Soal Try Out Prediksi UTBK - Matematika IPA
- (A) \(5\sqrt 3 \) cm
- (B) \(6\sqrt 2 \) cm
- (C) \(6\sqrt 3 \) cm
- (D) \(6\sqrt 6 \) cm
- (E) \(7\sqrt 3 \) cm
Pembahasan:
Segitiga
ATC adalah sama kaki (AT = CT)
\(A{T^2} = A{E^2} + E{T^2}\)
\( = {9^2} + {\left( {\frac{9}{2}\sqrt 2 }
\right)^2}\)
\( = 81 + \frac{{81}}{2} = \frac{{243}}{2}\)
\(AT = \frac{9}{2}\sqrt 6 \)
· Perbandingan luas segitiga
\(\frac{1}{2}ACTO = \frac{1}{2}AR \cdot TC\)
\(9\sqrt 2
\times 9 = AR \cdot \frac{9}{2}\sqrt 6 \)
\(AR = \frac{{9\sqrt 2 \times 9}}{{\frac{9}{2}\sqrt 6 }} = \frac{{18\sqrt
2 }}{{\sqrt 6 }}\)
💥 Kunci Jawaban: C
- (A) 7
- (B) 5
- (C) 3
- (D) -5
- (E) -7
Pembahasan:
Titik
(x, y) di cerminkan terhadap garis x = 4 menjadi (2(4) – x, y) = (8 – x, y)
Benda dicerminkan terhadap garis x = 4 tinggal diganti x nya menjadi 8 – x \(y = {\left( {8 - x} \right)^2} - 2\left( {8 - x} \right) - 3\)
\(y = 64 - 16x + {x^2} - 16 + 2x - 3\)
\(y = {x^2} - 14x + 45\) dan dari soal diketahui bayangannya maka
14 = 9 + a
a = 5
💥 Kunci Jawaban: B
- (A) -4
- (B) -2
- (C) -1
- (D) 2
- (E) 4
Pembahasan:
\(\mathop {\lim }\limits_{x \to 0} \frac{{g\left( x
\right)}}{x} = \frac{1}{2}\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{g\left( x \right)}}{{\left( {\sqrt {1 - x} - 1} \right)}} = \frac{{\sqrt {1 - x} + 1}}{{\left( {\sqrt {1 - x} + 1} \right)}}\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{g\left( x
\right)\left( {\sqrt {1 - x} + 1}
\right)}}{{1 - x - 1}}\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{g\left( x \right)}}{x} \cdot \frac{{\sqrt {1 - x} + 1}}{{\left( { - 1} \right)}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{{g\left( x \right)}}{x} \cdot \mathop {\lim }\limits_{x \to 0} - \left( {\sqrt {1 - x} + 1} \right)\)
\( = \left( {\frac{1}{2}} \right)\left( { - \left( {\sqrt {1 - 0} + 1} \right)} \right)\)
\( = \frac{1}{2}\left( { - 2} \right) = - 1\)
💥 Kunci Jawaban: C
- (A) 720
- (B) 840
- (C) 960
- (D) 1080
- (E) 1200
Pembahasan:
Diketahui
barisan aritmetika \({U_1} = a\) dan beda = 2a
\({u_1} + {u_2} + ... + {u_5} = 100\)
\({s_5} = \frac{5}{2}\left( {2{u_1} + \left( {5 - 1}
\right)b} \right)\)
\(100 = \frac{5}{2}\left( {2a + 4\left( {2a} \right)}
\right)\)
\(100 = 5\left( {a + 4a} \right)\)
\(100 = 25a \to a = 4\)
\({u_2} + {u_4} + .... + {u_{20}} = {s_{10}}\) (genap)
\({s_{10}} = \frac{{10}}{2}\left( {2{u_2} + \left(
{10 - 1} \right)2b} \right)\)
\({s_{10}} = \frac{{10}}{2}\left( {2\left( {a + b}
\right) + 9\left( {2b} \right)} \right)\)
\( = 5\left( {2\left( {a + 2a} \right) + 18\left(
{2a} \right)} \right)\)
\( = 5\left( {6a + 36a} \right)\)
\( = 5\left( {42\left( 4 \right)} \right) = 840\)
💥 Kunci Jawaban: B
- (A) 14
- (B) 15
- (C) 16
- (D) 17
- (E) 18
Pembahasan:
\(f\left( x \right) = 4\cos + P\sin x + 8\)
\(f\left( x \right) = K\cos \left( {x - x} \right) +
8\)
\( \Rightarrow K = \sqrt {{4^2} + {p^2}} = \sqrt {10 + {p^2}} \)
\( \Rightarrow f\left( x \right)\) minimum = 2
