Pembahasan Soal UM UGM Matematika IPA Tahun 2019. Ujian tulis masuk UGM yang sering disingkat Utul UGM atau UM UGM merupakan salah satu harapan untuk bisa melanjutkan study di salah satu universitas terbaik Indonesia. Setelah mengalami kegagalan dalam beberapa test masuk Perguruan Tinggi Negri, masih ada kesempatan buat adik-adik untuk menjajal Soal UM UGM tentunya. Untuk itu, tetap semangat dan pelajari Soal-soal dan Pembahasannya.

Soal nomor 1:

Sebuah kotak memuat 6 bola merah dan 4 bola hitam. Tiga bola diambil satu per satu tanpa pengembalian. Jika bola ketiga terambil merah, maka banyak kemungkinannya adalah . . . .
A. 234
B. 243
C. 324
D. 342
E. 432

Formasi:
MMM → 6.5.4 = 120 cara
MPM → 6.4.5 = 120 cara
PMM → 4.6.5 = 120 cara
PPM → 4.3.6 = 72 cara
Total = 432 cara
jawab: E.

Soal nomor 2:

Diketahui penyelesaian dari pertidaksamaan

\frac{3^{x + 3} + 3^{x} - 36}{9^{x} - 9} \leq 3

$\dfrac{3^{x + 3} + 3^x - 36}{9^x - 9} \leq 3$ adalah

a \leq x < b

$a \leq x < b$ atau

x \geq c

$x \geq c$ . Nilai

a + 2 b + c = \dots

$a + 2b + c = \cdots$
A. 0
B. 1
C. 2
D. 3
E. 4

\frac{3^{x + 3} + 3^{x} - 36}{9^{x} - 9} \leq 3

$\dfrac{3^{x + 3} + 3^x - 36}{9^x - 9} \leq 3$

\frac{3^{3} {.3}^{x} + 3^{x} - 36}{3^{2 x} - 9} - 3 \leq 0

$\dfrac{3^3.3^x + 3^x - 36}{3^{2x} - 9} - 3 \leq 0$

\frac{{27.3}^{x} + 3^{x} - 36 - 3 (3^{2 x} - 9)}{3^{2 x} - 9} \leq 0

$\dfrac{27.3^x + 3^x - 36 - 3(3^{2x} - 9)}{3^{2x} - 9}\leq 0$

\frac{{27.3}^{x} + 3^{x} - 36 - {3.3}^{2 x} + 27}{3^{2 x} - 9} \leq 0

$\dfrac{27.3^x + 3^x - 36 - 3.3^{2x} + 27}{3^{2x} - 9}\leq 0$

\frac{- {3.3}^{2 x} + {28.3}^{x} - 9}{3^{2 x} - 9} \leq 0

$\dfrac{-3.3^{2x} + 28.3^x - 9}{3^{2x} - 9}\leq 0$

\frac{{3.3}^{2 x} - {28.3}^{x} + 9}{3^{2 x} - 9} \geq 0

$\dfrac{3.3^{2x} - 28.3^x + 9}{3^{2x} - 9}\geq 0$

Misalkan

3^{x} = p

$3^x = p$

\frac{3 p^{2} - 28 p + 9}{p^{2} - 9} \geq 0

$\dfrac{3p^2 - 28p + 9}{p^2 - 9} \geq 0$

\frac{(3 p - 1) (p - 9)}{(p + 3) (p - 3)} \geq 0

$\dfrac{(3p - 1)(p - 9)}{(p + 3)(p - 3)} \geq 0$ →

p \neq 3

$p \ne 3$

(p + 3) (3 p - 1) (p - 3) (p - 9) \geq 0

$(p + 3)(3p - 1)(p - 3)(p - 9) \geq 0$

p \leq - 3

$p \leq -3$ atau

\frac{1}{3} \leq p < 3

$\dfrac13 \leq p < 3$ atau

p \geq 9

$p \geq 9$

3^{x} \leq - 3

$3^x \leq -3$ → tidak memenuhi

\frac{1}{3} \leq 3^{x} < 3

$\dfrac13 \leq 3^x < 3$

3^{- 1} \leq 3^{x} < 3^{1}

$3^{-1} \leq 3^x < 3^1$

- 1 \leq x < 1

$-1 \leq x < 1$

3^{x} \geq 9

$3^x \geq 9$

3^{x} \geq 3^{2}

$3^x \geq 3^2$

x \geq 2

$x \geq 2$

a = - 1, b = 1, c = 2

$a = -1,\ b = 1,\ c = 2$

a + 2 b + c = - 1 + 2.1 + 2 = 3

$a + 2b + c = -1 + 2.1 + 2 = 3$
jawab: D.

Soal nomor 3:

Jika

a < x < b

$a < x < b$ adalah solusi pertidaksamaan

1 + 2^{x} + 2^{2 x} + 2^{3 x} + \dots > 2

$1 + 2^x + 2^{2x} + 2^{3x} + \cdots > 2$ , dengan

x \neq 1

$x \ne 1$ , maka

a + b = \dots

$a + b = \cdots$

A . - 1

$A.\ -1$

B . - 2

$B.\ -2$

C . - 3

$C.\ -3$

D . - 4

$D.\ -4$

E . - 5

$E.\ -5$

Deret geometri tak hingga:

S_{\infty} = \frac{a}{1 - r}

$S_{\infty} = \dfrac{a}{1 - r}$

a = 1, r = 2^{x}

$a = 1,\ r = 2^x$

S_{\infty} > 2

$S_{\infty} > 2$

\frac{1}{1 - 2^{x}} > 2

$\dfrac{1}{1 - 2^x} > 2$

\frac{1}{1 - 2^{x}} - 2 > 0

$\dfrac{1}{1 - 2^x} - 2 > 0$

\frac{1 - 2 (1 - 2^{x})}{1 - 2^{x}} > 0

$\dfrac{1 - 2(1 - 2^x)}{1 - 2^x} > 0$

\frac{{2.2}^{x} - 1}{1 - 2^{x}} > 0

$\dfrac{2.2^x - 1}{1 - 2^x} > 0$

\frac{{2.2}^{x} - 1)}{2^{x} - 1} < 0

$\dfrac{2.2^x - 1)}{2^x - 1} < 0$

Misalkan

p = 2^{x}

$p = 2^x$

\frac{2 p - 1}{p - 1} < 0

$\dfrac{2p - 1}{p - 1} < 0$

(2 p - 1) (p - 1) < 0

$(2p - 1)(p - 1) < 0$

\frac{1}{2} < p < 1

$\dfrac12 < p < 1$

\frac{1}{2} < 2^{x} < 1

$\dfrac12 < 2^x < 1$

2^{- 1} < 2^{x} < 2^{0}

$2^{-1} < 2^x < 2^0$

- 1 < x < 0

$-1 < x < 0$

a = - 1, b = 0

$a = -1,\ b = 0$

a + b = - 1 + 0 = - 1

$a + b = -1 + 0 = -1$
jawab: A.

