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Dear Students,
Welcome to today’s quiz! This is your opportunity to demonstrate what you’ve learned so far, so do your best. Please keep in mind that you have a maximum of 40 minutes to complete all the questions. Make sure to manage your time wisely and answer each question thoughtfully.
Good luck!
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Pertanyaan 1 dari 7
1. Pertanyaan
1 pointsJika rata-rata dari \(a, b, c\) dan \(a^2, b^2, c^2\) berturut-turut adalah 2 dan 4, maka rata-rata dari \(ab, bc, ca\) adalah …
Benar
Rata-rata dari \(a, b, c\) adalah 2
\(\dfrac{a + b + c}{3} = 2\)
\(a + b + c = 6\dotso\color{blue}(1)\)
Rata-rata dari \(a^2, b^2, c^2\) adalah 4
\(\dfrac{a^2 + b^2 + c^2}{3} = 4\)
\(a^2 + b^2 + c^2 = 12\dotso\color{blue}(2)\)
Rata-rata dari \(ab, bc, ca\) adalah \(\dfrac{ab + bc + ca}{3}\)
\(\color{blue} a^2 + b^2 + c^2 = (a + b + c)^2\:-\:2(ab + ac + bc)\)
\(12 = 6^2\:-\:2(ab + ac + bc)\)
\(12 = 36\:-\:2(ab + ac + bc)\)
\(-24 = -2(ab + ac + bc)\)
\(ab + ac + bc = 12\)
Rata-rata dari \(ab, bc, ca\) adalah \(\dfrac{12}{3} = 4\)
Salah
Rata-rata dari \(a, b, c\) adalah 2
\(\dfrac{a + b + c}{3} = 2\)
\(a + b + c = 6\dotso\color{blue}(1)\)
Rata-rata dari \(a^2, b^2, c^2\) adalah 4
\(\dfrac{a^2 + b^2 + c^2}{3} = 4\)
\(a^2 + b^2 + c^2 = 12\dotso\color{blue}(2)\)
Rata-rata dari \(ab, bc, ca\) adalah \(\dfrac{ab + bc + ca}{3}\)
\(\color{blue} a^2 + b^2 + c^2 = (a + b + c)^2\:-\:2(ab + ac + bc)\)
\(12 = 6^2\:-\:2(ab + ac + bc)\)
\(12 = 36\:-\:2(ab + ac + bc)\)
\(-24 = -2(ab + ac + bc)\)
\(ab + ac + bc = 12\)
Rata-rata dari \(ab, bc, ca\) adalah \(\dfrac{12}{3} = 4\)
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Pertanyaan 2 dari 7
2. Pertanyaan
1 pointsDIketahui \(f(x) = x^2 + 1\) dan \(g(x) = ax + 2\), dengan \(a \neq 0\). Jika \((f \circ g^{-1})(1) = 5\), maka \(4a^2\:-\:3 = \dotso\)
Benar
\(g(x) = ax + 2\)
\(y = ax + 2\)
\(x = \dfrac{y\:-\:2}{a}\)
\(g^{-1}(x) = \dfrac{x\:-\:2}{a}\)
\(g^{-1}(1) = \dfrac{1\:-\:2}{a}\)
\(g^{-1}(1) = -\dfrac{1}{a}\)
\((f \circ g^{-1})(1) = 5\)
\(f[g^{-1}(1)] = 5\)
\(f(-\dfrac{1}{a}) = 5\)
\((-\dfrac{1}{a})^2 + 1 = 5\)
\(\dfrac{1}{a^2} \:-\:4 = 0\)