2 =
-K + 8
K =
6
\( \Rightarrow f\left( x \right)\) maksimum = k + 8
= 6 + 8
= 14
💥 Kunci Jawaban: A
- (A) pencerminan terhadap sumbu x
- (B) pencerminan terhadap sumbu y
- (C) perputaran + p terhadap O
- (D) perputaran - p terhadap O
- (E) perputaran + p terhadap O
Pembahasan:
P =
Matriks refleksi y = x
=\(\left(\begin{array}{l}\begin{array}{*{20}{c}}0&1\end{array}\\\begin{array}{*{20}{c}}1&0\end{array}\end{array}
\right)\)
Q =
Matriks rotasi seperempat putaran
=\(\left(\begin{array}{l}\begin{array}{*{20}{c}}{\cos90}&{ - \sin 90}\end{array}\\\begin{array}{*{20}{c}}{\sin 90}&{\cos
90}\end{array}\end{array} \right) = \left(
\begin{array}{l}\begin{array}{*{20}{c}}0&{-1}\end{array}\\\begin{array}{*{20}{c}}1&0\end{array}\end{array} \right)\)
PR =
Q -> R = \({P^{ - 1}}Q\)
\(R=\frac{1}{{0-1}}\left(\begin{array}{l}\begin{array}{*{20}{c}}0{-1}\end{array}\\\begin{array}{*{20}{c}}{ - 1}&0\end{array}\end{array}
\right)\left( \begin{array}{l}\begin{array}{*{20}{c}}0&{ - 1}\end{array}\\\begin{array}{*{20}{c}}1&{\,\,\,0}\end{array}\end{array}
\right)\)
\(R=\frac{1}{{0-1}}\left(\begin{array}{l}\begin{array}{*{20}{c}}0&1\end{array}\\\begin{array}{*{20}{c}}1&0\end{array}\end{array}
\right)\left( \begin{array}{l}\begin{array}{*{20}{c}}0&1\end{array}\\\begin{array}{*{20}{c}}1&0\end{array}\end{array}
\right) = \left(
\begin{array}{l}\begin{array}{*{20}{c}}1&{\,\,\,0}\end{array}\\\begin{array}{*{20}{c}}0&{
- 1}\end{array}\end{array} \right)\)
R =
Matriks refleksi terhadap sumbu x
💥 Kunci Jawaban: A
- (A) \(\sqrt {{b^2} - 4ac} \)
- (B) - \(\sqrt {{b^2} - 4ac} \)
- (C) - b + 2 \(\sqrt {ac} \)
- (D) b + 2 \(\sqrt {ac} \)
- (E) b2 4ac
Pembahasan:
\(a{x^2} + bx + c = 0\)
· akar – akarnya p & q dan p > q
\(p = \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\)
· gradient garis singgung
\(m
= {y^1} = 2ax + b\)
\( = 2a\left( {\frac{{ - b + \sqrt {{b^2} - }
4ac}}{{2a}}} \right) + b\)
\( = \sqrt {{b^2} - 4ac} \)
💥 Kunci Jawaban: A
- (A) - 59
- (B) - 5
- (C) 0
- (D) 5
- (E) 59
Pembahasan:
\(p\left( x \right) = {x^5} + a{x^3} + bx\)
· dibagi x – 2 bersisa 5
\(5 = p\left( 2 \right) = {2^5} + a\left( {{2^3}}
\right) + b\left( 2 \right)\)
5 =
32 + 8a + 2b
8a +
2b = -27
· Dibagi x + 2 bersisa p (-2)
\(p\left( { - 2} \right) = {\left( { - 2} \right)^5}
+ a{\left( { - 2} \right)^3} + b\left( { - 2} \right)\)
=
-32 – (8a +2b)
p
(-2) = -32 – (-27)
=
-32 + 27 = -5
💥 Kunci Jawaban: B
- (A) 10
- (B) 12
- (C) 14
- (D) 16
- (E) 18
Pembahasan:
Banyak
cara menempatkan penumpang :
Mobil
I Mobil II Banyak Cara
(1
sopir) (1 sopir)
1 3 \(C_1^4 = 4\)
2 2 \(C_2^4 = 6\)
3 1 \(C_3^4 = 4\)
Total = 14 cara
💥 Kunci Jawaban: C
- (A) 1 : 4 cos q
- (B) 4 : cos q
- (C) 4 cos q : 1
- (D) 1 : cos q
- (E) cos q : 4
Pembahasan:
\(\overline
x = 4\left| {\overline \upsilon } \right|\)
\(\overline x =
\left| u \right|\cos \theta \)
\(4\left| \upsilon \right| = \left| \upsilon \right|\cos \theta \)
