7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Mini Exam 5\",\"description\":\"AP Calculus BC Timed Mini Exams Mini Exam 5\",\"url\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-timed-mini-exams-mini-exam-5\",\"timeRequired\":\"PT25M\",\"numberOfQuestions\":9,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Timed Mini Exams\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Mini Exam 5\",\"item\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-timed-mini-exams-mini-exam-5\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","div",null,{"className":"container mx-auto px-4 py-8 md:py-12","children":[["$","div",null,{"className":"mb-12","children":["$undefined",["$","div",null,{"className":"mb-8","children":["$","h1",null,{"className":"text-4xl md:text-5xl lg:text-6xl font-bold text-gray-900 mb-4","children":"Mini Exam 5"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","div",null,{"className":"text-lg md:text-xl text-gray-700 mb-8 leading-relaxed prose prose-lg max-w-none","dangerouslySetInnerHTML":{"__html":"

AP Calculus BC Timed Mini Exams Mini Exam 5

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AP Calculus BC Timed Mini Exams Mini Exam 5

"},"excerpt":{"rendered":"

AP Calculus BC Timed Mini Exams Mini Exam 5

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":25,"niveau_difficulte":"$undefined","categorie":"Timed Mini Exams","nombre_questions":9,"score_minimum":80,"ordre_questions":"Aleatoire","questions":[{"id":"cmpz9r6p3060djgllnfv2ra0j","texte_question":"

If $$x=\\ln t$$ and $$y=e^{3t}$$, then $$\\frac{dy}{dx}=$$

","type_question":"QCM","explication":"

For parametric equations, $$\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}$$ Compute: $$\\frac{dy}{dt}=3e^{3t}$$ $$\\frac{dx}{dt}=\\frac{1}{t}$$ Therefore, $$\\frac{dy}{dx}=\\frac{3e^{3t}}{1/t}=3te^{3t}$$ Final answer: $$3te^{3t}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$3e^{3t}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{3e^{3t}}{t}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$3te^{3t}$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"

$$\\frac{te^{3t}}{3}$$

","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6p6060jjgllwp6ke5oe","texte_question":"

The function $$y=\\frac{(x-1)^2}{x^2-4x+3}$$ has a local minimum at $$x=$$

","type_question":"QCM","explication":"

Factor the denominator: $$x^2-4x+3=(x-1)(x-3)$$ So for $$x\\ne1,3$$, $$y=\\frac{x-1}{x-3}$$ Differentiate: $$y'=\\frac{(x-3)-(x-1)}{(x-3)^2}=\\frac{-2}{(x-3)^2}$$ This derivative is always negative where defined, so the function is strictly decreasing on each interval of its domain and has no local minimum.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"

$$2$$

","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"

The function has no local minimum.

","correcte":false,"explication":"

","imageUrl":"$undefined"}]},{"id":"cmpz9r6p9060pjgll9z6khwes","texte_question":"

Evaluate $$\\frac{d}{dx}\\left(e^{2x}\\ln(\\sin e^x)\\right)$$

","type_question":"QCM","explication":"

Use the product rule. Let $$u=e^{2x},\\quad v=\\ln(\\sin e^x)$$ Then $$u'=2e^{2x}$$ and by the chain rule, $$v'=\\frac{1}{\\sin e^x}\\cos e^x\\cdot e^x=e^x\\cot(e^x)$$ So $$\\frac{d}{dx}\\left(e^{2x}\\ln(\\sin e^x)\\right)=2e^{2x}\\ln(\\sin e^x)+e^{2x}\\cdot e^x\\cot(e^x)$$ $$=2e^{2x}\\ln(\\sin e^x)+e^{3x}\\cot(e^x)$$ Final answer: $$2e^{2x}\\ln(\\sin e^x)+e^{3x}\\cot(e^x)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2e^{2x}\\ln(\\sin e^x)+e^{3x}\\cot(e^x)$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$e^{2x}\\ln(\\sin e^x)+e^{3x}\\cot(e^x)$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$2e^{2x}\\ln(\\sin e^x)+e^{2x}\\cot(e^x)$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"

$$e^{3x}\\cot(e^x)$$

","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6pc060vjgll7rw4dknd","texte_question":"