\(\left(\dfrac{1}{a} + 2\right)\left(\dfrac{1}{a} \:-\: 2\right) = 0\)
\(\dfrac{1}{a} + 2 = 0\)
\(a = -\dfrac{1}{2}\)
\(4a^2 \:-\:3 = 4\left(-\dfrac{1}{2}\right)^2\:-\:3\)
\(4a^2 \:-\:3 = 1\:-\:3\)
\(4a^2 \:-\:3 = -2\)
Salah
\(g(x) = ax + 2\)
\(y = ax + 2\)
\(x = \dfrac{y\:-\:2}{a}\)
\(g^{-1}(x) = \dfrac{x\:-\:2}{a}\)
\(g^{-1}(1) = \dfrac{1\:-\:2}{a}\)
\(g^{-1}(1) = -\dfrac{1}{a}\)
\((f \circ g^{-1})(1) = 5\)
\(f[g^{-1}(1)] = 5\)
\(f(-\dfrac{1}{a}) = 5\)
\((-\dfrac{1}{a})^2 + 1 = 5\)
\(\dfrac{1}{a^2} \:-\:4 = 0\)
\(\left(\dfrac{1}{a} + 2\right)\left(\dfrac{1}{a} \:-\: 2\right) = 0\)
\(\dfrac{1}{a} + 2 = 0\)
\(a = -\dfrac{1}{2}\)
\(4a^2 \:-\:3 = 4\left(-\dfrac{1}{2}\right)^2\:-\:3\)
\(4a^2 \:-\:3 = 1\:-\:3\)
\(4a^2 \:-\:3 = -2\)
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Pertanyaan 3 dari 7
3. Pertanyaan
1 pointsNilai \(\lim\limits_{x \rightarrow 1} \dfrac{x\:-\:1 + \sqrt[3]{1\:-\:x}}{\sqrt[3]{1\:-\:x^2}}\) adalah…
Benar
\(\lim\limits_{x \rightarrow 1} \dfrac{x\:-\:1 + \sqrt[3]{1\:-\:x}}{\sqrt[3]{1\:-\:x^2}}\)
\(\lim\limits_{x \rightarrow 1} \dfrac{\sqrt[3]{1\:-\:x}\cdot [-(1\:-\:x)^{\frac{2}{3}} + 1]}{\sqrt[3]{1\:-\:x}\cdot \sqrt[3]{1 + x}}\)
\(\lim\limits_{x \rightarrow 1} \dfrac{\cancel{\sqrt[3]{1\:-\:x}}\cdot [-(1\:-\:x)^{\frac{2}{3}} + 1]}{\cancel{\sqrt[3]{1\:-\:x}}\cdot \sqrt[3]{1 + x}}\)
\(\lim\limits_{x \rightarrow 1} \dfrac{-(1\:-\:x)^{\frac{2}{3}} + 1}{\sqrt[3]{1 + x}}\)
\(\dfrac{-(1\:-\:1)^{\frac{2}{3}} + 1}{\sqrt[3]{1 + 1}}\)
\(\dfrac{1}{\sqrt[3]{2}}\)
Salah
\(\lim\limits_{x \rightarrow 1} \dfrac{x\:-\:1 + \sqrt[3]{1\:-\:x}}{\sqrt[3]{1\:-\:x^2}}\)
\(\lim\limits_{x \rightarrow 1} \dfrac{\sqrt[3]{1\:-\:x}\cdot [-(1\:-\:x)^{\frac{2}{3}} + 1]}{\sqrt[3]{1\:-\:x}\cdot \sqrt[3]{1 + x}}\)
\(\lim\limits_{x \rightarrow 1} \dfrac{\cancel{\sqrt[3]{1\:-\:x}}\cdot [-(1\:-\:x)^{\frac{2}{3}} + 1]}{\cancel{\sqrt[3]{1\:-\:x}}\cdot \sqrt[3]{1 + x}}\)
\(\lim\limits_{x \rightarrow 1} \dfrac{-(1\:-\:x)^{\frac{2}{3}} + 1}{\sqrt[3]{1 + x}}\)
\(\dfrac{-(1\:-\:1)^{\frac{2}{3}} + 1}{\sqrt[3]{1 + 1}}\)
\(\dfrac{1}{\sqrt[3]{2}}\)
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Pertanyaan 4 dari 7
4. Pertanyaan
1 pointsDiberikan \(f(x) = (ax^2 + bx + c)(x^2 + x)\). Jika \(f'(0) = 3\) dan \(f'(-1) = 10\), maka \(f'(-\frac{1}{2}) = \dotso\)