Soal nomor 4:

Diberikan lingkaran pada bidang koordinat dengan titik pusat

(a, b)

$(a,\ b)$ dan memotong sumbu X di titik

(3, 0)

$(3,\ 0)$ dan

(9, 0)

$(9,\ 0)$ . Jika garis yang melalui titik

(0, 3)

$(0,\ 3)$ menyinggung lingkaran di titik

(3, 0)

$(3,\ 0)$ maka nilai dari

a^{2} - b^{2}

$a^2 - b^2$ adalah . . . .
A. 9
B. 18
C. 27
D. 36
E. 45

Lihat Gambar !

a = \frac{9 + 3}{2} = \frac{12}{2} = 6

$a = \dfrac{9 + 3}{2} = \dfrac{12}{2} = 6$ .
Jarak antara titik

(6, b)

$(6,\ b)$ dengan garis

x + y - 3 = 0

$x + y - 3 = 0$ sama dengan jarak antara titik

(6, b)

$(6,\ b)$ dengan titik

(9, 0)

$(9,\ 0)$ .

| \begin{matrix} \frac{6 + b - 3}{\sqrt{1^{2} + 1^{2}}} \end{matrix} | = \sqrt{(6 - 9)^{2} + (b - 0)^{2}}

$\begin{vmatrix}\dfrac{6 + b - 3}{\sqrt{1^2 + 1^2}} \end{vmatrix} = \sqrt{(6 - 9)^2 + (b - 0)^2}$

| \begin{matrix} \frac{b + 3}{\sqrt{2}} \end{matrix} | = \sqrt{9 + b^{2}}

$\begin{vmatrix}\dfrac{b + 3}{\sqrt{2}} \end{vmatrix} = \sqrt{9 + b^2}$

(b + 3)^{2} = 2 (9 + b^{2})

$(b + 3)^2 = 2(9 + b^2)$

b^{2} + 6 b + 9 = 18 + 2 b^{2}

$b^2 + 6b + 9 = 18 + 2b^2$

b^{2} - 6 b + 9 = 0

$b^2 - 6b + 9 = 0$

(b - 3)^{2} = 0

$(b - 3)^2 = 0$

b - 3 = 0

$b - 3 = 0$

b = 3

$b = 3$

a^{2} - b^{2} = 6^{2} - 3^{2} = 27

$a^2 - b^2 = 6^2 - 3^2 = 27$
jawab: C.

Soal nomor 5:

Jika

(x - 2)^{2}

$(x - 2)^2$ membagi habis

x^{4} - a x^{3} + b x^{2} + 4 x - 4

$x^4 - ax^3 + bx^2 + 4x - 4$ , maka

a b = \dots

$ab = \cdots$
A. 9
B. 12
C. 16
D. 20
E. 25

x^{4} - a x^{3} + b x^{2} + 4 x - 4

$x^4 - ax^3 + bx^2 + 4x - 4$

= (x - 2)^{2} (x^{2} + p x - 1)

$= (x - 2)^2(x^2 + px - 1)$

= (x^{2} - 4 x + 4) (x^{2} + p x - 1)

$= (x^2 - 4x + 4)(x^2 + px - 1)$

= x^{4} + (p - 4) x^{3} + (3 - 4 p) x^{2} + (4 + 4 p) x - 4

$= x^4 + (p - 4)x^3 + (3 - 4p)x^2 + (4 + 4p)x - 4$

Kesamaan:

4 + 4 p = 4 \to p = 0

$4 + 4p = 4 → p = 0$

- a = p - 4

$-a = p - 4$

- a = 0 - 4 \to a = 4

$- a = 0 - 4 → a = 4$

3 - 4 p = b

$3 - 4p = b$

3 - 4.0 = b \to b = 3

$3 - 4.0 = b → b = 3$

a b = 4.3 = 12

$ab = 4.3 = 12$
jawab: B.

Soal nomor 6:

Diberikan empat matriks

A, B, C, D

$A,\ B,\ C,\ D$ berukuran

2 \times 2

$2 \times 2$ dengan

A + C B^{T} = C D

$A + CB^T = CD$ . Jika

A

$A$ mempunyai invers,

d e t (D^{T} - B) = m

$det(D^T - B) = m$ dan

d e t (C) = n

$det(C) = n$ , maka

d e t (2 A^{- 1}) = \dots

$det(2A^{-1}) = \cdots$

A . \frac{4}{m n}

$A.\ \dfrac{4}{mn}$

B . \frac{m n}{4}

$B.\ \dfrac{mn}{4}$

C . \frac{4 m}{n}

$C.\ \dfrac{4m}{n}$

D . 4 m n

$D.\ 4mn$

E . \frac{m + n}{4}

$E.\ \dfrac{m + n}{4}$

d e t (B) = d e t (B^{T})

$det(B) = det(B^T)$

d e t (D^{T}) = d e t (D)

$det(D^T) = det(D)$

d e t (D - B^{T}) = d e t (D^{T} - B)

$det(D - B^T) = det(D^T - B)$

A + C B^{T} = C D

$A + CB^T = CD$

A = C D - C B^{T}

$A = CD - CB^T$

A = C (D - B^{T})

$A = C(D - B^T)$

d e t (A) = d e t (C (D - B^{T}))

$det(A) = det(C(D - B^T))$

d e t (A) = d e t (C) . d e t (D - B^{T})

$det(A) = det(C).det(D - B^T)$

d e t (A) = d e t (C) . d e t (D^{T} - B)

$det(A) = det(C).det(D^T - B)$

d e t (A) = n . m = m n

$det(A) = n.m = mn$

\begin{aligned} d e t (2 A^{- 1}) & = 2^{2} . \frac{1}{d e t (A)} \\ = 4. \frac{1}{m n} \\ = \frac{4}{m n} \end{aligned}

$\begin{align} det(2A^{-1}) &= 2^2.\dfrac{1}{det(A)}\\ &= 4.\dfrac{1}{mn}\\ &= \dfrac{4}{mn}\\ \end{align}$
jawab: A.

Soal nomor 7:

Jika

- \frac{π}{2} < x < \frac{π}{2}

$-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$ dan

x

$x$ memenuhi

5 c o s^{2} x + 3 s i n x c o s x \geq 1

$5cos^2\ x + 3sin\ xcos\ x \geq 1$ , maka himpunan semua

y = t a n x

$y = tan\ x$ adalah . . . .

A . {y \in R : - 1 \leq y \leq 4}

$A.\ \{y \in R: -1 \leq y \leq 4\}$

B . {y \in R : - 4 \leq y \leq 1}

$B.\ \{y \in R: -4 \leq y \leq 1\}$

C . {y \in R : - 4 \leq y \leq - 1}

$C.\ \{y \in R: -4 \leq y \leq -1\}$

D . {y \in R : 1 \leq y \leq 4}

$D.\ \{y \in R: 1 \leq y \leq 4\}$

E . R

$E.\ R$

5 c o s^{2} x + 3 s i n x c o s x \geq 1

$5cos^2\ x + 3sin\ xcos\ x \geq 1$

5 c o s^{2} x + 3 s i n x c o s x - 1 \geq 0

$5cos^2\ x + 3sin\ xcos\ x - 1 \geq 0$ → bagi dengan

c o s^{2} x

$cos^2\ x$

5 + 3 t a n x - s e c^{2} x \geq 0

$5 + 3tan\ x - sec^2\ x \geq 0$

5 + 3 t a n x - (t a n^{2} x + 1) \geq 0

$5 + 3tan\ x - (tan^2\ x + 1) \geq 0$

- t a n^{2} x + 3 t a n x + 4 \geq 0

$-tan^2\ x + 3tan\ x + 4 \geq 0$

t a n^{2} x - 3 t a n x - 4 \leq 0

$tan^2\ x - 3tan\ x - 4 \leq 0$

(t a n x + 1) (t a n x - 4) \leq 0

$(tan\ x + 1)(tan\ x - 4) \leq 0$

- 1 \leq t a n x \leq 4

$-1 \leq tan\ x \leq 4$

- 1 \leq y \leq 4

$-1 \leq y \leq 4$
jawab: A.

Soal nomor 8:

Jika suku banyak

x^{4} + 3 x^{3} + A x^{2} + 5 x + B

$x^4 + 3x^3 + Ax^2 + 5x + B$ dibagi

x^{2} + 2 x + 2

$x^2 + 2x + 2$ bersisa

7 x + 14

$7x + 14$ , maka jika dibagi

x^{2} + 4 x + 4

$x^2 + 4x + 4$ akan bersisa . . . .

A . x + 1

$A.\ x + 1$

B . x + 2

$B.\ x + 2$

C . x + 3

$C.\ x + 3$

D . 2 x + 1

$D.\ 2x + 1$

E . 2 x + 4

$E.\ 2x + 4$

Lakukan pembagian langsung !

s i s a = 7 x + 14

$sisa = 7x + 14$

s i s a = (11 - 2 A) x + B + 8 - 2 A

$sisa = (11 - 2A)x + B + 8 - 2A$
Kesamaan:

11 - 2 A = 7

$11 - 2A = 7$

4 = 2 A \to A = 2

$4 = 2A → A = 2$

B + 8 - 2 A = 14

$B + 8 - 2A = 14$

B + 8 - 2.2 = 14

$B + 8 - 2.2 = 14$

B = 10

$B = 10$

Suku baanyak menjadi:

x^{4} + 3 x^{3} + 2 x^{2} + 5 x + 10

$x^4 + 3x^3 + 2x^2 + 5x + 10$
Bagi dengan

x^{2} + 4 x + 4

$x^2 + 4x + 4$ dengan pembagian langsung !

