Soal 1
Diketahui suku banyak f(x) dibagi x² + x − 2 bersisa ax + b dan dibagi x² − 4x + 3 bersisa 2bx + a − 1. Jika f(−2) = 7, maka a² + b² = …
(A) 12
(B) 10
(C) 9
(D) 8
(E) 5
Jawaban: B
\(f(x) = (x^2 + x \:-\:2)\cdot \text{H(x)} + ax + b\)
\(f(x) = (x + 2)(x\:-\:1)\cdot \text{H(x)} + ax + b\)
\(f(-2) = 0 + -2a + b \)
\(7 = -2a + b\dotso\dotso \color{yellow} (1)\)
\(f(1) = a + b\dotso\dotso \color{yellow} (2)\)
\(f(x) = (x^2 \:-\:4x + 3)\cdot \text{H(x)} + 2bx + a \:-\:1\)
\(f(x) = (x \:-\: 3)(x\:-\:1)\cdot \text{H(x)} + 2bx + a \:-\:1\)
\(f(1) = 0 + 2b + a \:-\:1\)
\(f(1) = 2b + a \:-\:1\dotso\dotso \color{yellow} (3)\)
Substitusikan persamaan (2) ke persamaan (3)
\(a + b = 2b + a \:-\:1\)
\(b = 1\)
Substitusikan nilai \(b = 1\) ke persamaan (1)
\(7 = -2a + b\)
\(7 = -2a + 1\)
\(-2a = 6\)
\(a = -3\)
\(a^2 + b^2 = (-3)^2 + 1^2\)
\(a^2 + b^2 = 9 + 1\)
\(a^2 + b^2 = 10\)
Soal 2
Jika b > a, nilai x yang memenuhi |x − 2a| + a ≤ b adalah…
(A) 3a ≤ x ≤ 2b + a
(B) x ≥ −b + 3a
(C) x ≤ b + a
(D) b − 3a ≤ x ≤ −b + a
(E) −b + 3a ≤ x ≤ b + a
Jawaban: E
\(|x \:-\:2a| \leq b \:-\:a\)
\(-(b\:-\:a) \leq x \:-\:2a \leq b \:-\:a\)
\(-b + a + 2a \leq x \leq b \:-\:a + 2a\)
\(-b + 3a \leq x \leq b + a\)
Soal 3
Jika \(x_1\) dan \(x_2\) memenuhi persamaan \(2 \sin^2 x \:-\:\cos x = 1\), \(0 \leq x \leq \pi\), nilai \(x_1 + x_2\) adalah…
(A) \(\dfrac{\pi}{3}\)
(B) \(\dfrac{2\pi}{3}\)
(C) \(\pi\)
(D) \(\dfrac{4\pi}{3}\)
(E) \(2\pi\)
Jawaban: D
\(2 \sin^2 x \:-\:\cos x = 1\)
\(2(1\:-\:\cos^2x \:-\:\cos x \:-\: 1 = 0\)
\(2 \:-\:2\cos^2x \:-\:\cos x \:-\:1 = 0\)
\(-2\cos^2x \:-\:\cos x + 1 = 0\)
\(2\cos^2x + \cos x \:-\: 1 = 0\)
\((2\cos x \:-\: 1)(\cos x + 1) = 0\)
\(2\cos x \:-\: 1 = 0\)
\(\cos x = \dfrac{1}{2}\)
\(x_1 = \dfrac{\pi}{3}\)
\(\cos x + 1 = 0\)
\(\cos x = -1\)
\(x_2 = \pi\)
\(x_1 + x_2 = \dfrac{\pi}{3} + \pi\)
\(x_1 + x_2 = \dfrac{4\pi}{3}\)
Soal 4
\(\lim\limits_{x \rightarrow 9} \dfrac{\sqrt{x}\:-\:\sqrt{4\sqrt{x}\:-\:3}}{x^2 \:-\:81}= \dotso\)
(A) \(\dfrac{1}{18}\)
(B) \(\dfrac{1}{48}\)
(C) \(\dfrac{1}{124}\)
(D) \(\dfrac{1}{324}\)
(E) \(\dfrac{1}{400}\)
Jawaban: D
