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

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AP Calculus BC Timed Mini Exams Mini Exam 6

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

AP Calculus BC Timed Mini Exams Mini Exam 6

"},"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":"cmpz9r6qs061yjgllgicfadp8","texte_question":"

Which of the following sequences converge? I. $$\\left\\{\\frac{5n}{2n^2+1}\\right\\}$$ II. $$\\left\\{\\frac{4}{n}\\right\\}$$ III. $$\\left\\{\\frac{2n^3}{n^2+1}\\right\\}$$

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

Check each sequence as $$n\\to\\infty$$. For I: $$\\frac{5n}{2n^2+1}\\sim\\frac{5n}{2n^2}=\\frac{5}{2n}\\to0$$ So I converges. For II: $$\\frac{4}{n}\\to0$$ So II converges. For III: $$\\frac{2n^3}{n^2+1}\\sim\\frac{2n^3}{n^2}=2n\\to\\infty$$ So III does not converge. Therefore the convergent sequences are I and II. Final answer: I and II

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

II only

","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"I and II","correcte":true,"explication":"

Correct.

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

I, II, and III

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

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6qw0624jgll2am9mz5v","texte_question":"

Evaluate $$\\lim_{x\\to2}\\frac{\\int_1^x t^2\\,dt}{x-2}$$

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

Compute the numerator as $$x\\to2$$: $$\\int_1^x t^2\\,dt\\to\\int_1^2 t^2\\,dt=\\left[\\frac{t^3}{3}\\right]_1^2=\\frac{8}{3}-\\frac{1}{3}=\\frac{7}{3}$$ The denominator satisfies $$x-2\\to0$$ So the expression behaves like $$\\frac{7/3}{x-2}$$ which goes to $$+\\infty$$ from the right and $$-\\infty$$ from the left. Therefore the limit does not exist. Final answer: It does not exist

","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":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"It does not exist","correcte":true,"explication":"

Correct.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6qy062ajgll67ia7g11","texte_question":"

Evaluate $$\\int\\frac{x^2+1}{x}\\,dx$$

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

Simplify the integrand: $$\\frac{x^2+1}{x}=x+\\frac{1}{x}$$ Integrate term by term: $$\\int x\\,dx+\\int\\frac{1}{x}\\,dx=\\frac{x^2}{2}+\\ln|x|+C$$ Final answer: $$\\frac{x^2}{2}+\\ln|x|+C$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{x^3}{3}+\\ln|x|+C$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{x^2}{2}+\\ln|x|+C$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$\\frac{x^3}{3}+x+C$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$x^2+\\ln|x|+C$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6r0062gjgllgo1ucas7","texte_question":"

What is the instantaneous rate of change of the derivative of the function $$f(x)=\\ln(x^3)$$ when $$x=1$$?

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

The instantaneous rate of change of the derivative means the second derivative. Simplify: $$f(x)=\\ln(x^3)=3\\ln x$$ Differentiate: $$f'(x)=\\frac{3}{x}$$ Differentiate again: $$f''(x)=-\\frac{3}{x^2}$$ Evaluate at $$x=1$$: $$f''(1)=-3$$ Final answer: $$-3$$

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

Incorrect.

","imageUrl":"$undefined"},{"texte":"$$$3$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6r3062mjgllbzotiw8e","texte_question":"

Evaluate $$\\lim_{x\\to\\infty}\\frac{3x^2+5x-1}{(x+1)(2x-1)}$$

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

Expand the denominator: $$(x+1)(2x-1)=2x^2+x-1$$ So the limit is $$\\lim_{x\\to\\infty}\\frac{3x^2+5x-1}{2x^2+x-1}$$ Since numerator and denominator have the same degree, the limit is the ratio of leading coefficients: $$\\frac{3}{2}$$ Final answer: $$\\frac{3}{2}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{2}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{3}{2}$$","correcte":true,"explication":"

Correct.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6r5062sjgllqi3128o5","texte_question":"