Sisa Pembagian adalah

x + 2

$x + 2$ jawab: B.

Soal nomor 9:

Jika

(^{2} l o g x)^{2} - (^{2} l o g y)^{2} =^{2} l o g 256

$(^2log\ x)^2 - (^2log\ y)^2 =\ ^2log\ 256$ dan

^{2} l o g x^{2} -^{2} l o g y^{2} =^{2} l o g 16

$^2log\ x^2 -\ ^2log\ y^2 =\ ^2log\ 16$ , maka nilai dari

^{2} l o g x^{6} y^{- 2}

$^2log\ x^6y^{-2}$ adalah . . . .
A. 24
B. 20
C. 16
D. 8
E. 4

Misalkan:

^{2} l o g x = p

$^2log\ x = p$

^{2} l o g y = q

$^2log\ y = q$

(^{2} l o g x)^{2} - (^{2} l o g y)^{2} =^{2} l o g 256

$(^2log\ x)^2 - (^2log\ y)^2 =\ ^2log\ 256$

p^{2} - q^{2} =^{2} l o g 256

$p^2 - q^2 =\ ^2log\ 256$

p^{2} - q^{2} =^{2} l o g 2^{8}

$p^2 - q^2 =\ ^2log\ 2^8$

p^{2} - q^{2} = 8

$p^2 - q^2 = 8$

(p + q) (p - q) = 8

$(p + q)(p - q) = 8$ . . . . (1)

^{2} l o g x^{2} -^{2} l o g y^{2} =^{2} l o g 16

$^2log\ x^2 -\ ^2log\ y^2 =\ ^2log\ 16$

{2.}^{2} l o g x - {2.}^{2} l o g y =^{2} l o g 2^{4}

$2.^2log\ x - 2.^2log\ y =\ ^2log\ 2^4$

{2.}^{2} l o g x - {2.}^{2} l o g y = 4

$2.^2log\ x - 2.^2log\ y = 4$

2 p - 2 q = 4

$2p - 2q = 4$

p - q = 2

$p - q = 2$ . . . . (2)

dari (1) dan (2):

(p + q) .2 = 8

$(p + q).2 = 8$

p + q = 4

$p + q = 4$ . . . . (3)

Eliminasi persamaan (2) dan (3) !

p - q = 2

$p - q = 2$

p + q = 4

$p + q = 4$
---------------- +

2 p = 6

$2p = 6$

p = 3

$p = 3$

q = 1

$q = 1$

\begin{aligned} ^{2} l o g x^{6} y^{- 2} & =^{2} l o g x^{6} +^{2} l o g y^{- 2} \\ = {6.}^{2} l o g x - {2.}^{2} l o g y \\ = 6 p - 2 q \\ = 6.3 - 2.1 \\ = 16 \end{aligned}

$\begin{align} ^2log\ x^6y^{-2} &=\ ^2log\ x^6 +\ ^2log\ y^{-2}\\ &= 6.^2log\ x -\ 2.^2log\ y\\ &= 6p - 2q\\ &= 6.3 - 2.1\\ &= 16\\ \end{align}$
jawab: C.

Soal nomor 10:

Diberikan kubus ABCD.EFGH dan P adalah titik tengah BC. Perbandingan luas segitiga APG dan luas segitiga DPG adalah . . . .

A . 1 : 1

$A.\ 1 : 1$

B . \sqrt{3} : \sqrt{2}

$B.\ \sqrt{3} : \sqrt{2}$

C . \sqrt{2} : 1

$C.\ \sqrt{2} : 1$

D . 3 : 2

$D.\ 3 : 2$

E . \sqrt{3} : 1

$E.\ \sqrt{3} : 1$

Misalkan panjang sisi kubus adalah

a

$a$ .

\begin{aligned} G P^{2} & = C G^{2} + C P^{2} \\ = a^{2} + {(\frac{1}{2} a)}^{2} \\ = a^{2} + \frac{1}{4} a^{2} \\ = \frac{5}{4} a^{2} \\ G P & = \frac{1}{2} a \sqrt{5} \\ A G & = a \sqrt{3} \to d i a g o n a l r u a n g \\ D G & = a \sqrt{2} \to d i a g o n a l s i s i \end{aligned}

$\begin{align} GP^2 &= CG^2 + CP^2\\ &= a^2 + \left(\dfrac12a\right)^2\\ &= a^2 + \dfrac14a^2\\ &= \dfrac54a^2\\ GP &= \dfrac12a\sqrt{5}\\ AG &= a\sqrt{3} → diagonal\ ruang\\ DG &= a\sqrt{2} → diagonal\ sisi\\ \end{align}$

L u a s Δ A P G = \frac{1}{2} . a \sqrt{3} . \frac{1}{2} a \sqrt{2} = \frac{1}{4} a^{2} \sqrt{6}

$Luas\ \Delta APG = \dfrac12.a\sqrt{3}.\dfrac12a\sqrt{2} = \dfrac14a^2\sqrt{6}$

L u a s Δ D P G = \frac{1}{2} . a \sqrt{2} . \frac{1}{2} a \sqrt{3} = \frac{1}{4} a^{2} \sqrt{6}

$Luas\ \Delta DPG = \dfrac12.a\sqrt{2}.\dfrac12a\sqrt{3} = \dfrac14a^2\sqrt{6}$

L u a s Δ A P D : L u a s Δ D P G = 1 : 1

$Luas\ \Delta APD : Luas\ \Delta DPG = 1 : 1$
jawab: A.

Soal nomor 11:

Misalkan

U_{n}

$U_n$ menyatakan suku

k e - n

$ke-n$ dari barisan aritmetika. Diketahui

U_{1} \times U_{2} = 10

$U_1 \times U_2 = 10$ dan

U_{1} \times U_{3} = 16

$U_1 \times U_3 = 16$ . Jika suku-suku dari barisan aritmetika tersebut merupakan bilangan positif, maka

U_{10} = \dots

$U_{10} = \cdots$
A. 21
B. 23
C. 25
D. 27
E. 29

U_{n} = a + (n - 1) b

$U_n = a + (n - 1)b$

U_{1} \times U_{2} = 10

$U_1 \times U_2 = 10$

a (a + b) = 10

$a(a + b) = 10$ . . . . (1)

U_{1} \times U_{3} = 16

$U_1 \times U_3 = 16$

a (a + 2 b) = 16

$a(a + 2b) = 16$ . . . . (2)

\frac{a (a + b)}{a (a + 2 b)} = \frac{10}{16}

$\dfrac{a(a + b)}{a(a + 2b)} = \dfrac{10}{16}$

\frac{(a + b)}{(a + 2 b)} = \frac{5}{8}

$\dfrac{(a + b)}{(a + 2b)} = \dfrac{5}{8}$

8 (a + b) = 5 (a + 2 b)

$8(a + b) = 5(a + 2b)$

8 a + 8 b = 5 a + 10 b

$8a + 8b = 5a + 10b$

3 a = 2 b

$3a = 2b$ . . . . (3)

dari pers (2) dan (3)

a (a + 3 a) = 16

$a(a + 3a) = 16$

4 a^{2} = 16

$4a^2 = 16$

a^{2} = 4

$a^2 = 4$

a = 2

$a = 2$ → suku-suku positif

b = 3

$b = 3$

U_{10} = a + 9 b

$U_{10} = a + 9b$

= 2 + 9.3

$= 2 + 9.3$

= 29

$= 29$
jawab: E.