\(\lim\limits_{x \rightarrow 9} \dfrac{\sqrt{x}\:-\:\sqrt{4\sqrt{x}\:-\:3}}{x^2 \:-\:81}\times \color{yellow} \dfrac{\sqrt{x} + \sqrt{4\sqrt{x}\:-\:3}}{\sqrt{x} + \sqrt{4\sqrt{x}\:-\:3}}\)
\(\lim\limits_{x \rightarrow 9}\dfrac {x \:-\: (4\sqrt{x}\:-\:3)}{(x + 9)(x\:-\:9)(\sqrt{x} + \sqrt{4\sqrt{x}\:-\:3})}\)
\(\lim\limits_{x \rightarrow 9}\dfrac {x \:-\: 4\sqrt{x} + 3}{(x + 9)((\sqrt{x})^2\:-\:3^2)(\sqrt{x} + \sqrt{4\sqrt{x}\:-\:3})}\)
\(\lim\limits_{x \rightarrow 9}\dfrac {(\sqrt{x}\:-\:3)(\sqrt{x} \:-\:1)}{(x + 9)((\sqrt{x} + 3)(\sqrt{x}\:-\:3)(\sqrt{x} + \sqrt{4\sqrt{x}\:-\:3})}\)
\(\lim\limits_{x \rightarrow 9}\dfrac {\cancel{(\sqrt{x}\:-\:3)}(\sqrt{x} \:-\:1)}{(x + 9)((\sqrt{x} + 3)\cancel{(\sqrt{x}\:-\:3)}(\sqrt{x} + \sqrt{4\sqrt{x}\:-\:3})}\)
\(\lim\limits_{x \rightarrow 9}\dfrac {\sqrt{x} \:-\:1}{(x + 9)(\sqrt{x} + 3)(\sqrt{x} + \sqrt{4\sqrt{x}\:-\:3})}\)
\(\dfrac {\sqrt{9} \:-\:1}{(9 + 9)(\sqrt{9} + 3)(\sqrt{9} + \sqrt{4\sqrt{9}\:-\:3})}\)
\(\dfrac {3 \:-\:1}{18(3+ 3)(3 + \sqrt{12\:-\:3})}\)
\(\dfrac {2}{18(6)(3 + 3)}\)
\(\dfrac {2}{18(6)(6)}\)
\(\dfrac {2}{648}\)
\(\dfrac {1}{324}\)
Soal 5
Jika \(\int_{-2}^{0} \left(\cos (-\pi k x) + \dfrac{6x^2\:-\:10x + 7}{k + 2}\right) \text{d}x = (k\:-\:2)(k + 7)\) untuk nilai \(k\) bilangan bulat, maka \(k + 5 = \dotso\)
(A) 10
(B) 9
(C) 8
(D) 7
(E) 6
Jawaban: C
Langkah 1: Menghitung integral
\(\int_{-2}^{0} \left(\cos (-\pi k x) \text{d}x + \int_{-2}^{0}\dfrac{6x^2\:-\:10x + 7}{k + 2}\right) \text{d}x\)
\(\int_{-2}^{0} \cos (-\pi k x) \text{d}x + \dfrac{1}{k + 2}\int_{-2}^{0} 6x^2\:-\:10x + 7 \text{d}x\)
\(\left.-\pi x \sin (-\pi k x)\right|_{-2}^0 + \dfrac{1}{k + 2} \left. 2x^3\:-\:5x^2 + 7x\right|_{-2}^0\)
\(0 + \dfrac{1}{k + 2} (0\:-\: (2(-2)^3\:-\:5(-2)^2 + 7(-2))\)
\(\dfrac{1}{k + 2} (0\:-\: (-16\:-\:20 \:-\:14)\)
\(\dfrac{1}{k + 2} (0\:-\: (-50)\)
\(\dfrac{50}{k + 2}\)
Langkah 2: Menghitung nilai k
\(\dfrac{50}{k + 2} = (k\:-\:2)(k + 7)\)
\(50 = (k\:-\:2)(k + 2)(k + 7)\)
\(k = 3\)
\(k + 5 = 3 + 5\)
\(k + 5 = 8\)
Soal 6
Pada balok ABCD.EFGH, dengan AB = 6, BC = 3, dan CG = 2, titik M, N, dan O masing-masing terletak pada rusuk EH, FG, dan AD. Jika 3EM = EH, FN = 2NG, 3DO = 2DA, dan α adalah bidang irisan balok yang melalui M, N, O, perbandingan luas bidang α dengan luas permukaan balok adalah…
