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

Multiple choice non-calculator section

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Multiple choice non-calculator section

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AP Calculus BC Full Practice Exam 4 Section I Part A

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

Multiple choice non-calculator section

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":60,"niveau_difficulte":"$undefined","categorie":"Full Practice Exam 4","nombre_questions":30,"score_minimum":80,"ordre_questions":"Fixe","questions":[{"id":"cmnhlz1mr0000jy04iza56toe","texte_question":"

Evaluate $$\\int_0^1 e^{3x}\\,dx$$.

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

Compute the antiderivative: $$\\int e^{3x}\\,dx=\\frac{1}{3}e^{3x}+C$$ Now evaluate from $$0$$ to $$1$$: $$\\int_0^1 e^{3x}\\,dx=\\left[\\frac{e^{3x}}{3}\\right]_0^1=\\frac{e^3-1}{3}$$ Final answer: $$\\frac{e^3-1}{3}$$

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

Correct

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms0005jy047psupzms","texte_question":"

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

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

Use the chain rule repeatedly. Outer derivative: $$\\frac{d}{dx}\\tan(u)=\\sec^2(u)\\cdot u'$$ Here $$u=e^{\\cos x}$$, so $$u'=e^{\\cos x}\\cdot(-\\sin x)$$ Therefore, $$f'(x)=-e^{\\cos x}\\sin x\\sec^2(e^{\\cos x})$$ Final answer: $$-e^{\\cos x}\\sin x\\sec^2(e^{\\cos x})$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms000ajy045qdar1gb","texte_question":"

If $$F(x)=\\int_1^{x^3} t^2\\,dt$$, then $$F'(x)=$$

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

By the Fundamental Theorem of Calculus with chain rule, $$F'(x)=\\left((x^3)^2\\right)\\cdot\\frac{d}{dx}(x^3)=x^6\\cdot 3x^2=3x^8$$ Final answer: $$3x^8$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$3x^2(x^3)^2$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$3x^2$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$x^6$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$3x^8$$","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms000fjy04dqc1lwst","texte_question":"

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

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

Differentiate: $$f'(x)=2\\tan x\\sec^2 x-\\sin x$$ Evaluate at $$x=\\frac{\\pi}{4}$$: $$\\tan\\frac{\\pi}{4}=1,\\quad \\sec^2\\frac{\\pi}{4}=2,\\quad \\sin\\frac{\\pi}{4}=\\frac{\\sqrt2}{2}$$ So $$f'\\left(\\frac{\\pi}{4}\\right)=2(1)(2)-\\frac{\\sqrt2}{2}=4-\\frac{\\sqrt2}{2}$$ Final answer: $$4-\\frac{\\sqrt2}{2}$$ Note: the source answer choices appear inconsistent with the actual derivative, so the answer set was corrected.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$4-\\frac{\\sqrt2}{2}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$4-\\frac{\\sqrt2}{2}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$1+\\frac{\\sqrt2}{2}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1-\\frac{\\sqrt2}{2}$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms000kjy04weze9w6o","texte_question":"

At which point is the graph of $$f(x)=x^4-4x^3+3x^2+2$$ decreasing and concave down?

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

Find the first and second derivatives: $$f'(x)=4x^3-12x^2+6x$$ $$f''(x)=12x^2-24x+6$$ Check the given $$x$$-values. At $$x=0$$: $$f'(0)=0$$ so not decreasing. At $$x=1$$: $$f'(1)=4-12+6=-2<0$$ and $$f''(1)=12-24+6=-6<0$$ So the graph is decreasing and concave down at $$x=1$$. Since $$f(1)=1-4+3+2=2$$, the point is $$(1,2)$$. Final answer: $$(1,2)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(0,2)$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$(1,2)$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$(2,2)$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$(3,2)$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms000pjy04dx7lvp4w","texte_question":"

Which of the following are antiderivatives of $$f(x)=\\sin x\\cos^2 x$$? I. $$F(x)=-\\frac{\\cos^3 x}{3}$$ II. $$F(x)=\\frac{\\sin^2 x}{2}-\\frac{\\sin^4 x}{4}$$ III. $$F(x)=\\frac{1-\\cos^3 x}{3}$$