If $$f(x)=\\frac{\\cos x}{x^2}$$ then $$f'(\\pi)=$$

","type_question":"QCM","explication":"

Differentiate using the product rule: $$f(x)=\\cos x\\cdot x^{-2}$$ $$f'(x)=-\\sin x\\cdot x^{-2}-2\\cos x\\cdot x^{-3}$$ Evaluate at $$x=\\pi$$: $$\\sin\\pi=0,\\quad \\cos\\pi=-1$$ So $$f'(\\pi)=0-\\frac{2(-1)}{\\pi^3}=\\frac{2}{\\pi^3}$$ Final answer: $$\\frac{2}{\\pi^3}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-\\frac{1}{\\pi^2}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{\\pi^2}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$-\\frac{2}{\\pi^3}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"

$$\\frac{2}{\\pi^3}$$

","correcte":true,"explication":"

Correct.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6pf0611jgll54zbl5pg","texte_question":"

Evaluate $$\\int_2^{\\infty}\\frac{\\ln x}{x^3}\\,dx$$

","type_question":"QCM","explication":"

Use integration by parts. Let $$u=\\ln x,\\quad dv=x^{-3}\\,dx$$ Then $$du=\\frac{1}{x}\\,dx,\\quad v=-\\frac{1}{2x^2}$$ So $$\\int\\frac{\\ln x}{x^3}\\,dx=-\\frac{\\ln x}{2x^2}+\\frac{1}{2}\\int x^{-3}\\,dx$$ $$=-\\frac{\\ln x}{2x^2}-\\frac{1}{4x^2}$$ Evaluate from $$2$$ to $$\\infty$$: $$\\int_2^{\\infty}\\frac{\\ln x}{x^3}\\,dx=\\left[-\\frac{\\ln x}{2x^2}-\\frac{1}{4x^2}\\right]_2^{\\infty}$$ The limit at infinity is $$0$$, so $$=0-\\left(-\\frac{\\ln2}{8}-\\frac{1}{16}\\right)=\\frac{\\ln2}{8}+\\frac{1}{16}$$ $$=\\frac{2\\ln2+1}{16}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{\\ln2+1}{4}$$","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"$$$\\frac{\\ln2}{4}$$","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"

$$\\frac{2\\ln2+1}{16}$$

","correcte":true,"explication":"

","imageUrl":"$undefined"},{"texte":"

It diverges

","correcte":false,"explication":"

","imageUrl":"$undefined"}]},{"id":"cmpz9r6pi0617jgllqoep4rg1","texte_question":"

Evaluate $$\\int x\\tan^2x\\,dx$$

","type_question":"QCM","explication":"

Use the identity $$\\tan^2x=\\sec^2x-1$$ Then $$\\int x\\tan^2x\\,dx=\\int x\\sec^2x\\,dx-\\int x\\,dx$$ For the first integral, use integration by parts: $$u=x,\\quad dv=\\sec^2x\\,dx$$ $$du=dx,\\quad v=\\tan x$$ So $$\\int x\\sec^2x\\,dx=x\\tan x-\\int\\tan x\\,dx$$ Since $$\\int\\tan x\\,dx=-\\ln|\\cos x|$$ we get $$\\int x\\sec^2x\\,dx=x\\tan x+\\ln|\\cos x|$$ Thus $$\\int x\\tan^2x\\,dx=x\\tan x+\\ln|\\cos x|-\\frac{x^2}{2}+C$$.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$x\\tan x+\\ln|\\cos x|+C$$","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"

$$x\\tan x+\\ln|\\cos x|-\\frac{x^2}{2}+C$$

","correcte":true,"explication":"

