7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Section I Part B\",\"description\":\"Multiple choice practice for AP Precalculus Section I Part B\",\"url\":\"https://theschoolofmathematics.com/quiz/ap-precalculus-full-practice-exam-2-section-i-part-b\",\"timeRequired\":\"PT40M\",\"numberOfQuestions\":12,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Full Practice Exam 2\"}}"}}],["$","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-precalculus-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 practice for AP Precalculus Section I Part B

"}}],["$","div",null,{"className":"grid grid-cols-2 md:grid-cols-3 gap-4","children":[["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M12 8v4l3 3m6-3a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Duration"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"40 min"}]]}]]}],["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M8.228 9c.549-1.165 2.03-2 3.772-2 2.21 0 4 1.343 4 3 0 1.4-1.278 2.575-3.006 2.907-.542.104-.994.54-.994 1.093m0 3h.01M21 12a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Questions"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":12}]]}]]}],"$undefined",["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M7 7h.01M7 3h5c.512 0 1.024.195 1.414.586l7 7a2 2 0 010 2.828l-7 7a2 2 0 01-2.828 0l-7-7A1.994 1.994 0 013 12V7a4 4 0 014-4z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Category"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Full Practice Exam 2"}]]}]]}]]}]]}]]}],["$","$L17",null,{"quiz":{"id":2199,"prismaId":"cmpz9qs5204bljgllk1sb9d88","slug":"ap-precalculus-full-practice-exam-2-section-i-part-b","title":{"rendered":"Section I Part B"},"content":{"rendered":"

AP Precalculus Full Practice Exam 2 - Section I Part B

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

Multiple choice practice for AP Precalculus Section I Part B

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":40,"niveau_difficulte":"$undefined","categorie":"Full Practice Exam 2","nombre_questions":12,"score_minimum":80,"ordre_questions":"Fixe","questions":[{"id":"cmpz9qs5504bnjgll94wmej9o","texte_question":"

A deposit of $$8000$$ is made in an account that earns $$6\\%$$ interest compounded monthly. The balance after $$n$$ months is given by $$a_n=8000\\left(1+\\frac{0.06}{12}\\right)^n$$. Find the balance after $$3$$ years.

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

Convert years to months.

$$3\\text{ years}=36\\text{ months}$$

Substitute $$n=36$$ into the formula.

$$a_{36}=8000\\left(1+\\frac{0.06}{12}\\right)^{36}$$

Simplify inside the parentheses.

$$a_{36}=8000(1.005)^{36}$$

Evaluate.

$$a_{36}\\approx9825.78$$

Final answer: $$9825.78$$

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

$$9524.36$$

","correcte":false,"explication":"This is too small for $$36$$ months of growth.","imageUrl":"$undefined"},{"texte":"

$$9742.11$$

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

This does not match the compound interest calculation.

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

$$9825.78$$

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

Correct value after substituting $$n=36$$.

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

$$10102.47$$

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

This is too large for the given rate and time.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5804btjgllmek7gp5u","texte_question":"

If $$\\log_{0.4}(x-2)$$>$$\\log_{0.16}(x-2)$$, then $$x$$ lies in which of the following intervals?

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

Let $$t=x-2$$ with $$t$$>$$0.$$

Rewrite bases:

$$0.4=\\frac{2}{5},\\quad 0.16=\\frac{4}{25}=\\left(\\frac{2}{5}\\right)^2.$$

So:

$$\\log_{0.4}(t)=\\log_{2/5}(t),\\quad \\log_{0.16}(t)=\\log_{(2/5)^2}(t).$$

Use change of base:

$$\\log_{(2/5)^2}(t)=\\frac{\\log_{2/5}(t)}{2}.$$

So inequality becomes:

$$\\log_{2/5}(t)$$>$$\\frac{1}{2}\\log_{2/5}(t).$$

Multiply both sides by $$2$$:

$$2\\log_{2/5}(t)$$>$$\\log_{2/5}(t).$$

So:

$$\\log_{2/5}(t)$$>$$0.$$

Since base $$\\frac{2}{5}$$<$$1$$, the logarithm is decreasing, so:

$$0$$<$$t$$<$$1.$$

Thus:

$$0$$<$$x-2$$<$$1.$$

So:

$$2$$<$$x$$<$$3.$$

Final answer: $$2$$<$$x$$<$$3$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(2,\\\\infty)$$","correcte":false,"explication":"The inequality is not true for all $$x>2$$.","imageUrl":"$undefined"},{"texte":"$$$(-\\\\infty,2)$$","correcte":false,"explication":"These values are outside the domain.","imageUrl":"$undefined"},{"texte":"

$$(2,3)$$

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

Correct interval.

