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-precalculus-full-practice-exam-1-section-i-part-b\",\"timeRequired\":\"PT40M\",\"numberOfQuestions\":12,\"about\":{\"@type\":\"Thing\",\"name\":\"Full Practice Exam 1\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://www.theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://www.theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Section I Part B\",\"item\":\"https://www.theschoolofmathematics.com/quiz/ap-precalculus-full-practice-exam-1-section-i-part-b\"}]}"}}],["$","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 Precalculus Full Practice Exam 1 Section I Part B

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

Multiple-choice calculator section

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":40,"niveau_difficulte":"$undefined","categorie":"Full Practice Exam 1","nombre_questions":12,"score_minimum":80,"ordre_questions":"Aleatoire","questions":[{"id":"cmo1c764800ygjggb6e5ve6fi","texte_question":"

Given the function $$f(x)=(x^2-4)(x-1)(2x-3)$$, at what value of $$x$$ does the absolute maximum of $$f(x)$$ occur on the interval $$[-2,3]$$?

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

Check the answer choices by evaluating the function. At $$x=-2$$: $$f(-2)=(4-4)(-3)(-7)=0.$$ At $$x=-1$$: $$f(-1)=(1-4)(-2)(-5)=-30.$$ At $$x=1$$: $$f(1)=(1-4)(0)(-1)=0.$$ At $$x=3$$: $$f(3)=(9-4)(2)(3)=30.$$ The largest value is $$30$$. Final answer: $$x=3$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-2$$","correcte":false,"explication":"At $$x=-2$$ the function value is $$0$$, not the maximum.","imageUrl":"$undefined"},{"texte":"$$$-1$$","correcte":false,"explication":"At $$x=-1$$ the function value is negative.","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"At $$x=1$$ the function value is $$0$$.","imageUrl":"$undefined"},{"texte":"

$$3$$

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

Correct because this gives the largest function value among the listed choices.

","imageUrl":"$undefined"}]},{"id":"cmo1c764800yljggbg9alijub","texte_question":"

Approximate $$\\log_a 18$$, given that $$\\log_a 2\\approx 0.4307$$ and $$\\log_a 3\\approx 0.6826$$.

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

Factor $$18$$: $$18=2cdot 3^2.$$ Use logarithm properties: $$log_a 18=log_a 2+log_a 3^2.$$ $$log_a 18=log_a 2+2log_a 3.$$ Substitute the given values: $$log_a 18approx 0.4307+2(0.6826).$$ $$log_a 18approx 0.4307+1.3652=1.7959.$$ Final answer: $$1.7959$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1.1133$$","correcte":false,"explication":"This does not include the full contribution of $$2\\log_a 3$$.","imageUrl":"$undefined"},{"texte":"$$$1.7959$$","correcte":true,"explication":"Correct approximation.","imageUrl":"$undefined"},{"texte":"$$$2.0459$$","correcte":false,"explication":"This is too large.","imageUrl":"$undefined"},{"texte":"

$$0.6826$$

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

This is only $$\\log_a 3$$.

","imageUrl":"$undefined"}]},{"id":"cmo1c764800yqjggbzik6rhjo","texte_question":"

Solve for $$x$$: $$2^{x+1}=4^{x-2}$$.

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

Rewrite $$4$$ as $$2^2$$: $$2^{x+1}=(2^2)^{x-2}=2^{2x-4}.$$ Since the bases are equal, set exponents equal: $$x+1=2x-4.$$ Solve: $$1=x-4$$ $$x=5.$$ Final answer: $$5$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2$$","correcte":false,"explication":"This does not satisfy the exponential equation.","imageUrl":"$undefined"},{"texte":"$$$3$$","correcte":false,"explication":"This is too small.","imageUrl":"$undefined"},{"texte":"$$$4$$","correcte":false,"explication":"This does not make the exponents equal.","imageUrl":"$undefined"},{"texte":"

$$5$$

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

Correct solution after rewriting with a common base.

","imageUrl":"$undefined"}]},{"id":"cmo1c764900yvjggb4rt6cao5","texte_question":"

Which of the following choices is equivalent to the complex number $$2-2\\sqrt{3}i$$?

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

Find the modulus: $$r=sqrt{2^2+(-2sqrt{3})^2}=sqrt{4+12}=4.$$ Find the argument: $$\tan\theta=\frac{-2sqrt{3}}{2}=-sqrt{3}.$$ The point is in Quadrant IV, so: $$\theta=-\frac{pi}{3}.$$ Therefore the polar form is: $$4left(cosleft(-\frac{pi}{3} ight)+isinleft(-\frac{pi}{3} ight) ight).$$ Final answer: $$4left(cosleft(-\frac{pi}{3} ight)+isinleft(-\frac{pi}{3} ight) ight)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$4\\left(\\cos\\frac{\\pi}{3}+i\\sin\\frac{\\pi}{3}\\right)$$","correcte":false,"explication":"This gives a positive imaginary part.","imageUrl":"$undefined"},{"texte":"$$$4\\left(\\cos\\left(-\\frac{\\pi}{3}\\right)+i\\sin\\left(-\\frac{\\pi}{3}\\right)\\right)$$","correcte":true,"explication":"Correct polar form.","imageUrl":"$undefined"},{"texte":"$$$2\\left(\\cos\\frac{\\pi}{6}+i\\sin\\frac{\\pi}{6}\\right)$$","correcte":false,"explication":"The modulus is incorrect.","imageUrl":"$undefined"},{"texte":"

$$2\\left(\\cos\\left(-\\frac{\\pi}{6}\\right)+i\\sin\\left(-\\frac{\\pi}{6}\\right)\\right)$$

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

Both the modulus and angle are incorrect.

