Soal dan Pembahasan SBMPTN Matematika Dasar 2016 (Kode 320)
- (A) -5
- (B) -4
- (C) 0
- (D) 4
- (E) 5
Pembahasan:
Dari bentuk umum persamaan kuadrat: \(a{x^2}
+ bx + c = 0\) nilai: \({x_1} + {x_2} =
\frac{{ - b}}{a},\,\,\,dan\,\,{x_1} \times {x_2} = \frac{c}{a}\)
Diketahui dari soal: \({x^2} + ax + b = 0\)
dimana
b (negatif ) dan a bilangan bulat
\( \Rightarrow
{x_1} + {x_2} = \frac{{ - b}}{a} \equiv
- a\,\,dan\,\,{x_1} \times {x_2} = \frac{c}{a} \equiv b\)
\(\begin{array}{l}{x_1} + {x_2} = \left( {7 - \sqrt 7 } \right) + \left( {k + \sqrt 7 } \right)\\7 + k \equiv - a\end{array}\)
\({x_1} \times {x_2} = \left( {7 - \sqrt 7
} \right) \times \left( {k + \sqrt 7 } \right) \equiv b\) dan b adalah
bilangan real negatif,sedangkan \(\left( {7 - \sqrt 7 } \right)\) sudah positif
maka \(\left( {k + \sqrt 7 } \right)\) harus negative.
\(k + \sqrt 7 < 0\,\,\, \to \,\,\,k < - \sqrt 7 \)
Supaya nilai a paling kecil maka nilai k
haruslah yang paling besar pada persamaan:
\(\begin{array}{l} - a = 7 + k\\a = - \left( {7 + \left( { - 3} \right)}
\right)\\a = - 4\end{array}\)
💥 Kunci Jawaban: C
- (A) \(\frac{{31}}{{18}}\)
- (B) \(\frac{{31}}{9}\)
- (C) \(\frac{{32}}{{18}}\)
- (D) \(\frac{{33}}{9}\)
- (E) \(\frac{{33}}{{18}}\)
Pembahasan:
Diketahui
\({A^{2x}} = 2\) maka
\(\begin{array}{c}\frac{{{A^{5x}} - {A^{ -
5x}}}}{{{A^{3x}} + {A^{ - 3x}}}} = \frac{{{{\left( {\sqrt 2 } \right)}^5} -
{{\left( {\sqrt 2 } \right)}^{ - 5}}}}{{{{\left( {\sqrt 2 } \right)}^3} +
{{\left( {\sqrt 2 } \right)}^{ - 3}}}}\\ = \frac{{4\sqrt 2 - \frac{1}{{4\sqrt 2 }}}}{{2\sqrt 2 + \frac{1}{{2\sqrt 2 }}}} \times \left(
{\frac{{4\sqrt 2 }}{{4\sqrt 2 }}} \right)\\ = \frac{{32 - 1}}{{16 + 2}}\\ =
\frac{{31}}{{18}}\end{array}\)
💥 Kunci Jawaban: A
- (A) \(\frac{1}{2}\)
- (B) 1
- (C) 2
- (D) \(\frac{{12}}{5}\)
- (E) 3
Pembahasan:
\(\begin{array}{l}{L_{arsiran}} = \frac{1}{2}L.P.panjang\\\frac{1}{2}\left[ {4 \times \left( {5m + m} \right)} \right] = \frac{1}{2}\left[ {4 \times 12} \right]\\4 \times 6m = 4 \times 12\\m = 2\end{array}\)
💥 Kunci Jawaban: C
- (A) \(x \le - \frac{9}{2}\) atau \(x > 3\)
- (B) \(x \le - \frac{9}{2}\) atau \( - 2 < x < 3\)
- (C) \( - \frac{9}{2}\, < x < - 2\) atau \( - 2 < x < 3\)
- (D) \( - \frac{9}{2}\, \le x < 3\)
- (E) x<-3 atau -2<x<3
Pembahasan
