// Numbas version: exam_results_page_options {"name": "Hyperbolic #2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "variable_groups": [], "functions": {}, "ungrouped_variables": ["k", "r", "a", "b"], "extensions": [], "name": "Hyperbolic #2", "variables": {"k": {"definition": "random(30..42#1)", "group": "Ungrouped variables", "name": "k", "templateType": "randrange", "description": ""}, "r": {"definition": "random(2..8#1)", "group": "Ungrouped variables", "name": "r", "templateType": "randrange", "description": ""}, "a": {"definition": "random(10..16#1)", "group": "Ungrouped variables", "name": "a", "templateType": "randrange", "description": ""}, "b": {"definition": "random(2..9#1)", "group": "Ungrouped variables", "name": "b", "templateType": "randrange", "description": ""}}, "statement": "

Solve the following equation for \\(x\\):

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\\(\\simplify{2*{a}}Cosh(\\var{r}x)-\\simplify{2*{b}}Sinh(\\var{r}x)=\\var{k}\\)

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Enter the greater of your two answers  \\(x=\\) [[0]]

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Enter the lesser of your two answers   \\(x=\\) [[1]]

"}], "advice": "

\\(\\simplify{2*{a}}Cosh(\\var{r}x)-\\simplify{2*{b}}Sinh(\\var{r}x)=\\var{k}\\)

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\\(\\simplify{2*{a}}.\\frac{1}{2}\\left(e^{\\var{r}x}+e^{-\\var{r}x}\\right)-\\simplify{2*{b}}.\\frac{1}{2}\\left(e^{\\var{r}x}-e^{-\\var{r}x}\\right)=\\var{k}\\)

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\\(\\var{a}e^{\\var{r}x}+\\var{a}e^{-\\var{r}x}-\\var{b}e^{\\var{r}x}+\\var{b}e^{-\\var{r}x}=\\var{k}\\)

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\\(\\simplify{{a}-{b}}e^{\\var{r}x}+\\simplify{{a}+{b}}e^{-\\var{r}x}=\\var{k}\\)

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Multiply across by \\(e^{\\var{r}x}\\)

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\\(\\simplify{{a}-{b}}\\left(e^{\\var{r}x}\\right)^2+\\simplify{{a}+{b}}=\\var{k}e^{\\var{r}x}\\)

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\\(\\simplify{{a}-{b}}\\left(e^{\\var{r}x}\\right)^2-\\var{k}e^{\\var{r}x}+\\simplify{{a}+{b}}=0\\)

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This is in the form of a quadratic equation, hence

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\\(e^{\\var{r}x}=\\frac{\\var{k}\\pm \\sqrt{\\var{k}^2-4*\\simplify{{a}-{b}}*\\simplify{{a}+{b}}}}{2*\\simplify{{a}-{b}}}\\)

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\\(e^{\\var{r}x}=\\frac{\\var{k}\\pm \\sqrt{\\simplify{{k}^2-4*({a}-{b})*({a}+{b})}}}{\\simplify{2*({a}-{b})}}\\)

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\\(\\var{r}x=ln\\left(\\frac{\\var{k}\\pm \\sqrt{\\simplify{{k}^2-4*({a}-{b})*({a}+{b})}}}{\\simplify{2*({a}-{b})}}\\right)\\)

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\\(x=\\frac{1}{\\var{r}}ln\\left(\\frac{\\var{k}\\pm \\sqrt{\\simplify{{k}^2-4*({a}-{b})*({a}+{b})}}}{\\simplify{2*({a}-{b})}}\\right)\\)

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\\(x=\\frac{1}{\\var{r}}ln\\left(\\frac{\\var{k}+\\sqrt{\\simplify{{k}^2-4*({a}-{b})*({a}+{b})}}}{\\simplify{2*({a}-{b})}}\\right)\\)    or    \\(x=\\frac{1}{\\var{r}}ln\\left(\\frac{\\var{k}-\\sqrt{\\simplify{{k}^2-4*({a}-{b})*({a}+{b})}}}{\\simplify{2*({a}-{b})}}\\right)\\)

\n

", "tags": [], "preamble": {"css": "", "js": ""}, "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}