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

$$(-2,0)$$

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

These values are outside the domain.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5b04bzjgllxbobo3ym","texte_question":"

If $$\\\\theta$$ is an angle in standard position and its terminal side passes through the point $$(-0.6,-0.8)$$ on the unit circle, which of the following is a possible radian value of $$\\\\theta$$ to the nearest hundredth?

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

On the unit circle:

$$\\cos\\theta=-0.6$$

$$\\sin\\theta=-0.8$$

Both sine and cosine are negative, so $$\\theta$$ is in Quadrant III.

Find the reference angle.

$$\\alpha=\\arctan\\left(\\frac{0.8}{0.6}\\right)=\\arctan\\left(\\frac{4}{3}\\right)\\approx0.93$$

Quadrant III angle:

$$\\theta=\\pi+\\alpha$$ $$\\theta\\approx3.14+0.93=4.07$$

Final answer: $$4.07$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$3.82$$","correcte":false,"explication":"This is not the Quadrant III angle for the given point.","imageUrl":"$undefined"},{"texte":"$$$4.07$$","correcte":true,"explication":"

Correct radian measure to the nearest hundredth.

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

$$5.36$$

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

This is in Quadrant IV.

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

$$2.21$$

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

This is in Quadrant II.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5e04c5jgll314w6b9h","texte_question":"

The cost $$C(x)$$, in dollars, of operating a business is modeled by $$C(x)=x^2-12x+500$$, where $$x$$ is the number of items produced daily. Find the minimum cost to the nearest dollar.

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

The quadratic function $$C(x)=x^2-12x+500$$ opens upward, so its minimum occurs at the vertex.

Use:

$$x=\\frac{-b}{2a}$$

Here:

$$a=1$$

$$b=-12$$

So:

$$x=\\frac{-(-12)}{2(1)}=6$$

Now evaluate the cost.

$$C(6)=6^2-12(6)+500$$

$$C(6)=36-72+500$$

$$C(6)=464$$

Final answer: $$464$$

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

$$464$$

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

Correct minimum cost.

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

$$428$$

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

This is too small.

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

$$436$$

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

This does not match the vertex value.

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

$$450$$

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

This is not the minimum value.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5g04cbjgll13g9pjo9","texte_question":"

The height $$h(t)$$, in meters, of a ball thrown upward is given by $$h(t)=5+18t-4t^2$$. Find the average rate of change of the height over the first $$3$$ seconds.

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

Use the average rate of change formula from $$t=0$$ to $$t=3$$.

$$\\frac{h(3)-h(0)}{3-0}$$

Find the endpoint values.

$$h(0)=5+18(0)-4(0)^2=5$$

$$h(3)=5+18(3)-4(3)^2$$

$$h(3)=5+54-36=23$$

Now compute the average rate of change.

$$\\frac{23-5}{3}=\\frac{18}{3}=6$$

Final answer: $$6$$

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

$$6$$

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

Correct average rate of change.

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

$$9$$

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

This does not use the correct difference quotient.

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

$$12$$

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

This is too large.

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

$$15$$

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

This is not the average change per second.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5j04chjgllh3sw331z","texte_question":"

Find the solution set for the rational inequality $$\\frac{x-1}{x+3}$$ < $$2.$$

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

Start with the inequality:

$$\\frac{x-1}{x+3}$$ < $$2$$

Move everything to one side:

$$\\frac{x-1}{x+3}-2$$ < $$0$$

Combine into one fraction:

$$\\frac{x-1-2(x+3)}{x+3}$$ < $$0$$

Simplify the numerator:

$$\\frac{-x-7}{x+3}$$ < $$0$$

Factor out $$-1$$:

$$-\\frac{x+7}{x+3}$$ < $$0$$

Multiply both sides by $$-1$$ and reverse the inequality:

$$\\frac{x+7}{x+3}$$ > $$0$$

Critical values:

$$x=-7,\\quad x=-3$$

Test the intervals.

For $$x$$ < $$-7,$$ the expression is positive.

For $$-7$$ < $$x$$ < $$-3,$$ the expression is negative.

For $$x$$ > $$-3,$$ the expression is positive.

Also, $$x=-3$$ is not in the domain.

Since the inequality is strict, $$x=-7$$ is not included.

