Kategori: Matematika IPA

\(\lim\limits_{x \to 0} \dfrac{\sin{(\pi + 2x)} + \tan {4x}}{3x+\sin{(\frac{\pi}{2} + x)}}\)

\(\dfrac{\sin{(\pi + 0)} + \tan {0}}{0 +\sin{(\frac{\pi}{2} +0)}}\)

\(\dfrac{0 + 0}{0 + 1}\)

\(\dfrac{0}{1}\)

\(0\)

Bagi pembilang dan penyebut dengan \(x\)

\(\lim\limits_{x \to 0} \dfrac{\frac{\sin {4x}}{x} + \frac{\tan {6x}}{x}\:-\:\frac{2x}{x}}{\frac{4x}{x}\:-\:\frac{\tan {3x}}{x} \:-\:\frac{\sin {3x}}{x}}\)

\(\dfrac{4 + 6 \:-\:2}{4\:-\:3\:-\:3}\)

\(\dfrac{8}{-2}\)

\(-4\)

Lakukan pemfaktoran terlebih dahulu,

\(\lim\limits_{x \to p} \dfrac{(x + p)(x\:-\:p)}{p\cdot \sin 4(x\:-\:p)\:-\:p(x\:-\:p)}\)

Selanjutnya, bagi pembilang dan penyebut dengan \((x\:-\:p)\)

\(\lim\limits_{x \to p} \dfrac{\frac{(x + p)(x\:-\:p)}{(x\:-\:p)}}{p\cdot \frac{\sin 4(x\:-\:p)}{(x\:-\:p)}\:-\:p\frac{(x\:-\:p)}{(x\:-\:p)}}\)

\(\lim\limits_{x \to p} \dfrac{x + p}{4p\:-\:p}\)

\(\dfrac{p + p}{3p}\)

\(\dfrac{2p}{3p}\)

\(\dfrac{2}{3}\)

Kali sekawan,

\(\lim\limits_{x \to 0} \dfrac{1\:-\:\sqrt{\cos 2x}}{16x^2}\times \color{red} \dfrac{1 + \sqrt{\cos 2x}}{1 + \sqrt{\cos 2x}}\)

\(\lim\limits_{x \to 0} \dfrac{1\:-\:\cos 2x}{16x^2(1+\sqrt{\cos 2x})}\)

Untuk menghilangkan angka 1, ubah \(\color{blue}\cos 2x = 1\:-\:2\sin^2{x}\)

\(\lim\limits_{x \to 0} \dfrac{1\:-\:(1\:-\:2\sin^2{x})}{16x^2(1+\sqrt{\cos 2x})}\)

\(\lim\limits_{x \to 0} \dfrac{2\sin^2{x}}{16x^2(1+\sqrt{\cos 2x})}\)

\(\lim\limits_{x \to 0} \dfrac{2\sin^2{x}}{16x^2}\times \dfrac{1}{(1+\sqrt{\cos 2x})}\)

\(\dfrac{2}{16}\times \dfrac{1}{(1+\sqrt{\cos 0})}\)

\(\dfrac{1}{8}\times \dfrac{1}{(1 + 1)}\)

\(\dfrac{1}{16}\)

\(\lim\limits_{x \to 0} \dfrac{2[\sin(x + \frac{\pi}{3})\:-\:\frac{\sqrt{3}}{2}]}{4x}\)

\(\lim\limits_{x \to 0} \dfrac{2[\sin(x + \frac{\pi}{3})\:-\:\sin \frac{\pi}{3}]}{4x}\)

Gunakan rumus:

\(\color{blue}\sin\text{A}\:-\:\sin \text{B} = 2\cos \frac{1}{2}(\text{A + B})\sin \frac{1}{2}(\text{A − B})\)

\(\lim\limits_{x \to 0} \dfrac{2[2\cos \frac{1}{2}(x + \frac{2\pi}{3})\sin \frac{1}{2}x]}{4x}\)

\(\lim\limits_{x \to 0} 4\cdot\cos \frac{1}{2}(x + \frac{2\pi}{3}) \times \frac{\sin \frac{1}{2}x}{4x}\)

\( 4\cdot\cos \frac{1}{2}(0 + \frac{2\pi}{3}) \times \frac{1}{8}\)

\( 4\cdot\cos \frac{\pi}{3} \times \frac{1}{8}\)

\( 4\cdot \frac{1}{2} \cdot \frac{1}{8}\)

\(\dfrac{1}{4}\)

\(\lim\limits_{x \to 0} \dfrac{\sqrt{1 + \sin 6x}\:-\:\sqrt{1\:-\:\sin 6x}}{\tan 3x}\times \color{red} \dfrac{\sqrt{1 + \sin 6x} + \sqrt{1\:-\:\sin 6x}}{\sqrt{1 + \sin 6x} + \sqrt{1\:-\:\sin 6x}}\)

