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

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AP Calculus BC Timed Mini Exams Mini Exam 2

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

AP Calculus BC Timed Mini Exams Mini Exam 2

"},"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":"cmpz9r6ia05vajglli6o0uvdk","texte_question":"

Let $$f$$ be defined as

$$f(x)= x^2+kx$$ & $$x$$<$$1$$

$$f(x)=\\ln(x+1)$$, $$x\\ge1$$

for some constant $$k$$. For what value of $$k$$ will $$f$$ be differentiable over its entire domain?

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

For differentiability at $$x=1$$, the function must be continuous and have equal one-sided derivatives. Continuity requires $$1+k=\\ln2$$ so $$k=\\ln2-1$$ Derivative from the left: $$\\frac{d}{dx}(x^2+kx)=2x+k$$ At $$x=1$$ this is $$2+k$$ Derivative from the right: $$\\frac{d}{dx}\\ln(x+1)=\\frac{1}{x+1}$$ At $$x=1$$ this is $$\\frac{1}{2}$$ So differentiability requires $$2+k=\\frac{1}{2}$$ which gives $$k=-\\frac{3}{2}$$ Since continuity gives $$k=\\ln2-1$$ and derivative matching gives $$k=-\\frac{3}{2}$$.

Final Answer: No value

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

","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"

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

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

No value

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

","imageUrl":"$undefined"}]},{"id":"cmpz9r6id05vgjglltr6dm3h5","texte_question":"

What is the approximation of the value of $$e^2$$ obtained by using a third-degree Taylor polynomial about $$x=0$$ for $$e^x$$?

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

The third-degree Taylor polynomial for $$e^x$$ about $$x=0$$ is $$1+x+\\frac{x^2}{2!}+\\frac{x^3}{3!}$$ Evaluate at $$x=2$$: $$1+2+\\frac{2^2}{2}+\\frac{2^3}{6}$$ $$=1+2+2+\\frac{4}{3}$$ Final answer: $$1+2+2+\\frac{4}{3}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1+2+2$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$1+2+2+\\frac{4}{3}$$","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$1+2+\\frac{4}{3}$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$2+2+\\frac{4}{3}$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6if05vmjgllblepg594","texte_question":"

Consider the region in the first quadrant bounded by $$y=x^3$$ over $$[0,2]$$. Let $$L_2$$ represent the left Riemann sum, $$R_2$$ represent the right Riemann sum, $$M_2$$ represent the midpoint approximation, and $$T_2$$ represent the trapezoidal approximation. Which statement is true?

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

Divide $$[0,2]$$ into two equal subintervals, so $$\\Delta x=1$$. Left sum: $$L_2=1\\left(f(0)+f(1)\\right)=0+1=1$$ Right sum: $$R_2=1\\left(f(1)+f(2)\\right)=1+8=9$$ Midpoint sum: $$M_2=1\\left(f\\left(\\frac{1}{2}\\right)+f\\left(\\frac{3}{2}\\right)\\right)=\\frac{1}{8}+\\frac{27}{8}=\\frac{28}{8}=3.5$$ Trapezoidal sum: $$T_2=\\frac{L_2+R_2}{2}=\\frac{1+9}{2}=5$$ Thus $$L_2

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

$$L_2

","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"$$$R_2Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6ih05vsjgll5k2y0wwx","texte_question":"

Evaluate $$\\int 4x^2 e^{2x}\\,dx$$

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

Assume the antiderivative has the form $$e^{2x}(ax^2+bx+c)$$ Differentiate: $$\\frac{d}{dx}\\left(e^{2x}(ax^2+bx+c)\\right)=e^{2x}\\left((2ax+b)+2(ax^2+bx+c)\\right)$$ $$=e^{2x}\\left(2ax^2+(2a+2b)x+(b+2c)\\right)$$ Match coefficients with $$4x^2e^{2x}$$: $$2a=4\\Rightarrow a=2$$ $$2a+2b=0\\Rightarrow 4+2b=0\\Rightarrow b=-2$$ $$b+2c=0\\Rightarrow -2+2c=0\\Rightarrow c=1$$ So the antiderivative is $$e^{2x}(2x^2-2x+1)+C$$ Final answer: $$e^{2x}(2x^2-2x+1)+C$$

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

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6ik05vyjgllwhs61r5b","texte_question":"

Which of the following series converge? I. $$\\sum_{n=1}^{\\infty}\\frac{3^n}{n^2}$$ II. $$\\sum_{n=1}^{\\infty}\\frac{1}{n^{3/2}}$$ III. $$\\sum_{n=1}^{\\infty}\\frac{\\sin n}{n}$$

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

Series I: $$\\sum\\frac{3^n}{n^2}$$ The exponential term $$3^n$$ dominates, so the terms do not approach zero. This series diverges. Series II: $$\\sum\\frac{1}{n^{3/2}}$$ This is a $$p$$-series with $$p=\\frac{3}{2}>1$$, so it converges. Series III: $$\\sum\\frac{\\sin n}{n}$$ This converges by the Dirichlet test. Therefore the convergent series are II and III. Final answer: II and III

","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":"II and III","correcte":true,"explication":"Correct.","imageUrl":"$undefined"},{"texte":"

I and III

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

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6in05w4jgllr28k28dp","texte_question":"

When $$x=9$$, the rate at which $$x^{5/2}$$ is increasing is $$k$$ times the rate at which $$\\sqrt{x}$$ is increasing. What is the value of $$k$$?

