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

Multiple-choice no calculator section

\",\"url\":\"https://www.theschoolofmathematics.com/quiz/ap-precalculus-full-practice-exam-1-section-i-part-a\",\"timeRequired\":\"PT80M\",\"numberOfQuestions\":27,\"about\":{\"@type\":\"Thing\",\"name\":\"Full Practice Exam 1\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://www.theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://www.theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Section I Part A\",\"item\":\"https://www.theschoolofmathematics.com/quiz/ap-precalculus-full-practice-exam-1-section-i-part-a\"}]}"}}],["$","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":["$","h1",null,{"className":"text-4xl md:text-5xl lg:text-6xl font-bold text-gray-900 mb-4","children":"Section I Part A"}]}],["$","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 no calculator section

"}}],["$","div",null,{"className":"grid grid-cols-2 md:grid-cols-3 gap-4","children":[["$","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":"1h 20min"}]]}]]}],["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","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"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":27}]]}]]}],"$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"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Full Practice Exam 1"}]]}]]}]]}]]}]]}],["$","$L17",null,{"quiz":{"id":2242,"prismaId":"59863b95-d959-477d-a505-a1f61204dc34","slug":"ap-precalculus-full-practice-exam-1-section-i-part-a","title":{"rendered":"Section I Part A"},"content":{"rendered":"

AP Precalculus Full Practice Exam 1 Section I Part A

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

Multiple-choice no calculator section

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":80,"niveau_difficulte":"$undefined","categorie":"Full Practice Exam 1","nombre_questions":27,"score_minimum":80,"ordre_questions":"Fixe","questions":[{"id":"cmo1ahzzp00smjggbjd0c10on","texte_question":"

What is the numerical coefficient of the term containing $$x^2y^3$$ in the expansion of $$(x+3y)^5$$?

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

Use the binomial theorem: $$(x+3y)^5=sum_{k=0}^{5}\binom{5}{k}x^{5-k}(3y)^k.$$ To get $$x^2y^3$$: $$5-k=2$$ $$k=3.$$ Now compute the coefficient: $$\binom{5}{3}(3)^3=10cdot 27=270.$$ Final answer: $$270$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$90$$","correcte":false,"explication":"This does not include the correct binomial coefficient and power of $$3$$.","imageUrl":"$undefined"},{"texte":"

$$135$$

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

This misses part of the coefficient calculation.

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

$$270$$

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

Correct coefficient from $$\\binom{5}{3}(3)^3$$.

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

$$405$$

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

This is too large for the required term.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzp00srjggb5kw28e1v","texte_question":"

If $$f(x)=x^3+Ax^2+Bx+2$$ and $$f(1)=6$$ and $$f(-1)=-2$$, what is the value of $$2A+B$$?

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

Use $$f(1)=6$$: $$1+A+B+2=6$$ $$A+B=3.$$ Use $$f(-1)=-2$$: $$-1+A-B+2=-2$$ $$A-B=-3.$$ Solve the system: $$(A+B)+(A-B)=3+(-3)$$ $$2A=0$$ $$A=0.$$ Then: $$B=3.$$ Now compute: $$2A+B=2(0)+3=3.$$ Final answer: $$3$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2$$","correcte":false,"explication":"This does not satisfy the two conditions together.","imageUrl":"$undefined"},{"texte":"

$$3$$

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

Correct after solving for $$A$$ and $$B$$.

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

$$4$$

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

This comes from an incorrect system solution.

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

$$6$$

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

This is too large for the computed value.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzp00swjggbovdyzt10","texte_question":"

If $$f(x)=x^3-3x+2$$, then $$f(-1)=$$

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

Substitute $$x=-1$$: $$f(-1)=(-1)^3-3(-1)+2.$$ Simplify: $$f(-1)=-1+3+2=4.$$ Final answer: $$4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"This is not the value after substitution.","imageUrl":"$undefined"},{"texte":"

$$2$$

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

This misses part of the evaluation.

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

$$4$$

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

Correct value of $$f(-1)$$.

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

$$6$$

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

This is too large.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzp00t1jggb6y0qbht5","texte_question":"

What is the domain of $$f(x)=\\frac{x+2}{x^2+16}$$?

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

The denominator cannot be zero. Set the denominator equal to zero: $$x^2+16=0$$ $$x^2=-16.$$ This has no real solutions. So there are no real restrictions on $$x$$. Final answer: All real numbers

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"All real numbers except $$x=-2$$","correcte":false,"explication":"The numerator can be zero. That does not restrict the domain.","imageUrl":"$undefined"},{"texte":"

All real numbers

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

Correct because $$x^2+16$$ is never zero for real $$x$$.