\(\lim\limits_{x \to 0} \dfrac{1 + \sin 6x \:-\: (1 \:-\:\sin 6x)}{\tan 3x(\sqrt{1 + \sin 6x} + \sqrt{1\:-\:\sin 6x})}\)

\(\lim\limits_{x \to 0} \dfrac{2\sin 6x}{\tan 3x(\sqrt{1 + \sin 6x} + \sqrt{1\:-\:\sin 6x})}\)

\(\lim\limits_{x \to 0} \dfrac{2\sin 6x}{\tan 3x}\times \dfrac{1}{(\sqrt{1 + \sin 6x} + \sqrt{1\:-\:\sin 6x})}\)

\(2\cdot \frac{6}{3}\times \dfrac{1}{(\sqrt{1 + \sin 0} + \sqrt{1\:-\:\sin 0})}\)

\(4\times \dfrac{1}{(\sqrt{1 + 0} + \sqrt{1\:-\:0})}\)

\(4\times\dfrac{1}{2}\)

\(2\)

\(\lim\limits_{x \to 0} \dfrac{2(1\:-\:\cos 4x)}{\cos 3x\:-\:\cos 5x}\)

Gunakan rumus berikut:

\(\color{blue}\cos 4x = 1\:-\:2\sin^2 {2x}\)

\(\color{blue}\cos \text{A}\:-\:\cos \text{B} = -2\sin \frac{1}{2}(\text{A + B}) \sin \frac{1}{2} (\text{A − B})\)

\(\lim\limits_{x \to 0} \dfrac{2[1\:-\:(1\:-\:2\sin^2 {2x})]}{-2\sin \frac{1}{2} (8x)\cdot \sin \frac{1}{2}(-2x)}\)

\(\lim\limits_{x \to 0} \dfrac{2(2\sin^2 {2x})}{-2\sin 4x\cdot \sin (-x)}\)

\(-\frac{4}{2}\cdot \lim\limits_{x \to 0}\dfrac{\sin {2x}}{\sin 4x}\cdot \dfrac{\sin 2x}{\sin {(-x)}}\)

\(-2\cdot \frac{2}{4}\cdot (-2)\)

\(2\)

\(\lim\limits_{x \to 0} \dfrac{96x^3}{2\sin 2x\cdot \cos 2x\:-\:2\sin 2x}\)

\(\lim\limits_{x \to 0} \dfrac{96x^3}{2\sin 2x(\cos 2x\:-\:1)}\)

\(\lim\limits_{x \to 0} \dfrac{96x}{2\sin 2x}\times \dfrac{x^2}{(\cos 2x\:-\:1)}\)

\(\dfrac{96}{4}\cdot \lim\limits_{x \to 0}\dfrac{x^2}{\cos 2x\:-\:1}\)

\(\dfrac{96}{4}\cdot \lim\limits_{x \to 0}\dfrac{x^2}{1\:-\:2\sin^2 x\:-\:1}\)

\(\dfrac{96}{4}\cdot \lim\limits_{x \to 0}\dfrac{x^2}{-2\sin^2 x}\)

\(\dfrac{96}{4}\cdot \dfrac{1}{-2}\)

\(-12\)

\(\lim\limits_{x \to 0} \dfrac{2\:-\:\cos^2 4x \:-\:\cos^2 8x}{2\:-\:\cos 2x \:-\:\cos 4x}\)

\(\lim\limits_{x \to 0} \dfrac{2\:-\:(1\:-\:\sin^2 {4x}) \:-\:(1\:-\:\sin^2 {8x})}{2\:-\:(1\:-\:2\sin^2 x) \:-\:(1\:-\:2\sin^2 2x)}\)

\(\lim\limits_{x \to 0} \dfrac{\sin^2 {4x} + \sin^2 {8x}}{2\sin^2 x + 2\sin^2 2x}\)

Bagi pembilang dan penyebut dengan \(x^2\)

\(\lim\limits_{x \to 0} \dfrac{\frac{\sin^2 {4x}}{x^2} + \frac{\sin^2 {8x}}{x^2}}{2\frac{\sin^2 x}{x^2} + 2\frac{\sin^2 2x}{x^2}}\)

\(\dfrac{4^2 + 8^2}{2\cdot 1^2 + 2\cdot 2^2}\)

\(\dfrac{80}{10}\)

\(8\)

\(\lim\limits_{x \to \frac{\pi}{2}} \dfrac{\frac{\sin x\:-\:1}{\sin x}[\sin \frac{3}{2}x + \sin (\frac{\pi}{2}\:-\:\frac{3}{2}x)]}{1+(2\cos^2 x\:-\:1)}\)

\(\lim\limits_{x \to \frac{\pi}{2}}\dfrac{(\sin x \:-\:1)[2\sin \frac{1}{2}(\frac{\pi}{2})\cos \frac{1}{2}(3x\:-\:\frac{\pi}{2})]}{\sin x \cdot 2\cos^2 x}\)