If $$f(x)=\\tan^2x$$, then $$f'\\left(\\frac{\\pi}{4}\\right)=$$

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

Differentiate: $$f(x)=\\tan^2x$$ $$f'(x)=2\\tan x\\sec^2x$$ Evaluate at $$x=\\frac{\\pi}{4}$$: $$\\tan\\left(\\frac{\\pi}{4}\\right)=1$$ $$\\sec^2\\left(\\frac{\\pi}{4}\\right)=2$$ So $$f'\\left(\\frac{\\pi}{4}\\right)=2(1)(2)=4$$ Final answer: $$4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$4$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$8$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6r8062yjglll6vpdu9e","texte_question":"

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

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

Use the chain rule: $$\\frac{d}{dx}\\ln(\\sin x^2)=\\frac{1}{\\sin x^2}\\cdot\\frac{d}{dx}(\\sin x^2)$$ $$=\\frac{1}{\\sin x^2}\\cdot\\cos x^2\\cdot2x$$ $$=2x\\cot(x^2)$$ Final answer: $$2x\\cot(x^2)$$

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

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6ra0634jgllqdxxwcyo","texte_question":"

What is the approximation of the area under $$y=x^2+1$$ for $$0\\le x\\le2$$ using the trapezoidal rule with $$2$$ subintervals?

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

Using the trapezoidal rule on $$[0,2]$$ with $$n=2$$: $$\\Delta x=\\frac{2-0}{2}=1$$ Evaluate: $$f(0)=1$$ $$f(1)=2$$ $$f(2)=5$$ Then $$T_2=\\frac{\\Delta x}{2}\\left[f(0)+2f(1)+f(2)\\right]$$ $$T_2=\\frac{1}{2}(1+4+5)=5$$ Final answer: $$5$$

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

$$\\frac{14}{3}$$

","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$5$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"

$$6$$

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

Incorrect.

","imageUrl":"$undefined"},{"texte":"$$$7$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6rd063ajgll3qh82faq","texte_question":"

Let $$f$$ be the function given by the first four nonzero terms of the Maclaurin polynomial used to approximate $$\\sin x$$ Determine the area bounded by the graph and the $$x$$-axis for $$0\\le x\\le1$$.

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

The first four nonzero terms of the Maclaurin polynomial for $$\\sin x$$ are $$x-\\frac{x^3}{3!}+\\frac{x^5}{5!}-\\frac{x^7}{7!}$$ So the required area is $$\\int_0^1\\left(x-\\frac{x^3}{6}+\\frac{x^5}{120}-\\frac{x^7}{5040}\\right)\\,dx$$ Integrate term by term: $$\\left[\\frac{x^2}{2}-\\frac{x^4}{24}+\\frac{x^6}{720}-\\frac{x^8}{40320}\\right]_0^1$$ $$=\\frac{1}{2}-\\frac{1}{24}+\\frac{1}{720}-\\frac{1}{40320}$$ $$=\\frac{18511}{40320}\\approx0.459$$ This value is not listed among the choices, so the source appears inconsistent.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{1}{2}$$","correcte":false,"explication":"

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

$$\\frac{5}{12}$$

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

","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{3}$$","correcte":false,"explication":"

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

$$\\frac{18511}{40320}$$

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

","imageUrl":"$undefined"}]},{"id":"cmpz9r6rf063gjgllor5nemag","texte_question":"

Evaluate $$\\int_0^4\\frac{e^{\\sqrt{x}}}{\\sqrt{x}}\\,dx$$

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

Use the substitution $$u=\\sqrt{x}$$ Then $$x=u^2,\\quad dx=2u\\,du$$ Also $$\\sqrt{x}=u$$ So $$\\int_0^4\\frac{e^{\\sqrt{x}}}{\\sqrt{x}}\\,dx=\\int_0^2\\frac{e^u}{u}(2u)\\,du$$ $$=2\\int_0^2 e^u\\,du$$ $$=2\\left[e^u\\right]_0^2$$ $$=2(e^2-1)$$ Final answer: $$2(e^2-1)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2(e^2-1)$$","correcte":true,"explication":"

Correct.

","imageUrl":"$undefined"},{"texte":"$$$e^2-1$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"

$$4(e^2-1)$$

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

Incorrect.

","imageUrl":"$undefined"},{"texte":"$$$2(e^4-1)$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]}]},"categories":["Timed Mini Exams"],"date":"2026-06-04T09:04:29.665Z","modified":"2026-06-04T09:04:29.665Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"ap-calculus-bc-timed-mini-exams-mini-exam-6"}]}]]}]}]]

Mini Exam 6

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