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

All real numbers except $$x=4$$

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

There is no restriction at $$x=4$$.

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

All real numbers except $$x=\\pm 4$$

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

The denominator is never zero for real numbers.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzp00t6jggbx24xvbob","texte_question":"

What is the domain of $$g(x)=\\frac{\\sqrt{x-2}}{x^2-4x}$$?

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

The square root requires: $$x-2ge 0$$ $$xge 2.$$ The denominator cannot be zero: $$x^2-4x=x(x-4) e 0.$$ So: $$x e 0, x e 4.$$ Because $$xge 2$$ already, the restriction $$x e 0$$ is automatic. So the domain is: $$xge 2, x e 4.$$ Final answer: $$xge 2, x e 4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"All real numbers except $$x=0,4$$","correcte":false,"explication":"This ignores the square root restriction.","imageUrl":"$undefined"},{"texte":"

$$x\\ge 2,\\ x\\ne 0,4$$

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

The restriction $$x\\ne 0$$ is unnecessary once $$x\\ge 2$$.

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

$$x\\ge 2,\\ x\\ne 4$$

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

Correct domain.

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

$$x>4$$

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

Values from $$2$$ up to but not including $$4$$ are allowed.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzq00tbjggbo0c1i2iz","texte_question":"

Which of the following is equivalent to the expression $$\\sqrt[3]{\\frac{a^3b^2}{c}}$$?

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

Rewrite the cube root using exponents: $$sqrt[3]{\frac{a^3b^2}{c}}=left(\frac{a^3b^2}{c} ight)^{1/3}.$$ Apply the exponent to each factor: $$a^{3/3}b^{2/3}c^{-1/3}=ab^{2/3}c^{-1/3}.$$ Rewrite without negative exponents: $$\frac{ab^{2/3}}{c^{1/3}}.$$ Final answer: $$\frac{ab^{2/3}}{c^{1/3}}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{ab^{2/3}}{c^{1/3}}$$","correcte":true,"explication":"Correct simplification of the cube root.","imageUrl":"$undefined"},{"texte":"

$$\\frac{a^{1/3}b^{2/3}}{c}$$

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

The exponent on $$a$$ should simplify to $$1$$, not $$\\frac{1}{3}$$.

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

$$\\frac{ab^2}{c^{1/3}}$$

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

The exponent on $$b$$ should be $$\\frac{2}{3}$$.

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

$$\\frac{a^{1/3}b^{2/3}}{c^{1/3}}$$

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

The exponent on $$a$$ is incorrect.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzq00tgjggb883h2bwf","texte_question":"

Identify the vertical asymptote(s) for the function $$f(x)=\\frac{x^2-9}{x^3-3x^2}$$.

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

Factor the numerator and denominator: $$x^2-9=(x-3)(x+3),$$ $$x^3-3x^2=x^2(x-3).$$ Cancel the common factor $$x-3$$: $$f(x)=\frac{x+3}{x^2}, x e 3.$$ The factor $$x=3$$ creates a hole, not a vertical asymptote. The denominator is zero at: $$x=0.$$ So the only vertical asymptote is: $$x=0.$$ Final answer: $$x=0$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$x=0$$","correcte":true,"explication":"Correct vertical asymptote.","imageUrl":"$undefined"},{"texte":"

$$x=0,\\ x=3$$

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

$$x=3$$ is a hole after cancellation, not a vertical asymptote.

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

$$x=3$$

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

This is not a vertical asymptote.

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

$$x=-3,\\ x=3$$

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

Neither of these is the vertical asymptote.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzq00tljggbsoc2dfdw","texte_question":"

Given $$f(x)=\\begin{cases}x^2+4,&x\\le 0\\\\(x+1)^2,&x>0\\end{cases}$$, find $$f(-2)$$.

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

Since $$-2le 0$$, use the first piece: $$f(x)=x^2+4.$$ Substitute $$x=-2$$: $$f(-2)=(-2)^2+4=4+4=8.$$ Final answer: $$8$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"This is not the result of substitution.","imageUrl":"$undefined"},{"texte":"

$$2$$

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

This is too small.

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

$$8$$

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

Correct value from the first piece.

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

$$12$$

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

This does not match the evaluation.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzq00tqjggbk7pzhhon","texte_question":"

Find the slant asymptote of $$f(x)=\\frac{x^2+3x+2}{x-1}$$.

