right) - \left( { - 1 + \frac{1}{3}} \right)\\\,\,\,\,\, = 1 - \frac{1}{3} + 1 - \frac{1}{3}\\\,\,\,\,\, = \frac{4}{3}\\Kunci\,\,C\end{array}\)
- (A) 48
- (B) 54
- (C) 60
- (D) 64
- (E) 72
\(\begin{array}{l}{\rm{Angka}}\,\,{\rm{di}}\,\,{\rm{C}}\,\,hanya\,\,{\rm{terisi}}\,\,\,{\rm{dengan}}\,\,\,{\rm{angka}}\,\,\,{\rm{6}}\,\,{\rm{atau}}\,\,{\rm{8}}\\{\rm{jika}}\,\,\,C = 6\,\,\,\, \Rightarrow \left[ 9 \right] \times \left[ 2 \right] \times \left[ 1 \right]\,\,\,\, = \,\,\,18\\{\rm{jika}}\,\,\,C = 8\,\,\,\, \Rightarrow \left[ 9 \right] \times \underline {\left[ 4 \right] \times \left[ 1 \right]\,\,\,\, = \,\,\,36} \,{\,_{\, + }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,54\\Kunci\,\,\,B\end{array}\)
- (A) \(\sqrt[2]{3}\)
- (B) \(\sqrt 3 \)
- (C) \(\frac{1}{2}\sqrt 3 \)
- (D) \(\frac{1}{3}\sqrt 3 \)
- (E) \(\frac{1}{4}\sqrt 3 \)
\(\begin{array}{l}{\rm{Gradien garis singgung kurva}}\,\\y = 3 - {x^2}\,\,adalah\,\,\,m = y' = 2x\\{\rm{I}}{\rm{.}}\,\,{\rm{Persamaan garis singgung}}\,\,{\rm{di }}\left( { - a,b} \right)\\*\,m = - 2a\\\, \Rightarrow y - b = - 2a\left( {x + a} \right)\\\,\,\,\,\,\,\,y = - 2a\,\,x - 2{a^2} + b\\{\rm{II}}{\rm{.}}\,\,{\rm{Persamaan garis singgung}}\,\,{\rm{di }}\left( {a,b} \right)\\*\,m = 2a\\\, \Rightarrow y - b = 2a\left( {x + a} \right)\\\,\,\,\,\,\,\,y = 2a\,\,x + 2{a^2} + b\\{\rm{Kita eliminasi kedua persamaan tersebut}}\\{\rm{untuk mendapatkan titik R yaitu R}}\,\,{\rm{ = }}\,\,{\rm{(0}}{\rm{,2}}{{\rm{a}}^{\rm{2}}}{\rm{ + b)}}\\{\rm{Jadi}}\,\,RS = (2{a^2} + b - b) = 2{a^2}\,\,dan\,\,\,SQ = a\\\tan \,\,Q = \frac{{RS}}{{SQ}} = \tan \,\,60\,(sama\,\,sisi)\\\,\frac{{2{a^2}}}{a} = \sqrt 3 \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,a = \frac{1}{2}\sqrt 3 \\Kunci jawaban\,\,C\end{array}\)
- (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