","imageUrl":"$undefined"}]},{"id":"cmo1c764900z0jggbr4id10q2","texte_question":"

Find all solutions in the interval $$[0,2\\pi]$$ for the trigonometric equation $$4\\cos^2 x-4=0$$.

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

Solve the equation: $$4cos^2 x-4=0.$$ Factor out $$4$$: $$cos^2 x-1=0.$$ Then: $$cos^2 x=1$$ $$cos x=pm 1.$$ On the interval $$[0,2pi]$$: $$cos x=1$$ at $$x=0,2pi$$ $$cos x=-1$$ at $$x=pi$$ Final answer: $$0,pi,2pi$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0,\\pi$$","correcte":false,"explication":"This misses the endpoint $$2\\pi$$.","imageUrl":"$undefined"},{"texte":"$$$\\frac{\\pi}{2},\\frac{3\\pi}{2}$$","correcte":false,"explication":"These make cosine equal to $$0$$.","imageUrl":"$undefined"},{"texte":"$$$0,\\frac{\\pi}{2},\\pi,\\frac{3\\pi}{2}$$","correcte":false,"explication":"This includes points where cosine is not $$\\pm1$$.","imageUrl":"$undefined"},{"texte":"

$$0,\\pi,2\\pi$$

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

Correct solutions on the closed interval.

","imageUrl":"$undefined"}]},{"id":"cmo1c764900z5jggbe6ca6s01","texte_question":"

Give an algebraic expression for $$\\sin(\\cos^{-1} x)$$.

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

Let: $$\theta=cos^{-1}x.$$ Then: $$cos\theta=x.$$ Since $$\theta$$ is in the range of arccos, $$0le\thetalepi$$. So $$sin\thetage 0$$. Use the identity: $$sin^2\theta+cos^2\theta=1.$$ Then: $$sin\theta=sqrt{1-cos^2\theta}=sqrt{1-x^2}.$$ Final answer: $$sqrt{1-x^2}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\sqrt{1-x^2}$$","correcte":true,"explication":"Correct algebraic expression.","imageUrl":"$undefined"},{"texte":"$$$-\\sqrt{1-x^2}$$","correcte":false,"explication":"Sine is nonnegative on the range of arccos.","imageUrl":"$undefined"},{"texte":"$$$1-x^2$$","correcte":false,"explication":"The square root is missing.","imageUrl":"$undefined"},{"texte":"

$$\\sqrt{1+x^2}$$

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

The identity is applied incorrectly.

","imageUrl":"$undefined"}]},{"id":"cmo1c764900zajggbfhr93xq3","texte_question":"

How long will it take for $$500$$ dollars to double in an investment when interest is compounded continuously at the rate of $$6\\%$$ per annum? Round to the nearest year.

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

Use the continuous compounding formula: $$A=Pe^{rt}.$$ To double the investment: $$2P=Pe^{0.06t}.$$ Divide by $$P$$: $$2=e^{0.06t}.$$ Take the natural logarithm: $$ln 2=0.06t.$$ Solve for $$t$$: $$t=\frac{ln 2}{0.06}approx rac{0.6931}{0.06}approx 11.55.$$ Round to the nearest year: $$12$$ years. Final answer: $$12$$ years

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$10$$ years","correcte":false,"explication":"This is too small.","imageUrl":"$undefined"},{"texte":"$$$11$$ years","correcte":false,"explication":"This is closer, but the nearest whole year is $$12$$.","imageUrl":"$undefined"},{"texte":"$$$12$$ years","correcte":true,"explication":"Correct rounded value.","imageUrl":"$undefined"},{"texte":"

$$13$$ years

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

This is too large.

","imageUrl":"$undefined"}]},{"id":"cmo1c764900zfjggbnl0stnsu","texte_question":"

An object is launched straight up from ground level with an initial velocity of $$40$$ ft/s. The height is given by $$h(t)=-16t^2+40t$$. At approximately what value of $$t$$ will the object be at a height of $$24$$ feet and moving downward?

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

Set the height equal to $$24$$: $$-16t^2+40t=24.$$ Move all terms to one side: $$-16t^2+40t-24=0.$$ Divide by $$-8$$: $$2t^2-5t+3=0.$$ Factor: $$(2t-3)(t-1)=0.$$ So: $$t=1$$ or $$t=\frac{3}{2}=1.5.$$ The object is moving downward after it reaches its maximum height. The later time is the downward time. Final answer: $$1.5$$ seconds

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0.8$$ seconds","correcte":false,"explication":"

This is before the object reaches $$24$$ feet.