\(\begin{array}{l}\frac{x}{{x - 3}} - \frac{{x + 3}}{{x + 2}} \le 0\\\frac{{x\left( {x + 2} \right) - \left( {x - 3} \right)\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 2} \right)}} \le 0\\\frac{{x2 + 2x - x2 + 9}}{{\left( {x - 3} \right)\left( {x + 2} \right)}} \le 0\\\frac{{2x + 9}}{{\left( {x - 3} \right)\left( {x + 2} \right)}} \le 0\end{array}\)
💥 Kunci Jawaban: B
- (A) 7
- (B) 5
- (C) 3
- (D) -5
- (E) -7
Pembahasan:
\(\begin{array}{l}(x,y)\,\,di\,\,refleksi\,\,x = h\,\,menjadi:\,\,\left( {2h - x,y} \right)\\\left( {x,y} \right)\,\,\,di\,\,refleksi\,\,x = 4\,\,\,menjadi:\,\left( {8 - x,y} \right)\\x' = 8 - x \to x = 8 - x\\y' = y\\Benda:\,\,\,y = {x^2} - 2x - 3\\Bayangan:\,\,\,y' = {\left( {8 - x} \right)^2} - 2\left( {8 - x} \right) - 3\\y = {x^2} - 14x + 45 \equiv y = {x^2} - \left( {9 + a} \right)x + 9a\\maka:45 = 9a\,\,\,\, \Rightarrow \,\,\,\,a = 5\end{array}\)
💥 Kunci Jawaban: B
- (A) 144
- (B) 108
- (C) 72
- (D) 36
- (E) 35
Pembahasan:
- 7 Finalis terdiri atas : 4 pria (P) dan 3 wanita (W)
- Banyak urutan tampil selang seling = 4!\( \times \)3! = 24\( \times \)6 = 144 cara
- Terdapat 2 finalis yang berasal dari SMA "A" (1 Pria = PA dan 1 Wanita = WA ) maka ada 6 buah kemungkinan finalis dari SMA A tampil berurutan yaitu seperti pada ilustrasi berikut
- sehingga, banyaknya susunan yang mungkin agar pria dan wanita dari SMA "A" tampil scara brurutan adalah enam kali dari banyaknya cara menyusun 3 pria selain SMA "A" secara permutasi dan banyaknya cara menyusun 2 wanita selain SMA "A" secara permutasi yaitu:\(6.3{P_3}.2{P_2} = 6 \times \left( {3 \times 2 \times 1} \right) \times \left( {2 \times 1} \right) = 72\)
- Banyaknya cara finalis dari SMA "A" tidak tampil berurutan adalah 144-72 =72 cara
- (A) 2
- (B) 1
- (C) 0
- (D) -1
- (E) -2
Pembahasan:
\(\begin{array}{c}f\left( x \right) = ax - 1\\g\left( x \right) = x + 1\\\left( {fog} \right)\left( {x + 1} \right) = \left( {gof} \right)\left( {x + 1} \right)\\f\left( {g\left( x \right)} \right) = g\left( {f\left( x \right)} \right)\\a\left( {x + 1} \right) - 1 = \left( {ax - 1} \right) + 1\\ax + a - 1 = ax\end{array}\)
Perhatikan koefisiennya yang akan kita samakan:
Koefisien x: a = 1, maka
\(\begin{array}{c}f(2) - g(1) = \left( {1(2) - 1} \right) - \left( {1 + 1} \right)\\ = 1 - 2 = - 1\end{array}\)
💥 Kunci Jawaban: D
- (A) \({g^{ - 1}}(x) + 11\)
- (B) \({g^{ - 1}}(x) + 9\)