(A) \(\dfrac{\sqrt{35}}{36}\)
(B) \(\dfrac{\sqrt{37}}{36}\)
(C) \(\dfrac{\sqrt{38}}{36}\)
(D) \(\dfrac{\sqrt{39}}{36}\)
(E) \(\dfrac{\sqrt{41}}{36}\)
Jawaban: B
Bidang \(\alpha\) = bidang MOPN
Langkah 1: Menghitung panjang OP
\(\text{OP}^2 = \text{OR}^2 + \text{RP}^2\)
\(\text{OP}^2 = 6^2 + 1^2\)
\(\text{OP}^2 = 37\)
\(\text{OP} = \sqrt{37}\)
Langkah 2: Menghitung luas bidang α
\(\text{Luas} = \text{OP} \times \text{PN}\)
\(\text{Luas} = \sqrt{37}\times 2\)
\(\text{Luas} = 2\sqrt{37}\)
Langkah 3: Menghitung luas permukaan balok
\(\text{LP} = 2(pl + pt + lt)\)
\(\text{LP} = 2(6(3) + 6(2) + 3(2))\)
\(\text{LP} = 2(18 + 12 + 6)\)
\(\text{LP} = 2(36)\)
\(\text{LP} = 72\)
Langkah 4: Menghitung perbandingan luas bidang α dengan luas permukaan balok
Perbandingan luas bidang α dengan luas permukaan balok adalah:
\(2\sqrt{37} : 72\)
\(\sqrt{37} : 36\)
Soal 7
Jika θ adalah sudut antara bidang BEG dan DEG pada kubus ABCD.EFGH, maka \(\sin 2 \theta = \dotso\)
(A) \(\dfrac{\sqrt{2}}{9}\)
(B) \(\dfrac{2\sqrt{2}}{9}\)
(C) \(\dfrac{4\sqrt{2}}{9}\)
(D) \(\dfrac{5\sqrt{2}}{9}\)
(E) \(0\)
Jawaban: C
Sudut yang dibentuk antara bidang BEG dengan bidang DEG adalah sudut DTB (θ).
Langkah 1: Menghitung panjang BT
\(\text{BT}^2 = \text{BF}^2 + \text{FT}^2\)
\(\text{BT}^2 = 1^2 + (\frac{1}{2}\sqrt{2})^2\)
\(\text{BT}^2 = 1 + \frac{1}{2}\)
\(\text{BT} = \sqrt{\frac{3}{2}}\)
Langkah 2: Menghitung cos θ
\(\text{DB}^2 = \text{TB}^2 + \text{TD}^2 \:-\:2\cdot\text{TB}\cdot\text{TD}\cos \theta\)
\((\sqrt{2})^2 = \left(\sqrt\frac{3}{2}\right)^2 + \left(\sqrt\frac{3}{2}\right)^2 \:-\:2\cdot \sqrt\frac{3}{2} \cdot \sqrt\frac{3}{2}\cos \theta\)
\(2 = \frac{3}{2} + \frac{3}{2} \:-\:2\cdot \frac{3}{2}\cos \theta\)
\(2 \:-\:\frac{3}{2} \:-\: \frac{3}{2} = – 3\cos \theta\)
\(-1 = – 3\cos \theta\)
\(\cos \theta = \frac{1}{3}\)
Langkah 2: Menghitung sin 2θ
\(\cos \theta = \frac{1}{3}\)
\(\sin \theta = \frac{2\sqrt{2}}{3}\)
\(\color{yellow} \sin 2\theta = 2\sin \theta \cdot \cos \theta\)
\(\sin 2\theta = 2 \cdot \frac{2\sqrt{2}}{3} \cdot \frac{1}{3}\)
\(\sin 2\theta = \frac{4\sqrt{2}}{9}\)
