7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Mini Exam 3\",\"description\":\"AP Calculus BC Timed Mini Exams Mini Exam 3\",\"url\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-timed-mini-exams-mini-exam-3\",\"timeRequired\":\"PT25M\",\"numberOfQuestions\":10,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Timed Mini Exams\"}}"}}],["$","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 3\",\"item\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-timed-mini-exams-mini-exam-3\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","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 3"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","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 3

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AP Calculus BC Timed Mini Exams Mini Exam 3

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

AP Calculus BC Timed Mini Exams Mini Exam 3

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":25,"niveau_difficulte":"$undefined","categorie":"Timed Mini Exams","nombre_questions":10,"score_minimum":80,"ordre_questions":"Aleatoire","questions":[{"id":"cmpz9r6jt05x1jgll8fvmctih","texte_question":"

Evaluate $$\\int_1^{\\infty}\\frac{x}{e^x}\\,dx$$.

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

Rewrite the integral as $$\\int_1^{\\infty}x e^{-x}\\,dx$$ Use integration by parts: Let $$u=x,\\ dv=e^{-x}\\,dx$$ Then $$du=dx,\\ v=-e^{-x}$$ So $$\\int x e^{-x}\\,dx=-x e^{-x}+\\int e^{-x}\\,dx=-x e^{-x}-e^{-x}$$ $$=-(x+1)e^{-x}$$ Evaluate from $$1$$ to $$\\infty$$: $$\\lim_{b\\to\\infty}\\left[-(x+1)e^{-x}\\right]_1^b$$ As $$b\\to\\infty$$, $$-(b+1)e^{-b}\\to0$$ So the value is $$0-\\left(-\\frac{2}{e}\\right)=\\frac{2}{e}$$ Final answer: $$\\frac{2}{e}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{2}{e}$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$\\frac{3}{e}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{e}{2}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$e$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6jw05x7jglly89y0euk","texte_question":"

The population $$P(t)$$ satisfies the logistic differential equation $$\\frac{dP}{dt}=\\frac{1}{2}P\\left(4-\\frac{P}{200}\\right)$$ What is $$\\lim_{t\\to\\infty}P(t)$$

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

For a logistic differential equation of the form $$\\frac{dP}{dt}=kP\\left(M-P\\right)$$ or an equivalent form, the population approaches the carrying capacity. Here, $$\\frac{dP}{dt}=\\frac{1}{2}P\\left(4-\\frac{P}{200}\\right)$$ Set the second factor equal to zero: $$4-\\frac{P}{200}=0$$ $$\\frac{P}{200}=4$$ $$P=800$$ So the carrying capacity is $$800$$, and therefore $$\\lim_{t\\to\\infty}P(t)=800$$ Final answer: $$800$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$200$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$400$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$600$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$800$$","correcte":true,"explication":"

Correct.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6jy05xdjglldv8kiym9","texte_question":"

If $$\\sum_{n=0}^{\\infty}a_n(x-c)^n$$ is a Taylor series that converges to $$f(x)$$, then $$f'(c)=$$

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

For a Taylor series centered at $$x=c$$, $$f(x)=a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\\cdots$$ Differentiate: $$f'(x)=a_1+2a_2(x-c)+3a_3(x-c)^2+\\cdots$$ Now evaluate at $$x=c$$: $$f'(c)=a_1$$ Final answer: $$a_1$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$a_0$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$a_1$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$2a_2$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$3a_3$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6k105xjjgllje93bgbm","texte_question":"

The graph of the function represented by the Taylor series centered at $$x=0$$, $$1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\cdots$$ intersects the graph of $$y=2$$ at $$x=$$

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

The series $$1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\cdots$$ is the Taylor series for $$e^x$$ So we solve $$e^x=2$$ Take natural logarithms: $$x=\\ln2\\approx0.693$$ Final answer: $$0.693$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0.567$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$0.693$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$1.000$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$1.386$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6k305xpjgll9m9zja2y","texte_question":"

If $$f$$ is a vector-valued function defined by $$f(t)=\\langle\\sin t,\\ e^t\\rangle$$ then $$f'(t)=$$

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

Differentiate each component: $$\\frac{d}{dt}(\\sin t)=\\cos t$$ $$\\frac{d}{dt}(e^t)=e^t$$ So $$f'(t)=\\langle\\cos t,\\ e^t\\rangle$$ Final answer: $$\\langle\\cos t,\\ e^t\\rangle$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\langle\\cos t,\\ e^t\\rangle$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$\\langle-\\cos t,\\ e^t\\rangle$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\langle\\cos t,\\ t e^t\\rangle$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\langle\\sin t,\\ e^t\\rangle$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6k805xvjgll09fv5lij","texte_question":"

If $$f$$ is continuous over the set of real numbers and $$f(x)=\\frac{x^2-4}{x-2},\\quad x\\ne2$$ then $$f(2)=$$