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

Check each candidate by differentiating. I. $$\\frac{d}{dx}\\left(-\\frac{\\cos^3 x}{3}\\right)=-\\frac13\\cdot 3\\cos^2 x(-\\sin x)=\\sin x\\cos^2 x$$ So I works. II. $$\\frac{d}{dx}\\left(\\frac{\\sin^2 x}{2}-\\frac{\\sin^4 x}{4}\\right)=\\sin x\\cos x-\\sin^3 x\\cos x$$ $$=\\sin x\\cos x(1-\\sin^2 x)=\\sin x\\cos^3 x$$ So II does not work. III. $$\\frac{d}{dx}\\left(\\frac{1-\\cos^3 x}{3}\\right)=\\frac13\\cdot 3\\cos^2 x\\sin x=\\sin x\\cos^2 x$$ So III works. Final answer: I 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":"$undefined"},{"texte":"I and III","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"I, II, and III","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms000ujy04pedpjb61","texte_question":"

Determine $$\\frac{dy}{dx}$$ for the curve defined by $$x^2+y^3=3xy$$.

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

Differentiate implicitly: $$2x+3y^2y'=3(xy'+y)$$ Rearrange: $$3y^2y'-3xy'=3y-2x$$ Factor: $$y'(3y^2-3x)=3y-2x$$ So $$\\frac{dy}{dx}=\\frac{3y-2x}{3y^2-3x}$$ Final answer: $$\\frac{3y-2x}{3y^2-3x}$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms000zjy04qhlednz1","texte_question":"

Evaluate $$\\int_0^{\\pi/6}\\cos 2x\\,dx$$.

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

Compute the antiderivative: $$\\int \\cos 2x\\,dx=\\frac{1}{2}\\sin 2x+C$$ Now evaluate: $$\\int_0^{\\pi/6}\\cos 2x\\,dx=\\left[\\frac{1}{2}\\sin 2x\\right]_0^{\\pi/6}=\\frac12\\sin\\frac{\\pi}{3}=\\frac12\\cdot\\frac{\\sqrt3}{2}=\\frac{\\sqrt3}{4}$$ Final answer: $$\\frac{\\sqrt3}{4}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac12$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac14$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac{\\sqrt3}{4}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\frac{\\sqrt3}{2}$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms0014jy043bs8x2kk","texte_question":"

Find the average value of $$f(x)=\\cos 3x$$ over $$\\left[0,\\frac{\\pi}{3}\\right]$$.

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

Average value on $$[a,b]$$ is $$\\frac{1}{b-a}\\int_a^b f(x)\\,dx$$ So $$\\frac{1}{\\pi/3}\\int_0^{\\pi/3}\\cos 3x\\,dx$$ Compute the integral: $$\\int \\cos 3x\\,dx=\\frac13\\sin 3x$$ Thus $$\\int_0^{\\pi/3}\\cos 3x\\,dx=\\left[\\frac13\\sin 3x\\right]_0^{\\pi/3}=0$$ Therefore the average value is $$0$$ Final answer: $$0$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\frac13$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac23$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{\\pi}$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms0019jy04ndd4ey7s","texte_question":"

Evaluate $$\\lim_{x\\to\\infty}\\frac{4x^2-5x+1}{2x^2+3x-7}$$.

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

Divide numerator and denominator by $$x^2$$: $$\\lim_{x\\to\\infty}\\frac{4-\\frac5x+\\frac1{x^2}}{2+\\frac3x-\\frac7{x^2}}$$ As $$x\\to\\infty$$ all fractional terms go to zero: $$=\\frac{4}{2}=2$$ Final answer: $$2$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\frac12$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms001ejy04fn1ogbec","texte_question":"

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

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

Use substitution: Let $$u=\\ln x$$ Then $$du=\\frac1x dx$$ Integral becomes $$\\int_0^1 u\\,du=\\frac12u^2\\Big|_0^1=\\frac12$$ Final answer: $$\\frac12$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac12$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\frac13$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1ms001jjy04ie4a9jeq","texte_question":"

A particle's position is given by $$s(t)=\\cos t+t$$. The average velocity over $$[0,\\pi]$$ is

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

Average velocity: $$\\frac{s(\\pi)-s(0)}{\\pi}$$ Compute: $$s(\\pi)=\\cos\\pi+\\pi=-1+\\pi$$ $$s(0)=1$$ So $$\\frac{(-1+\\pi)-1}{\\pi}=\\frac{\\pi-2}{\\pi}$$ Final answer: $$\\frac{\\pi-2}{\\pi}$$

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

$$\\frac1\\pi$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmnhlz1mt001ojy047uo6zd6x","texte_question":"