Soal 8
Jika \(3^x + 5^y = 18\), nilai maksimum \(3^x\cdot 5^y\) adalah…
(A) 72
(B) 80
(C) 81
(D) 86
(E) 88
Jawaban: C
\(3^x + 5^y = 18\)
\(5^y = 18 \:-\:3^x\)
\(\text{L} = 3^x\cdot 5^y\)
\(\text{L} = 3^x(18 \:-\:3^x)\)
\(\text{L} = 18\cdot 3^x \:-\: (3^x)^2\)
Misal \(3^x = p\)
\(\text{L} = 18p \:-\: p^2\)
Agar L maksimum maka L’ = 0
\(18\:-\:2p = 0\)
\(-2p = -18\)
\(p = 9\)
\(3^x = 9\)
\(x = 2\)
\(5^y = 18 \:-\:3^2\)
\(5^y = 9\)
\(y = \:^5 \log 9\)
\(\text{L}_{\text{max}} = 3^2\cdot 5^{^5 \log 9}\)
\(\text{note:} \color{yellow} a^{^a\log b} = b\)
\(\text{L}_{\text{max}} = 9\cdot 9\)
\(\text{L}_{\text{max}} = 81\)
Soal 9
Diketahui \(sx\:-\:y = 0\) adalah garis singgung sebuah lingkaran yang titik pusatnya berada di kuadran ketiga dan berjarak 1 satuan ke sumbu-x. Jika lingkaran tersebut menyinggung sumbu-x dan titik pusatnya dilalui garis \(x = -2\), nilai \(3s\) adalah…
(A) \(\dfrac{1}{6}\)
(B) \(\dfrac{4}{3}\)
(C) \(3\)
(D) \(4\)
(E) \(6\)
Jawaban: D
Karena pusat lingkaran dilalui garis x = −2, berjarak 1 satuan ke sumbu-x, dan terletak di kuadran III, maka pusat lingkaran berada di titik (−2, −1) dan jari-jari lingkarannya adalah 1.
Jarak pusat lingkaran (−2, −1) ke garis singgung \(sx\:-\:y = 0\) sama dengan jari-jari lingkaran.
\(r = \left|\dfrac{sx\:-\:y}{\sqrt{s^2 + (-1)^2}}\right|\)
\(1 = \left|\dfrac{-2s + 1}{\sqrt{s^2 + 1}}\right|\)
kuadratkan kedua ruas,
\(1 = \dfrac{4s^2 \:-\: 4s + 1}{s^2 + 1}\)
Kali silang,
\(s^2 + 1 = 4s^2 \:-\: 4s + 1\)
\(0 = 3s^2 \:-\: 4s\)
\(0 = s(3s \:-\: 4)\)
\(s = 0 \text{ atau } s = \frac{4}{3}\)
\(3s = 3\cdot \frac{4}{3}\)
\(3s = 4\)
Soal 10
Jika kurva \(y = (a\:-\:2)x^2 + \sqrt{3}(1\:-\:a)x + (a\:-\:2)\) selalu berada di atas sumbu-x, bilangan bulat terkecil \(a\:-\:2\) yang memenuhi adalah…
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Jawaban: B
Kurva selalu berada di atas sumbu-x, maka fungsi tersebut definit positif (selalu bernilai positif)
Syarat fungsi definit positif:
(1) \(\color{cyan} a > 0\)
(2) \(\color{cyan} b^2 \:-\:4ac < 0\)
Langkah 1
\(a\:-\:2 > 0\)
\(a > 2\)
Langkah 2
\(b^2 \:-\:4ac < 0\)
\((\sqrt{3}(1\:-\:a))^2 \:-\:4(a\:-\:2)(a\:-\:2) < 0\)
