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We note first that substituting \\(x=\\var{b}\\) into the function results in a \"value\" of \\(\\frac{\\simplify{({b}-{a})*({b}-{c})}}{0}\\), hence the function is not defined at \\(x=\\var{b}\\). Because this value is not of the form \\(\\frac00\\) we can say that the limit does not exist.

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Evaluate \\(\\displaystyle{\\lim_{x\\to\\var{b}}\\frac{\\simplify{x^2-{a+c}x+{a*c}}}{\\simplify{x^2-{a+b}x+{a*b}}}}\\). Show full working on your handwritten working. Type the answer as a fraction in the box below. If the limit does not exist type infinity in the box below.

\\(\\displaystyle{\\lim_{x\\to\\var{b}}\\frac{\\simplify{x^2-{a+c}x+{a*c}}}{\\simplify{x^2-{a+b}x+{a*b}}}}=\\)[[0]]

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We note first that substituting \\(x=\\var{c}\\) into the function results in a value of \\(\\frac{0}{\\simplify{({c}-{a})*({c}-{b})}}\\), hence the value of the function at \\(x=\\var{c}\\) is \\(0\\). Using the rule which says that \\[\\lim_{x\\to c}\\frac{f(x)}{g(x)}=\\frac{f(c)}{g(c)}\\] we can conclude that the value of the limit is also \\(0\\).

Evaluate \\(\\displaystyle{\\lim_{x\\to\\var{c}}\\frac{\\simplify{x^2-{a+c}x+{a*c}}}{\\simplify{x^2-{a+b}x+{a*b}}}}\\). Show full working on your handwritten working. Type the answer as a fraction in the box below. If the limit does not exist type infinity in the box below.

\\(\\displaystyle{\\lim_{x\\to\\var{c}}\\frac{\\simplify{x^2-{a+c}x+{a*c}}}{\\simplify{x^2-{a+b}x+{a*b}}}}=\\)[[0]]

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We note first that substituting \\(x=\\var{a}\\) into the function results in a \"value\" of \\(\\frac00\\), hence the function is not defined at \\(x=\\var{a}\\). To evaluate the limit we try factorising the numerator and denominator and looking for common factors.

\\begin{align} \\lim_{x\\to\\var{a}}\\frac{\\simplify{x^2-{a+c}x+{a*c}}}{\\simplify{x^2-{a+b}x+{a*b}}}&=\\lim_{x\\to\\var{a}}\\frac{\\simplify{(x-{a})(x-{c})}}{\\simplify{(x-{a})(x-{b})}}\\\\&=\\lim_{x\\to\\var{a}}\\frac{\\simplify{x-{c}}}{\\simplify{x-{b}}}\\\\&=\\simplify{{a-c}/{a-b}}\\end{align}

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Evaluate \\(\\displaystyle{\\lim_{x\\to\\var{a}}\\frac{\\simplify{x^2-{a+c}x+{a*c}}}{\\simplify{x^2-{a+b}x+{a*b}}}}\\). Show full working on your handwritten working. Type the answer as a fraction in the box below. If the limit does not exist type infinity in the box below.

\\(\\displaystyle{\\lim_{x\\to\\var{a}}\\frac{\\simplify{x^2-{a+c}x+{a*c}}}{\\simplify{x^2-{a+b}x+{a*b}}}}=\\)[[0]]

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Note first that the function \\(\\displaystyle{\\simplify{y={function}}}\\) can be written as \\(\\displaystyle{y=\\simplify{{a}*x^{b}+{c}*x^{-1*d}}+\\ln({\\var{f}x}})\\).

Using the differentiation rules (where \\(u\\) and \\(v\\) are functions of \\(x\\)):

\\(\\frac{d}{dx}\\left(x^n\\right)=nx^{n-1}\\) (The power rule)
\\(\\frac{d}{dx}(cu)=c\\frac{du}{dx}\\)
\\(\\frac{d}{dx}(u\\pm v)=\\frac{du}{dx}\\pm\\frac{dv}{dx}\\)
\\(\\frac{d}{dx}\\left(\\ln(u)\\right)=\\frac1u\\frac{du}{dx}\\)

we have \\(\\frac{d}{dx}\\left(\\simplify{{a}*x^{b}+{c}*x^{-1*d}}+\\ln({\\var{f}x})\\right)=\\simplify{{a*b}x^{b-1}-{c*d}x^{-d-1}}+\\frac{1}{x}\\)

Find the first derivative of the function \\(\\displaystyle{\\simplify{y={function}}}\\). Show any necessary working on your handwritten working.