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

Divide $$x^2+3x+2$$ by $$x-1$$. Long division gives: $$\frac{x^2+3x+2}{x-1}=x+4+\frac{6}{x-1}.$$ As $$x\topminfty$$, the fraction term approaches $$0$$. So the slant asymptote is: $$y=x+4.$$ Final answer: $$y=x+4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$y=x+1$$","correcte":false,"explication":"This is not the quotient from division.","imageUrl":"$undefined"},{"texte":"

$$y=x+3$$

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

The constant term is incorrect.

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

$$y=x+4$$

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

Correct slant asymptote.

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

$$y=2x+1$$

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

The slope is incorrect.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzq00tvjggbuakemu9t","texte_question":"

In polar coordinates, which of the following choices is not equivalent to $$(3,\\frac{2\\pi}{3})$$?

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

Equivalent polar coordinates can be found by: Adding multiples of $$2pi$$ to the angle when $$r$$ stays positive or adding $$pi$$ and changing the sign of $$r$$. Check the choices: $$(-3,\frac{5pi}{3})$$ is equivalent. $$left(3,-\frac{4pi}{3} ight)$$ is equivalent. $$left(-3,-\frac{pi}{3} ight)$$ is equivalent. $$left(3,\frac{pi}{3} ight)$$ is not equivalent because $$\frac{pi}{3} e rac{2pi}{3}+2kpi$$. Final answer: $$left(3,\frac{pi}{3} ight)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(-3,\\frac{5\\pi}{3})$$","correcte":false,"explication":"

Changing the sign of $$r$$ and adding $$\\pi$$ makes this equivalent.

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

$$(3,-\\frac{4\\pi}{3})$$

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

This differs by $$-2\\pi$$, so it is equivalent.

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

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

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

Correct non-equivalent choice.

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

$$(-3,-\\frac{\\pi}{3})$$

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

This is equivalent after adding $$\\pi$$ to the angle.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzq00u0jggbjsyv0ig8","texte_question":"

Which of the following represents zeros of $$r=3-3\\cos 2\\theta$$?

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

Set $$r=0$$: $$3-3cos 2\theta=0.$$ Solve: $$1-cos 2\theta=0$$ $$cos 2\theta=1.$$ This happens when: $$2\theta=2kpi$$ $$\theta=kpi.$$ In the interval $$[0,2pi)$$, the solutions are: $$0,pi.$$ Final answer: $$0,pi$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0,\\pi$$","correcte":true,"explication":"Correct zeros in one full rotation.","imageUrl":"$undefined"},{"texte":"

$$\\frac{\\pi}{4},\\frac{3\\pi}{4}$$

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

These do not make $$\\cos 2\\theta=1$$.

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

$$\\frac{\\pi}{2},\\frac{3\\pi}{2}$$

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

These make $$\\cos 2\\theta=-1$$.

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

$$\\frac{\\pi}{6},\\frac{5\\pi}{6}$$

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

These are not zeros of the function.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzq00u5jggbd6p082ks","texte_question":"

Evaluate $$\\sin\\left[\\arctan\\left(\\frac{4}{3}\\right)\\right]$$.

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

Let: $$\theta=arctanleft(\frac{4}{3} ight).$$ Then: $$\tan\theta=\frac{4}{3}.$$ Draw a right triangle: Opposite $$=4$$ Adjacent $$=3$$ Hypotenuse $$=5$$ Now compute sine: $$sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{4}{5}.$$ Final answer: $$\frac{4}{5}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{4}{5}$$","correcte":true,"explication":"Correct value from the reference triangle.","imageUrl":"$undefined"},{"texte":"

$$\\frac{3}{5}$$

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

This is the cosine, not the sine.

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

$$\\frac{5}{4}$$

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

Sine cannot be greater than $$1$$ here.

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

$$\\frac{3}{4}$$

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

This is not the sine value.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzq00uajggbqc786liv","texte_question":"

Which of the following is equivalent to $$\\sin 4x\\cos 2x-\\cos 4x\\sin 2x$$?

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

Use the sine difference identity: $$sin Acos B-cos Asin B=sin(A-B).$$ Let: $$A=4x,$$ $$B=2x.$$ Then: $$sin 4xcos 2x-cos 4xsin 2x=sin(4x-2x).$$ Simplify: $$sin 2x.$$ Final answer: $$sin 2x$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\sin 6x$$","correcte":false,"explication":"

This would come from incorrectly using the angle-sum identity.