Benar
\(f(x) = (ax^2 + bx + c)(x^2 + x)\)
\(u = ax^2 + bx + c \rightarrow u’ = 2ax + b\)
\(v = x^2 + x \rightarrow v’ = 2x + 1\)
\(f'(x) = u’\cdot v + u\cdot v’\)
\(f'(x) = (2ax + b)(x^2 + x) + (ax^2 + bx + c)(2x + 1)\)
\(f'(0) = (2a(0) + b)(0^2 + 0) + (a(0)^2 + b(0) + c)(2(0) + 1)\)
\(3 = 0 + c\)
\(c = 3\)
\(f'(-1) = (-2a + b)((-1)^2 \:-\:1) + (a(-1)^2 \:-\:b + c)(-2 + 1)\)
\(f'(-1) = 0 + (a \:-\:b + c)(-1)\)
\(10 = -a + b \:-\:c\)
\(10 = -a + b \:-\:3\)
\(-a + b = 13\)
\(f'(-\frac{1}{2}) = (2a(-\frac{1}{2}) + b)((-\frac{1}{2})^2\:-\:\frac{1}{2}) + (a(-\frac{1}{2})^2 \:-\:\frac{1}{2}b + c)(2(-\frac{1}{2}) + 1)\)
\(f'(-\frac{1}{2}) = (-a + b)(\frac{1}{4}\:-\:\frac{1}{2}) + 0\)
\(f'(-\frac{1}{2}) = 13(-\frac{1}{4})\)
\(f'(-\frac{1}{2}) =-\frac{13}{4}\)
Salah
\(f(x) = (ax^2 + bx + c)(x^2 + x)\)
\(u = ax^2 + bx + c \rightarrow u’ = 2ax + b\)
\(v = x^2 + x \rightarrow v’ = 2x + 1\)
\(f'(x) = u’\cdot v + u\cdot v’\)
\(f'(x) = (2ax + b)(x^2 + x) + (ax^2 + bx + c)(2x + 1)\)
\(f'(0) = (2a(0) + b)(0^2 + 0) + (a(0)^2 + b(0) + c)(2(0) + 1)\)
\(3 = 0 + c\)
\(c = 3\)
\(f'(-1) = (-2a + b)((-1)^2 \:-\:1) + (a(-1)^2 \:-\:b + c)(-2 + 1)\)
\(f'(-1) = 0 + (a \:-\:b + c)(-1)\)
\(10 = -a + b \:-\:c\)
\(10 = -a + b \:-\:3\)
\(-a + b = 13\)
\(f'(-\frac{1}{2}) = (2a(-\frac{1}{2}) + b)((-\frac{1}{2})^2\:-\:\frac{1}{2}) + (a(-\frac{1}{2})^2 \:-\:\frac{1}{2}b + c)(2(-\frac{1}{2}) + 1)\)
\(f'(-\frac{1}{2}) = (-a + b)(\frac{1}{4}\:-\:\frac{1}{2}) + 0\)
\(f'(-\frac{1}{2}) = 13(-\frac{1}{4})\)
\(f'(-\frac{1}{2}) =-\frac{13}{4}\)
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Pertanyaan 5 dari 7
5. Pertanyaan
1 pointsDiberikan bilangan real \(r\), dengan \(0 < r < 1\). Jika jumlah deret geometri tak hingga dengan suku pertama 2 dan rasio \(\dfrac{1}{1 + r}\) adalah 8, maka jumlah deret geometri tak hingga dengan suku pertama 8 dan rasio \(r\) adalah…
Benar
Jumlah deret geometri tak hingga \(\color{blue} s_{\infty} = \dfrac{a}{1\:-\:r}\)
\(8 = \dfrac{2}{1\:-\:(\frac{1}{1 + r})}\)
\(8 = \dfrac{2}{\frac{r}{1 + r}}\)
\(8 = \dfrac{2(1 + r)}{r}\)
\(8r = 2 + 2r\)
\(6r = 2\)
\(r = \dfrac{1}{3}\)
\(s_{\infty} = \dfrac{a}{1\:-\:r}\)
\(s_{\infty} = \dfrac{8}{1\:-\:\frac{1}{3}}\)
