7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Section I Part B\",\"description\":\"Multiple choice calculator section\",\"url\":\"https://theschoolofmathematics.com/quiz/ap-calculus-ab-full-practice-exam-2-section-i-part-b\",\"timeRequired\":\"PT45M\",\"numberOfQuestions\":15,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Full Practice Exam 2\"}}"}}],["$","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\":\"Section I Part B\",\"item\":\"https://theschoolofmathematics.com/quiz/ap-calculus-ab-full-practice-exam-2-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 AB Full Practice Exam 2 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 2","nombre_questions":15,"score_minimum":80,"ordre_questions":"Fixe","questions":[{"id":"cmpz9rcau09lsjgll0tuetwtk","texte_question":"

Let $$f$$ be a function such that the rate of change of the derivative of $$f$$ is constant and equal to $$3$$ for all $$x$$. If $$f'(1)=2$$ and $$f(1)=4$$, find $$f(3)$$.

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

The rate of change of the derivative is constant and equal to $$3$$, so $$f''(x)=3$$ Integrate: $$f'(x)=3x+C$$ Use $$f'(1)=2$$: $$3(1)+C=2$$ $$C=-1$$ So $$f'(x)=3x-1$$ Integrate again: $$f(x)=\\frac{3}{2}x^2-x+C$$ Use $$f(1)=4$$: $$\\frac{3}{2}-1+C=4$$ $$\\frac{1}{2}+C=4$$ $$C=\\frac{7}{2}$$ Thus $$f(x)=\\frac{3}{2}x^2-x+\\frac{7}{2}$$ Evaluate at $$x=3$$: $$f(3)=\\frac{3}{2}(9)-3+\\frac{7}{2}$$ $$=\\frac{27}{2}-3+\\frac{7}{2}=17-3=14$$ Final answer: $$14$$ Note: the source answer choices did not include the correct value, so the answer set was corrected.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$10$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$12$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$13$$","correcte":false,"explication":"

Incorrect

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

$$14$$

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

Correct

","imageUrl":"$undefined"}]},{"id":"cmpz9rcax09lyjgll8vqbwdiq","texte_question":"

Evaluate $$\\lim_{x\\to a}\\frac{x-a}{x^2-a^2}$$.

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

Factor the denominator: $$x^2-a^2=(x-a)(x+a)$$ So for $$x\\ne a$$, $$\\frac{x-a}{x^2-a^2}=\\frac{1}{x+a}$$ Now take the limit: $$\\lim_{x\\to a}\\frac{1}{x+a}=\\frac{1}{2a}$$ Final answer: $$\\frac{1}{2a}$$

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

Incorrect

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

$$0$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcaz09m4jgllcxc72d4s","texte_question":"

Population $$y$$ grows according to the equation $$\\frac{dy}{dt}=ky$$ where $$k$$ is a constant and $$t$$ is measured in years. If the population doubles every $$4$$ years, then $$k=$$

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

The general solution is $$y=Ce^{kt}$$ Doubling every $$4$$ years means $$e^{4k}=2$$ Take the natural logarithm of both sides: $$4k=\\ln 2$$ $$k=\\frac{\\ln 2}{4}$$ Using $$\\ln 2\\approx0.693$$: $$k\\approx\\frac{0.693}{4}\\approx0.173$$ Final answer: $$0.173$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0.173$$","correcte":true,"explication":"

Correct

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

$$0.250$$

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

Incorrect

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

$$0.346$$

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

Incorrect

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

$$0.693$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcb109majgllexphdanr","texte_question":"

The radius of a circle is increasing at a rate of $$3$$ inches per minute. When the radius is $$5$$ inches, how fast is the area increasing in square inches per minute?

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

Area of a circle is $$A=\\pi r^2$$ Differentiate with respect to time: $$\\frac{dA}{dt}=2\\pi r\\frac{dr}{dt}$$ Substitute $$r=5$$ and $$\\frac{dr}{dt}=3$$: $$\\frac{dA}{dt}=2\\pi(5)(3)=30\\pi$$ Final answer: $$30\\pi$$

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

$$15\\pi$$

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

Incorrect

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

$$20\\pi$$

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

Incorrect

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

$$30\\pi$$

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

Correct

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

$$45\\pi$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcb409mgjglln43n8d26","texte_question":"

Let $$f(x)=\\int_0^{3x}e^t\\,dt$$. What value on $$[0,2]$$ satisfies the Mean Value Theorem for $$f$$?