","imageUrl":"$undefined"},{"texte":"$$$1.2$$ seconds","correcte":false,"explication":"This is not one of the solutions.","imageUrl":"$undefined"},{"texte":"$$$1.5$$ seconds","correcte":true,"explication":"

Correct downward time at height $$24$$ feet.

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

$$2.2$$ seconds

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

This is too large.

","imageUrl":"$undefined"}]},{"id":"cmo1c764900zkjggbauckfniy","texte_question":"

Suppose a ball is dropped from a height of $$10$$ feet. After each bounce, it reaches $$80\\%$$ of its previous height. How high will the ball rebound after the third bounce, to the nearest tenth?

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

This is a geometric pattern. Initial height: $$10$$ After the first bounce: $$10(0.8)=8.0$$ After the second bounce: $$8.0(0.8)=6.4$$ After the third bounce: $$6.4(0.8)=5.12$$ Round to the nearest tenth: $$5.1$$ feet Final answer: $$5.1$$ feet

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$4.1$$ feet","correcte":false,"explication":"This is too small for the third bounce height.","imageUrl":"$undefined"},{"texte":"$$$5.1$$ feet","correcte":true,"explication":"Correct rebound height after rounding.","imageUrl":"$undefined"},{"texte":"$$$6.4$$ feet","correcte":false,"explication":"This is the height after the second bounce.","imageUrl":"$undefined"},{"texte":"

$$8.0$$ feet

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

This is the height after the first bounce.

","imageUrl":"$undefined"}]},{"id":"cmo1c764a00zpjggbce7oumcu","texte_question":"

Let $$f(x)=-x^5+10x^4$$. For what values of $$x$$ does the graph of $$f$$ have a point of inflection?\\n\\nI. $$x=0$$\\nII. $$x=2$$\\nIII. $$x=6$$

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

Find the second derivative. First derivative: $$f'(x)=-5x^4+40x^3$$ Second derivative: $$f''(x)=-20x^3+120x^2=-20x^2(x-6).$$ Set the second derivative equal to zero: $$-20x^2(x-6)=0$$ Critical values for concavity: $$x=0, 6.$$ Now test for a sign change. For $$x<0$$: $$f''(x)>0$$ For $$00$$ For $$x>6$$: $$f''(x)<0$$ There is no sign change at $$x=0$$. There is a sign change at $$x=6$$. So the only point of inflection is: $$x=6$$ Final answer: III only

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"I only","correcte":false,"explication":"The concavity does not change at $$x=0$$.","imageUrl":"$undefined"},{"texte":"II only","correcte":false,"explication":"$$$x=2$$ is not a zero of the second derivative.","imageUrl":"$undefined"},{"texte":"III only","correcte":true,"explication":"Correct because concavity changes only at $$x=6$$.","imageUrl":"$undefined"},{"texte":"

I, II, and III

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

Only one listed value gives an inflection point.

","imageUrl":"$undefined"}]},{"id":"cmo1c764a00zujggbu6uq616o","texte_question":"

After performing analyses on a set of data, a student examined the scatterplot of residual values. Which scatterplot indicates the best linear fit?

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

For a good linear fit, the residuals should have no visible pattern. A best-fit residual plot should: Be randomly scattered Be centered around zero Show no curvature or trend Curved patterns, trends, or clustering suggest the linear model is missing structure in the data. Final answer: Residuals are randomly scattered around zero

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Residuals form a curved pattern","correcte":false,"explication":"A curved pattern suggests a nonlinear relationship.","imageUrl":"$undefined"},{"texte":"Residuals show increasing trend","correcte":false,"explication":"A trend means the linear model is biased across the data range.","imageUrl":"$undefined"},{"texte":"Residuals are randomly scattered around zero","correcte":true,"explication":"Correct indicator of a good linear fit.","imageUrl":"$undefined"},{"texte":"

Residuals show strong clustering above axis

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

Clustering above zero suggests systematic error.

","imageUrl":"$undefined"}]},{"id":"cmo1c764a00zzjggbj77sxy5u","texte_question":"$18","type_question":"QCM","explication":"

Use the average rate of change formula: $$\frac{f(b)-f(a)}{b-a}.$$ Here: $$a=0,quad f(0)=0$$ $$b=4,quad f(4)=16$$ Substitute: $$\frac{16-0}{4-0}=\frac{16}{4}=4.$$ Final answer: $$4$$

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

","imageUrl":"$undefined"},{"texte":"$$$4$$","correcte":true,"explication":"

","imageUrl":"$undefined"},{"texte":"$$$8$$","correcte":false,"explication":"

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

$$16$$

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

","imageUrl":"$undefined"}]}]},"categories":["Full-Practice-Exam-1"],"date":"2026-04-16T02:42:10.934Z","modified":"2026-04-16T10:29:02.263Z"}}]]}]}]]