Soal nomor 12:

Diketahui fungsi

f

$f$ dan

g

$g$ dengan

f (x) = (2 x + 1)^{5}

$f(x) = (2x + 1)^5$ dan

h = f o g

$h = f\ o\ g$ . Jika

g (5) = - 1

$g(5) = -1$ dan

g^{'} (\frac{x + 1}{x - 1}) = 2 x + 2

$g'\left(\dfrac{x + 1}{x - 1}\right) = 2x + 2$ , maka

h^{'} (x) = \dots

$h'(x) = \cdots$
A. 10
B. 25
C. 50
D. 60
E. 120

g^{'} (\frac{x + 1}{x - 1}) = 2 x + 2

$g'\left(\dfrac{x + 1}{x - 1}\right) = 2x + 2$
Jika

\frac{x + 1}{x - 1} = 5

$\dfrac{x + 1}{x - 1} = 5$ , maka:

x + 1 = 5 (x - 1)

$x + 1 = 5(x - 1)$

x + 1 = 5 x - 5

$x + 1 = 5x - 5$

6 = 4 x

$6 = 4x$

x = \frac{3}{2}

$x = \dfrac32$
Dengan demikian:

g^{'} (5) = 2. \frac{3}{2} + 2 = 5

$g'(5) = 2.\dfrac32 + 2 = 5$

h = f o g

$h = f\ o\ g$

h (x) = (2 g (x) + 1)^{5}

$h(x) = (2g(x) + 1)^5$

h^{'} (x) = 5. (2 g (x) + 1)^{4} .2 g^{'} (x)

$h'(x) = 5.(2g(x) + 1)^4.2g'(x)$

h^{'} (5) = 5. (2 g (5) + 1)^{4} .2 g^{'} (5)

$h'(5) = 5.(2g(5) + 1)^4.2g'(5)$

= 5. (2. (- 1) + 1)^{4} .2 .5

$= 5.(2.(-1) + 1)^4.2.5$

= 5.1.2.5

$= 5.1.2.5$

= 50

$= 50$
jawab: C.

Soal nomor 13:

Jika

p > 0

$p > 0$ dan

lim_{x \to p} \frac{x^{3} + p x^{2} + q x}{x - p} = 12

$\displaystyle \lim_{x \to p} \dfrac{x^3 + px^2 + qx}{x - p} = 12$ , maka nilai

p - q

$p - q$ adalah . . . .
A. 14
B. 10
C. 8
D. 5
E. 3

Limit adalah limit tak tentu

\frac{0}{0}

$\dfrac 00$
Dengan demikian:

p^{3} + p . p^{2} + q p = 0

$p^3 + p.p^2 + qp = 0$

2 p^{3} + q p = 0

$2p^3 + qp = 0$

p (2 p^{2} + q) = 0

$p(2p^2 + q) = 0$

p = 0 a t a u q = - 2 p^{2}

$p = 0\ atau\ q = -2p^2$ →

p > 0

$p > 0$

lim_{x \to p} \frac{x^{3} + p x^{2} + q x}{x - p} = 12

$\displaystyle \lim_{x \to p} \dfrac{x^3 + px^2 + qx}{x - p} = 12$
Gunakan L'Hospital !

lim_{x \to p} \frac{3 x^{2} + 2 p x + q}{1} = 12

$\displaystyle \lim_{x \to p} \dfrac{3x^2 + 2px + q}{1} = 12$

3 p^{2} + 2 p . p - 2 p^{2} = 12

$3p^2 + 2p.p - 2p^2 = 12$

3 p^{2} = 12

$3p^2 = 12$

p^{2} = 4

$p^2 = 4$

p = 2

$p = 2$

q = - 2 p^{2} = - {2.2}^{2} = - 8

$q = -2p^2 = -2.2^2 = -8$

p - q = 2 - (- 8) = 2 + 8 = 10

$p - q = 2 - (-8) = 2 + 8 = 10$
jawab: B.

Soal nomor 14:

Jika

s i n x + s i n 2 x + s i n 3 x = 0

$sin\ x + sin\ 2x + sin\ 3x = 0$ untuk

\frac{π}{2} < x < π

$\dfrac{\pi}{2} < x < \pi$ , maka

t a n 2 x = \dots

$tan\ 2x = \cdots$

A . - \sqrt{2}

$A.\ -\sqrt{2}$

B . - 1

$B.\ -1$

C . - \frac{1}{3} \sqrt{3}

$C.\ -\dfrac13\sqrt{3}$

D . \frac{1}{3} \sqrt{3}

$D.\ \dfrac13\sqrt{3}$

E . \sqrt{3}

$E.\ \sqrt{3}$

s i n A + s i n B = 2 s i n \frac{1}{2} (A + B) c o s \frac{1}{2} (A - B)

$sin\ A + sin\ B = 2sin\dfrac12(A + B)cos\dfrac12(A - B)$

c o s (- A) = c o s A

$cos\ (-A) = cos\ A$

t a n (180^{o} + A) = t a n A

$tan\ (180^o + A) = tan\ A$

s i n x + s i n 2 x + s i n 3 x = 0

$sin\ x + sin\ 2x + sin\ 3x = 0$

s i n x + s i n 3 x + s i n 2 x = 0

$sin\ x + sin\ 3x + sin\ 2x = 0$

2 s i n \frac{1}{2} (x + 3 x) c o s \frac{1}{2} (x - 3 x) + s i n 2 x = 0

$2sin\dfrac12(x + 3x)cos\dfrac12(x - 3x) + sin\ 2x = 0$

2 s i n 2 x c o s (- x) + s i n 2 x = 0

$2sin\ 2xcos(-x) + sin\ 2x = 0$

2 s i n 2 x c o s x + s i n 2 x = 0

$2sin\ 2xcos\ x + sin\ 2x = 0$

s i n 2 x (2 c o s x + 1) = 0

$sin\ 2x(2cos\ x + 1) = 0$

s i n 2 x = 0 a t a u c o s x = - \frac{1}{2}

$sin\ 2x = 0\ atau\ cos\ x = -\dfrac12$

s i n 2 x = 0

$sin\ 2x = 0$

x = \frac{π}{2}, π

$x = \dfrac{\pi}{2},\ \pi$ → tidak memenuhi syarat

c o s x = - \frac{1}{2}

$cos\ x = -\dfrac12$

x = \frac{2 π}{3} = 120^{o}

$x = \dfrac{2\pi}{3} = 120^o$

t a n 2 x = t a n ({2.120}^{o})

$tan\ 2x = tan\ (2.120^o)$

= t a n 240^{o}

$= tan\ 240^o$

= t a n (180 + 60)^{o}

$= tan\ (180 + 60)^o$

= t a n 60^{o}

$= tan\ 60^o$

= \sqrt{3}

$= \sqrt{3}$
jawab: E.

Soal nomor 15:

Diketahui

x^{2} + 2 x y + 4 x = - 3

$x^2 + 2xy + 4x = -3$ dan

9 y^{2} + 4 x y + 12 y = - 1

$9y^2 + 4xy + 12y = -1$ . Nilai dari

x + 3 y

$x + 3y$ adalah . . . .

A . 2

$A.\ 2$

B . 1

$B.\ 1$

C . 0

$C.\ 0$

D . - 1

$D.\ -1$

E . - 2

$E.\ -2$

x^{2} + 2 x y + 4 x = - 3

$x^2 + 2xy + 4x = -3$

9 y^{2} + 4 x y + 12 y = - 1

$9y^2 + 4xy + 12y = -1$
----------------------------------- +

x^{2} + 9 y^{2} + 6 x y + 4 x + 12 y = - 4

$x^2 + 9y^2 + 6xy + 4x + 12y = -4$

(x + 3 y)^{2} - 6 x y + 6 x y + 4 (x + 3 y) + 4 = 0

$(x + 3y)^2 - 6xy + 6xy + 4(x + 3y) + 4 = 0$

(x + 3 y)^{2} + 4 (x + 3 y) + 4 = 0

$(x + 3y)^2 + 4(x + 3y) + 4 = 0$
Misalkan

x + 3 y = p

$x + 3y = p$

p^{2} + 4 p + 4 = 0

$p^2 + 4p + 4 = 0$

(p + 2)^{2} = 0

$(p + 2)^2 = 0$

p + 2 = 0

$p + 2 = 0$

p = - 2

$p = -2$

x + 3 y = - 2

$x + 3y = -2$
jawab: E.