\(3(1 \:-\:2a + a^2) \:-\:4(a^2\:-\:4a + 4) < 0\)
\(3 \:-\:6a + 3a^2 \:-\:4a^2 + 16a \:-\:16 < 0\)
\(-a^2 + 10a \:-\:13 < 0\)
pembuat nol:
\(x_1 = 5\:-\:2\sqrt{3}\)
\(x_2 = 5 + 2\sqrt{3}\)
Langkah 3
Irisan dari solusi langkah 1 dan 2 adalah:
\(\lbrace x | x > 5 + 2\sqrt{3}, \: x \in R \rbrace\)
\(\lbrace x | x > 8,46, \: x \in R \rbrace\)
Bilangan bulat terkecil \(a\) yang memenuhi adalah 9
\(a\:-\:2 = 9\:-\:2\)
\(a\:-\:2 = 7\)
Soal 11
Jika diberikan \(\sqrt{3}a + b \:-\:c = 2\), \(bc = -1,5a^2\), dan \(b^2 + c^2 = \sqrt{3}a\), nilai \(a\) adalah…
(A) \(\dfrac{2\sqrt{3}}{15}\)
(B) \(\dfrac{4\sqrt{3}}{15}\)
(C) \(\dfrac{7\sqrt{3}}{15}\)
(D) \(\dfrac{8\sqrt{3}}{15}\)
(E) \(\dfrac{11\sqrt{3}}{15}\)
Jawaban: B
\(\sqrt{3}a + b \:-\:c = 2\)
\(b \:-\:c = 2\:-\:\sqrt{3}a\)
Kuadratkan kedua ruas,
\((b \:-\:c)^2 = (2\:-\:\sqrt{3}a)^2\)
\(b^2 \:-\:2bc + c^2 = 4 \:-\:4\sqrt{3}a + 3a^2\)
\(b^2 + c^2 \:-\:2bc = 4 \:-\:4\sqrt{3}a + 3a^2\)
\(\sqrt{3}a\:-\:2(-1,5a^2) = 4 \:-\:4\sqrt{3}a + 3a^2\)
\(\sqrt{3}a + \cancel{3a^2} = 4 \:-\:4\sqrt{3}a + \cancel{3a^2}\)
\(5\sqrt{3}a = 4\)
\(a = \dfrac{4}{5\sqrt{3}}\)
\(a = \dfrac{4}{5\sqrt{3}} \times \color{yellow} \dfrac{\sqrt{3}}{\sqrt{3}}\)
\(a = \dfrac{4\sqrt{3}}{15}\)
Soal 12
Diketahui sebuah barisan \(0, \dfrac{5}{6}, \dfrac{5}{36}, \dfrac{35}{216}, \dotso\). Suku ke-12 dari barisan tersebut adalah…
(A) \(\dfrac{1}{2^{11}}\:-\:\dfrac{1}{3^{11}}\)
(B) \(\dfrac{1}{2^{11}}\:-\:\dfrac{2}{3^{11}}\)
(C) \(\dfrac{3}{2^{11}}\:-\:\dfrac{1}{3^{11}}\)
(D) \(\dfrac{1}{2^{11}} + \dfrac{1}{3^{11}}\)
(E) \(\dfrac{2}{2^{11}} + \dfrac{3}{3^{11}}\)
Jawaban: D
\(\text{U}_1 = 0 = \dfrac{1}{2^0}\:-\:\dfrac{1}{3^0} = \dfrac{1}{2^{1\:-\:1}} + (-1)^1\cdot \dfrac{1}{3^{1\:-\:1}}\)
\(\text{U}_2 = \dfrac{5}{6} = \dfrac{1}{2^1} + \dfrac{1}{3^1} = \dfrac{1}{2^{2\:-\:1}} + (-1)^2\cdot \dfrac{1}{3^{2\:-\:1}}\)
\(\text{U}_3 = \dfrac{5}{36} = \dfrac{1}{2^2}\:-\:\dfrac{1}{3^2} = \dfrac{1}{2^{3\:-\:1}} + (-1)^3\cdot \dfrac{1}{3^{3\:-\:1}}\)
\(\text{U}_4 = \dfrac{35}{216} = \dfrac{1}{2^3}+\dfrac{1}{3^3} = \dfrac{1}{2^{4\:-\:1}} + (-1)^4\cdot \dfrac{1}{3^{4\:-\:1}}\)
\(\dotso\)
\(\text{U}_n = \dfrac{1}{2^{n\:-\:1}} + (-1)^n \cdot \dfrac{1}{3^{n\:-\:1}}\)