If $$f(x)=\\begin{cases} e^{2x}, & x<\\ln3 \\\\ 3, & x\\ge\\ln3 \\end{cases}$$ then $$\\lim_{x\\to\\ln3} f(x)=$$

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

Left limit: $$\\lim_{x\\to\\ln3^-} e^{2x}=e^{2\\ln3}=3^2=9$$ Right limit: $$\\lim_{x\\to\\ln3^+} 3=3$$ Since limits are not equal, the limit does not exist. Final answer: Limit does not exist

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$3$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\ln3$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Limit does not exist","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"}]},{"id":"b4bb980b-9c7f-4284-9fba-7d88536fdb4b","texte_question":"Evaluate $$\\lim_{h\\to0}\\frac{\\sin\\left(\\frac{\\pi}{6}+h\\right)-\\sin\\frac{\\pi}{6}}{h}$$.","type_question":"QCM","explication":"This is the definition of the derivative of $$\\sin x$$ at $$x=\\frac{\\pi}{6}$$:\r\n\r\n$$\\lim_{h\\to0}\\frac{\\sin\\left(\\frac{\\pi}{6}+h\\right)-\\sin\\frac{\\pi}{6}}{h}=\\cos\\frac{\\pi}{6}$$\r\n\r\nSince\r\n\r\n$$\\cos\\frac{\\pi}{6}=\\frac{\\sqrt3}{2}$$\r\n\r\nFinal answer: $$\\frac{\\sqrt3}{2}$$","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{\\sqrt3}{2}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\frac12$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\sqrt3$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]},{"id":"a1584131-b41c-447e-9396-e4d04f7956ed","texte_question":"The graph of $$f(x)=(x-2)^3(x+1)^2$$ has a local minimum at $$x=$$","type_question":"QCM","explication":"Differentiate:\r\n\r\n$$f'(x)=3(x-2)^2(x+1)^2+2(x-2)^3(x+1)$$\r\n\r\nFactor:\r\n\r\n$$f'(x)=(x-2)^2(x+1)(5x-1)$$\r\n\r\nCritical points are\r\n\r\n$$x=2,\\ -1,\\ \\frac15$$\r\n\r\nSince $$(x-2)^2\\ge0$$, the sign of $$f'(x)$$ is determined by $$(x+1)(5x-1)$$.\r\n\r\nAround $$x=-1$$, the derivative changes from positive to negative, so that is a local maximum.\r\n\r\nAround $$x=\\frac15$$, the derivative changes from negative to positive, so that is a local minimum.\r\n\r\nFinal answer: $$\\frac15$$\r\n\r\nNote: the source answer choices did not include the correct value, so the answer set was corrected.","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-1$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac15$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"}]},{"id":"1d955c55-92c3-4597-9a98-10a9ae3d27a7","texte_question":"Let $$f$$ and $$g$$ be twice differentiable functions such that $$f'(x)<0$$ for all $$x$$. If $$h(x)=f(g(x))$$ and $$h'(2)>0$$, then at $$x=2$$","type_question":"QCM","explication":"Since $$f'(x)<0$$ for all $$x$$, the function $$f$$ is concave down everywhere.\r\n\r\nTherefore the true statement is about $$f$$, independent of additional information about $$g$$ or $$h$$.\r\n\r\nFinal answer: $$f$$ is concave down","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$g$$ is increasing","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$g$$ is decreasing","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$f$$ is concave down","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$g$$ is concave up","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]},{"id":"3a60728d-02ad-4d4c-b08c-8c3add3acaeb","texte_question":"What is the area of the region bounded by the curves $$y=x^2+1$$ and $$y=x$$ from $$x=0$$ to $$x=2$$?","type_question":"QCM","explication":"On $$[0,2]$$,\r\n\r\n$$x^2+1-x=x^2-x+1>0$$\r\n\r\nso the area is\r\n\r\n$$\\int_0^2\\big((x^2+1)-x\\big)\\,dx$$\r\n\r\n$$=\\int_0^2(x^2-x+1)\\,dx$$\r\n\r\n$$=\\left[\\frac{x^3}{3}-\\frac{x^2}{2}+x\\right]_0^2$$\r\n\r\n$$=\\frac{8}{3}-2+2=\\frac{8}{3}$$\r\n\r\nFinal answer: $$\\frac{8}{3}$$","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{4}{3}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac{8}{3}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac{10}{3}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"}]}]},"categories":["full-practice-exam-4"],"date":"2026-04-02T14:41:47.789Z","modified":"2026-04-02T15:07:15.843Z"}}]]}]}]]