\(s_{\infty} = \dfrac{8}{\frac{2}{3}}\)
\(s_{\infty} = 12\)
Salah
Jumlah deret geometri tak hingga \(\color{blue} s_{\infty} = \dfrac{a}{1\:-\:r}\)
\(8 = \dfrac{2}{1\:-\:(\frac{1}{1 + r})}\)
\(8 = \dfrac{2}{\frac{r}{1 + r}}\)
\(8 = \dfrac{2(1 + r)}{r}\)
\(8r = 2 + 2r\)
\(6r = 2\)
\(r = \dfrac{1}{3}\)
\(s_{\infty} = \dfrac{a}{1\:-\:r}\)
\(s_{\infty} = \dfrac{8}{1\:-\:\frac{1}{3}}\)
\(s_{\infty} = \dfrac{8}{\frac{2}{3}}\)
\(s_{\infty} = 12\)
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Pertanyaan 6 dari 7
6. Pertanyaan
1 pointsJika \(\textbf {A} = \begin{bmatrix}1 & 2 \\0 & 1\\1 & 1 \end{bmatrix}\), \(\textbf {B} = \begin{bmatrix}1 \\ 2 \end{bmatrix}\), maka determinan dari \(\textbf{A}^{\text{t}}\textbf{A} + \textbf{B}\textbf{B}^{\text{t}}\) adalah…
Benar
\(\textbf{A}^{\text{t}}\textbf{A} + \textbf{B}\textbf{B}^{\text{t}}\)
\(\begin{bmatrix}1 & 0 & 1\\2 & 1 & 1 \end{bmatrix}\cdot \begin{bmatrix}1 & 2\\0 & 1 \\1 & 1\end{bmatrix} + \begin{bmatrix}1 \\ 2 \end{bmatrix}\cdot \begin{bmatrix}1 & 2\end{bmatrix}\)
\(\begin{bmatrix}2 & 3\\3 & 6\end{bmatrix} + \begin{bmatrix}1 & 2\\2 & 4\end{bmatrix}\)
\(\begin{bmatrix}3 & 5\\5 & 10\end{bmatrix}\)
Determinan matriks = 3(10) − 5(5)
Determinan matriks = 30 − 25 = 5
Salah
\(\textbf{A}^{\text{t}}\textbf{A} + \textbf{B}\textbf{B}^{\text{t}}\)
\(\begin{bmatrix}1 & 0 & 1\\2 & 1 & 1 \end{bmatrix}\cdot \begin{bmatrix}1 & 2\\0 & 1 \\1 & 1\end{bmatrix} + \begin{bmatrix}1 \\ 2 \end{bmatrix}\cdot \begin{bmatrix}1 & 2\end{bmatrix}\)
\(\begin{bmatrix}2 & 3\\3 & 6\end{bmatrix} + \begin{bmatrix}1 & 2\\2 & 4\end{bmatrix}\)
\(\begin{bmatrix}3 & 5\\5 & 10\end{bmatrix}\)
Determinan matriks = 3(10) − 5(5)
Determinan matriks = 30 − 25 = 5
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Pertanyaan 7 dari 7
7. Pertanyaan
1 pointsJika \(\tan x = 2\), maka \(\dfrac{\sin x + \cos x}{\sin x \:-\:\cos x} = \dotso\)
Benar
\(\dfrac{\sin x + \cos x}{\sin x \:-\:\cos x}\)
Bagi bagian pembilang dan penyebut dengan \(\cos x\)
\(\dfrac{\tan x + 1}{\tan x \:-\:1}\)
\(\dfrac{2 + 1}{2 \:-\:1}\)
\(3\)
Salah
\(\dfrac{\sin x + \cos x}{\sin x \:-\:\cos x}\)
Bagi bagian pembilang dan penyebut dengan \(\cos x\)
\(\dfrac{\tan x + 1}{\tan x \:-\:1}\)
\(\dfrac{2 + 1}{2 \:-\:1}\)
\(3\)