\\(\\displaystyle{\\frac{dy}{dx}}=\\)[[0]]

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Note first that the function \\(\\displaystyle{\\simplify{y={function}}}\\) can be written as \\(\\displaystyle{y=\\simplify{{a}*x^{b}+{c}*sin({d}*x)+{f}}\\ln(x)}\\).

Using the differentiation rules (where \\(u\\) and \\(v\\) are functions of \\(x\\)):

\\(\\frac{d}{dx}\\left(x^n\\right)=nx^{n-1}\\) (The power rule)
\\(\\frac{d}{dx}\\left(\\sin (u)\\right)=\\cos(u)\\frac{du}{dx}\\)
\\(\\frac{d}{dx}(cu)=c\\frac{du}{dx}\\)
\\(\\frac{d}{dx}(u\\pm v)=\\frac{du}{dx}\\pm\\frac{dv}{dx}\\)
\\(\\frac{d}{dx}\\left(\\ln u\\right)=\\frac{1}{u}\\frac{du}{dx}\\)

we have \\(\\frac{d}{dx}\\left(\\simplify{{a}*x^{b}+{c}*sin({d}*x)+{f}ln (x)}\\right)=\\simplify{{a*b}x^{b-1}+{c*d}cos({d}*x)+{f}/x}\\)

Find the first derivative of the function \\(\\displaystyle{\\simplify{y={function}}}\\). Show any necessary working on your handwritten working.

\\(\\displaystyle{\\frac{dy}{dx}}=\\)[[0]]

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We use the product rule:

Let \\(u=\\simplify{{function_u}}\\) and \\(v=\\simplify{{function_v}}\\).

Hence we have

\\(\\frac{du}{dx}=\\simplify{{deriv_u}}\\) and \\(\\frac{dv}{dx}=\\simplify{{deriv_v}}\\)

Now by the product rule we have \\begin{align}\\frac{dy}{dx}&=u\\frac{dv}{dx}+v\\frac{du}{dx}\\\\&=\\simplify{{function_u}}\\times\\simplify{{deriv_v}}+\\simplify{{function_v}}\\times\\simplify{{deriv_u}}\\\\&=\\simplify{{function_u}*{deriv_v}+{function_v}*{deriv_u}}\\end{align}

Find \\(\\displaystyle{\\frac{dy}{dx}}\\) given that \\(\\displaystyle{\\simplify{y={function}}}\\). Show full working on your handwritten working.

\\(\\displaystyle{\\frac{dy}{dx}=}\\)[[0]]

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We use the product rule:

Let \\(u=\\simplify{{function_u}}\\) and \\(v=\\simplify{{function_v}}\\).

Hence we have

\\(\\frac{du}{dx}=\\simplify{{deriv_u}}\\) and \\(\\frac{dv}{dx}=\\simplify{{deriv_v}}\\)

Find \\(\\displaystyle{\\frac{dy}{dx}}\\) given that \\(\\displaystyle{\\simplify{y={function}}}\\). Show full working on your handwritten working. (Note to enter a function such as \\(\\sec^2x\\) type sec x^2 in the answer box.)

\\(\\displaystyle{\\frac{dy}{dx}=}\\)[[0]]

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We use the quotient rule:

Let \\(u=\\simplify{{function_u}}\\) and \\(v=\\simplify{{function_v}}\\).