\(\lim\limits_{x \to \frac{\pi}{2}} \dfrac{(\sin x\:-\:\sin \frac{\pi}{2})[\sqrt{2}\cos (\frac{3}{2}x\:-\:\frac{\pi}{4})]}{2\sin x \cdot \cos x \cdot \cos x}\)

\(\lim\limits_{x \to \frac{\pi}{2}}\dfrac{\sqrt{2}[2\cos \frac{1}{2}(x + \frac{\pi}{2})\cancelto{-\frac{1}{2}}{\sin \frac{1}{2}(x \:-\:\frac{\pi}{2})}]\sin[\frac{\pi}{2}\:-\:(\frac{3}{2}x\:-\:\frac{\pi}{4})]}{2\sin\frac{\pi}{2}\cdot \cancel{\sin (\frac{\pi}{2}\:-\:x)}\cdot \sin (\frac{\pi}{2}\:-\:x)}\)

\(\lim\limits_{x \to \frac{\pi}{2}}\dfrac{-\sqrt{2}\cos(\frac{1}{2}x + \frac{\pi}{4})\cancelto{\frac{3}{2}}{\sin (\frac{3\pi}{4}\:-\:\frac{3}{2}x)}}{2\cancel{\sin(\frac{\pi}{2}\:-\:x)}}\)

\(\dfrac{-\sqrt{2}\cos (\frac{\pi}{4} + \frac{\pi}{4})}{2}\)

\(\dfrac{-\sqrt{2}\cos \frac{\pi}{2}}{2}\)

\(\dfrac{-\sqrt{2}\cdot 0}{2}\)

\(0\)

Turunan pertama dari \(y = \sec x\) adalah \(y’ = \sec x \cdot \tan x\)

\(f(x) = 5\sec^3 2x\)

\(f'(x) = 15\cdot \sec^2 2x \cdot 2\cdot \sec 2x \cdot \tan 2x\)

\(f'(x) = 30\cdot \sec^3 2x \cdot \tan 2x\)

Misal:

\(u = 1 + \sin 2x \rightarrow u’ = 2\cos 2x\)

\(v = \cos 2x \rightarrow v’ = -2\sin 2x\)

Turunan pertama dari \(y = \dfrac{u}{v}\) adalah \(y’ = \dfrac{u’\cdot v\:-\:u\cdot v’}{v^2}\)

\(y’ = \dfrac{2\cos 2x\cdot \cos 2x\:-\:(1 + \sin 2x )\cdot ( -2\sin 2x)}{(\cos 2x )^2}\)

\(y’ = \dfrac{2\cos^2 2x + 2\sin 2x + 2\sin^2 2x}{\cos^2 2x}\)

\(y’ = \dfrac{2 + 2\sin 2x}{\cos^2 2x}\)

\(y’ = \dfrac{2(1 + \sin 2x)}{1\:-\:\sin^2 2x}\)

\(y’ = \dfrac{2\cancel{(1 + \sin 2x)}}{\cancel{(1 + \sin 2x)}(1\:-\:\sin 2x)}\)

\(y’ = \dfrac{2}{1\:-\:\sin 2x}\)

Turunan pertama dari \(y = \ln u\) adalah \(y’ = \dfrac{1}{u}\times u’\)

Turunan pertama dari \(y = \cot x\) adalah \(y’ = -\csc^2 x\)

\(y = \ln(\cot 2x)\)

\(y’ = \dfrac{1}{\cot 2x}\cdot -2\csc^2 2x\)

\(y’ = -2\dfrac{\csc^2 2x}{\cot 2x}\)

\(y’ = -2\dfrac{(\frac{1}{\sin 2x})^2}{\frac {\cos 2x}{\sin 2x}}\)

\(y’ = -2\dfrac{\frac{1}{\sin^2 2x}}{\frac {\cos 2x}{\sin 2x}}\)

\(y’ = -2\dfrac{\frac{1}{\sin 2x}}{\frac {\cos 2x}{1}}\)

\(y’ = -2\dfrac{1}{\sin 2x\cdot \cos 2x}\:\:\:\:\:\color{blue} (\sin 4x = 2\cdot \sin 2x \cdot \cos 2x)\)

\(y’ = -2\dfrac{1}{\frac{1}{2}\sin 4x}\)

\(y’ = -4\csc 4x\)

\(f(\tan 2x) = \sin^4 2x\)

Misal \(u = \tan 2x\)

\(\sin 2x = \dfrac{u}{\sqrt{1 + u^2}}\)

\(f(u) = \left[\dfrac{u}{\sqrt{1 + u^2}}\right]^4\)

\(f(u) = \dfrac{u^4}{\sqrt{(1 + u^2)^2}}\)

\(f(u) = \dfrac{u^4}{\sqrt{u^4 + 2u^2 + 1}}\)

\(f'(u) = \dfrac{4u^3(u^4 + 2u^2 + 1)\:-\:u^4(4u^3 + 4u)}{(u^4 + 2u^2 + 1)^2}\)