Demikianlah pembahasan soal UM UGM 2019 Matematika IPA, semoga bermanfaat dan bisa membantu. Selamat belajar !
Disusun oleh:
Joslin Sibarani
Alumni Teknik Sipil ITB

Another Januari 14, 2022 JawabCara seo Banjarmasin, Kalimatan Selatan

Home » UM UGM MIPA » Soal dan Pembahasan UM UGM 2019 Matematika IPA

Soal dan Pembahasan UM UGM 2019 Matematika IPA

Posted by Caraku on Jumat, 14 Januari 2022

Formasi:
MMM → 6.5.4 = 120 cara
MPM → 6.4.5 = 120 cara
PMM → 4.6.5 = 120 cara
PPM → 4.3.6 = 72 cara
Total = 432 cara
jawab: E.

Soal nomor 2:

Diketahui penyelesaian dari pertidaksamaan

\frac{3^{x + 3} + 3^{x} - 36}{9^{x} - 9} \leq 3

$\dfrac{3^{x + 3} + 3^x - 36}{9^x - 9} \leq 3$ adalah

a \leq x < b

$a \leq x < b$ atau

x \geq c

$x \geq c$ . Nilai

a + 2 b + c = \dots

$a + 2b + c = \cdots$
A. 0
B. 1
C. 2
D. 3
E. 4

\frac{3^{x + 3} + 3^{x} - 36}{9^{x} - 9} \leq 3

$\dfrac{3^{x + 3} + 3^x - 36}{9^x - 9} \leq 3$

\frac{3^{3} {.3}^{x} + 3^{x} - 36}{3^{2 x} - 9} - 3 \leq 0

$\dfrac{3^3.3^x + 3^x - 36}{3^{2x} - 9} - 3 \leq 0$

\frac{{27.3}^{x} + 3^{x} - 36 - 3 (3^{2 x} - 9)}{3^{2 x} - 9} \leq 0

$\dfrac{27.3^x + 3^x - 36 - 3(3^{2x} - 9)}{3^{2x} - 9}\leq 0$

\frac{{27.3}^{x} + 3^{x} - 36 - {3.3}^{2 x} + 27}{3^{2 x} - 9} \leq 0

$\dfrac{27.3^x + 3^x - 36 - 3.3^{2x} + 27}{3^{2x} - 9}\leq 0$

\frac{- {3.3}^{2 x} + {28.3}^{x} - 9}{3^{2 x} - 9} \leq 0

$\dfrac{-3.3^{2x} + 28.3^x - 9}{3^{2x} - 9}\leq 0$

\frac{{3.3}^{2 x} - {28.3}^{x} + 9}{3^{2 x} - 9} \geq 0

$\dfrac{3.3^{2x} - 28.3^x + 9}{3^{2x} - 9}\geq 0$

Misalkan

3^{x} = p

$3^x = p$

\frac{3 p^{2} - 28 p + 9}{p^{2} - 9} \geq 0

$\dfrac{3p^2 - 28p + 9}{p^2 - 9} \geq 0$

\frac{(3 p - 1) (p - 9)}{(p + 3) (p - 3)} \geq 0

$\dfrac{(3p - 1)(p - 9)}{(p + 3)(p - 3)} \geq 0$ →

p \neq 3

$p \ne 3$

(p + 3) (3 p - 1) (p - 3) (p - 9) \geq 0

$(p + 3)(3p - 1)(p - 3)(p - 9) \geq 0$

p \leq - 3

$p \leq -3$ atau

\frac{1}{3} \leq p < 3

$\dfrac13 \leq p < 3$ atau

p \geq 9

$p \geq 9$

3^{x} \leq - 3

$3^x \leq -3$ → tidak memenuhi

\frac{1}{3} \leq 3^{x} < 3

$\dfrac13 \leq 3^x < 3$

3^{- 1} \leq 3^{x} < 3^{1}

$3^{-1} \leq 3^x < 3^1$

- 1 \leq x < 1

$-1 \leq x < 1$

3^{x} \geq 9

$3^x \geq 9$

3^{x} \geq 3^{2}

$3^x \geq 3^2$

x \geq 2

$x \geq 2$

a = - 1, b = 1, c = 2

$a = -1,\ b = 1,\ c = 2$

a + 2 b + c = - 1 + 2.1 + 2 = 3

$a + 2b + c = -1 + 2.1 + 2 = 3$
jawab: D.

Soal nomor 3:

Jika

a < x < b

$a < x < b$ adalah solusi pertidaksamaan

1 + 2^{x} + 2^{2 x} + 2^{3 x} + \dots > 2

$1 + 2^x + 2^{2x} + 2^{3x} + \cdots > 2$ , dengan

x \neq 1

$x \ne 1$ , maka

a + b = \dots

$a + b = \cdots$

A . - 1

$A.\ -1$

B . - 2

$B.\ -2$

C . - 3

$C.\ -3$

D . - 4

$D.\ -4$

E . - 5

$E.\ -5$

Deret geometri tak hingga:

S_{\infty} = \frac{a}{1 - r}

$S_{\infty} = \dfrac{a}{1 - r}$

a = 1, r = 2^{x}

$a = 1,\ r = 2^x$

S_{\infty} > 2

$S_{\infty} > 2$

\frac{1}{1 - 2^{x}} > 2

$\dfrac{1}{1 - 2^x} > 2$

\frac{1}{1 - 2^{x}} - 2 > 0

$\dfrac{1}{1 - 2^x} - 2 > 0$

\frac{1 - 2 (1 - 2^{x})}{1 - 2^{x}} > 0

$\dfrac{1 - 2(1 - 2^x)}{1 - 2^x} > 0$

\frac{{2.2}^{x} - 1}{1 - 2^{x}} > 0

$\dfrac{2.2^x - 1}{1 - 2^x} > 0$

\frac{{2.2}^{x} - 1)}{2^{x} - 1} < 0

$\dfrac{2.2^x - 1)}{2^x - 1} < 0$

Misalkan

p = 2^{x}

$p = 2^x$

\frac{2 p - 1}{p - 1} < 0

$\dfrac{2p - 1}{p - 1} < 0$

(2 p - 1) (p - 1) < 0

$(2p - 1)(p - 1) < 0$

\frac{1}{2} < p < 1

$\dfrac12 < p < 1$

\frac{1}{2} < 2^{x} < 1

$\dfrac12 < 2^x < 1$

2^{- 1} < 2^{x} < 2^{0}

$2^{-1} < 2^x < 2^0$

- 1 < x < 0

$-1 < x < 0$

a = - 1, b = 0

$a = -1,\ b = 0$

a + b = - 1 + 0 = - 1

$a + b = -1 + 0 = -1$
jawab: A.

Soal nomor 4:

Diberikan lingkaran pada bidang koordinat dengan titik pusat

(a, b)

$(a,\ b)$ dan memotong sumbu X di titik

(3, 0)

$(3,\ 0)$ dan

(9, 0)

$(9,\ 0)$ . Jika garis yang melalui titik

(0, 3)

$(0,\ 3)$ menyinggung lingkaran di titik

(3, 0)

$(3,\ 0)$ maka nilai dari

a^{2} - b^{2}

$a^2 - b^2$ adalah . . . .
A. 9
B. 18
C. 27
D. 36
E. 45

Lihat Gambar !

a = \frac{9 + 3}{2} = \frac{12}{2} = 6

$a = \dfrac{9 + 3}{2} = \dfrac{12}{2} = 6$ .
Jarak antara titik

(6, b)

$(6,\ b)$ dengan garis

x + y - 3 = 0

$x + y - 3 = 0$ sama dengan jarak antara titik

(6, b)

$(6,\ b)$ dengan titik

(9, 0)

$(9,\ 0)$ .

| \begin{matrix} \frac{6 + b - 3}{\sqrt{1^{2} + 1^{2}}} \end{matrix} | = \sqrt{(6 - 9)^{2} + (b - 0)^{2}}

$\begin{vmatrix}\dfrac{6 + b - 3}{\sqrt{1^2 + 1^2}} \end{vmatrix} = \sqrt{(6 - 9)^2 + (b - 0)^2}$

| \begin{matrix} \frac{b + 3}{\sqrt{2}} \end{matrix} | = \sqrt{9 + b^{2}}

$\begin{vmatrix}\dfrac{b + 3}{\sqrt{2}} \end{vmatrix} = \sqrt{9 + b^2}$

(b + 3)^{2} = 2 (9 + b^{2})

$(b + 3)^2 = 2(9 + b^2)$

b^{2} + 6 b + 9 = 18 + 2 b^{2}

$b^2 + 6b + 9 = 18 + 2b^2$

b^{2} - 6 b + 9 = 0

$b^2 - 6b + 9 = 0$

(b - 3)^{2} = 0

$(b - 3)^2 = 0$

b - 3 = 0

$b - 3 = 0$

b = 3

$b = 3$

a^{2} - b^{2} = 6^{2} - 3^{2} = 27

$a^2 - b^2 = 6^2 - 3^2 = 27$
jawab: C.