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

Since $$f$$ is continuous everywhere, its value at $$x=2$$ must equal the limit. Factor: $$x^2-4=(x-2)(x+2)$$ So for $$x\\ne2$$, $$f(x)=x+2$$ Therefore, $$f(2)=\\lim_{x\\to2}(x+2)=4$$ Final answer: $$4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$4$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"None of the above","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6kb05y1jgll6be8sw4p","texte_question":"

If $$0\\le k\\le3$$ and the area between the curves $$y=x^2\\quad\\text{and}\\quad y=x$$ from $$x=0$$ to $$x=k$$ is $$1$$, then $$k=$$

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

The curves intersect where $$x^2=x$$ $$x(x-1)=0$$ So they intersect at $$x=0$$ and $$x=1$$. For $$0\\le x\\le1$$, the upper curve is $$y=x$$. For $$1\\le x\\le k$$, the upper curve is $$y=x^2$$. Since the answer lies in $$[0,3]$$, check $$k=2$$: Area from $$0$$ to $$1$$: $$\\int_0^1(x-x^2)\\,dx=\\left[\\frac{x^2}{2}-\\frac{x^3}{3}\\right]_0^1=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$$ Area from $$1$$ to $$2$$: $$\\int_1^2(x^2-x)\\,dx=\\left[\\frac{x^3}{3}-\\frac{x^2}{2}\\right]_1^2$$ $$=\\left(\\frac{8}{3}-2\\right)-\\left(\\frac{1}{3}-\\frac{1}{2}\\right)=\\frac{2}{3}+\\frac{1}{6}=\\frac{5}{6}$$ Total area: $$\\frac{1}{6}+\\frac{5}{6}=1$$ So $$k=2$$ Final answer: $$2.000$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1.145$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$1.442$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"},{"texte":"$$$1.732$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$2.000$$","correcte":true,"explication":"

Correct.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6kd05y7jgll7hh8t7gu","texte_question":"

Determine $$\\frac{dy}{dx}$$ for the curve defined by $$x\\cos y=1$$

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

Differentiate implicitly: $$\\frac{d}{dx}(x\\cos y)=0$$ Use the product rule: $$\\cos y+x(-\\sin y)\\frac{dy}{dx}=0$$ $$\\cos y-x\\sin y\\frac{dy}{dx}=0$$ Solve for $$\\frac{dy}{dx}$$: $$x\\sin y\\frac{dy}{dx}=\\cos y$$ $$\\frac{dy}{dx}=\\frac{\\cos y}{x\\sin y}$$ Final answer: $$\\frac{\\cos y}{x\\sin y}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{\\sin y}{x}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"

$$\\frac{\\sin y}{x}$$

","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{\\cos y}{x\\sin y}$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$\\frac{\\cos y}{x}$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6kh05ydjgllcosszul6","texte_question":"

If $$f(x)=p(x)+q(x)$$ for $$0\\le x\\le5$$, then $$\\int_0^5 (f(x)-p(x)+2)\\,dx=$$

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

Substitute $$f(x)=p(x)+q(x)$$ Then $$f(x)-p(x)+2=p(x)+q(x)-p(x)+2$$ Simplify $$=q(x)+2$$ So $$\\int_0^5 (f(x)-p(x)+2)\\,dx=\\int_0^5 (q(x)+2)\\,dx$$ Split the integral $$=\\int_0^5 q(x)\\,dx+\\int_0^5 2\\,dx$$ $$=\\int_0^5 q(x)\\,dx+10$$ Final answer: $$\\int_0^5 q(x)\\,dx+10$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\int_0^5 (q(x)+2)\\,dx$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\int_0^5 (q(x)-p(x))\\,dx+10$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\int_0^5 q(x)\\,dx+10$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$\\int_0^5 (q(x)+2)\\,dx+5$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6kk05yjjglllmwp6xpv","texte_question":"

Use a fourth-degree Taylor polynomial centered at $$x=0$$ to estimate $$e^{1.5}$$

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

The fourth-degree Taylor polynomial for $$e^x$$ at $$x=0$$ is $$1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}+\\frac{x^4}{4!}$$ Evaluate at $$x=1.5$$ $$1+1.5+\\frac{1.5^2}{2}+\\frac{1.5^3}{6}+\\frac{1.5^4}{24}$$ Compute each term $$1+1.5+1.125+0.5625+0.2109375$$ $$=4.3984375$$ Rounded $$4.398$$ Final answer: $$4.398$$

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

$$4.375$$

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

Incorrect.

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

$$4.398$$

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

Correct.

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

$$4.462$$

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

Incorrect.

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

$$4.500$$

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

Incorrect.

","imageUrl":"$undefined"}]}]},"categories":["Timed Mini Exams"],"date":"2026-06-04T09:04:29.414Z","modified":"2026-06-04T09:04:29.414Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"ap-calculus-bc-timed-mini-exams-mini-exam-3"}]}]]}]}]]

Mini Exam 3

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