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

$$\\sin 2x$$

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

Correct by the sine difference identity.

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

$$\\cos 6x$$

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

The expression simplifies to a sine expression, not cosine.

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

$$\\cos 2x$$

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

This does not match the identity.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00ufjggb50hq86ij","texte_question":"

Given $$\\cos x=\\frac{3}{5}$$ and $$\\tan x>0$$, find $$\\sin 2x$$.

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

Use the identity: $$sin 2x=2sin xcos x.$$ Since $$cos x=\frac{3}{5}$$ and $$\tan x>0$$, sine and cosine have the same sign. Because cosine is positive, sine is also positive. Find $$sin x$$: $$sin x=sqrt{1-cos^2 x}=sqrt{1-left(\frac{3}{5} ight)^2}=sqrt{\frac{16}{25}}=\frac{4}{5}.$$ Now compute: $$sin 2x=2left(\frac{4}{5} ight)left(\frac{3}{5} ight)=\frac{24}{25}.$$ Final answer: $$\frac{24}{25}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{24}{25}$$","correcte":true,"explication":"Correct value of $$\\sin 2x$$.","imageUrl":"$undefined"},{"texte":"

$$\\frac{12}{25}$$

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

This misses the factor of $$2$$ in the double-angle identity.

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

$$\\frac{6}{25}$$

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

This is too small.

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

$$\\frac{18}{25}$$

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

This does not come from the correct substitution.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00ukjggbv0lgnlvw","texte_question":"

Determine the period of the function $$y=3\\sin\\left(\\frac{1}{4}x+\\pi\\right)$$.

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

For a function of the form: $$y=Asin(Bx+C),$$ its period is: $$\frac{2pi}{|B|}.$$ Here: $$B=\frac{1}{4}.$$ So the period is: $$\frac{2pi}{1/4}=8pi.$$ Final answer: $$8pi$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2\\pi$$","correcte":false,"explication":"This would be the period if $$B=1$$.","imageUrl":"$undefined"},{"texte":"

$$4\\pi$$

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

This is only half of the actual period.

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

$$8\\pi$$

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

Correct period.

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

$$\\pi$$

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

This is too small.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00upjggbzs0o6aqf","texte_question":"

Solve the equation $$\\log_a 5-\\log_a b=c$$ for $$b$$.

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

Use the quotient property of logarithms: $$log_a 5-log_a b=log_aleft(\frac{5}{b} ight).$$ So: $$log_aleft(\frac{5}{b} ight)=c.$$ Rewrite in exponential form: $$\frac{5}{b}=a^c.$$ Solve for $$b$$: $$5=ba^c$$ $$b=\frac{5}{a^c}.$$ Final answer: $$\frac{5}{a^c}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{5}{a^c}$$","correcte":true,"explication":"Correct solution for $$b$$.","imageUrl":"$undefined"},{"texte":"

$$5a^c$$

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

This inverts the relationship incorrectly.

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

$$\\frac{a^c}{5}$$

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

This solves for the reciprocal instead of $$b$$.

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

$$a^5c$$

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

This is not algebraically valid.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00uujggbtmt7b2vb","texte_question":"

Which of the following statements is true about the function $$g(x)=x^5+x^3+\\sin x$$?

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

Check whether each term is odd or even. For $$x^5$$: $$(-x)^5=-x^5$$ So $$x^5$$ is odd. For $$x^3$$: $$(-x)^3=-x^3$$ So $$x^3$$ is odd. For $$sin x$$: $$sin(-x)=-sin x$$ So $$sin x$$ is odd. The sum of odd functions is odd. Therefore: $$g(-x)=-g(x).$$ Final answer: The function is odd and symmetric about the origin.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"The function is even and symmetric about the y-axis.","correcte":false,"explication":"Each term is odd, not even.","imageUrl":"$undefined"},{"texte":"

The function is odd and symmetric about the origin.

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

Correct description.

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

The function is neither even nor odd.

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

The function is odd.

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

The function is even and symmetric about the origin.

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

Even functions are symmetric about the y-axis, not the origin.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00uzjggbzkjq0pxm","texte_question":"

Which of these is an equation for the linear function $$f$$ that satisfies $$f(-1)=3$$ and $$f(3)=-5$$?

