// Numbas version: exam_results_page_options {"name": "Week 1 homework test", "question_groups": [{"pickingStrategy": "all-ordered", "name": "Group", "pickQuestions": 1, "questions": [{"name": "The domain and codomain of a function", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Daniel Mansfield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/743/"}], "functions": {}, "ungrouped_variables": ["x1", "y1"], "tags": [], "preamble": {"css": "", "js": ""}, "advice": "", "rulesets": {}, "parts": [{"maxAnswers": 0, "displayColumns": 0, "prompt": "

Which combinations of domain and codomain make the formula $g(x) = 2x$ into a function?

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Domain $\\left\\{0,1,2\\right\\}$ and codomain $\\left\\{0,2,4\\right\\}$

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Domain $\\left\\{0,1,2\\right\\}$ and codomain $\\left\\{0,2\\right\\}$

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Domain $\\left\\{0,1,2\\right\\}$ and codomain $\\left\\{0,2,4,5\\right\\}$

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Domain $\\mathbb N$ and codomain $\\mathbb N$.

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Which of the following forumulas make $h:\\mathbb Z \\mapsto \\mathbb Z$ into a function?

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$h(x) = -x$

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$h(x) = \\pm x$

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$h(x) = \\sqrt{x}$

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$h(x) = x^2$

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$h(x) = \\frac12$

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$h(x) = 1$

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Informally, a function tells you how to transform elements of one set into another. The parts of a function are the: name, domain, codomain and formula.

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• The name of a function can be anything. This is usually is just a letter such as $f,g$ or $h$.
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• The domain is the set of elements to be transformed, otherwise known as the inputs to the function.
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• The codomain is the set containing the transformed elements, otherwise known as the outputs to the function.
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• The formula is an expression which tells you exactly how to transform an element of the domain into an element of the codomain.
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We use the notation $f: X\\mapsto Y$ to mean that the function named $f$ has domain $X$ and codomain $Y$. The formula is expressed separately, for example

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$f: \\left\\{-1,0,1\\right\\} \\mapsto \\left\\{0,1\\right\\}, f(x) = x^2.$

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The distinctive aspect of a function is that for any input, the formula determines exactly one output.

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A short question explaining the domain of a function.

Enter the value for \$$A\$$ as an exact fraction.      \$$A=\$$ [[0]]

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Enter the value for \$$B\$$ as an exact fraction.      \$$B=\$$ [[1]]

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Enter the value for \$$C\$$ as an exact fraction.      \$$C=\$$ [[2]]

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Enter the value for \$$D\$$.                                           \$$D=\$$ [[3]]

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The diagram below shows a typical mass-spring-damper system as might apply to the suspension of a car.

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(Masses have mass M, springs with stiffness k and dampers having damping coefficient B).

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The associated variables are displacement x(t) and force F(t).

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Initially the mass is at a distance \$$\\var{i0}cm\$$ from the equilibrium point and is moving at \$$\\var{i1}cm/s\$$.

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If \$$B=\\simplify{2*{a1}}\$$  and  \$$k=\\simplify{{a1}^2}\$$ and the system is subjected to an external applied force \$$F(t)=\\var{c1}e^{-\\var{d1}t}+\\var{b}\\delta(t-\\var{f})\$$

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then from Newton's law we get the differential equation:

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\$$\\frac{d^2x}{dt^2}+\\simplify{{a1}+{a1}}\\frac{dx}{dt}+\\simplify{{a1}*{a1}}x(t)=\\var{c1}e^{-\\var{d1}t}+\\var{b}\\delta(t-\\var{f})\$$  where \$$x(0)=\\var{i0} \\,\\, and \\,\\, x'(0)=\\var{i1}\$$

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The solution of the equation is given by:

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\$$x(t)=Ae^{-\\var{d1}t}+Be^{-\\var{a1}t}+Cte^{-\\var{a1}t}+Du(t-\\var{f})(t-\\var{f})e^{-\\var{a1}(t-\\var{f})}\$$

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.