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

First evaluate $$f(x)$$. By the Fundamental Theorem of Calculus, $$f(x)=\\int_0^{3x}e^t\\,dt=e^{3x}-1$$ Differentiate: $$f'(x)=3e^{3x}$$ By the Mean Value Theorem, there exists $$c$$ in $$(0,2)$$ such that $$f'(c)=\\frac{f(2)-f(0)}{2-0}$$ Compute the average rate of change: $$f(2)=e^6-1$$ $$f(0)=0$$ $$\\frac{f(2)-f(0)}{2}=\\frac{e^6-1}{2}$$ Set equal: $$3e^{3c}=\\frac{e^6-1}{2}$$ $$e^{3c}=\\frac{e^6-1}{6}$$ Take natural logarithms: $$3c=\\ln\\left(\\frac{e^6-1}{6}\\right)$$ $$c=\\frac{1}{3}\\ln\\left(\\frac{e^6-1}{6}\\right)$$ Using a calculator, $$c\\approx1.402$$ The source answer choices did not include the exact value, so the closest corrected choice is used. Final answer: $$1.402$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1.210$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1.402$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$1.732$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$1.910$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcb609mmjgllr7nqldm3","texte_question":"

The position for a particle is $$s(t)=-2t^3+3t^2+1$$. At what time $$t$$ on $$[0,2]$$ is the instantaneous velocity equal to the average velocity?

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

Average velocity on $$[0,2]$$ is $$\\frac{s(2)-s(0)}{2-0}$$ Compute: $$s(2)=-2(8)+3(4)+1=-16+12+1=-3$$ $$s(0)=1$$ So $$\\text{average velocity}=\\frac{-3-1}{2}=-2$$ Instantaneous velocity: $$v(t)=s'(t)=-6t^2+6t$$ Set equal to $$-2$$: $$-6t^2+6t=-2$$ $$6t^2-6t-2=0$$ $$3t^2-3t-1=0$$ Using the quadratic formula: $$t=\\frac{3\\pm\\sqrt{21}}{6}$$ The value in $$[0,2]$$ is $$t=\\frac{3+\\sqrt{21}}{6}\\approx1.264$$ Final answer: $$1.264$$ Note: the source answer choices did not include the exact correct value, so the answer set was corrected.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0.422$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$0.667$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1.215$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$1.264$$

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

Correct

","imageUrl":"$undefined"}]},{"id":"cmpz9rcb909msjgllhdda3v9l","texte_question":"

Let $$f(x)=x^2+2x+1$$. Using the trapezoidal rule with $$n=4$$, approximate $$\\int_0^2 f(x)\\,dx$$.

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

For $$[0,2]$$ with $$n=4$$, $$\\Delta x=\\frac{2-0}{4}=0.5$$ The partition points are $$x_0=0,\\ x_1=0.5,\\ x_2=1,\\ x_3=1.5,\\ x_4=2$$ Evaluate the function: $$f(0)=1$$ $$f(0.5)=0.25+1+1=2.25$$ $$f(1)=1+2+1=4$$ $$f(1.5)=2.25+3+1=6.25$$ $$f(2)=4+4+1=9$$ Apply the trapezoidal rule: $$T_4=\\frac{\\Delta x}{2}\\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)\\right]$$ $$T_4=\\frac{0.5}{2}\\left[1+2(2.25)+2(4)+2(6.25)+9\\right]$$ $$T_4=0.25\\left[1+4.5+8+12.5+9\\right]$$ $$T_4=0.25(35)=8.75$$ Final answer: $$8.75$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$8.25$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$8.50$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$8.75$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

$$9.00$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcbb09myjgllkct2013m","texte_question":"

Let $$f(x)=4\\ln(3x)$$ and $$g(x)=x^3+x$$. At what value of $$x$$ do the graphs have parallel tangent lines?

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

Differentiate: $$f'(x)=\\frac{4}{x}$$ $$g'(x)=3x^2+1$$ Set equal: $$\\frac{4}{x}=3x^2+1$$ Multiply: $$4=3x^3+x$$ Rearrange: $$3x^3+x-4=0$$ Test $$x=1$$: $$3(1)^3+1-4=0$$ Thus: $$x=1$$ Final answer: $$1$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0.412$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$0.693$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

$$0.882$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcbe09n4jgllot9izt57","texte_question":"

The derivative of a function is $$f'(x)=\\frac{e^{-x}}{x}-\\cos x$$. How many critical values does $$f$$ have on $$(0,8)$$?

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

Critical points occur where: $$f'(x)=0$$ Solve: $$\\frac{e^{-x}}{x}-\\cos x=0$$ $$\\frac{e^{-x}}{x}=\\cos x$$ This equation must be solved graphically. On $$(0,8)$$: - $$\\frac{e^{-x}}{x}$$ decreases - $$\\cos x$$ oscillates They intersect multiple times. Checking graphically gives: $$3 \\text{ solutions}$$ Final answer: $$3$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"One","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Two","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Three","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

Four

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcbg09najglllzxamz4r","texte_question":"

Evaluate $$\\frac{d}{dx}\\left(\\int_x^4 f'(t)\\,dt\\right)$$

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

By the Fundamental Theorem of Calculus: $$\\frac{d}{dx}\\int_x^4 f'(t)\\,dt=-f'(x)$$ Final answer: $$-f'(x)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$f'(x)$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$-f'(x)$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$f'(4)$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$-f'(4)$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcbj09ngjgllonfbs4ur","texte_question":"

Let

$$f(x)=\\sqrt{x}+k$$, $$x$$<$$1$$.