- (C) \({g^{ - 1}}(x) + 6\)
- (D) \({g^{ - 1}}(\frac{x}{2}) + 6\)
- (E) \({g^{ - 1}}(2x) + 6\)
Pembahasan:
\(\begin{array}{l}f\left( {x + 5} \right) = g\left( {2x - 1} \right)\\f\left( x \right) = g\left( {2\left( {x - 5} \right) - 1} \right)\\f\left( x \right) = g\left( {2x - 11} \right)\\{f^{ - 1}}\left( x \right) = {\left( {goh} \right)^{ - 1}}\left( x \right)\\h\left( x \right) = 2x - 11 \to {h^{ - 1}}\left( x \right) = \frac{{x + 11}}{2}\\{f^{ - 1}}\left( x \right) = \left( {{h^{ - 1}}o{g^{ - 1}}} \right)\left( x \right)\\{f^{ - 1}}\left( x \right) = {h^{ - 1}}\left( {{g^{ - 1}}\left( x \right)} \right)\\{f^{ - 1}}\left( x \right) = \frac{{g - 1\left( x \right) + 11}}{2}\\2.{f^{ - 1}}\left( x \right) = {g^{ - 1}}\left( x \right) + 11\end{array}\)
💥 Kunci Jawaban: A
- (A) 0
- (B) \(\frac{1}{3}\)
- (C) \(\frac{1}{2}\)
- (D) 1
- (E) 3
Pembahasan:
\(AB = \left[ {\begin{array}{*{20}{c}}{2a}\\a\end{array}\begin{array}{*{20}{c}}{}&{}\end{array}\begin{array}{*{20}{c}}{b + 1}\\{2b}\end{array}} \right]\)
\(\left| {AB} \right| = 0 \to \) tidak punya invest
\(4ab - a\left( {b + 1} \right) = 0\)
\(4ab - ab - a = 0\)
\(3ab = a\)
\( = \frac{1}{3}\)
💥 Kunci Jawaban: B
- (A) 29
- (B) 55
- (C) 66
- (D) 95
- (E) 121
Pembahasan:
\(\begin{array}{l}\log \left( {{a^3}{b^7}} \right) - \log \left( {a{b^4}} \right) = \log \left( {{a^6}{b^9}} \right) - \log \left( {{a^3}{b^7}} \right)\\\log \left[ {\frac{{{a^3}{b^7}}}{{a{b^4}}}} \right] = \log \left[ {\frac{{{a^6}{b^9}}}{{{a^3}{b^7}}}} \right]\\{a^2}{b^3} = {a^3}{b^2}\\a = b\\ \Rightarrow \log {a^5},\log {a^{10}},\log {a^{15}}\\{u_1} = \log {a^5}\\beda = {u_2} - {u_1} = \log {a^{10}}\log {a^5} = \log {a^5}\\ \Rightarrow {u_{11}} = a + 10(beda) = \log {a^5} + 10.\log {a^5}\\\log {a^p} = \log {a^{55}}\\p = 55\end{array}\)
💥 Kunci Jawaban: B
- (A) a + b + c
- (B) a – b = c
- (C) a + b - c
- (D) -a – b + c
- (E) b + c – a
\(\begin{array}{l}p + q = c\\{\underline {q + r = a} _ + }\\p - r = c - a\\{\underline {p + r = b} _ + }\\2p = b + c - a\\2AY = b + c - a\end{array}\)
💥 Kunci Jawaban E
💦 Soal No 57
- (A) 1
- (B) 2
- (C) 3
- (D) 4
- (E) 5
Pembahasan:
\(\begin{array}{l}\frac{{{x_1} + {x_n}}}{2} = 7\\{x_1} + {x_n} = 14\end{array}\)
Angka yang memenuhi untuk dipasangkan adalah
👉 10 dan 4
👉 9 dan 5
👉 8 dan 6 yang lainnya tidak memenuhi lagi.