Soal nomor 5:

Jika

(x - 2)^{2}

$(x - 2)^2$ membagi habis

x^{4} - a x^{3} + b x^{2} + 4 x - 4

$x^4 - ax^3 + bx^2 + 4x - 4$ , maka

a b = \dots

$ab = \cdots$
A. 9
B. 12
C. 16
D. 20
E. 25

x^{4} - a x^{3} + b x^{2} + 4 x - 4

$x^4 - ax^3 + bx^2 + 4x - 4$

= (x - 2)^{2} (x^{2} + p x - 1)

$= (x - 2)^2(x^2 + px - 1)$

= (x^{2} - 4 x + 4) (x^{2} + p x - 1)

$= (x^2 - 4x + 4)(x^2 + px - 1)$

= x^{4} + (p - 4) x^{3} + (3 - 4 p) x^{2} + (4 + 4 p) x - 4

$= x^4 + (p - 4)x^3 + (3 - 4p)x^2 + (4 + 4p)x - 4$

Kesamaan:

4 + 4 p = 4 \to p = 0

$4 + 4p = 4 → p = 0$

- a = p - 4

$-a = p - 4$

- a = 0 - 4 \to a = 4

$- a = 0 - 4 → a = 4$

3 - 4 p = b

$3 - 4p = b$

3 - 4.0 = b \to b = 3

$3 - 4.0 = b → b = 3$

a b = 4.3 = 12

$ab = 4.3 = 12$
jawab: B.

Soal nomor 6:

Diberikan empat matriks

A, B, C, D

$A,\ B,\ C,\ D$ berukuran

2 \times 2

$2 \times 2$ dengan

A + C B^{T} = C D

$A + CB^T = CD$ . Jika

A

$A$ mempunyai invers,

d e t (D^{T} - B) = m

$det(D^T - B) = m$ dan

d e t (C) = n

$det(C) = n$ , maka

d e t (2 A^{- 1}) = \dots

$det(2A^{-1}) = \cdots$

A . \frac{4}{m n}

$A.\ \dfrac{4}{mn}$

B . \frac{m n}{4}

$B.\ \dfrac{mn}{4}$

C . \frac{4 m}{n}

$C.\ \dfrac{4m}{n}$

D . 4 m n

$D.\ 4mn$

E . \frac{m + n}{4}

$E.\ \dfrac{m + n}{4}$

d e t (B) = d e t (B^{T})

$det(B) = det(B^T)$

d e t (D^{T}) = d e t (D)

$det(D^T) = det(D)$

d e t (D - B^{T}) = d e t (D^{T} - B)

$det(D - B^T) = det(D^T - B)$

A + C B^{T} = C D

$A + CB^T = CD$

A = C D - C B^{T}

$A = CD - CB^T$

A = C (D - B^{T})

$A = C(D - B^T)$

d e t (A) = d e t (C (D - B^{T}))

$det(A) = det(C(D - B^T))$

d e t (A) = d e t (C) . d e t (D - B^{T})

$det(A) = det(C).det(D - B^T)$

d e t (A) = d e t (C) . d e t (D^{T} - B)

$det(A) = det(C).det(D^T - B)$

d e t (A) = n . m = m n

$det(A) = n.m = mn$

\begin{aligned} d e t (2 A^{- 1}) & = 2^{2} . \frac{1}{d e t (A)} \\ = 4. \frac{1}{m n} \\ = \frac{4}{m n} \end{aligned}

$\begin{align} det(2A^{-1}) &= 2^2.\dfrac{1}{det(A)}\\ &= 4.\dfrac{1}{mn}\\ &= \dfrac{4}{mn}\\ \end{align}$
jawab: A.

Soal nomor 7:

Jika

- \frac{π}{2} < x < \frac{π}{2}

$-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$ dan

x

$x$ memenuhi

5 c o s^{2} x + 3 s i n x c o s x \geq 1

$5cos^2\ x + 3sin\ xcos\ x \geq 1$ , maka himpunan semua

y = t a n x

$y = tan\ x$ adalah . . . .

A . {y \in R : - 1 \leq y \leq 4}

$A.\ \{y \in R: -1 \leq y \leq 4\}$

B . {y \in R : - 4 \leq y \leq 1}

$B.\ \{y \in R: -4 \leq y \leq 1\}$

C . {y \in R : - 4 \leq y \leq - 1}

$C.\ \{y \in R: -4 \leq y \leq -1\}$

D . {y \in R : 1 \leq y \leq 4}

$D.\ \{y \in R: 1 \leq y \leq 4\}$

E . R

$E.\ R$

5 c o s^{2} x + 3 s i n x c o s x \geq 1

$5cos^2\ x + 3sin\ xcos\ x \geq 1$

5 c o s^{2} x + 3 s i n x c o s x - 1 \geq 0

$5cos^2\ x + 3sin\ xcos\ x - 1 \geq 0$ → bagi dengan

c o s^{2} x

$cos^2\ x$

5 + 3 t a n x - s e c^{2} x \geq 0

$5 + 3tan\ x - sec^2\ x \geq 0$

5 + 3 t a n x - (t a n^{2} x + 1) \geq 0

$5 + 3tan\ x - (tan^2\ x + 1) \geq 0$

- t a n^{2} x + 3 t a n x + 4 \geq 0

$-tan^2\ x + 3tan\ x + 4 \geq 0$

t a n^{2} x - 3 t a n x - 4 \leq 0

$tan^2\ x - 3tan\ x - 4 \leq 0$

(t a n x + 1) (t a n x - 4) \leq 0

$(tan\ x + 1)(tan\ x - 4) \leq 0$

- 1 \leq t a n x \leq 4

$-1 \leq tan\ x \leq 4$

- 1 \leq y \leq 4

$-1 \leq y \leq 4$
jawab: A.

Soal nomor 8:

Jika suku banyak

x^{4} + 3 x^{3} + A x^{2} + 5 x + B

$x^4 + 3x^3 + Ax^2 + 5x + B$ dibagi

x^{2} + 2 x + 2

$x^2 + 2x + 2$ bersisa

7 x + 14

$7x + 14$ , maka jika dibagi

x^{2} + 4 x + 4

$x^2 + 4x + 4$ akan bersisa . . . .

A . x + 1

$A.\ x + 1$

B . x + 2

$B.\ x + 2$

C . x + 3

$C.\ x + 3$

D . 2 x + 1

$D.\ 2x + 1$

E . 2 x + 4

$E.\ 2x + 4$

Lakukan pembagian langsung !

s i s a = 7 x + 14

$sisa = 7x + 14$

s i s a = (11 - 2 A) x + B + 8 - 2 A

$sisa = (11 - 2A)x + B + 8 - 2A$
Kesamaan:

11 - 2 A = 7

$11 - 2A = 7$

4 = 2 A \to A = 2

$4 = 2A → A = 2$

B + 8 - 2 A = 14

$B + 8 - 2A = 14$

B + 8 - 2.2 = 14

$B + 8 - 2.2 = 14$

B = 10

$B = 10$

Suku baanyak menjadi:

x^{4} + 3 x^{3} + 2 x^{2} + 5 x + 10

$x^4 + 3x^3 + 2x^2 + 5x + 10$
Bagi dengan

x^{2} + 4 x + 4

$x^2 + 4x + 4$ dengan pembagian langsung !