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

Use the two points: $$(-1,3)$$ and $$(3,-5).$$ Find the slope: $$m=\frac{-5-3}{3-(-1)}=\frac{-8}{4}=-2.$$ Use point-slope form with point $$(-1,3)$$: $$y-3=-2(x+1).$$ Final answer: $$y-3=-2(x+1)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$y-3=-2(x+1)$$","correcte":true,"explication":"Correct point-slope equation.","imageUrl":"$undefined"},{"texte":"

$$y+5=-2(x-3)$$

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

Also correct; it uses the point $$(3,-5)$$.

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

$$y-3=-2(x-3)$$

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

This uses the wrong point with the same slope.

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

$$y+5=-2(x+1)$$

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

This does not pass through either required point correctly.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00v4jggbiyfyu8hs","texte_question":"

Given the functions $$f(x)=x^2+1$$ and $$g(x)=\\sqrt{x-1}$$, determine the composition $$(g\\circ f)(x)$$.

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

Use the composition rule: $$(gcirc f)(x)=g(f(x)).$$ Substitute $$f(x)=x^2+1$$ into $$g(x)=sqrt{x-1}$$: $$g(f(x))=sqrt{(x^2+1)-1}.$$ Simplify: $$g(f(x))=sqrt{x^2}.$$ Final answer: $$sqrt{x^2}$$

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

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

","correcte":true,"explication":"Correct composition after substitution and simplification.","imageUrl":"$undefined"},{"texte":"

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

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

This subtracts $$1$$ outside the radical instead of inside.

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

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

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

The extra $$+1$$ should cancel inside the radical.

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

$$x^2$$

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

The square root is missing.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00v9jggbukyc4ex9","texte_question":"

Given the function $$g(x)=-(5x-2)^2(2x+3)^3$$, determine $$\\lim_{x\\to-\\infty} g(x)$$.

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

Look at the leading behavior of each factor. As $$x\to-infty$$: $$(5x-2)^2sim 25x^2,$$ $$(2x+3)^3sim 8x^3.$$ So: $$g(x)sim -(25x^2)(8x^3)=-200x^5.$$ Since $$x^5\to-infty$$ as $$x\to-infty$$: $$-200x^5\toinfty.$$ Final answer: $$infty$$

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

The outside negative changes the sign of the odd-degree leading term.

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

$$0$$

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

The function grows without bound in magnitude.

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

$$1$$

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

Polynomial behavior does not approach a constant here.

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

$$\\infty$$

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

Correct end behavior.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00vejggbad77r0i4","texte_question":"

Evaluate $$\\lim_{x\\to\\infty} 2^{-x}+5$$.

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

As $$x\toinfty$$: $$2^{-x}=\frac{1}{2^x}\to 0.$$ So: $$2^{-x}+5\to 0+5=5.$$ Final answer: $$5$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"Only the exponential term approaches $$0$$.","imageUrl":"$undefined"},{"texte":"

$$2$$

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

This is not the limit of the full expression.

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

$$5$$

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

Correct limit.

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

$$\\infty$$

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

The expression approaches a horizontal value, not infinity.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00vjjggbjul54az7","texte_question":"

What is the function whose graph is a reflection over the $$y$$-axis of the graph of $$f(x)=2^x+1$$?

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

A reflection over the $$y$$-axis replaces $$x$$ with $$-x$$. So: $$g(x)=f(-x)=2^{-x}+1.$$ Final answer: $$g(x)=2^{-x}+1$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$g(x)=2^{-x}+1$$","correcte":true,"explication":"Correct reflection over the $$y$$-axis.","imageUrl":"$undefined"},{"texte":"

$$g(x)=2^x-1$$

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

This changes the vertical shift, not the horizontal reflection.

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

$$g(x)=1-2^x$$

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

This reflects over the $$x$$-axis and shifts.

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

$$g(x)=2^{x+1}$$

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

This is a horizontal shift/scaling effect, not reflection over the $$y$$-axis.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzr00vojggbmkzygchx","texte_question":"

Which of the following functions does not have an inverse function?

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

A function has an inverse only if it is one-to-one. Check each choice. For $$y=cos x, 0le xle pi$$: This is one-to-one on the interval. For $$y=x^3-1$$: This is one-to-one for all real $$x$$. For $$y=\frac{x^2}{x+1}$$: This is not one-to-one in its domain. For example, many horizontal lines intersect its graph more than once. For $$y=e^x$$: This is one-to-one. Final answer: $$y=\frac{x^2}{x+1}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$y=\\cos x,\\ 0\\le x\\le \\pi$$","correcte":false,"explication":"This is one-to-one on the restricted interval.","imageUrl":"$undefined"},{"texte":"

$$y=x^3-1$$

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

This cubic is one-to-one.