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

For what value of $$k$$ is $$f$$ continuous at $$x=1$$?

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

Continuity requires:

$$\\lim_{x\\to1^-}f(x)=\\lim_{x\\to1^+}f(x)$$

Left side:

$$\\sqrt{1}+k=1+k$$

Right side:

$$\\ln(2)$$

Set equal:

$$1+k=\\ln2$$

$$k=\\ln2-1$$

Final answer: $$\\ln2-1$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-1$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\ln2-1$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

$$1$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcbl09nmjgll6dr9noja","texte_question":"

Let $$f(x)=x^2+2x-3$$. The tangent line to the graph at $$x=1$$ is used to approximate values of $$f$$. For what value(s) of $$x$$ is the tangent line approximation twice that of $$f$$?

I. $$-1$$

II. $$0$$

III. $$2$$

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

First find the tangent line at $$x=1$$. $$f(1)=1+2-3=0$$ $$f'(x)=2x+2$$ $$f'(1)=4$$ So the tangent line is $$L(x)=f(1)+f'(1)(x-1)=4(x-1)=4x-4$$ We want the tangent line approximation to be twice the function value: $$L(x)=2f(x)$$ Substitute: $$4x-4=2(x^2+2x-3)$$ $$4x-4=2x^2+4x-6$$ $$0=2x^2-2$$ $$x^2=1$$ $$x=\\pm1$$ From the listed values, this means I. $$-1$$ works II. $$0$$ does not work III. $$2$$ does not work Final answer: I only

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

I and III

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcbp09nsjgllzfjmuze3","texte_question":"$18","type_question":"QCM","explication":"

To compare function values, use accumulated area under the graph of $$f''$$. From the graph: - On $$[0,2]$$, the net area is slightly negative, so $$f(2)f(0)$$. - Therefore the ordering is $$f(2)

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$f(0)$$f(2)

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

Correct

","imageUrl":"$undefined"}]},{"id":"cmpz9rcc109nyjgllxwufypxg","texte_question":"$19","type_question":"QCM","explication":"

By the Fundamental Theorem of Calculus, $$g''(x)=f(x)$$ A relative minimum of $$g$$ occurs where - $$g''(x)=0$$ - and $$g''(x)$$ changes from negative to positive Since $$g''(x)=f(x)$$, we look for where the graph of $$f$$ crosses the x-axis from below to above. From the graph: - At $$a$$, $$f$$ changes from positive to negative, so $$g$$ has a relative maximum there. - At $$c$$, $$f$$ changes from negative to positive, so $$g$$ has a relative minimum there. - At $$d$$, the graph ends at the axis and there is no change from negative to positive followed by continuation. Final answer: $$c$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$a$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$b$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$c$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

$$d$$

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

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9rcca09o4jgllpvrs46da","texte_question":"$1a","type_question":"QCM","explication":"

Check each statement from the graph. (A) At $$x=a$$, the graph is continuous but has a sharp corner, so it is not differentiable there. This is true. (B) At $$x=a$$, the left-hand and right-hand derivatives are different because of the corner. This is true. (C) At $$x=a$$, the function is defined, but at $$x=c$$ there is an open circle, so $$f(c)$$ is not defined. This is true. (D) At $$x=b$$, the graph has a smooth local minimum, so $$f''(b)=0$$. This is true. (E) The graph approaches the same open-circle value from both sides of $$x=c$$, so $$\\lim_{x\\to c^-}f(x)=\\lim_{x\\to c^+}f(x)$$ Therefore statement (E), which says those two one-sided limits are not equal, is false. Final answer: (E)

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$f$$ is continuous but not differentiable at $$x=a$$.","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"The left-hand and right-hand derivatives at $$x=a$$ are not equal.","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$f(a)$$ is defined, but $$f(c)$$ is not.","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$\\lim_{x\\to c^-}f(x)\\ne\\lim_{x\\to c^+}f(x)$$

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

Correct

","imageUrl":"$undefined"}]}]},"categories":["Full Practice Exam 2"],"date":"2026-06-04T09:04:36.867Z","modified":"2026-06-04T09:04:36.867Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L1b",null,{"targetType":"quiz","targetSlug":"ap-calculus-ab-full-practice-exam-2-section-i-part-b"}]}]]}]}]]

Section I Part B

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