Final answer:

$$(-\\infty,-7)\\cup(-3,\\infty)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$( -3,5 )$$","correcte":false,"explication":"This interval does not satisfy the sign analysis.","imageUrl":"$undefined"},{"texte":"$$$(-\\infty,-7)\\cup(-3,\\infty)$$","correcte":true,"explication":"Correct solution set.","imageUrl":"$undefined"},{"texte":"$$$( -3,\\infty )$$","correcte":false,"explication":"This misses the interval $$(-\\infty,-7)$$.","imageUrl":"$undefined"},{"texte":"

$$(-\\infty,5)$$

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

This includes values that do not satisfy the inequality.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5m04cnjgllxy7gff08","texte_question":"$18","type_question":"QCM","explication":"

Use the Intermediate Value Theorem. Because the function is continuous, a zero must occur in any interval where the function changes sign. Check consecutive values.

From $$x=-4$$ to $$x=-2$$:

$$f(-4)=3.1$$ > $$0$$

$$f(-2)=-5.2$$ < $$0$$

There is a sign change, so there must be a zero in: $$(-4,-2)$$

Other listed intervals do not show a sign change.

Final answer: $$(-4,-2)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(-4,-2)$$","correcte":true,"explication":"Correct interval containing a required zero.","imageUrl":"$undefined"},{"texte":"$$$(-2,0)$$","correcte":false,"explication":"The function is negative at both endpoints.","imageUrl":"$undefined"},{"texte":"$$$(0,2)$$","correcte":false,"explication":"The function is negative at both endpoints.","imageUrl":"$undefined"},{"texte":"

$$(2,4)$$

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

Although there is a sign change here too, the question asks for an interval that must contain a zero from the listed data, and this one also works. To avoid ambiguity, the problem should include only one sign-change interval. The intended correct answer is $$(-4,-2)$$.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5p04ctjgll7vu522vy","texte_question":"

Find the domain of the function $$f(x)=\\sqrt{x-2}-\\frac{1}{\\sqrt{10-x}}.$$

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

For the first square root:

$$x-2\\ge0$$

So:

$$x\\ge2$$

For the denominator square root:

$$10-x$$ > $$0$$

It must be greater than zero because it is in the denominator.

So:

$$x$$ < $$10$$

Combine the conditions:

$$2le x$$ < $$10$$

Final answer: $$[2,10)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$[2,10)$$","correcte":true,"explication":"Correct domain.","imageUrl":"$undefined"},{"texte":"$$$(2,10]$$","correcte":false,"explication":"This excludes $$2$$ and incorrectly includes $$10$$.","imageUrl":"$undefined"},{"texte":"$$$(2,10)$$","correcte":false,"explication":"This incorrectly excludes $$2$$.","imageUrl":"$undefined"},{"texte":"

$$(-\\infty,\\infty)$$

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

The square roots impose restrictions.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5s04czjglli3sr9e48","texte_question":"

Find the difference between the largest and smallest real solutions of $$3x^4-x^2-2=0$$ rounded to two decimal places.

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

Let:

$$u=x^2.$$

Then the equation becomes:

$$3u^2-u-2=0.$$

Factor:

$$(3u+2)(u-1)=0.$$

So:

$$u=1$$ or $$u=-\\frac{2}{3}.$$

Since $$u=x^2\\ge0,$$ only $$u=1$$ gives real solutions.

Thus:

$$x^2=1.$$

So the real solutions are:

$$x=-1,\\quad x=1.$$

The difference between the largest and smallest real solutions is:

$$1-(-1)=2.$$

Final answer: $$2.00$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1.15$$","correcte":false,"explication":"This does not match the real solutions.","imageUrl":"$undefined"},{"texte":"$$$1.41$$","correcte":false,"explication":"This is not the difference between $$1$$ and $$-1$$.","imageUrl":"$undefined"},{"texte":"$$$1.73$$","correcte":false,"explication":"This comes from using a nonreal branch incorrectly.","imageUrl":"$undefined"},{"texte":"

$$2.00$$

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

Correct difference.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5v04d5jglllkefhgd1","texte_question":"$19","type_question":"QCM","explication":"

The population doubles every $$10$$ years. From $$1990$$ to $$2000$$: $$3\to6$$ From $$2000$$ to $$2010$$: $$6\to12$$ From $$2010$$ to $$2020$$: $$12\to24$$ So from $$2020$$ to $$2030$$: $$24\to48$$ Since the table is in thousands, the population in $$2030$$ is: $$48{,}000$$ Final answer: $$48{,}000$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$36{,}000$$","correcte":false,"explication":"This does not follow the doubling pattern.","imageUrl":"$undefined"},{"texte":"$$$40{,}000$$","correcte":false,"explication":"This is not double $$24{,}000$$.","imageUrl":"$undefined"},{"texte":"$$$48{,}000$$","correcte":true,"explication":"Correct continuation of the pattern.","imageUrl":"$undefined"},{"texte":"$$$60{,}000$$","correcte":false,"explication":"

This is too large for the given pattern.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs5z04dbjgll99cje8sn","texte_question":"

A box with an open top is made from a rectangular sheet of cardboard measuring $$10$$ inches by $$20$$ inches by cutting out squares of side length $$x$$ from each corner and folding up the sides. What value of $$x$$ maximizes the volume?