jadi jangkauan yang bisa terjadi hanya dari pasangan tersebut ada sebanyak 3
💥 Kunci Jawaban: C
- (A) -4
- (B) -2
- (C) 0
- (D) 2
- (E) 8
Pembahasan:
\( \Rightarrow \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - 4}}{{x + 2}} = \mathop {\lim }\limits_{x \to 1} a{x^2} + b\)
kita selesaikan kedua ruas tersebut, diperoleh
\(\begin{array}{l}\mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} a{x^2} + b\\\mathop {\lim }\limits_{x \to - 2} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 1} a{x^2} + b\\\left( { - 2 - 2} \right) = a{(1)^2} + b\\a + b = - 4\end{array}\)
\(\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 2} a{x^2} + b = 2\\4a + b = 2...\,2)\\a + b = - 4\\{\underline {4a + b = 2} _ + }\\ - 3a = - 6\\a = 2\,\, \to b = - 6\\maka:\\a - b = 8\end{array}\)
💥 Kunci Jawaban: E
- (A) 7
- (B) 6
- (C) 6\(\frac{1}{2}\)
- (D) -6
- (E) - 7
Pembahasan:
\(\begin{array}{l} \Rightarrow 3x + y = b\\\left( {1,1} \right):3\left( 1 \right) + 1 = b\\b = 4\\\left( {a, - 2} \right):3\left( a \right) - 2 = 4\\a = 2\\ \Rightarrow cx - dy = 1\\\left( {1,1} \right):c - d = 1\,\,...\,\,i)\\a + b + c - d = 2 + 4 + 1 = 7\end{array}\)
💥 Kunci Jawaban: A
- (A) x < 1
- (B) x > 1
- (C) x < 2
- (D) x < 1 atau\(\frac{3}{2}\) < x < 2
- (E) \(1 < x < \frac{3}{2}\,\,\,atau\,\,x > 2\)
Pembahasan:
\(\begin{array}{l}\frac{1}{{\left| {x - 2} \right|}} < \frac{1}{{1 - x}}\\ \Rightarrow untuk\,\,x < 1\,\,...\,\,i)\\ \Rightarrow \frac{{ - 1}}{{x - 2}} < \frac{1}{{1 - x}}\\\frac{{ - 1}}{{x - 2}} - \frac{1}{{1 - x}} < 0\\\frac{{ - 1 + x - x + 2}}{{\left( {x - 2} \right)\left( {1 - x} \right)}} < 0\\\frac{1}{{\left( {x - 2} \right)\left( {1 - x} \right)}} > 0\\x < 1\,\,atau\,\,x > 2\,\,...\,\,ii)\\irisan\,\,i)\,\,dan\,\,ii)\\HP:\,\,\,x < 1\end{array}\)
\(\begin{array}{l} \Leftrightarrow untuk\,\,1 < x < 2\,\,...\,\,i)\\ \Rightarrow \frac{{ - 1}}{{x - 2}} < \frac{{ - 1}}{{1 - x}}\\\frac{{ - 1}}{{x - 2}} + \frac{1}{{1 - x}} < 0\\\frac{{ - 1 + x + x + 2}}{{\left( {x - 2} \right)\left( {1 - x} \right)}} < 0\\\frac{{2x - 3}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} > 0\\x < 1\,\,atau\,\,\frac{3}{2} < x > 2\,\,...\,\,ii)\\irisan\,\,i)\,\,dan\,\,ii)\\HP:\,\,\,\frac{3}{2} < x > 2\end{array}\)
\(\begin{array}{l} \Leftrightarrow untuk\,\,x > 2\,\,...\,\,i)\\ \Rightarrow \frac{1}{{x - 2}} < \frac{{ - 1}}{{1 - x}}\\\frac{1}{{x - 2}} + \frac{1}{{1 - x}} < 0\\\frac{{1 - x + x - 2}}{{\left( {x - 2} \right)\left( {1 - x} \right)}} < 0\\\frac{{ - 1}}{{\left( {x - 2} \right)\left( {1 - x} \right)}} > 0\\\frac{1}{{\left( {x - 2} \right)\left( {x - 1} \right)}} < 0\\1 < x < 2\,\,...\,\,ii)\\irisan\,\,i)\,\,dan\,\,ii)\\HP:\,\,\,\left\{ {} \right\}\end{array}\)
Jadi himpunan penyelesaiannya adalah:
\(x < 1\,\,atau\,\,\frac{3}{2} < x < 2\)
Kunci Jawaban D