\(\text{U}_{12} = \dfrac{1}{2^{12\:-\:1}} + (-1)^{12} \cdot \dfrac{1}{3^{12\:-\:1}}\)
\(\text{U}_{12} = \dfrac{1}{2^{11}} + \dfrac{1}{3^{11}}\)
Soal 13
Jika vektor \(\textbf{a} = (3, -2, -5)\), \(\textbf{b} = (1, 4, -4)\), dan \(\textbf{c} = (0, 3, 2)\), maka…
(1) \(\textbf{a, b, c}\) membentuk jajargenjang
(2) \(\textbf{a}\cdot (\textbf{b} \times \textbf{c}) = (\textbf{b} \times \textbf{c})\cdot \textbf{a}\)
(3) Volume jajargenjang = 49
(4) \(\textbf{a} \times \textbf{b} = -(\textbf{b} \times \textbf{a})\)
Jawaban: C
\(\textbf{b} \times \textbf{c} = \begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\1 & 4 & -4\\0 & 3 & 2\end{vmatrix}\)
\(\textbf{b} \times \textbf{c} =\begin{vmatrix}4 & -4\\ 3 & 2\end{vmatrix}\textbf{i}\:-\:\begin{vmatrix}1 & 0\\ -4 & 2\end{vmatrix}\textbf{j} + \begin{vmatrix}1 & 4\\ 0 & 3\end{vmatrix}\textbf{k}\)
\(\textbf{b} \times \textbf{c} =(4\cdot 2\:-\:(-4)\cdot 3)\textbf{i}\:-\:(1\cdot 2\:-\:(-4)\cdot 0)\textbf{j} + (1\cdot 3\:-\:4\cdot 0)\textbf{k}\)
\(\textbf{b} \times \textbf{c} =20\textbf{i}\:-\:2\textbf{j} + 3\textbf{k}\)
\(\textbf{a}\cdot (\textbf{b} \times \textbf{c}) = (3, -2, -5) \cdot (20, -2, 3)\)
\(\textbf{a}\cdot (\textbf{b} \times \textbf{c}) = 3(20) + (-2)(-2) + (-5)(3)\)
\(\textbf{a}\cdot (\textbf{b} \times \textbf{c}) = 60 + 4 \:-\:15 = 49\)
\((\textbf{b} \times \textbf{c})\cdot \textbf{a} = (20, -2, 3) \cdot (3, -2, -5)\)
\((\textbf{b} \times \textbf{c})\cdot \textbf{a} = 20(3) + (-2)(-2) + 3(-5)\)
\((\textbf{b} \times \textbf{c})\cdot \textbf{a} = 60 + 4 \:-\:15 = 49\)
Pernyataan (2) benar
Pernyataan (4) benar
\(\textbf{a} \times \textbf{b} = -(\textbf{b} \times \textbf{a})\)
Jadi, pernyataan (2) dan (4) benar
Soal 14
Jika \(f(x) = (2x\:-\:3)^7\:-\:(2x\:-\:3)^5 + (2x\:-\:3)^3\), maka…
(1) \(f\) selalu naik pada R
(2) \(f\) tidak pernah turun
(3) \(f\) tidak memiliki maksimum relatif
(4) \(f\) minimum relatif pada \(x = \dfrac{3}{2}\)
Jawaban: A
\(f(x) = (2x\:-\:3)^7\:-\:(2x\:-\:3)^5 + (2x\:-\:3)^3\)
Menentukan titik stasioner, \(f'(x) = 0\)
\(f'(x) = 7(2)(2x\:-\:3)^6\:-\:5(2)(2x\:-\:3)^4 + 3(2)(2x\:-\:3)^2\)
\(0 = 14(2x\:-\:3)^6\:-\:10(2x\:-\:3)^4 + 6(2x\:-\:3)^2\)
