7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Mini Exam 1\",\"description\":\"AP Calculus BC Timed Mini Exams Mini Exam 1\",\"url\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-timed-mini-exams-mini-exam-1\",\"timeRequired\":\"PT25M\",\"numberOfQuestions\":9,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Timed Mini Exams\"}}"}}],["$","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 1\",\"item\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-timed-mini-exams-mini-exam-1\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","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 1"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","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 1

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AP Calculus BC Timed Mini Exams Mini Exam 1

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

AP Calculus BC Timed Mini Exams Mini Exam 1

"},"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":"cmpz9r6gu05tpjgllgm7c5278","texte_question":"

The function $$f$$ is given by $$f(x)=2x^4-4x^3+3x^2-1$$. On which of the following intervals is $$f'$$ decreasing?

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

Differentiate: $$f'(x)=24x^2-24x+6$$ Factor: $$f'(x)=6(2x-1)^2$$ Since $$f'(x)\\ge0$$ for all $$x$$, the derivative $$f'$$ is never negative, so $$f'$$ is not decreasing on any of the listed intervals.

Final Answer: None

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

$$(-\\infty,\\infty)$$

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

","imageUrl":"$undefined"},{"texte":"$$$(-\\infty,0)$$","correcte":false,"explication":"

","imageUrl":"$undefined"},{"texte":"$$$(1,\\infty)$$","correcte":false,"explication":"

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

None

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

","imageUrl":"$undefined"}]},{"id":"cmpz9r6gx05tvjgllcp14pax0","texte_question":"

What is the area under the curve described by the parametric equations $$x=\\cos t,\\ y=\\sin^2 t$$, for $$0\\le t\\le\\frac{\\pi}{2}$$?

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

Since $$x=\\cos t$$ and $$y=\\sin^2 t=1-\\cos^2 t$$, the curve is $$y=1-x^2$$ from $$x=1$$ to $$x=0$$. The area under the curve is $$\\int_0^1(1-x^2)\\,dx=\\left[x-\\frac{x^3}{3}\\right]_0^1=1-\\frac{1}{3}=\\frac{2}{3}$$ So the correct area is $$\\frac{2}{3}$$.

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

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

$$\\frac{2}{3}$$

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

","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{2}$$","correcte":false,"explication":"

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

","imageUrl":"$undefined"}]},{"id":"cmpz9r6h005u1jgll4slaqgcg","texte_question":"

The function $$f$$ is given by $$f(x)=x^3+6x^2+12x+8$$. All of these statements are true EXCEPT

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

Factor: $$f(x)=x^3+6x^2+12x+8=(x+2)^3$$ So $$-2$$ is a zero of $$f$$. Differentiate: $$f'(x)=3(x+2)^2$$ Thus $$-2$$ is a zero of $$f'$$. Second derivative: $$f''(x)=6x+12=6(x+2)$$ So $$-2$$ is a point of inflection. However, because $$f'(x)=3(x+2)^2\\ge0$$ and does not change from positive to negative or negative to positive at $$x=-2$$, $$-2$$ is not a local extremum. Final answer: $$-2$$ is a local extremum is the false statement.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-2$$ is a zero of $$f$$","correcte":false,"explication":"True statement.","imageUrl":"$undefined"},{"texte":"$$$-2$$ is a point of inflection","correcte":false,"explication":"True statement.","imageUrl":"$undefined"},{"texte":"$$$-2$$ is a local extremum","correcte":true,"explication":"Correct. This is the false statement.","imageUrl":"$undefined"},{"texte":"$$$-2$$ is a zero of $$f'$$","correcte":false,"explication":"

True statement.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6h205u7jgllno9z40hy","texte_question":"

Evaluate $$\\int x e^x\\,dx$$

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

Use integration by parts. Let $$u=x,\\ dv=e^x\\,dx$$ Then $$du=dx,\\ v=e^x$$ So $$\\int x e^x\\,dx=x e^x-\\int e^x\\,dx=x e^x-e^x+C$$ Final answer: $$x e^x-e^x+C$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$x e^x+e^x+C$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"

$$x e^x-e^x+C$$

","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$e^x+x+C$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$x^2 e^x+C$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6h505udjgll7upijfzf","texte_question":"

Let $$h(x)=\\ln(f(x))$$. If $$f$$ is increasing for all $$x$$ in its domain, then

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

The natural logarithm function is increasing on its domain. Since $$f$$ is increasing and $$\\ln(x)$$ is increasing for $$x>0$$, the composition $$h(x)=\\ln(f(x))$$ is strictly increasing wherever it is defined. Final answer: $$h$$ is strictly increasing.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$h$$ is strictly increasing","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$h$$ is strictly decreasing","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"},{"texte":"$$$h$$ has a relative maximum","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"none of the above","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6h705ujjgllcwiipubs","texte_question":"

Evaluate $$\\int_0^1\\left(\\frac{x^2+2}{x+1}\\right)\\,dx$$

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

Divide: $$\\frac{x^2+2}{x+1}=x-1+\\frac{3}{x+1}$$ So $$\\int_0^1\\left(\\frac{x^2+2}{x+1}\\right)\\,dx=\\int_0^1\\left(x-1+\\frac{3}{x+1}\\right)\\,dx$$ Integrate: $$\\left[\\frac{x^2}{2}-x+3\\ln(x+1)\\right]_0^1=\\frac{1}{2}-1+3\\ln2$$ Thus the value is $$3\\ln2-\\frac{1}{2}\\approx1.579$$.