\(f'(u) = \dfrac{\cancel{4u^7} + 8u^5 + 4u^3\:-\:\cancel{4u^7}\:-\:4u^5}{(u^4 + 2u^2 + 1)^2}\)

\(f'(1) = \dfrac{8 + 4\:-\:4}{(1 + 2 + 1)^2}\)

\(f'(1) = \dfrac{8}{16} = \dfrac{1}{2}\)

Langkah 1: Menentukan titik singgung kurva

Titik singgung kurva berada di absis \(-\frac{\pi}{8}\)

\(f(-\frac{\pi}{8}) = \tan (\frac{\pi}{8} + \frac{\pi}{8})\)

\(f(-\frac{\pi}{8}) = \tan (\frac{\pi}{4})\)

\(f(-\frac{\pi}{8}) = 1\)

Jadi, titik singgung kurvanya adalah \((-\frac{\pi}{8},\: 1)\)

Langkah 2: Menentukan gradien garis singgung kurva

Gradien garis singgung kurva adalah turunan pertama kurva tersebut di titik berabsis \(-\frac{\pi}{8}\)

\(f'(x) = -\sec^2 {(-x + \frac{\pi}{8})}\)

\(\text{m}_{\text{gs}} = f'(-\frac{\pi}{8})\)

\(\text{m}_{\text{gs}} = -\sec^2 {(\frac{\pi}{8} + \frac{\pi}{8})}\)

\(\text{m}_{\text{gs}} = -\sec^2 {\frac{\pi}{4}}\)

\(\text{m}_{\text{gs}} = -(\sqrt{2})^2\)

\(\text{m}_{\text{gs}} = -2\)

Langkah 3: Menentukan gradien garis normal

Garis normal adalah garis yang tegak lurus dengan garis singgung di titik berabsis \(-\frac{\pi}{8}\) pada kurva

Syarat dua garis tegak lurus adalah:

\(\text{m}_{\text{gs}}\times \text{m}_{\text{gn}} = -1\)

\(-2\times \text{m}_{\text{gn}} = -1\)

Jadi, \(\text{m}_{\text{gn}} = \frac{1}{2}\)

Langkah 4: Menentukan persamaan garis normal

Persamaan garis normal yang melalui titik \((-\frac{\pi}{8},\: 1)\) dan bergradien \(\text{m}_{\text{gn}} = 1\) adalah:

\(y\:-\:y_{1} = \text{m}_{\text {gn}}(x\:-\:x_{1})\)

\(y\:-\:1 = \frac{1}{2}(x + \frac{\pi}{8})\)

Luas trapezium = jumlah sisi sejajar kali tinggi bagi 2

\(\text{x} = 20\cdot \cos 60^{\circ}\)

\(\text{t} = 20\cdot \sin60^{\circ}\)

\(\text{L} = \dfrac{(20 + 20 + \text{x} + \text{x})\times \text{t}}{2}\)

\(\text{L} = \dfrac{(20 + 20 + 20\cos \theta + 20 \cos \theta)\times 20\sin \theta}{2}\)

\(\text{L} = \dfrac{(40 + 40\cos \theta)\times 20\sin \theta}{2}\)

\(\text{L} = (20 + 20\cos \theta)\times 20\sin \theta\)

\(\text{L} = 400\sin \theta + 400\sin \theta \cos \theta\)

\(\text{L} = 400\sin \theta + 200\cdot 2\sin \theta \cos \theta\)

\(\text{L} = 400\sin \theta + 200\cdot \sin 2\theta\)

\(\text{L}’ = 0\)

\(400\cos \theta + 200\cdot 2\cdot \cos 2\theta = 0\)

\(400\cos \theta + 400\cos 2\theta = 0\)

\(400\cos \theta  = – 400\cos 2\theta\)

\(\cos \theta  = -\cos 2\theta\)

\(\cos \theta  = -(2\cos^2 \theta \:-\:1)\)

\(\cos \theta  = -2\cos^2 \theta + 1\)

\(2\cos^2 \theta + \cos \theta \:-\: 1 = 0\)

\((2\cos \theta \:-\: 1)(\cos \theta + 1) = 0\)

\(2\cos \theta \:-\: 1 = 0\)

\(\cos \theta = \frac{1}{2}\)

\(\theta = 60^{\circ}\)

Fungsi \(f(x) = \sqrt{3}\cos x \:-\:\sin x\) naik jika \(f'(x) > 0\)

\(-\sqrt{3}\sin x \:-\:\cos x > 0\)

\(-\sqrt{3}\sin x \:-\:\cos x\) dapat diubah ke bentuk \(r\cos (x\:-\:\alpha)\)

\(r = \sqrt{(-\sqrt{3})^2 + (-1)^2}\)

\(r = \sqrt{3 + 1}\)

\(r = \sqrt{4} = 2\)

\(\tan \alpha = \dfrac{-\sqrt{3}}{-1}\)

\(\tan \alpha = \sqrt{3}\:\:\:\:\:\color{blue}\text{ kuadran III}\)