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

$$y=\\frac{x^2}{x+1}$$

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

Correct choice; it is not one-to-one.

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

$$y=e^x$$

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

This exponential is one-to-one.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzs00vtjggbw9yf7aa5","texte_question":"$18","type_question":"QCM","explication":"

Look for a pattern in the sequence: $$12,24,36,48,60.$$ Each term is $$12$$ times the term number: $$12(1)=12,$$ $$12(2)=24,$$ $$12(3)=36.$$ So the explicit formula is: $$a_n=12n.$$ Final answer: $$a_n=12n$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$a_n=12n$$","correcte":true,"explication":"

Correct explicit formula.

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

$$a_n=n+12$$

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

This does not match the table values.

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

$$a_n=6n$$

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

This gives values that are too small.

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

$$a_n=12n+12$$

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

This shifts every term too high.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzs00vyjggbq6zbgy36","texte_question":"$19","type_question":"QCM","explication":"

Read the amplitude from the graph: The maximum is $$3$$ and the minimum is $$-3$$. So the amplitude is $$3$$. Read the period: From $$0$$ to $$2pi$$, the graph completes $$2$$ full cycles. So the period is: $$\frac{2pi}{2}=pi.$$ For $$y=Asin(Bx)$$: Amplitude is $$|A|$$ Period is $$\frac{2pi}{|B|}$$ So we need: $$|A|=3$$ and $$\frac{2pi}{|B|}=piRightarrow |B|=2.$$ Final answer: $$y=3sin 2x$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$y=3\\sin 2x$$","correcte":true,"explication":"Correct amplitude and period.","imageUrl":"$undefined"},{"texte":"

$$y=2\\sin 3x$$

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

Both the amplitude and period are incorrect.

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

$$y=3\\sin x$$

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

The amplitude is correct, but the period is $$2\\pi$$, not $$\\pi$$.

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

$$y=2\\sin 4x$$

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

The amplitude is too small and the period is too short.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzs00w3jggb9ui7106e","texte_question":"

What are the points where the graph of the polynomial $$f(x)=4(x-3)(x+2)^2$$ passes through the $$x$$-axis?

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

Set the polynomial equal to zero: $$4(x-3)(x+2)^2=0.$$ The zeros are: $$x=3$$ and $$x=-2.$$ Now check multiplicity: $$x=3$$ has multiplicity $$1$$, so the graph crosses the $$x$$-axis there. $$x=-2$$ has multiplicity $$2$$, so the graph touches the $$x$$-axis and turns around there. Since the question asks where the graph passes through the $$x$$-axis, we want the crossing point only. Final answer: $$x=3$$ only

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$x=3$$ only","correcte":true,"explication":"

Correct because the root at $$x=3$$ has odd multiplicity.

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

$$x=-2$$ only

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

At $$x=-2$$ the graph only touches the axis because the multiplicity is even.

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

$$x=-2$$ and $$x=3$$

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

These are both zeros, but only $$x=3$$ is where the graph passes through.

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

Nowhere

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

The graph does cross the axis at $$x=3$$.

","imageUrl":"$undefined"}]},{"id":"cmo1ahzzs00w8jggbki2n857d","texte_question":"

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

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

The numerator and denominator have the same degree. For a rational function with equal degrees, the limit as $$x\toinfty$$ is the ratio of the leading coefficients. Leading coefficient of the numerator: $$2$$ Leading coefficient of the denominator: $$4$$ So: $$lim_{x\toinfty}\frac{2x^2-5x+1}{4x^2+3x-7}=\frac{2}{4}=\frac{1}{2}.$$ Final answer: $$\frac{1}{2}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Does not exist","correcte":false,"explication":"The limit exists because the numerator and denominator are polynomials of the same degree.","imageUrl":"$undefined"},{"texte":"

$$0$$

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

This would happen if the denominator had larger degree.

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

$$\\frac{1}{2}$$

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

Correct ratio of leading coefficients.

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

$$\\frac{2}{4}$$

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

This is equivalent to the correct answer, but the simplified choice is $$\\frac{1}{2}$$.

","imageUrl":"$undefined"}]}]},"categories":["Full-Practice-Exam-1"],"date":"2026-04-16T01:56:09.864Z","modified":"2026-04-16T09:41:28.308Z"}}]]}]}]]