Sisa Pembagian adalah

x + 2

$x + 2$ jawab: B.

Soal nomor 9:

Jika

(^{2} l o g x)^{2} - (^{2} l o g y)^{2} =^{2} l o g 256

$(^2log\ x)^2 - (^2log\ y)^2 =\ ^2log\ 256$ dan

^{2} l o g x^{2} -^{2} l o g y^{2} =^{2} l o g 16

$^2log\ x^2 -\ ^2log\ y^2 =\ ^2log\ 16$ , maka nilai dari

^{2} l o g x^{6} y^{- 2}

$^2log\ x^6y^{-2}$ adalah . . . .
A. 24
B. 20
C. 16
D. 8
E. 4

Misalkan:

^{2} l o g x = p

$^2log\ x = p$

^{2} l o g y = q

$^2log\ y = q$

(^{2} l o g x)^{2} - (^{2} l o g y)^{2} =^{2} l o g 256

$(^2log\ x)^2 - (^2log\ y)^2 =\ ^2log\ 256$

p^{2} - q^{2} =^{2} l o g 256

$p^2 - q^2 =\ ^2log\ 256$

p^{2} - q^{2} =^{2} l o g 2^{8}

$p^2 - q^2 =\ ^2log\ 2^8$

p^{2} - q^{2} = 8

$p^2 - q^2 = 8$

(p + q) (p - q) = 8

$(p + q)(p - q) = 8$ . . . . (1)

^{2} l o g x^{2} -^{2} l o g y^{2} =^{2} l o g 16

$^2log\ x^2 -\ ^2log\ y^2 =\ ^2log\ 16$

{2.}^{2} l o g x - {2.}^{2} l o g y =^{2} l o g 2^{4}

$2.^2log\ x - 2.^2log\ y =\ ^2log\ 2^4$

{2.}^{2} l o g x - {2.}^{2} l o g y = 4

$2.^2log\ x - 2.^2log\ y = 4$

2 p - 2 q = 4

$2p - 2q = 4$

p - q = 2

$p - q = 2$ . . . . (2)

dari (1) dan (2):

(p + q) .2 = 8

$(p + q).2 = 8$

p + q = 4

$p + q = 4$ . . . . (3)

Eliminasi persamaan (2) dan (3) !

p - q = 2

$p - q = 2$

p + q = 4

$p + q = 4$
---------------- +

2 p = 6

$2p = 6$

p = 3

$p = 3$

q = 1

$q = 1$

\begin{aligned} ^{2} l o g x^{6} y^{- 2} & =^{2} l o g x^{6} +^{2} l o g y^{- 2} \\ = {6.}^{2} l o g x - {2.}^{2} l o g y \\ = 6 p - 2 q \\ = 6.3 - 2.1 \\ = 16 \end{aligned}

$\begin{align} ^2log\ x^6y^{-2} &=\ ^2log\ x^6 +\ ^2log\ y^{-2}\\ &= 6.^2log\ x -\ 2.^2log\ y\\ &= 6p - 2q\\ &= 6.3 - 2.1\\ &= 16\\ \end{align}$
jawab: C.

Soal nomor 10:

Diberikan kubus ABCD.EFGH dan P adalah titik tengah BC. Perbandingan luas segitiga APG dan luas segitiga DPG adalah . . . .

A . 1 : 1

$A.\ 1 : 1$

B . \sqrt{3} : \sqrt{2}

$B.\ \sqrt{3} : \sqrt{2}$

C . \sqrt{2} : 1

$C.\ \sqrt{2} : 1$

D . 3 : 2

$D.\ 3 : 2$

E . \sqrt{3} : 1

$E.\ \sqrt{3} : 1$

Misalkan panjang sisi kubus adalah

a

$a$ .

\begin{aligned} G P^{2} & = C G^{2} + C P^{2} \\ = a^{2} + {(\frac{1}{2} a)}^{2} \\ = a^{2} + \frac{1}{4} a^{2} \\ = \frac{5}{4} a^{2} \\ G P & = \frac{1}{2} a \sqrt{5} \\ A G & = a \sqrt{3} \to d i a g o n a l r u a n g \\ D G & = a \sqrt{2} \to d i a g o n a l s i s i \end{aligned}

L u a s Δ A P G = \frac{1}{2} . a \sqrt{3} . \frac{1}{2} a \sqrt{2} = \frac{1}{4} a^{2} \sqrt{6}

$Luas\ \Delta APG = \dfrac12.a\sqrt{3}.\dfrac12a\sqrt{2} = \dfrac14a^2\sqrt{6}$

L u a s Δ D P G = \frac{1}{2} . a \sqrt{2} . \frac{1}{2} a \sqrt{3} = \frac{1}{4} a^{2} \sqrt{6}

$Luas\ \Delta DPG = \dfrac12.a\sqrt{2}.\dfrac12a\sqrt{3} = \dfrac14a^2\sqrt{6}$

L u a s Δ A P D : L u a s Δ D P G = 1 : 1

$Luas\ \Delta APD : Luas\ \Delta DPG = 1 : 1$
jawab: A.

Soal nomor 11:

Misalkan

U_{n}

$U_n$ menyatakan suku

k e - n

$ke-n$ dari barisan aritmetika. Diketahui

U_{1} \times U_{2} = 10

$U_1 \times U_2 = 10$ dan

U_{1} \times U_{3} = 16

$U_1 \times U_3 = 16$ . Jika suku-suku dari barisan aritmetika tersebut merupakan bilangan positif, maka

U_{10} = \dots

$U_{10} = \cdots$
A. 21
B. 23
C. 25
D. 27
E. 29

U_{n} = a + (n - 1) b

$U_n = a + (n - 1)b$

U_{1} \times U_{2} = 10

$U_1 \times U_2 = 10$

a (a + b) = 10

$a(a + b) = 10$ . . . . (1)

U_{1} \times U_{3} = 16

$U_1 \times U_3 = 16$

a (a + 2 b) = 16

$a(a + 2b) = 16$ . . . . (2)

\frac{a (a + b)}{a (a + 2 b)} = \frac{10}{16}

$\dfrac{a(a + b)}{a(a + 2b)} = \dfrac{10}{16}$

\frac{(a + b)}{(a + 2 b)} = \frac{5}{8}

$\dfrac{(a + b)}{(a + 2b)} = \dfrac{5}{8}$

8 (a + b) = 5 (a + 2 b)

$8(a + b) = 5(a + 2b)$

8 a + 8 b = 5 a + 10 b

$8a + 8b = 5a + 10b$

3 a = 2 b

$3a = 2b$ . . . . (3)

dari pers (2) dan (3)

a (a + 3 a) = 16

$a(a + 3a) = 16$

4 a^{2} = 16

$4a^2 = 16$

a^{2} = 4

$a^2 = 4$

a = 2

$a = 2$ → suku-suku positif

b = 3

$b = 3$

U_{10} = a + 9 b

$U_{10} = a + 9b$

= 2 + 9.3

$= 2 + 9.3$

= 29

$= 29$
jawab: E.