\(0 = 2(2x\:-\:3)^2(7(2x\:-\:3)^4\:-\:5(2x\:-\:3)^2 + 3)\)
Hanya terdapat satu buah titik stasioner di \(x = \dfrac{3}{2}\)
(1) \(f\) selalu naik pada R (benar)
(2) \(f\) tidak pernah turun (benar)
(3) \(f\) tidak memiliki maksimum relatif (benar)
(4) \(f\) minimum relatif pada \(x = \dfrac{3}{2}\) (salah) harusnya titik belok
Soal 15
Jika \(\alpha = \dfrac{\pi}{12}\), maka…
(1) \(\sin^4\alpha + \cos^4\alpha = \dfrac{7}{8}\)
(2) \(\sin^6\alpha + \cos^6\alpha = \dfrac{11}{16}\)
(3) \(\cos^4\alpha = \dfrac{7}{16}\:-\:\dfrac{1}{4}\sqrt{3}\)
(4) \(\sin^4 \alpha = \dfrac{3}{8} \:-\:\dfrac{1}{2}\sqrt{3}\)
Jawaban: B
\(\color{cyan}\sin \dfrac{\pi}{12} = \dfrac{\sqrt{6}\:-\:\sqrt{2}}{4}\)
\(\color{cyan}\cos \dfrac{\pi}{12} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\)
\(\sin^4\alpha + \cos^4\alpha\)
\((\sin^2 \alpha)^2 + (\cos^2 \alpha)^2\)
\((\sin^2 \alpha + \cos^2 \alpha)^2\:-\:2\sin^2 \alpha\cdot \cos^2\alpha\)
\(1^2\:-\:2(\sin\alpha\cdot \cos\alpha)^2\)
\(1\:-\:2\left[ \left(\dfrac{\sqrt{6}\:-\:\sqrt{2}}{4}\right)\left(\dfrac{\sqrt{6} + \sqrt{2}}{4}\right)\right]^2\)
\(1\:-\:2\left[ \dfrac{6\:-\:2}{16}\right]^2\)
\(1\:-\:2\left[ \dfrac{1}{4}\right]^2\)
\(1\:-\:\dfrac{1}{8}\)
\(\dfrac{7}{8}\)
Pernyataan 1 benar
\(\sin^6\alpha + \cos^6\alpha\)
\((\sin^2 \alpha)^3 + (\cos^2 \alpha)^3\)
\((\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha \:-\:\sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha)\)
\(1\cdot (\sin^4 \alpha + \cos^4 \alpha \:-\: (\sin\alpha \cos\alpha)^2)\)
\(\dfrac{7}{8} \:-\:\left (\dfrac{1}{4}\right)^2\)
\(\dfrac{14}{16} \:-\: \dfrac{1}{16}\)
\(\dfrac{13}{16}\)
Pernyataan 2 salah
\(\cos^4\alpha = \left(\dfrac{\sqrt{6} + \sqrt{2}}{4} \right)^4\)
\(\cos^4\alpha = \dfrac{36 + 4\cdot 6\sqrt{6}\cdot \sqrt{2} + 6\cdot 6 \cdot 2 + 4\cdot \sqrt{6}\cdot 2\sqrt{2} + 4}{256}\)
\(\cos^4\alpha = \dfrac{36 + 24\sqrt{12} + 72 + 8\sqrt{12} + 4}{256}\)
\(\cos^4\alpha = \dfrac{112 + 32\sqrt{12}}{256}\)
\(\cos^4\alpha = \dfrac{112 + 64\sqrt{3}}{256}\)
\(\cos^4\alpha = \dfrac{112}{256} + \dfrac{64\sqrt{3}}{256}\)
\(\cos^4\alpha = \dfrac{7}{16}\:-\:\dfrac{1}{4}\sqrt{3}\)
Pernyataan 3 benar