\(\alpha = \pi + \frac{\pi}{3} = \frac{4}{3}\pi\)

Jadi, \(-\sqrt{3}\sin x \:-\:\cos x\) dapat diubah ke bentuk \(2\cos (x\:-\:\frac{4}{3}\pi)\)

\(2\cos (x\:-\:\frac{4}{3}\pi) > 0\)

\(\cos (x\:-\:\frac{4}{3}\pi) > 0\)

Mencari pembuat nol fungsi dengan menyelesaikan persamaan \(\cos (x\:-\:\frac{4}{3}\pi) = 0\)

\(\cos (x\:-\:\frac{4}{3}\pi) = \cos \frac{\pi}{2}\)

Kemungkinan 1:

\(x\:-\:\frac{4}{3}\pi = \frac{\pi}{2} + k\cdot 2\pi\)

\(x = \frac{\pi}{2} + \frac{4}{3}\pi + k\cdot 2\pi\)

\(x = \frac{11}{6}\pi + k\cdot 2\pi\)

\(\text{Untuk } k = 0 \rightarrow  x = \frac{11}{6}\pi\)

Kemungkinan 2:

\(x\:-\:\frac{4}{3}\pi = -\frac{\pi}{2} + k\cdot 2\pi\)

\(x = -\frac{\pi}{2} + \frac{4}{3}\pi + k\cdot 2\pi\)

\(x = \frac{5}{6}\pi + k\cdot 2\pi\)

\(\text{Untuk } k = 0 \rightarrow  x = \frac{5}{6}\pi\)

\(\cos (x\:-\:\frac{4}{3}\pi) > 0\)

Jadi, fungsi \(f(x) = \sqrt{3}\cos x \:-\:\sin x\) naik pada interval \(\frac{5}{6}\pi < x < \frac{11}{6}\pi\)

\(\int \cos^3 2x\: \text{dx}\)

\(\int \cos^2 2x\cdot \cos 2x\: \text{dx}\)

\(\int (1\:-\:\sin^2 2x) \cdot \cos 2x\: \text{dx}\)

\(\int (1\:-\:\sin^2 2x) \cdot \cos 2x\: \text{d}(\dfrac{\sin 2x}{2\cos 2x})\)

\(\int (1\:-\:\sin^2 2x) \cdot \cancel{\cos 2x}\: \text{d}\left(\dfrac{\sin 2x}{2\cancel{\cos 2x}}\right)\)

\(\frac{1}{2} \int (1\:-\:\sin^2 2x)\:\text{d}(\sin 2x)\)

\(\frac{1}{2}[\sin 2x \:-\:\frac{1}{3}\sin^3 2x] + c\)

\(\frac{1}{2}\sin 2x \:-\:\frac{1}{6}\sin^3 2x + c\)

\(\int (\cos^2 x)^2 \:\text{dx}\)

\(\int \left(\dfrac{1 + \cos 2x}{2}\right)^2 \:\text{dx}\)

\(\frac{1}{4}\int (1 + 2\cos 2x + \cos^2 2x) \:\text{dx}\)

\(\frac{1}{4}x + \frac{1}{4}\sin 2x + \frac{1}{4}\int \cos^2 2x\:\text{dx}\)

\(\frac{1}{4}x + \frac{1}{4}\sin 2x + \frac{1}{4}\int \dfrac{1 + \cos 4x}{2} \:\text{dx}\)

\(\frac{1}{4}x + \frac{1}{4}\sin 2x + \frac{1}{8}(x + \frac{1}{4}\sin 4x) + c\)

\(\frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + c\)

\(\int \sec x \cdot \tan x \cdot \tan^2 x \text{ dx}\)

\(\int \sec x\cdot \tan x(\sec^2 x \:-\:1)\text{ dx}\)

\(\int \cancel{\sec x \cdot \tan x}(\sec^2 x \:-\:1)\text{ d}(\dfrac{\sec x}{\cancel{\sec x \cdot \tan x}})\)

\(\int (\sec^2 x\:-\:1)\text{ d}(\sec x)\)

\(\frac{1}{3}\sec^3 x \:-\:\sec x + c\)

Misal tiga bilangan berurutan yang membentuk barisan geometri tersebut adalah \(\dfrac{a}{r}, a, ar\)

Hasil kali ketiga bilangan = 125

\(\dfrac{a}{\cancel{r}}\times a \times a\cancel{r} = 125\)

\(a^3 = 125\)

\(a^3 = 5^3\)

\(a = 5\)

Tiga bilangan \(\dfrac{5}{r}, 5, 5r\) masing-masing merupakan suku pertama, suku ketiga, dan suku keenam barisan aritmetika.