","imageUrl":"$undefined"},{"texte":"

$$x\\tan x-\\ln|\\sec x+\\tan x|+C$$

","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"

$$x\\tan x+\\ln|\\sec x+\\tan x|+C$$

","correcte":false,"explication":"

","imageUrl":"$undefined"}]},{"id":"cmpz9r6pl061djgll085do2fi","texte_question":"

Evaluate $$\\lim_{x\\to1}\\frac{(\\ln x)^3}{x^2-1}$$

","type_question":"QCM","explication":"

As $$x\\to1$$, $$(\\ln x)^3\\to0$$ and $$x^2-1\\to0$$ so this is an indeterminate form $$\\frac{0}{0}$$. Apply L'Hospital's Rule: $$\\lim_{x\\to1}\\frac{(\\ln x)^3}{x^2-1}=\\lim_{x\\to1}\\frac{3(\\ln x)^2\\cdot\\frac{1}{x}}{2x}$$ $$=\\lim_{x\\to1}\\frac{3(\\ln x)^2}{2x^2}$$ Now as $$x\\to1$$, $$(\\ln x)^2\\to0$$ so the limit is $$0$$ Final answer: $$0$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{2}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"

It does not exist

","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6pn061jjgllpkz5bf5t","texte_question":"

What is the approximation of the value of $$\\sin(1)$$ obtained by using the fifth-degree Taylor polynomial about $$x=0$$ for $$\\sin x$$?

","type_question":"QCM","explication":"

The fifth-degree Taylor polynomial for $$\\sin x$$ centered at $$x=0$$ is $$x-\\frac{x^3}{3!}+\\frac{x^5}{5!}$$ Evaluate at $$x=1$$: $$1-\\frac{1}{6}+\\frac{1}{120}$$ Final answer: $$1-\\frac{1}{6}+\\frac{1}{120}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1-\\frac{1}{6}+\\frac{1}{120}$$","correcte":true,"explication":"

Correct.

","imageUrl":"$undefined"},{"texte":"$$$1-\\frac{1}{2}-\\frac{1}{24}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$1-\\frac{1}{6}-\\frac{1}{120}$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"},{"texte":"

$$1-\\frac{1}{2}+\\frac{1}{120}$$

","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6pq061pjglloz1ygidc","texte_question":"

A particle moves on a plane curve so that at any time $$t>0$$ its position is defined by $$x(t)=t^2+1,\\quad y(t)=\\frac{t^3+2}{t}$$ The acceleration vector of the particle at $$t=1$$ is

","type_question":"QCM","explication":"

Write the position vector: $$\\mathbf{r}(t)=\\langle x(t),y(t)\\rangle=\\left\\langle t^2+1,\\frac{t^3+2}{t}\\right\\rangle$$ Simplify $$y(t)$$: $$y(t)=t^2+\\frac{2}{t}$$ Velocity: $$\\mathbf{v}(t)=\\left\\langle\\frac{d}{dt}(t^2+1),\\frac{d}{dt}\\left(t^2+\\frac{2}{t}\\right)\\right\\rangle=\\left\\langle2t,2t-\\frac{2}{t^2}\\right\\rangle$$ Acceleration: $$\\mathbf{a}(t)=\\left\\langle2,2+\\frac{4}{t^3}\\right\\rangle$$ At $$t=1$$: $$\\mathbf{a}(1)=\\langle2,6\\rangle$$.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$\\langle2,1\\rangle$$

","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"

$$\\langle2,2\\rangle$$

","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"

$$\\langle2,6\\rangle$$

","correcte":true,"explication":"

","imageUrl":"$undefined"},{"texte":"

$$\\langle3,1\\rangle$$

","correcte":false,"explication":"

","imageUrl":"$undefined"}]}]},"categories":["Timed Mini Exams"],"date":"2026-06-04T09:04:29.603Z","modified":"2026-06-04T09:04:29.603Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"ap-calculus-bc-timed-mini-exams-mini-exam-5"}]}]]}]}]]

Mini Exam 5

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