Hence we have

\\(\\frac{du}{dx}=\\simplify{{deriv_u}}\\) and \\(\\frac{dv}{dx}=\\simplify{{deriv_v}}\\)

Now by the quotient rule we have \\begin{align}\\frac{dy}{dx}&=\\frac{v\\frac{du}{dx}-u\\frac{dv}{dx}}{v^2}\\\\&=\\frac{\\simplify{{function_v}}\\times\\simplify{{deriv_u}}-\\simplify{{function_u}}\\times\\simplify{{deriv_v}}}{\\left(\\simplify{{function_v}}\\right)^2}\\\\&=\\frac{\\simplify{{function_v}*{deriv_u}-{function_u}*{deriv_v}}}{\\left(\\simplify{{function_v}}\\right)^2}\\end{align}

\\(\\displaystyle{\\frac{dy}{dx}=}\\)[[0]]

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We use the quotient rule:

Let \\(u=\\simplify{{function_u}}\\) and \\(v=\\simplify{{function_v}}\\).

Hence we have

\\(\\frac{du}{dx}=\\simplify{{deriv_u}}\\) and \\(\\frac{dv}{dx}=\\simplify{{deriv_v}}\\)

Now by the quotient rule we have \\begin{align}\\frac{dy}{dx}&=\\frac{v\\frac{du}{dx}-u\\frac{dv}{dx}}{v^2}\\\\&=\\frac{\\simplify{{function_v}}\\times\\simplify{{deriv_u}}-\\simplify{{function_u}}\\times\\simplify{{deriv_v}}}{\\simplify{{function_v}^2}}\\\\&=\\frac{\\simplify{{function_v}*{deriv_u}-{function_u}*{deriv_v}}}{\\simplify{{function_v}^2}}\\end{align}

Find \\(\\displaystyle{\\frac{dy}{dx}}\\) given that \\(\\displaystyle{\\simplify{y={function}}}\\). Show full working on your handwritten working.

\\(\\displaystyle{\\frac{dy}{dx}=}\\)[[0]]

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To find the equation of the tangent at \\(x=0\\) we need to find the value of \\(y\\) and \\(\\frac{dy}{dx}\\) when \\(x=0\\).

We have: \\begin{align}y(0)&=\\sin^{-1}(\\var{c}\\times 0)e^{\\simplify[!collectNumbers]{{a}*0+{b}}}\\\\&=0\\times e^\\var{b}\\\\&=0\\end{align}

To find the gradient of the tangent we first need to find the first derivative of \\(y\\). Using the product rule we have: \\(\\frac{dy}{dx}=\\sin^{-1}(\\var{c}x)\\var{a}e^{\\simplify{{a}x +{b}}}+\\frac{\\var{c}}{\\sqrt{1-\\var{c^2}x^2}}e^{\\simplify{{a}x +{b}}}\\)

Hence: \\begin{align}\\left.\\frac{dy}{dx}\\right|_{x=0}&=\\sin^{-1}(0)\\times\\var{a}e^{\\var{b}}+\\var{c}\\times e^{\\var{b}}\\\\&=\\var{c}e^{\\var{b}}\\end{align}

The equation of the tangent can now be found using the point gradient formula which gives: \\begin{align}y-y_1&=m(x-x_1)\\\\y-0&=\\var{c}e^{\\var{b}}(x-0)\\\\y&=\\var{c}e^{\\var{b}}x\\\\&\\approx\\var{m}x\\end{align}

Thus the equation of the tangent to \\(y=\\sin^{-1} (\\var{c}x)e^{\\simplify{{a}x +{b}}}\\) at \\(x=0\\) is the line \\(y=\\var{c}e^{\\var{b}}x\\approx\\var{m}x\\).

Find the equation of the tangent to the curve \\[ y=\\sin^{-1} (\\var{c}x)e^{\\simplify{{a}x +{b}}} \\] at the point where \\( x=0 \\). Include full working for the problem in your handwritten working.

Equation of tangent: [[0]] (Round all numbers to 2 decimal places)

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To find the equation of the tangent at \\(x=0\\) we need to find the value of \\(y\\) and \\(\\frac{dy}{dx}\\) when \\(x=0\\).