Rumus suku ke-n barisan aritmetika:

\(\color{blue} \text{U}_{\text{n}} = \text{U}_{\text{1}} + (\text{n}\:-\:1)b\)

\(\text{U}_1 = \dfrac{5}{r}\)

\(\text{U}_3= 5\)

\(\text{U}_1 + 2b = 5\)

\(\dfrac{5}{r} + 2b = 5\)

\(b = \dfrac{5}{2} \:-\: \dfrac{5}{2r}\dotso \color{red} (1)\)

\(\text{U}_6 = 5r\)

\(\text{U}_1 + 5b = 5r\)

\(\dfrac{5}{r} \:-\: 5r + 5b = 0\dotso \color{red} (2)\)

Substitusikan persamaan (1) ke persamaan (2)

\(\dfrac{5}{r} \:-\: 5r + 5\left(\dfrac{5}{2} \:-\: \dfrac{5}{2r}\right) = 0\)

\(\dfrac{5}{r} \:-\: 5r + \dfrac{25}{2}\:-\:\dfrac{25}{2r} = 0\)

Kalikan kedua ruas dengan \(2r\)

\(10\:-\:10r^2 + 25r \:-\: 25 = 0\)

\(-10r^2 + 25r\:-\:15 = 0\)

Bagi kedua ruas dengan \(-5\)

\(2r^2 \:-\: 5r + 3 = 0\)

\((2r\:-\: 3)(r\:-\: 1) = 0\)

\(2r\:-\: 3 = 0 \rightarrow r = \dfrac{3}{2}\)

\(r\:-\: 1 = 0 \rightarrow r = 1\)

Untuk \(r = \dfrac{3}{2}\) maka ketiga bilangan tersebut adalah \(\dfrac{10}{3}, 5, \dfrac{15}{2}\)

Jumlah ketiga bilangan \(\dfrac{10}{3} + 5 + \dfrac{15}{2} = \dfrac{95}{6}\)

Langkah 1: Menentukan titik potong parabola dan garis 

Substitusikan persamaan parabola \(y = x^2 + 6x\) ke persamaan garis \(2x\:-\:y = 0\)

\(2x\:-\:(x^2 + 6x) = 0\)

\(-x^2\:-\:4x = 0\)

\(x^2 + 4x = 0\)

\(x(x + 4) = 0\)

\(x_1 = 0\rightarrow y = 0\)

\(x_2 = -4 \rightarrow y = -8\)

Titik potongnya adalah \((0, 0) \text{ dan } (-4, -8)\)

Langkah 2: Menentukan pusat dan jari-jari lingkaran

Pusat lingkaran terletak pada sumbu x

Misal pusat lingkaran \((p, 0)\)

Jarak pusat lingkaran ke titik \((0, 0)\) dan juga ke titik \((-4, -8)\) merupakan jari-jari lingkaran

Jarak titik \((p, 0)\) dan \((0, 0)\) = Jarak titik \((p, 0)\) dan \((-4, -8)\)

\(\sqrt{(0\:-\:p)^2 + (0\:-\:0)^2} = \sqrt{(-4\:-\:p)^2 + (-8\:-\:0)^2}\)

Kuadratkan kedua ruas

\(\cancel{p^2} = 16 + 8p + \cancel{p^2} + 64\)

\(-80 = 8p\)

\(p = -10\)

Pusat lingkaran berada di titik \((-10, 0)\)

Jari-jari lingkaran adalah 10

Langkah 3: Menentukan persamaan lingkaran

Persamaan lingkaran dengan pusat \((a, b)\) berjari-jari \(r\) adalah:

\(\color{blue} (x\:-\:a)^2 + (y\:-\:b)^2 = r^2\)

\((x\:-\:(-10))^2 + (y\:-\:0)^2 = 10^2\)

\(x^2 + 20x + 100 + y^2 = 100\)

\(x^2 + y^2 + 20x = 0\)

Jadi, persamaan lingkarannya adalah \(x^2 + y^2 + 20x = 0\)

Langlah 1: Menentukan turunan pertama fungsi g

\(g(x) = \sqrt{\dfrac{1 + x + f(x)}{f^2(x)}}\)

\(g'(x) = \dfrac{1}{2}\cdot \left(\dfrac{1 + x + f(x)}{f^2(x)}\right)^{-\frac{1}{2}}\cdot \dfrac{(1 + f'(x))\cdot f^2(x)\:-\:(1 + x + f(x))(2f(x)f'(x)}{f^4(x)}\)

\(g'(1) = \dfrac{1}{2}\cdot \left(\dfrac{1 + 1 + f(1)}{f^2(1)}\right)^{-\frac{1}{2}}\cdot \dfrac{(1 + f'(1))\cdot f^2(1)\:-\:(1 + 1 + f(1))(2f(1)f'(1)}{f^4(1)}\)

\(g'(1) = \dfrac{1}{2}\cdot 1\cdot \dfrac{8\:-\:16}{16}\)

\(g'(1) = \dfrac{1}{2}\cdot 1\cdot \dfrac{-1}{2}\)

\(g'(1) = -\dfrac{1}{4}\)

\(f(x) = x\:-\:2\sqrt{x + a}\) mencapai minimum jika \(f'(x) = 0\)