\(\left| \upsilon \right| = \frac{1}{4}\left| \upsilon \right|\cos \theta \)
maka
\(\frac{{\left|
{\overline 4 } \right|}}{{\left| \upsilon
\right|}} = \frac{{\left| \upsilon
\right|}}{{\frac{1}{4}\left| \upsilon
\right|\cos \theta }}\)
\( = \frac{4}{{\cos \theta
}}\)
\(\left| u \right|:\left|
\upsilon \right| = 4:\cos \theta \)
💥 Kunci Jawaban: B
- (A) \(\frac{7}{{12}}\)
- (B) \(\frac{4}{{33}}\)
- (C) \(\frac{3}{{30}}\)
- (D) \(\frac{2}{{33}}\)
- (E) \(\frac{1}{{12}}\)
Pembahasan:
Dalam
kotak terdapat 11 bola terdiri dari 6 merah, 2 putih, dan 3 biru
Kejadian A adalah memilih 7 bola dengan
ketentuan bola merah 3kali bola putih
\(n\left( A \right) = C_3^6 \times C_1^2 \times
C_3^3\)
\( = \frac{{6!}}{{3! \times 3!}} \times 2 \times 1\)
\( = 20 \times 2 \times 1 = 40\)
· \(n\left(
S \right) = C_7^{11} = \frac{{11!}}{{7!\,\, \cdot 4!}}\)
\( = \frac{{11 \cdot 10 \cdot 9 \cdot 8}}{{4 \cdot 3
\cdot 2}} = 330\)
\(p\left( A \right) = \frac{{40}}{{330}} =
\frac{4}{{33}}\)
💥 Kunci Jawaban: B
- (A) 144\(\pi \)
- (B) 288\(\pi \)
- (C) 360\(\pi \)
- (D) 384\(\pi \)
- (E) 484\(\pi \)
Pembahasan:
Diketahui
panjang AC = 12
· Dari rumus kesebangunan
\(\frac{{BA}}{{BO}} = \frac{{OD}}{{AC}}\)
\(\frac{R}{{2R}} = \frac{r}{{12}}\)
\(r = \frac{{12}}{2} = 6cm\)
r : R = 1 : 3
maka R = 3 x 6 = 18
Selisih luas kedua lingkaran
= \(\pi {R^2} - \pi {r^2} = \pi \left( {{R^2} - {r^2}}
\right)\_
= \(\pi \left( {{{18}^2} - {6^2}} \right) = 288\pi \)
💥 Kunci Jawaban: B
- (A) \(b \le 3\)
- (B) \(b \le - 3\)
- (C) \(b \le - 3\,\,atau\,\,b \ge 1\)
- (D) \( - 3\, \le \,b \le \,1\)
- (E) \(1\, \le \,b \le \,3\)
Pembahasan:
Titik
(2, 2) dimasukkan ke
\({\left(
{b - 3} \right)^2}{x^2} - {\left( {5 - b} \right)^2}{y^2} - b\left( {x + y}
\right) + 2 \le 0\)
\(4\left( {{b^2} - 6b + 9} \right) - 4\left( {25 -
10b + {b^2}} \right) - 6\left( 4 \right) + 28 \le 0\)
sama
– sama dibagi 4
\({b^2} - 6b + 9 - 25 + 10b - {b^2} - b + 7 \le 0\)
\(3b - 9 \le 0\)
\(b \le 3\)
💥 Kunci Jawaban: A
- (A) 5
- (B) 6
- (C) 7
- (D) 8
- (E) 9
Pembahasan:
\(\int\limits_0^2 {ax - b} \,\,dx = 4\)
\(\frac{1}{2}a{x^2} - \left. {bx} \right]_0^2 = 4\)
\(2a - 2b = 4\)
\(a - b = 2....\left( 1 \right)\)
· \(\frac{1}{3}{x^3}
+ 2\left. {bx} \right]_1^3 = 10\)
\(\left( {9 + 6b} \right) - \left( {\frac{1}{3} + 2b}
\right) = 10\)
\(8\frac{2}{3} + 4b = 10\)
\(4b = 1\frac{1}{3} = \frac{4}{3}\)
\(b = \frac{1}{3}\)
\(a - \frac{1}{3} = 2\)
\(a = 2\frac{1}{3} = \frac{7}{3}\)
maka 3a + 6b
=\(3\left( {\frac{7}{3}} \right) + 6\left( {\frac{1}{3}}
\right)\)
= 7
+ 2 = 9
💥 Kunci Jawaban: E
- (A) \(x < {a^2}\,\,{\rm{atau}}\,\,x > {a^{ - 1}}\)
- (B) \(x < {a^2}\,\,{\rm{atau}}\,\,x > {a^{ - 2}}\)
- (C) \({a^2} < x < {a^{ - 1}}\)
- (D) \({a^2} < x < {a^{ - 2}}\)
- (E) \({a^{ - 2}} < x < {a^2}\)
Pembahasan:
log
a X = \(^a\log x\)
misal \(^a\log x\) = p
\({p^2}
- p - 2 > 0\)
(p – 2) (p + 1) > 0
p < -1 atau p > 2
karena o < a < 1
< -1 atau > 2
x > atau x <
💥 Kunci Jawaban: A