Suku ke-n deret geometri adalah Un. Jika diketahui \(\dfrac{\text{U}_6}{\text{U}_8} = 3\) dan \(\text{U}_2 \cdot \text{U}_8 = \dfrac{1}{3}\), maka nilai \(\text{U}_{10} = \dotso\)
Jawaban: A
Rumus suku ke-n barisan geometri:
\(\color{blue} \text{U}_n = a\cdot r^{n\:-\:1}\)
\(\dfrac{\text{U}_6}{\text{U}_8} = 3\)
\(\dfrac{\cancel{a}\cdot r^5}{\cancel{a} \cdot r^7} = 3\)
\(\dfrac{1}{r^2} = 3\)
\(r^2 = \dfrac{1}{3}\)
\(r = \pm \dfrac{1}{\sqrt{3}}\)
\(\text{U}_2 \cdot \text{U}_8 = \dfrac{1}{3}\)
\(a\cdot r \cdot a \cdot r^7= \dfrac{1}{3}\)
\(a^2\cdot r^8= \dfrac{1}{3}\)
\(a^2\cdot (r^2)^4 = \dfrac{1}{3}\)
\(a^2\cdot \left( \dfrac{1}{3}\right)^4 = \dfrac{1}{3}\)
\(a^2\cdot \dfrac{1}{81}= \dfrac{1}{3}\)
\(a^2 = \dfrac{1}{3}\cdot \dfrac{81}{1}\)
\(a^2 = 27\)
\(a = \pm \sqrt{27}\)
\(a = \pm 3\sqrt{3}\)
\(\text{U}_{10} = a\cdot r^9\)
Untuk
\(r = \dfrac{1}{\sqrt{3}}\)
\(a = 3\sqrt{3}\)
\(\text{U}_{10} = 3\sqrt{3}\cdot \left(\dfrac{1}{\sqrt{3}}\right)^9\)
\(\text{U}_{10} = 3\cancel{\sqrt{3}}\cdot \dfrac{1}{3^4\cancel{\sqrt{3}}}\)
\(\text{U}_{10}= 3\cdot \dfrac{1}{3^4}\)
\(\text{U}_{10} = \dfrac{1}{3^3}\)
\(\text{U}_{10} = \dfrac{1}{27}\)
Jika \(x_1\) dan \(x_2\) memenuhi persamaan \(12 \cos^2 x \:-\: \cos x \:-\:1 = 0\), maka nilai \(\sec^2 x_1 + \sec^2 x_2 = \dotso\)
Jawaban: B
\(12 \cos^2 x \:-\: \cos x \:-\:1 = 0\)
\((4\cos x + 1)(3\cos x \:-\:1) = 0\)
\(4\cos x + 1 = 0\)
\(\cos x = – \dfrac{1}{4}\)
\(\color{blue} \sec x = \dfrac{1}{\cos x}\)
\(\sec x = – 4\)
\(3\cos x \:-\:1 = 0\)
\(\cos x = \dfrac{1}{3}\)
\(\sec x = – 3\)
\(\sec^2 x_1 + \sec^2 x_2 = (-4)^2 + (-3)^2\)
\(\sec^2 x_1 + \sec^2 x_2 = 16 + 9\)
\(\sec^2 x_1 + \sec^2 x_2 = 25\)
Panjang proyeksi vektor \((a, 5, -1)\) pada vektor \((1, 4, 8)\) adalah 2, maka \(a = \dotso\)
Jawaban: E
Panjang proyeksi disebut juga proyeksi skalar
Misal:
\(\vec{\text{a}} = (a, 5, -1)\)
\(\vec{\text{b}} = (1, 4, 8)\)
Proyeksi skalar \(\vec{\text{a}}\) pada \(\vec{\text{b}}\) adalah \(\text{p}\)
\(\color{blue} \text{p} = \dfrac{\vec{\text{a}} \cdot \vec{\text{b}}}{|\vec{\text{b}}|}\)
\(2 = \dfrac{(a, 5, -1) \cdot (1, 4, 8)}{\sqrt{1^2 + 4^2 + 8^2}}\)
\(2 = \dfrac{a(1) + 5(4) + (-1)(8)}{\sqrt{1 + 16 + 64}}\)
\(2 = \dfrac{a + 20 \:-\:8}{\sqrt{81}}\)
\(2 = \dfrac{a + 12}{9}\)
\(18 = a + 12\)
\(a = 18 \:-\:12\)
\(a = 2\)
Jika luas daerah yang dibatasi oleh kurva \(y = x^2\) dan garis \(y = (2m\:-\:1)x\) adalah 4½, maka \(m = \dotso\)
Jawaban:
Luas daerah yang dibatasi kurva dan garis dapat dihitung menggunakan rumus:
\(\color{blue} \text{L} = \dfrac{\text{D}\sqrt{\text{D}}}{6\text{a}^2}\)
dengan D = b² − 4ac
Substitusikan persamaan garis ke kurva,
\((2m\:-\:1)x = x^2\)
\(0 = x^2 \:-\:(2m\:-\:1)x\)
Nilai diskriminan:
\(\text{D} = (-(2m\:-\:1))^2\:-\:4(1)(0)\)
\(\text{D} = (2m\:-\:1)^2\)
\(\text{L} = \dfrac{\text{D} \sqrt{\text{D}}}{6\text{a}^2}\)
\(4\dfrac{1}{2} = \dfrac{(2m\:-\:1)^2 \sqrt{(2m\:-\:1)^2}}{6(1)^2}\)
\(4\dfrac{1}{2} = \dfrac{(2m\:-\:1)^3}{6}\)
\(\dfrac{9}{2}\times 6 = (2m\:-\:1)^3\)
\(27 = (2m\:-\:1)^3\)
\(3^3 = (2m\:-\:1)^3\)
\(3 = 2m\:-\:1\)
Perhatikan bahwa \(2m\:-\:1\) merupakan gradien garis \(y = (2m\:-\:1)x\). Garis ini memiliki gradien dengan kemungkinan (+) dan (-).
Kemungkinan 1: \(2m\:-\:1\) bernilai positif
\(3 = 2m\:-\:1\)
\(4 = 2m\)
\(m = 2\)
Kemungkinan 2: \(2m\:-\:1\) bernilai negatif
\(3 = -(2m\:-\:1)\)
\(3 = -2m + 1\)
\(-2m = 2\)
\(m = -1\)
Jadi, nilai \(m\) yang memenuhi adalah 2 dan −1