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\$$\\frac{d^2x}{dt^2}+\\simplify{{a1}+{a1}}\\frac{dx}{dt}+\\simplify{{a1}*{a1}}x(t)=\\var{c1}e^{-\\var{d1}t}+\\var{b}\\delta(t-\\var{f})\$$  where \$$x(0)=\\var{i0} \\,\\, and \\,\\, x'(0)=\\var{i1}\$$

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The Laplace transform of this is given by:

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\$$s^2X(s)-\\var{i0}s-\\var{i1}+\\simplify{{a1}+{a1}}(sX(s)-\\var{i0})+\\simplify{{a1}*{a1}}X(s)=\\frac{\\var{c1}}{s+\\var{d1}}+\\var{b}e^{-\\var{f}s}\$$

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Gathering only \$$X(s)\$$ terms on the left hand side and factoring gives:

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\$$(s^2+\\simplify{{a1}+{a1}}s+\\simplify{{a1}*{a1}})X(s)=\\frac{\\var{c1}}{s+\\var{d1}}+\\var{i0}s+\\simplify{{i1}+({a1}+{a1})*{i0}}+\\var{b}e^{-\\var{f}s}\$$

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\$$(s^2+\\simplify{{a1}+{a1}}s+\\simplify{{a1}*{a1}})X(s)=\\frac{\\simplify{{c1}+({i0}s+{i1}+({a1}+{a1})*{i0})*(s+{d1})}}{s+\\var{d1}}+\\var{b}e^{-\\var{f}s}\$$

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\$$X(s)=\\frac{\\simplify{{c1}+({i0}s+{i1}+({a1}+{a1})*{i0})*(s+{d1})}}{(s+\\var{d1})(s+\\var{a1})^2}+\\frac{\\var{b}}{(s+\\var{a1})^2}e^{-\\var{f}s}\$$

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Solving the first fraction gives:

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\$$X(s)=\\frac{A}{s+\\var{d1}}+\\frac{B}{s+\\var{a1}}+\\frac{C}{(s+\\var{a1})^2}\$$

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\$$\\simplify{{c1}+({i0}s+{i1}+({a1}+{a1})*{i0})*(s+{d1})}=A(s+\\var{a1})(s+\\var{a1})+B(s+\\var{d1})(s+\\var{a1})+C(s+\\var{d1})\$$

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Let \$$s=-\\var{d1}\$$

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\$$\\simplify{{c1}+({i0}*-{d1}+{i1}+({a1}+{a1})*{i0})*(-{d1}+{d1})}=\\simplify{(-{d1}+{a1})(-{d1}+{a1})}A\$$

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\$$A=\\simplify{({c1})/((-{d1}+{a1})(-{d1}+{a1}))}\$$

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Let \$$s=-\\var{a1}\$$

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\$$\\simplify{{c1}+({i0}*-{a1}+{i1}+({a1}+{a1})*{i0})*(-{a1}+{d1})}=\\simplify{(-{a1}+{d1})}C\$$

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\$$C=\\simplify{({c1}+({i0}*-{a1}+{i1}+({a1}+{a1})*{i0})*(-{a1}+{d1}))/((-{a1}+{d1}))}\$$

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Compare the coefficients of \$$s^2\$$

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\$$\\var{i0}=A+B\$$

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\$$B=\\simplify{{i0}-(({c1})/((-{d1}+{a1})(-{d1}+{a1})))}\$$

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\$$B=\\simplify{({i0}*(-{d1}+{a1})*(-{d1}+{a1})-{c1})/((-{d1}+{a1})*(-{d1}+{a1}))}\$$

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We can find the inverse Laplace transform of the second fraction without breaking it down:

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\$$\\frac{\\var{b}}{(s+\\var{a1})^2}e^{-\\var{f}s}\$$ changes to \$$\\var{b}u(t-\\var{f})(t-\\var{f})e^{-\\var{a1}(t-\\var{f})}\$$

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\$$D=\\var{b}\$$

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Solve a Differential equation with a repeated linear factor & a delta function

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description

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