\(f'(x) = 1\:-\:2\cdot \frac{1}{2}\cdot (x + a)^{-\frac{1}{2}} = 0\)

\(1\:-\:\dfrac{1}{\sqrt{x + a}} = 0\)

Substitusikan \(x = -4\)

\(1\:-\:\dfrac{1}{\sqrt{-4 + a}} = 0\)

\(1 = \dfrac{1}{\sqrt{-4 + a}}\)

\(-4 + a = 1\)

\(a = 1 + 4 \)

\(a = 5\)

\(f(x) = x\:-\:2\sqrt{x + 5}\) mempunyai nilai minimum \(b\) di titik \(x = -4\)

\(b = -4\:-\:2\sqrt{-4 + 5}\)

\(b = -4\:-\:2\)

\(b = -6\)

\(a + b = 5 \:-\:6\)

\(a + b = -1\)

Pada pencerminan, jarak titik C ke garis akan sama dengan jarak C’ ke garis.

Garis yang melewati titik C dan C’ akan tegak lurus dengan cerminnya yaitu garis \(ax + by + 6 = 0\)

Langkah 1: Menentukan gradien garis yang melewati titik C(-4, -2) dan C'(4, 10)

\(\color{blue} m = \dfrac{y_2\:-\:y_1}{x_2\:-\:x_1}\)

\(m_1 = \frac{10\:-\:(-2)}{4\:-\:(-4)}\)

\(m_1 = \frac{12}{8}\)

\(m_1 = \frac{3}{2}\)

Langkah 2: Menentukan gradien garis ax + by + 6 = 0

\(ax + by + 6 = 0\)

\(by = -ax\:-\:6\)

\(y = -\frac{a}{b}x\:-\:\frac{6}{b}\)

\(m_2 = -\frac{a}{b}\)

Langkah 3: Syarat kedua garis saling tegak lurus

\(m_1 \times m_2 = -1\)

\(\frac{3}{2} \times -\frac{a}{b} = -1\)

\(\frac{a}{b} = \frac{2}{3}\)

\(a = \frac{2}{3}b\)

Langkah 4: Menentukan titik tengah garis yang melewati titik C dan C’

Misal titik tengah garis yang melalui titik C(-4, -2) dan C'(4, 10) adalah T

\(\text{T}(\frac{-4 + 4}{2}, \frac{-2 + 10}{2})\)

\(\text{T}(0, 4)\)

Langkah 4: Menentukan nilai \(a\) dan \(b\)

Titik T (0, 4) dilalui oleh garis \(ax + by + 6 = 0\)

\(a(0) + b(4) + 6 = 0\)

\(4b = -6\)

\(b = -\frac{3}{2}\)

\(a = \frac{2}{3}b\)

\(a = \frac{2}{3}[-\frac{3}{2}]\)

\(a = -1\)

\(a + 2b = -1 + 2(-\frac{3}{2})\)

\(a + 2b = -1\:-\:3\)

\(a + 2b = -4\)

\(\color{blue}\cos 2x = 1\:-\:2\sin^2 x\)

\(f(x) = 2\sin x + \cos 2x\)

\(f(x) = 2\sin x + 1\:-\:2\sin^2 x\)

\(-1 \leq \sin x \leq 1\)

Untuk \(\sin x = -1\) maka \(f(x) = 2(-1) + 1\:-\:2(-1)^2 = -3\:\:\:\color{blue} \text{ min}\)

Untuk \(\sin x = 1\) maka \(f(x) = 2(1) + 1\:-\:2(1)^2 = 1\:\:\:\color{blue} \text{ max}\)

Jadi, nilai minimum fungsi \(f\) adalah \(-3\)

Persamaan garis singgung dengan gradien \(m\) pada lingkaran yang berpusat di \((a, b)\) dengan jari-jari \(r\) adalah:

\(\color{blue} y\:-\:b = m(x\:-\:a) \pm r\sqrt{1 + m^2}\)

Pusat lingkaran (1, p)

\(y\:-\:p = m(x\:-\:1) \pm r\sqrt{1 + m^2}\)

Garis-garis singgung lingkaran \(y + 2x  = -4\) dan \(y + 2x  = 6\) memiliki nilai gradien m = 2.

\(y = p + 2(x\:-\:1) \pm r\sqrt{1 + 2^2}\)

\(y = p + 2x + 2 \pm r\sqrt{5} + p\)

Dengan menyamakan koefisien persamaan garis-garis singgung, didapatkan persamaan:

\(p + 2\:-\: r\sqrt{5} = -4\dotso \color{blue} (1)\)

\(p + 2 +  r\sqrt{5} = 6\dotso \color{blue} (2)\)

Kurangkan persamaan (1) dan (2),

\(-2r\sqrt{5} = -10\)

\(r = \dfrac{5}{\sqrt{5}}\)

\(r = \sqrt{5}\)

Substitusikan nilai r ke persamaan (1)

\(p + 2\:-\: \sqrt{5}\cdot \sqrt{5} = -4\)

