// Numbas version: exam_results_page_options {"name": "Set Soal Matematika IIA - Bab 12 - 14", "metadata": {"description": "", "licence": "None specified"}, "duration": 3600, "percentPass": 0, "showQuestionGroupNames": false, "showstudentname": true, "question_groups": [{"name": "PG + Isian", "pickingStrategy": "all-shuffled", "pickQuestions": 1, "questionNames": ["", "", "", "", "", ""], "questions": [{"name": "Mat 2A - UAS - No 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Meong Meong Project", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4687/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

Pernyataan yang benar terkait limit dua peubah adalah $\\ldots$.

", "advice": "", "rulesets": {}, "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["Karena $1-\\dfrac{x^2y^2}{3}<\\dfrac{\\tan^{-1}(xy)}{xy}<1$, maka $\\displaystyle \\lim\\limits_{(x,y)\\to (0,0)}\\dfrac{\\tan^{-1}(xy)}{xy}=1$.", "Karena $\\left| \\sin\\left( \\dfrac{1}{x} \\right) \\right|<1$, maka $\\displaystyle \\lim\\limits_{(x,y)\\to (0,0)}y\\sin\\left( \\dfrac{1}{x} \\right)=0$.", "$\\displaystyle\\lim\\limits_{x\\to 0}\\left(\\lim\\limits_{y\\to 0}f(x,y)\\right)=\\lim\\limits_{y\\to 0}\\left(\\lim\\limits_{x\\to 0}f(x,y)\\right)$.", "Jika $x=r\\cos(\\theta)$ dan $y=r\\sin(\\theta)$, maka $\\displaystyle \\lim\\limits_{(x,y)\\to (0,0)}f(x,y)=\\lim\\limits_{r\\to 0}f(r,\\theta)$."], "matrix": ["1", 0, 0, 0], "distractors": ["", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Mat 2A - UAS - No 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Meong Meong Project", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4687/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

Nilai maksimum dari fungsi $f(x,y,z)=\\var{a}x+\\var{b}y+\\var{c}z$ pada elipsoida $\\var{a^2}x^2+\\var{b^2}y^2+\\var{c^2}z^2=\\var{d^2}$ adalah $a\\sqrt{b}$ dengan $a+b=\\ldots$.

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Luas daerah yang berada di dalam kardioid $r=\\var{a}+\\var{a}\\cos(\\theta)$ dan di luar lingkaran $r=\\var{a}$ adalah $A=k \\pi$ dengan $k=\\ldots$.

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$\\displaystyle \\int_0^{\\sqrt{3}}\\int_0^1 \\dfrac{\\var{4*a}x}{(x^2+y^2+1)^2}\\, dy\\, dx = a\\tan^{-1}(2)$ dengan $a=\\ldots$.

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Pilih semua solusi dari persamaan diferensial $y''+2y'+2y=1+x$.

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Solusi khusus dari masalah nilai awal $y''=\\sec(x)-y$, $y(0)=\\var{a}$, dan $y'(0)=\\var{b}$ untuk $x\\in\\left[0,\\dfrac{\\pi}{2}\\right)$ adalah $y(x)=\\cos(x)\\left(a+\\ln(\\cos(x))\\right)+\\sin(x)(b+x)$ dengan $a+b=\\ldots$.

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Diberikan lamina $D$ berbentuk cakram satuan dengan pusat $(0,\\var{a})$. Suhu lamina tersebut di titik $(x,y)$ dinyatakan oleh fungsi $T(x,y)=xy^2$.

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Titik terpanas lamina tersebut berada pada $\\left( \\dfrac{\\sqrt{5}}{c_1},\\dfrac{c_2}{3} \\right)$ dengan $c_1+c_2=\\ldots$.

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Suhu terdinginnya adalah $-\\dfrac{a^{5/2}}{b^3}$ dengan $a$ dan $b$ bilangan asli dan pecahan $\\dfrac{a^2}{b^3}$ sudah dalam bentuk paling sederhana. Maka $a+b=\\ldots$.

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Suhu rata-rata lamina $D$ dapat dihitung dengan menggunakan integral lipat, yaitu jumlahan seluruh suhu di seluruh titik pada $D$ dibagi dengan luas lamina tersebut, yakni

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$T_{\\text{avg}}=\\displaystyle \\dfrac{1}{A(D)} \\iint_D T(x,y)\\, dA.$

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Suhu rata-ratanya adalah $T_{\\text{avg}}=\\dfrac{\\simplify{{a}m^2}}{n\\pi}$ dengan $m$ dan $n$ bilangan asli dan pecahan $\\dfrac{m^2}{n}$ sudah dalam bentuk paling sederhana. Maka $m+n=\\ldots$.

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Waktunya 5 menit lagi ya.

"}}, "feedback": {"showactualmark": false, "showtotalmark": false, "showanswerstate": false, "allowrevealanswer": false, "advicethreshold": 0, "intro": "

Selamat datang!

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Silakan gunakan set soal ini dengan bijak. Kalian bisa mencoba set soal ini terus-menerus.

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Set soal ini dapat digunakan oleh siapapun secara gratis.

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Petunjuk pengerjaan soal:

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Jika ada kunci jawaban yang salah, kalian dapat memberitahu kami di Instagram @meongmeongproject atau OA Line @eog7710d.

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