We have: \\begin{align}y(0)&=\\tan^{-1}(\\var{c}\\times 0)e^{\\simplify[!collectNumbers]{{a}*0+{b}}}\\\\&=0\\times e^\\var{b}\\\\&=0\\end{align}

To find the gradient of the tangent we first need to find the first derivative of \\(y\\). Using the product rule we have: \\(\\frac{dy}{dx}=\\tan^{-1}(\\var{c}x)\\var{a}e^{\\simplify{{a}x +{b}}}+\\frac{\\var{c}}{1+\\var{c^2}x^2}e^{\\simplify{{a}x +{b}}}\\)

Hence: \\begin{align}\\left.\\frac{dy}{dx}\\right|_{x=0}&=\\tan^{-1}(0)\\times\\var{a}e^{\\var{b}}+\\var{c}\\times e^{\\var{b}}\\\\&=\\var{c}e^{\\var{b}}\\end{align}

Thus the equation of the tangent to \\(y=\\tan^{-1} (\\var{c}x)e^{\\simplify{{a}x +{b}}}\\) at \\(x=0\\) is the line \\(y=\\var{c}e^{\\var{b}}x\\approx\\var{m}x\\).

Find the equation of the tangent to the curve \\[ y=\\tan^{-1} (\\var{c}x)e^{\\simplify{{a}x +{b}}} \\] at the point where \\( x=0 \\). Include full working for the problem in your handwritten working.

Equation of tangent: [[0]] (Round all numbers to 2 decimal places).

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The position of a particle, \\(y\\), is given by \\(y(t)=\\simplify{t^3-{0.5*(a+3*b)}t^2+{a*b}*t+{c}}\\) where \\(t\\) represents time in seconds. On your written working find the values of the position and acceleration of the particle when its velocity is \\(0\\). Using these results sketch the graph of \\(y(t)\\) for \\(0\\le t\\le \\var{max(a,b)+2}\\).

", "advice": "

We know that the velocity of the particle is given by \\begin{align}\\frac{dy}{dt}&=\\frac{d}{dt}\\left(\\simplify{{function}}\\right)\\\\&=\\simplify{{deriv}}\\end{align}

This can be factorised as \\[\\frac{dy}{dt}=\\simplify{(3t-{a})(t-{b})}\\] and hence the velocity will be \\(0\\) when \\(t=\\simplify[collectNumbers,simplifyFractions,unitDenominator]{{a}/3}\\) and \\(t=\\var{b}\\).

The acceleration of the particle is given by \\begin{align}\\frac{d^2y}{dt^2}&=\\frac{d}{dt}\\left(\\simplify{{deriv}}\\right)\\\\&=\\simplify{6t-({a}+{3*b})}\\end{align}

Thus we have at \\(t=\\simplify[collectNumbers,simplifyFractions,unitDenominator]{{a}/3}\\), position \\(y=\\simplify{{rational(y_1)}}\\) metres, acceleration \\(\\left.\\frac{d^2y}{dt^2}\\right|_{t=\\simplify[collectNumbers,simplifyFractions,unitDenominator]{{a}/3}}=\\simplify{6*{a/3}-({a}+{3*b})}\\) ms\\(^{-2}\\)

and at \\(t=\\var{b}\\), position \\(y=\\simplify{{rational(y_2)}}\\) metres, acceleration \\(\\left.\\frac{d^2y}{dt^2}\\right|_{t=\\var{b}}=\\simplify{6*{b}-({a}+{3*b})}\\) ms\\(^{-2}\\).

Since velocity corresponds to the first derivative of position this means that we have stationary points at \\(\\left(\\simplify[collectNumbers,simplifyFractions,unitDenominator]{{a}/3},\\simplify{{rational(y_1)}}\\right)\\) and \\(\\left(\\var{b},\\simplify{{rational(y_2)}}\\right)\\).

Further, since acceleration corresponds to the second derivative of position, we know, using the second derivative test that these points correspond to a maximum minimum and minimum maximum respectively.