Soal nomor 12:

Diketahui fungsi

f

$f$ dan

g

$g$ dengan

f (x) = (2 x + 1)^{5}

$f(x) = (2x + 1)^5$ dan

h = f o g

$h = f\ o\ g$ . Jika

g (5) = - 1

$g(5) = -1$ dan

g^{'} (\frac{x + 1}{x - 1}) = 2 x + 2

$g'\left(\dfrac{x + 1}{x - 1}\right) = 2x + 2$ , maka

h^{'} (x) = \dots

$h'(x) = \cdots$
A. 10
B. 25
C. 50
D. 60
E. 120

g^{'} (\frac{x + 1}{x - 1}) = 2 x + 2

$g'\left(\dfrac{x + 1}{x - 1}\right) = 2x + 2$
Jika

\frac{x + 1}{x - 1} = 5

$\dfrac{x + 1}{x - 1} = 5$ , maka:

x + 1 = 5 (x - 1)

$x + 1 = 5(x - 1)$

x + 1 = 5 x - 5

$x + 1 = 5x - 5$

6 = 4 x

$6 = 4x$

x = \frac{3}{2}

$x = \dfrac32$
Dengan demikian:

g^{'} (5) = 2. \frac{3}{2} + 2 = 5

$g'(5) = 2.\dfrac32 + 2 = 5$

h = f o g

$h = f\ o\ g$

h (x) = (2 g (x) + 1)^{5}

$h(x) = (2g(x) + 1)^5$

h^{'} (x) = 5. (2 g (x) + 1)^{4} .2 g^{'} (x)

$h'(x) = 5.(2g(x) + 1)^4.2g'(x)$

h^{'} (5) = 5. (2 g (5) + 1)^{4} .2 g^{'} (5)

$h'(5) = 5.(2g(5) + 1)^4.2g'(5)$

= 5. (2. (- 1) + 1)^{4} .2 .5

$= 5.(2.(-1) + 1)^4.2.5$

= 5.1.2.5

$= 5.1.2.5$

= 50

$= 50$
jawab: C.

Soal nomor 13:

Jika

p > 0

$p > 0$ dan

lim_{x \to p} \frac{x^{3} + p x^{2} + q x}{x - p} = 12

$\displaystyle \lim_{x \to p} \dfrac{x^3 + px^2 + qx}{x - p} = 12$ , maka nilai

p - q

$p - q$ adalah . . . .
A. 14
B. 10
C. 8
D. 5
E. 3

Limit adalah limit tak tentu

\frac{0}{0}

$\dfrac 00$
Dengan demikian:

p^{3} + p . p^{2} + q p = 0

$p^3 + p.p^2 + qp = 0$

2 p^{3} + q p = 0

$2p^3 + qp = 0$

p (2 p^{2} + q) = 0

$p(2p^2 + q) = 0$

p = 0 a t a u q = - 2 p^{2}

$p = 0\ atau\ q = -2p^2$ →

p > 0

$p > 0$

lim_{x \to p} \frac{x^{3} + p x^{2} + q x}{x - p} = 12

$\displaystyle \lim_{x \to p} \dfrac{x^3 + px^2 + qx}{x - p} = 12$
Gunakan L'Hospital !

lim_{x \to p} \frac{3 x^{2} + 2 p x + q}{1} = 12

$\displaystyle \lim_{x \to p} \dfrac{3x^2 + 2px + q}{1} = 12$

3 p^{2} + 2 p . p - 2 p^{2} = 12

$3p^2 + 2p.p - 2p^2 = 12$

3 p^{2} = 12

$3p^2 = 12$

p^{2} = 4

$p^2 = 4$

p = 2

$p = 2$

q = - 2 p^{2} = - {2.2}^{2} = - 8

$q = -2p^2 = -2.2^2 = -8$

p - q = 2 - (- 8) = 2 + 8 = 10

$p - q = 2 - (-8) = 2 + 8 = 10$
jawab: B.

Soal nomor 14:

Jika

s i n x + s i n 2 x + s i n 3 x = 0

$sin\ x + sin\ 2x + sin\ 3x = 0$ untuk

\frac{π}{2} < x < π

$\dfrac{\pi}{2} < x < \pi$ , maka

t a n 2 x = \dots

$tan\ 2x = \cdots$

A . - \sqrt{2}

$A.\ -\sqrt{2}$

B . - 1

$B.\ -1$

C . - \frac{1}{3} \sqrt{3}

$C.\ -\dfrac13\sqrt{3}$

D . \frac{1}{3} \sqrt{3}

$D.\ \dfrac13\sqrt{3}$

E . \sqrt{3}

$E.\ \sqrt{3}$

s i n A + s i n B = 2 s i n \frac{1}{2} (A + B) c o s \frac{1}{2} (A - B)

$sin\ A + sin\ B = 2sin\dfrac12(A + B)cos\dfrac12(A - B)$

c o s (- A) = c o s A

$cos\ (-A) = cos\ A$

t a n (180^{o} + A) = t a n A

$tan\ (180^o + A) = tan\ A$

s i n x + s i n 2 x + s i n 3 x = 0

$sin\ x + sin\ 2x + sin\ 3x = 0$

s i n x + s i n 3 x + s i n 2 x = 0

$sin\ x + sin\ 3x + sin\ 2x = 0$

2 s i n \frac{1}{2} (x + 3 x) c o s \frac{1}{2} (x - 3 x) + s i n 2 x = 0

$2sin\dfrac12(x + 3x)cos\dfrac12(x - 3x) + sin\ 2x = 0$

2 s i n 2 x c o s (- x) + s i n 2 x = 0

$2sin\ 2xcos(-x) + sin\ 2x = 0$

2 s i n 2 x c o s x + s i n 2 x = 0

$2sin\ 2xcos\ x + sin\ 2x = 0$

s i n 2 x (2 c o s x + 1) = 0

$sin\ 2x(2cos\ x + 1) = 0$

s i n 2 x = 0 a t a u c o s x = - \frac{1}{2}

$sin\ 2x = 0\ atau\ cos\ x = -\dfrac12$

s i n 2 x = 0

$sin\ 2x = 0$

x = \frac{π}{2}, π

$x = \dfrac{\pi}{2},\ \pi$ → tidak memenuhi syarat

c o s x = - \frac{1}{2}

$cos\ x = -\dfrac12$

x = \frac{2 π}{3} = 120^{o}

$x = \dfrac{2\pi}{3} = 120^o$

t a n 2 x = t a n ({2.120}^{o})

$tan\ 2x = tan\ (2.120^o)$

= t a n 240^{o}

$= tan\ 240^o$

= t a n (180 + 60)^{o}

$= tan\ (180 + 60)^o$

= t a n 60^{o}

$= tan\ 60^o$

= \sqrt{3}

$= \sqrt{3}$
jawab: E.

Soal nomor 15:

Diketahui

x^{2} + 2 x y + 4 x = - 3

$x^2 + 2xy + 4x = -3$ dan

9 y^{2} + 4 x y + 12 y = - 1

$9y^2 + 4xy + 12y = -1$ . Nilai dari

x + 3 y

$x + 3y$ adalah . . . .

A . 2

$A.\ 2$

B . 1

$B.\ 1$

C . 0

$C.\ 0$

D . - 1

$D.\ -1$

E . - 2

$E.\ -2$

x^{2} + 2 x y + 4 x = - 3

$x^2 + 2xy + 4x = -3$

9 y^{2} + 4 x y + 12 y = - 1

$9y^2 + 4xy + 12y = -1$
----------------------------------- +

x^{2} + 9 y^{2} + 6 x y + 4 x + 12 y = - 4

$x^2 + 9y^2 + 6xy + 4x + 12y = -4$

(x + 3 y)^{2} - 6 x y + 6 x y + 4 (x + 3 y) + 4 = 0

$(x + 3y)^2 - 6xy + 6xy + 4(x + 3y) + 4 = 0$

(x + 3 y)^{2} + 4 (x + 3 y) + 4 = 0

$(x + 3y)^2 + 4(x + 3y) + 4 = 0$
Misalkan

x + 3 y = p

$x + 3y = p$

p^{2} + 4 p + 4 = 0

$p^2 + 4p + 4 = 0$

(p + 2)^{2} = 0

$(p + 2)^2 = 0$

p + 2 = 0

$p + 2 = 0$

p = - 2

$p = -2$

x + 3 y = - 2

$x + 3y = -2$
jawab: E.

Demikianlah pembahasan soal UM UGM 2019 Matematika IPA, semoga bermanfaat dan bisa membantu. Selamat belajar !
Disusun oleh:
Joslin Sibarani
Alumni Teknik Sipil ITB

Terima kasih telah membaca artikel ini & dipublikasikan oleh Caraku

Soal dan Pembahasan UM UGM 2019 Matematika IPA

Posted by Caraku on Jumat, 14 Januari 2022

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