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

","imageUrl":"$undefined"},{"texte":"$$$\\frac{3}{2}$$","correcte":false,"explication":"

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

$$1.58$$

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

","imageUrl":"$undefined"},{"texte":"$$$\\frac{5}{2}$$","correcte":false,"explication":"

","imageUrl":"$undefined"}]},{"id":"cmpz9r6ha05upjgll0s14m3lz","texte_question":"

The function $$f$$ given by $$f(x)=x^5-5x^3+4x$$ is increasing and concave down over which of these intervals?

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

Differentiate: $$f'(x)=5x^4-15x^2+4$$ Second derivative: $$f''(x)=20x^3-30x=10x(2x^2-3)$$ Check the intervals: On $$(-\\infty,-1)$$, $$f'(x)>0$$ and $$f''(x)<0$$, so $$f$$ is increasing and concave down there. The other listed intervals fail one of the two conditions over the entire interval. Final answer: $$(-\\infty,-1)$$

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

$$(-\\infty,-1)$$

","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$(-1,0)$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$(0,1)$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$(1,\\infty)$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6hc05uvjgll2nzsxwqo","texte_question":"

The length of the curve described by the parametric equations $$x=t^2,\\ y=t^3$$ where $$0\\le t\\le1$$ is

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

For a parametric curve, arc length is $$L=\\int_0^1\\sqrt{\\left(\\frac{dx}{dt}\\right)^2+\\left(\\frac{dy}{dt}\\right)^2}\\,dt$$ Compute derivatives: $$\\frac{dx}{dt}=2t$$ $$\\frac{dy}{dt}=3t^2$$ So $$L=\\int_0^1\\sqrt{4t^2+9t^4}\\,dt$$ $$L=\\int_0^1 t\\sqrt{4+9t^2}\\,dt$$ Let $$u=4+9t^2$$ Then $$du=18t\\,dt$$ $$t\\,dt=\\frac{1}{18}du$$ Thus $$L=\\frac{1}{18}\\int_4^{13}u^{1/2}\\,du$$ $$L=\\frac{1}{18}\\cdot\\frac{2}{3}u^{3/2}\\Big|_4^{13}$$ $$L=\\frac{1}{27}\\left(13^{3/2}-8\\right)$$ $$L\\approx1.440$$

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

$$\\frac{1}{2}$$

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

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

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

$$1.44$$

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

","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"

","imageUrl":"$undefined"}]},{"id":"cmpz9r6hf05v1jgllp5clf5pc","texte_question":"

What is the average value of $$f(x)=2\\sin^2x+\\cos^2x$$ over $$\\left[0,\\frac{\\pi}{2}\\right]$$?

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

The average value of $$f$$ on $$[a,b]$$ is $$\\frac{1}{b-a}\\int_a^b f(x)\\,dx$$ Here $$b-a=\\frac{\\pi}{2}$$ So $$f_{\\mathrm{avg}}=\\frac{2}{\\pi}\\int_0^{\\pi/2}\\left(2\\sin^2x+\\cos^2x\\right)\\,dx$$ Simplify: $$2\\sin^2x+\\cos^2x=\\sin^2x+\\sin^2x+\\cos^2x=1+\\sin^2x$$ Thus $$f_{\\mathrm{avg}}=\\frac{2}{\\pi}\\left(\\int_0^{\\pi/2}1\\,dx+\\int_0^{\\pi/2}\\sin^2x\\,dx\\right)$$ $$=\\frac{2}{\\pi}\\left(\\frac{\\pi}{2}+\\frac{\\pi}{4}\\right)$$ $$=\\frac{2}{\\pi}\\cdot\\frac{3\\pi}{4}$$ $$=\\frac{3}{2}$$ Final answer: $$\\frac{3}{2}$$

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

Incorrect.

","imageUrl":"$undefined"}]}]},"categories":["Timed Mini Exams"],"date":"2026-06-04T09:04:29.307Z","modified":"2026-06-04T09:04:29.307Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"ap-calculus-bc-timed-mini-exams-mini-exam-1"}]}]]}]}]]

Mini Exam 1

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