Thus the graph of \\(y=\\simplify{{function}}\\) is

{applet()}

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// , , , \n //app.setCoordSystem(-1,10,Math.min(-50,20,0),Math.max(-20,30,0)); // , , , \n //app.setAxisSteps(1,0.25*Math.PI,1)\n //app.refreshViews();\n //app.evalCommand(\"ZoomIn[-1,-1,1,1]\"); // , , , \n //app.evalCommand(\"SetAxesRatio(1,1/20)\");\n //app.evalCommand(\"CenterView((0, 0))\"); //comment out if coordinates are not centred on (0,0)\n app.evalCommand(\"ShowAxes(true)\");\n app.evalCommand(\"ShowGrid(true)\");\n \n var rsggb = question.scope.evaluate(\"function2\").value; \n //console.log(\"rsggb= \"+rsggb);\n // var ratio=\"\";\n //if (rsggb==0) {\n //ratio=\"sin\";\n //} else if (rsggb==1) {\n // ratio=\"cos\"; \n //} else if (rsggb==2) {\n // ratio=\"tan\";\n // }\n \n //app.evalCommand(\"f=Curve(t,\" + aggb + \"*\" + ratio + \"(t), t, -2*pi, 2*pi)\");\n //app.evalCommand(\"ShowLabel(f,false)\");\n //app.evalCommand(\"SetColor(f,0,0,1)\");\n \n app.evalCommand(\"f=Function(\"+rsggb+\", x, 0, 10)\");\n app.evalCommand(\"ShowLabel(f,false)\");\n app.evalCommand(\"SetColor(f,0,0,1)\");\n \n});\n\n// This function returns the result of `createGeogebraApplet` as an object \n// with the JME data type 'ggbapplet', which can be substituted into the question's content.\nreturn new Numbas.jme.types.ggbapplet(result);"}}, "preamble": {"js": "", "css": ""}, "parts": [], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}, {"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": [""], "variable_overrides": [[]], "questions": [{"name": "Implicit 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Don Shearman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/680/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "", "advice": "

We need to use implicit differentiation since the function is not written in terms of \\(y\\) as a function of \\(x\\).

Differentiating both sides and using the product rule on the first term we have \\begin{align}\\frac{d}{dx}\\left(\\simplify{x^{a}*y^{b}+{c}}\\cos y\\right)&=\\frac{d}{dx}\\left(y^\\var{d}\\right)\\\\\\simplify[!expandBrackets]{({b}x^{a}*y^{b-1})*(dy/dx)+{a}x^{a-1}*y^{b}+{c}*cos(y)*(dy/dx)}&=\\var{d}y^\\var{d-1}\\frac{dy}{dx}\\\\\\frac{dy}{dx}\\left(\\simplify{{b}x^{a}*y^{b-1}+{c}*cos(y)-{d}y^{d-1}}\\right)&=\\simplify{-{a}x^{a-1}*y^{b}}\\\\\\frac{dy}{dx}&=\\frac{\\simplify{-{a}x^{a-1}*y^{b}}}{\\simplify{{b}x^{a}*y^{b-1}+{c}*cos(y)-{d}y^{d-1}}}\\end{align}

Find the rate of change of \\(y\\) with respect to \\(x\\) if \\[\\simplify{x^{a}*y^{b}+{c}}\\sin (y)=\\simplify{y^{d}}\\]

\\(\\displaystyle{\\frac{dy}{dx}}=\\)[[0]]

Note: To display \\(\\sin x\\) above type sin (x) and similar for other trig functions.

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Time out. Upload your handwriting answers to vUWS.

"}, "timedwarning": {"action": "warn", "message": "

You have 5 minutes left

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

There are a total of 30 marks available for this test with most marks being given to your handwritten working for each question. Total marks for each question will be

Question 1 - 4 marks
Question 2 - 4 marks
Question 3 - 4 marks
Question 4 - 6 marks
Question 5 - 6 marks
Question 6 - 6 marks

You must write your working for each question on paper. This must be scanned and uploaded to vUWS as a single PDF file within 10 minutes of the end of the test. The first page of your working should include your name and student number.

", "reviewshowscore": true, "reviewshowfeedback": true, "reviewshowexpectedanswer": true, "reviewshowadvice": true, "feedbackmessages": []}, "diagnostic": {"knowledge_graph": {"topics": [], "learning_objectives": []}, "script": "diagnosys", "customScript": ""}, "contributors": [{"name": "Don Shearman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/680/"}, {"name": "Hanh Vo", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/12984/"}], "extensions": ["geogebra"], "custom_part_types": [], "resources": []}