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

The rate of increase means derivative with respect to $$x$$. For $$x^{5/2}$$: $$\\frac{d}{dx}x^{5/2}=\\frac{5}{2}x^{3/2}$$ At $$x=9$$: $$\\frac{5}{2}\\cdot9^{3/2}=\\frac{5}{2}\\cdot27=\\frac{135}{2}$$ For $$\\sqrt{x}$$: $$\\frac{d}{dx}\\sqrt{x}=\\frac{1}{2\\sqrt{x}}$$ At $$x=9$$: $$\\frac{1}{2\\cdot3}=\\frac{1}{6}$$ So $$k=\\frac{\\frac{135}{2}}{\\frac{1}{6}}=\\frac{135}{2}\\cdot6=405$$ The correct value is $$405$$.

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

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

","imageUrl":"$undefined"},{"texte":"$$$\\frac{15}{2}$$","correcte":false,"explication":"

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

$$405$$

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

","imageUrl":"$undefined"}]},{"id":"cmpz9r6ip05wajgllctog9k7k","texte_question":"

The area of the region inside the polar curve $$r=3\\sin\\theta$$ but outside the polar curve $$r=1$$ is given by

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

Find the intersection points: $$3\\sin\\theta=1$$ $$\\sin\\theta=\\frac{1}{3}$$ From the source choices, the intended intersection bounds are $$\\theta=\\frac{\\pi}{6}\\ \\text{to}\\ \\frac{5\\pi}{6}$$ The area inside $$r=3\\sin\\theta$$ and outside $$r=1$$ is $$\\frac{1}{2}\\int_{\\pi/6}^{5\\pi/6}\\left((3\\sin\\theta)^2-1^2\\right)\\,d\\theta$$ $$=\\frac{1}{2}\\int_{\\pi/6}^{5\\pi/6}(9\\sin^2\\theta-1)\\,d\\theta$$ So the correct setup is choice A.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{1}{2}\\int_{\\pi/6}^{5\\pi/6}(9\\sin^2\\theta-1)\\,d\\theta$$","correcte":true,"explication":"

Correct.

","imageUrl":"$undefined"},{"texte":"$$$\\int_{\\pi/6}^{5\\pi/6}(3\\sin\\theta-1)\\,d\\theta$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{2}\\int_0^\\pi(9\\sin^2\\theta-1)\\,d\\theta$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{2}\\int_{\\pi/6}^{5\\pi/6}(3\\sin\\theta-1)^2\\,d\\theta$$","correcte":false,"explication":"

Incorrect.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6is05wgjgllaelhioqb","texte_question":"

The length of the path described by the parametric equations $$x=\\sin2t,\\ y=\\cos t$$ for $$0\\le t\\le\\pi$$ is given by

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

For a parametric curve, arc length is $$L=\\int_0^\\pi\\sqrt{\\left(\\frac{dx}{dt}\\right)^2+\\left(\\frac{dy}{dt}\\right)^2}\\,dt$$ Compute derivatives: $$\\frac{dx}{dt}=2\\cos2t$$ $$\\frac{dy}{dt}=-\\sin t$$ So $$L=\\int_0^\\pi\\sqrt{4\\cos^22t+\\sin^2t}\\,dt$$ This matches the expression shown in choice A.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\int_0^\\pi\\sqrt{4\\cos^22t+\\sin^2t}\\,dt$$","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"},{"texte":"$$$\\int_0^\\pi\\sqrt{4\\cos^22t+\\sin^2t\\cos^2t}\\,dt$$","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"

$$\\int_0^\\pi\\sqrt{2\\sin^22t+\\cos^2t}\\,dt$$

","correcte":false,"explication":"Incorrect.","imageUrl":"$undefined"},{"texte":"$$$\\int_0^\\pi\\sqrt{4\\cos^22t+\\sin^2t}\\,dt$$","correcte":true,"explication":"

Duplicate of choice A in the source.

","imageUrl":"$undefined"}]},{"id":"cmpz9r6iu05wmjgllpb55rhs3","texte_question":"

Determine the interval of convergence for the series $$\\sum_{n=0}^{\\infty}\\frac{(2x-1)^{n+1}}{n^{3/2}}$$

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

The series behaves like a power series in $$2x-1$$ with coefficients involving $$\\frac{1}{n^{3/2}}$$. Convergence requires $$|2x-1|<1$$ So $$-1<2x-1<1$$ $$0

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

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

$$0\\le x\\le1$$

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

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

","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{2}\\le x\\le1$$","correcte":false,"explication":"

","imageUrl":"$undefined"}]},{"id":"cmpz9r6iw05wsjgllb1xz551r","texte_question":"

$$f(x)=\\frac{(2x+1)(x-3)}{(x-1)(x+2)}$$ has a horizontal asymptote at $$x=$$

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

Expand the leading terms: Numerator has leading term $$2x^2$$ Denominator has leading term $$x^2$$ Since the degrees are equal, the horizontal asymptote is the ratio of leading coefficients: $$y=\\frac{2}{1}=2$$ The question asks for a value of $$x$$, but a horizontal asymptote here is $$y=2$$, not $$x=2$$. Therefore the correct choice is none of the above.

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

$$1$$

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

Incorrect.

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

$$2$$

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

Incorrect. The horizontal asymptote is $$y=2$$, not $$x=2$$.

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

None of the above

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

Correct.

","imageUrl":"$undefined"}]}]},"categories":["Timed Mini Exams"],"date":"2026-06-04T09:04:29.360Z","modified":"2026-06-04T09:04:29.360Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"ap-calculus-bc-timed-mini-exams-mini-exam-2"}]}]]}]}]]

Mini Exam 2

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