7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Section I Part B\",\"description\":\"

Multiple choice calculator section

\",\"url\":\"https://www.theschoolofmathematics.com/quiz/ap-calculus-bc-full-practice-exam-4-section-i-part-b\",\"timeRequired\":\"PT45M\",\"numberOfQuestions\":12,\"about\":{\"@type\":\"Thing\",\"name\":\"Full Practice Exam 4\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://www.theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://www.theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Section I Part B\",\"item\":\"https://www.theschoolofmathematics.com/quiz/ap-calculus-bc-full-practice-exam-4-section-i-part-b\"}]}"}}],["$","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":["$","h1",null,{"className":"text-4xl md:text-5xl lg:text-6xl font-bold text-gray-900 mb-4","children":"Section I Part B"}]}],["$","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":"

Multiple choice calculator section

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AP Calculus BC Full Practice Exam 4 Section I Part B

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

Multiple choice calculator section

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":45,"niveau_difficulte":"$undefined","categorie":"Full Practice Exam 4","nombre_questions":15,"score_minimum":80,"ordre_questions":"Fixe","questions":[{"id":"cmnim2slq0000la049jki40ve","texte_question":"

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

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

Write $$f(x)=e^{2x}\\sec x$$ Differentiate using the product rule: $$f'(x)=2e^{2x}\\sec x+e^{2x}\\sec x\\tan x$$ Factor: $$f'(x)=e^{2x}\\sec x(2+\\tan x)$$ Rewrite with sine and cosine: $$f'(x)=\\frac{e^{2x}(2\\cos x+\\sin x)}{\\cos^2 x}$$ Final answer: $$\\frac{e^{2x}(2\\cos x+\\sin x)}{\\cos^2 x}$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnim2slq0005la047e1de4r3","texte_question":"

Let $$f$$ be a function such that $$\\lim_{h\\to0}\\frac{f(2+h)-f(2)}{h}=5$$. Which of the following must be true?

I. $$f'(2)=5$$

II. $$f$$ is continuous at $$x=2$$

III. $$f(2)=5$$

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

The given limit is the definition of the derivative at $$x=2$$, so $$f'(2)=5$$ is true as intended by the source notation. Differentiability implies continuity, so II is also true. III need not be true. Final answer: I and II

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"I only","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"I and II","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"II only","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"I, II, and III","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnim2slq000ala04nal2bq38","texte_question":"

The function $$f$$ whose derivative is $$f'(x)=3x^2-6x+2$$ has a local minimum at $$x=$$

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

To find where $$f$$ has a local minimum, solve $$f'(x)=3x^2-6x+2=0$$ Using the quadratic formula: $$x=\\frac{6\\pm\\sqrt{36-24}}{6}=\\frac{6\\pm2\\sqrt3}{6}=1\\pm\\frac{\\sqrt3}{3}$$ Since $$f'$$ is an upward-opening quadratic, the sign changes from negative to positive at the larger root, so the local minimum occurs at $$x=1+\\frac{\\sqrt3}{3}\\approx1.577$$ The closest listed choice is $$\\frac{5}{3}$$ if that was intended, but it is not shown in the screenshot. With the visible choices $$\\frac13,1,\\frac23,2$$, none match exactly. This item appears flawed in the source, so the answer set below uses the mathematically correct value.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac13$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1+\\frac{\\sqrt3}{3}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnim2slq000fla04qr7m74nu","texte_question":"

A car travels east at $$50$$ mph while another car travels north at $$40$$ mph. If both cars start from the same point at the same time, how fast is the distance between them changing after $$2$$ hours?

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

Let $$x=50t$$ and $$y=40t$$. Then the distance between the cars is $$d=\\sqrt{x^2+y^2}=\\sqrt{(50t)^2+(40t)^2}=10t\\sqrt{41}$$ Differentiate: $$\\frac{dd}{dt}=10\\sqrt{41}\\approx64.03$$ So after $$2$$ hours, the rate is still $$64.03\\text{ mph}$$ Final answer: $$64.03\\text{ mph}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$50\\text{ mph}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$64.03\\text{ mph}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

$$90\\text{ mph}$$

","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$70\\text{ mph}$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnim2slq000kla04fuz41bvy","texte_question":"

Which of the following is an equation for a line tangent to the graph of $$f(x)=e^{3x}$$ when $$f'(x)=12$$?

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

First solve for the point where the slope is $$12$$. $$f'(x)=3e^{3x}=12$$ $$e^{3x}=4$$ So $$x=\\frac{\\ln4}{3}$$ Then $$y=e^{3x}=4$$ Using point-slope form with slope $$12$$: $$y-4=12\\left(x-\\frac{\\ln4}{3}\\right)$$ $$y=12x-4\\ln4+4$$ Since $$-4\\ln4+4\\approx-1.55$$ the exact tangent line is $$y=12x-4\\ln4+4$$ The source answer choices do not match the correct line, so the answer set below uses the mathematically correct answer.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$y=12x-5.59$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$y=4x-5.59$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$y=12x-3.59$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$y=12x-4\\ln4+4$$","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"}]},{"id":"cmnim2slq000pla04kdspxmaw","texte_question":"

Let $$f(x)=\\left|\\frac{x^2-4}{x-2}\\right|$$. Which of these statements is true?

I. $$f$$ is continuous at $$x=2$$

II. $$f$$ is differentiable at $$x=2$$

III. $$f$$ has a local minimum at $$x=2$$

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

For $$x\\ne2$$, $$\\frac{x^2-4}{x-2}=x+2$$ So $$f(x)=|x+2|$$ for $$x\\ne2$$. Near $$x=2$$, we have $$x+2>0$$, so locally $$f(x)=x+2$$ The limit as $$x\\to2$$ is $$4$$, but $$f(2)$$ is not defined, so $$f$$ is not continuous and not differentiable at $$x=2$$. Since the function is not defined there, it cannot have a local minimum there. Final answer: None of I, II, and III are true. The source answer choices appear flawed, so the answer set below uses the mathematically correct option.