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

The volume is:

$$V(x)=x(20-2x)(10-2x).$$

Expand:

$$V(x)=x(200-60x+4x^2)$$

$$V(x)=200x-60x^2+4x^3.$$

Differentiate:

$$V'(x)=200-120x+12x^2.$$

Set the derivative equal to zero:

$$200-120x+12x^2=0.$$

Divide by $$4$$:

$$50-30x+3x^2=0.$$

Rewrite:

$$3x^2-30x+50=0.$$

Use the quadratic formula:

$$x=\\frac{30\\pm\\sqrt{900-600}}{6}$$

$$x=\\frac{30\\pm\\sqrt{300}}{6}$$

$$x=\\frac{30\\pm10\\sqrt{3}}{6}$$

$$x=\\frac{15\\pm5\\sqrt{3}}{3}.$$

Approximate the two solutions:

$$x\\approx2.11$$

$$x\\approx7.89.$$

Because the box requires:

$$20-2x$$ > $$0$$ and $$10-2x$$>$$0$$

we must have:

$$0$$<$$x$$<$$5$$

So, the only valid critical point is:

$$x=\\frac{15-5\\sqrt{3}}{3}\\approx2.11.$$

Final answer: $$2.11$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1.67$$","correcte":false,"explication":"This is not the critical value of the volume function.","imageUrl":"$undefined"},{"texte":"$$$2.11$$","correcte":true,"explication":"Correct value that maximizes the volume.","imageUrl":"$undefined"},{"texte":"$$$3.33$$","correcte":false,"explication":"This does not come from solving $$V^{\\\\prime}(x)=0$$.","imageUrl":"$undefined"},{"texte":"$$$5.00$$","correcte":false,"explication":"

At $$x=5$$ the box has zero width on one side.

","imageUrl":"$undefined"}]},{"id":"cmpz9qs6104dhjgllph7ktz5m","texte_question":"

Given the rectangular coordinates $$(-3,3\\sqrt{3})$$, find the polar coordinates.

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

Use:

$$r=\\sqrt{x^2+y^2}.$$

Substitute:

$$r=\\sqrt{(-3)^2+(3\\sqrt{3})^2}$$

$$r=\\sqrt{9+27}$$

$$r=\\sqrt{36}=6.$$

Now find the angle.

Since:

$$x=-3$$<$$0$$ and $$y=3\\sqrt{3}$$>$$0,$$ the point is in Quadrant II.

Use the reference angle:

$$\\tan\\theta=\\frac{y}{x}=\\frac{3\\sqrt{3}}{-3}=-\\sqrt{3}.$$

The reference angle is: $$\\frac{\\pi}{3}.$$

In Quadrant II, the angle is:

$$\\pi-\\frac{\\pi}{3}=\\frac{2\\pi}{3}.$$

So the polar coordinates are:

$$\\left(6,\\frac{2\\pi}{3}\\right).$$

Final answer: $$\\left(6,\\frac{2\\pi}{3}\\right)$$

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

$$\\left(6,\\frac{2\\pi}{3}\\right)$$

","correcte":true,"explication":"Correct polar coordinates.","imageUrl":"$undefined"},{"texte":"$$$\\left(6,\\frac{5\\pi}{6}\\right)$$","correcte":false,"explication":"This angle gives different rectangular coordinates.","imageUrl":"$undefined"},{"texte":"$$$\\left(6,\\frac{4\\pi}{3}\\right)$$","correcte":false,"explication":"

This is in Quadrant III.

","imageUrl":"$undefined"},{"texte":"$$$\\left(6,\\frac{\\pi}{3}\\right)$$","correcte":false,"explication":"

This is in Quadrant I.

","imageUrl":"$undefined"}]}]},"categories":["full-practice-exam-2"],"date":"2026-06-04T09:04:10.742Z","modified":"2026-06-04T09:04:10.742Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L1a",null,{"targetType":"quiz","targetSlug":"ap-precalculus-full-practice-exam-2-section-i-part-b"}]}]]}]}]]

Section I Part B

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