\(p + 2\:-\: 5 = -4\)

\(p\:-\:3= -4\)

\(p = -1\)

Jadi, pusat lingkaran berada di titik (1, -1) dan berjari-jari \(\sqrt{5}\)

Persamaan lingkaran:

\(\color{blue} (x\:-\:a)^2 + (y\:-\:b)^2 = r^2\)

\((x\:-\:1)^2 + (y + 1)^2 = (\sqrt{5})^2\)

\(x^2 \:-\:2x + 1 + y^2 + 2y + 1 = 5\)

\(x^2 + y^2\:-\:2x + 2y \:-\:3 = 0\)

\(\vec{p} \bot \vec{q}\rightarrow \vec{p} \cdot \vec{q} = 0\)

\(a + 2b + 2c = 0\dotso \color{blue} (1)\)

\(\vec{p} \bot \vec{r}\rightarrow \vec{p} \cdot \vec{r} = 0\)

\(3a + 6b + 2c = 0\dotso \color{blue} (2)\)

Eliminasi kedua persamaan sehingga didapatkan:

\(c = 0\)

\(b = -\dfrac{1}{2}a\)

\(\dfrac{a^2 + 4b^2}{ab}\)

\(\dfrac{a^2 + 4(-\frac{1}{2}a)^2}{a\cdot (-\frac{1}{2}a)}\)

\(\dfrac{a^2 + a^2}{-\frac{1}{2}a^2}\)

\(\dfrac{2\cancel{a^2}}{-\frac{1}{2}\cancel{a^2}}\)

\(\dfrac{2}{-\frac{1}{2}}\)

\(-4\)

\(|3x\:-\:4| = \begin{cases}3x\:-\:4,  & x \geq \frac{4}{3}\\-(3x\:-\:4), & x < \frac{4}{3}\end{cases}\)

\(\color{blue} x_1 + x_2 + x_3 = -\dfrac{b}{a}\)

\(9 + x_2 + x_3 = -\dfrac{-6}{1}\)

\( x_2 + x_3 = 6\:-\:9\)

\( x_2 + x_3 = -3\)

\(\color{blue} x_1x_2 + x_1x_3 + x_2x_3 = \dfrac{c}{a}\)

\(9x_2 + 9x_3 + x_2x_3 = \dfrac{c}{a}\)

\(9(x_2 + x_3) + x_2x_3 = \dfrac{-a}{1}\)

\(9(-3) + x_2x_3 = -a\)

\(x_2x_3 = 27\:-\:a\)

\(\color{blue} x_1 \cdot x_2 \cdot x_3 = -\dfrac{d}{a}\)

\(9 \cdot x_2 \cdot x_3 = -\dfrac{b}{1}\)

\(9 (27\:-\:a) = -b\)

\(243\:-\:9a = -b\)

\(b\:-\:9a = -243\)

Eliminasikan persamaan \(b\:-\:9a = -243\) dan \(b\:-\:a = 5\) sehingga didapatkan nilai \(a = 31\).

\(x_1 + x_2 + x_1\cdot x_2 = -3 + 27\:-\:a\)

\(x_1 + x_2 + x_1\cdot x_2 = -3 + 27\:-\:31\)

\(x_1 + x_2 + x_1\cdot x_2 = -7\)

\((3x)^{1 +\:^3\log 3x} > 81x^2\)

\((3x)^{1 +\:^3\log 3 + \:^3\log x} > 81x^2\)

\((3x)^{1 + 1 + \:^3\log x} > 81x^2\)

\((3x)^{2+ \:^3\log x} > 81x^2\)

\((3x)^2\cdot (3x)^{^3\log x}> 81x^2\)

\(9x^2\cdot (3x)^{^3\log x}\:-\: 81x^2 > 0\)

\(9x^2((3x)^{^3\log x}\:-\: 9) > 0\)

\(9x^2 = 0\)

\(x = 0\)

\((3x)^{^3\log x} = 9\)

\(^3\log x = \:^{3x}\log 9\)

\(^3\log x = \dfrac{1}{^9\log 3x}\)

\(^3\log x = \dfrac{1}{^9\log 3 + \:^9\log x}\)

\(^3\log x = \dfrac{1}{\frac{1}{2} + \frac{1}{2}\:^3\log x}\)

\(^3\log x = \dfrac{2}{1 + \:^3\log x}\)

\(^3\log x (1 + \:^3\log x = 2\)

\(^3\log x  + \:(^3\log x)^2\:-\:2 = 0\)

\((^3\log x)^2 + \:^3\log x\:-\:2 = 0\)

\((^3\log x + 2)(^3\log x \:-\:1) = 0\)

\(^3\log x + 2 = 0\)

\(x = 3^{-2} = \dfrac{1}{9}\)

\(^3\log x \:-\:1 = 0\)

\(^3\log x = 1\)

\(x = 3\)

\(9x^2((3x)^{^3\log x}\:-\: 9) > 0\)