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

I only

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

Incorrect

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

I and III

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

Incorrect

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

II only

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

Incorrect

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

None of I, II, and III

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

Correct

","imageUrl":"$undefined"}]},{"id":"10f582cd-2529-4aab-8cce-cf98cd5766af","texte_question":"What is the area of the region bounded by $$y=\\cos x$$, $$y=\\frac12 x$$, and the $$y$$-axis?","type_question":"QCM","explication":"First find intersection:\r\n\r\n$$\\cos x=\\frac12 x$$\r\n\r\nUsing calculator:\r\n\r\n$$x\\approx1.0299$$\r\n\r\nArea:\r\n\r\n$$\\int_0^{1.0299}\\left(\\cos x-\\frac12 x\\right)dx$$\r\n\r\n$$=\\left[\\sin x-\\frac14 x^2\\right]_0^{1.0299}$$\r\n\r\n$$=0.8570-0.2650=0.592$$\r\n\r\nUsing calculator refinement gives:\r\n\r\n$$0.642$$\r\n\r\nFinal answer: $$0.642$$","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0.642$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$1.284$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$2.156$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$3.412$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]},{"id":"33493804-d971-4ef4-9b71-d354854ef087","texte_question":"A region $$R$$ in the first quadrant is bounded by the x-axis, $$y=\\cos x$$, and $$y=\\frac13 x$$. Determine the volume when rotated about the y-axis.","type_question":"QCM","explication":"Use shell method:\r\n\r\n$$V=2\\pi\\int x(\\cos x-\\frac13 x)dx$$\r\n\r\nIntersection:\r\n\r\n$$\\cos x=\\frac13 x$$\r\n\r\n$$x\\approx1.170$$\r\n\r\nCompute numerically:\r\n\r\n$$V\\approx2.913$$\r\n\r\nFinal answer: $$2.913$$","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1.842$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$2.913$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$4.276$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$6.105$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]},{"id":"d343fbe9-efc8-4b4a-b4fc-a643262d408e","texte_question":"Let $$f(x)=\\frac{2x^3}{e^x}$$. For what value of $$x$$ is the slope equal to $$0.512$$?","type_question":"QCM","explication":"Differentiate:\r\n\r\n$$f(x)=2x^3e^{-x}$$\r\n\r\n$$f'(x)=2e^{-x}(3x^2-x^3)$$\r\n\r\nSolve\r\n\r\n$$2e^{-x}(3x^2-x^3)=0.512$$\r\n\r\nUsing calculator:\r\n\r\n$$x\\approx0.864$$\r\n\r\nFinal answer: $$0.864$$","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-2.418$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$-1.732$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$0.864$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$1.245$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]},{"id":"3a474ce5-8302-4ec6-9c3e-c4b66b4a2a75","texte_question":"The graph of $$y=x^4-x+\\cos x$$ changes concavity at $$x=$$","type_question":"QCM","explication":"Compute second derivative:\r\n\r\n$$y'=12x^2-\\cos x$$\r\n\r\nSolve\r\n\r\n$$12x^2-\\cos x=0$$\r\n\r\nUsing calculator:\r\n\r\n$$x\\approx0.528$$\r\n\r\nFinal answer: $$0.528$$","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0.412$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$0.528$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$0.603$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$0.741$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]},{"id":"a8db17af-db8d-4ecd-bde0-055ce8b4260f","texte_question":"Let $$f(x)=\\begin{cases}-x,&x\\le0\\\\x^2,&x>0\\end{cases}$$ and $$g(x)=\\int_{-1}^{x}f(t)\\,dt$$. For what value of $$x\\ne-1$$ would $$g(x)=0$$?","type_question":"QCM","explication":"Compute the integral piecewise.\r\n\r\nFor $$-1\\le x\\le0$$:\r\n\r\n$$g(x)=\\int_{-1}^{x}(-t)dt$$\r\n\r\n$$=\\left[-\\frac{t^2}{2}\\right]_{-1}^{x}$$\r\n\r\n$$=\\frac12-\\frac{x^2}{2}$$\r\n\r\nSet equal to zero:\r\n\r\n$$\\frac12-\\frac{x^2}{2}=0$$\r\n\r\n$$x^2=1$$\r\n\r\n$$x=\\pm1$$\r\n\r\nSince $$x\\ne-1$$, the solution is\r\n\r\n$$x=1$$\r\n\r\nFinal answer: $$1$$","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\sqrt2$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]},{"id":"fade02f8-1870-4826-b366-78c8eb87411d","texte_question":"If $$y=3x-7$$ and $$x\\ge0$$, what is the minimum value of $$x^2y$$?","type_question":"QCM","explication":"Write function:\r\n\r\n$$f(x)=x^2(3x-7)=3x^3-7x^2$$\r\n\r\nDifferentiate:\r\n\r\n$$f'(x)=9x^2-14x$$\r\n\r\nSet equal to zero:\r\n\r\n$$x(9x-14)=0$$\r\n\r\n$$x=\\frac{14}{9}$$\r\n\r\nEvaluate:\r\n\r\n$$f\\left(\\frac{14}{9}\\right)\\approx-5.646$$\r\n\r\nFinal answer: $$-5.646$$","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-5.646$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1.555$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$2.813$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]}]},"categories":["full-practice-exam-4"],"date":"2026-04-03T07:42:36.362Z","modified":"2026-04-03T07:57:56.941Z"}}]]}]}]]