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1"},"content":{"rendered":""},"excerpt":{"rendered":""},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":10,"niveau_difficulte":"Intermediate","categorie":"Functions","nombre_questions":10,"score_minimum":80,"ordre_questions":"Fixe","questions":[{"id":"cmpz9r9ld07jhjgllqq02nqty","texte_question":"

Find the domain of $$g(x)=\\frac{\\sqrt{7-x}}{x-2}.$$

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

Problem: Find the domain of $$g(x)=\\frac{\\sqrt{7-x}}{x-2}.$$

Solution:

Inside the square root:

$$7-x \\ge 0$$

$$x \\le 7$$

Denominator cannot be zero:

$$x-2 \\ne 0$$

$$x \\ne 2$$

The domain is all real x such that:

$$x \\le 7$$ and $$x \\ne 2$$

Final answer: $$(-\\infty,2)\\cup(2,7]$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(-7,2)\\cup(2,\\infty)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(-2,7)\\cup(7,\\infty)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(-\\infty,2)\\cup(2,7]$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$[-2,7)\\cup(7,\\infty)$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9lg07jnjgllj7bcvo0b","texte_question":"

If $$f(x)=x^2+6x-7$$ and $$g(x)=x-3$$, then find $$\\frac{f(x)}{g(x)}$$.

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

Problem: If $$f(x)=x^2+6x-7$$ and $$g(x)=x-3$$, then find $$\\frac{f(x)}{g(x)}$$.

Solution:

Write the expression:

$$\\frac{(x^2+6x-7)}{(x-3)}$$

Factor the numerator:

$$\\frac{(x+7)(x-1)}{(x-3)}$$

No common factors with $$x-3$$, so perform polynomial division:

$$\\frac{(x^2+6x-7)}{(x-3)}=x+9+(\\frac{20}{(x-3)}$$

Final answer: $$x+9+\\frac{20}{x-3}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{20}{x-3}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$x+9+\\frac{20}{x-3}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$7+\\frac{3}{x-3}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{6x-7}{x-3}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9lj07jtjglludy2x0al","texte_question":"

If $$f(x)=3x^2+2$$ and $$g(x)=\\sqrt{x+1}$$, find $$f(g(x))$$.

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

Problem: If $$f(x)=3x^2+2$$ and $$g(x)=\\sqrt{x+1}$$, find $$f(g(x))$$.

Solution:

Substitute g(x) into f(x):

$$f(g(x))=3(\\sqrt{x+1})^2+2$$

Simplify the square:

$$3(x+1)+2=3x+5$$

Final answer: $$3x+5$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$x+2$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$x+3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2x+3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$3x+5$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9lm07jzjgllmuikx0o7","texte_question":"

Find the inverse of the one-to-one function $$g(x)=\\sqrt[3]{x+5}$$.

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

Problem: Find the inverse of the one-to-one function $$g(x)=\\sqrt[3]{x+5}$$.

Solution:

Let

$$y=\\sqrt[3]{x+5}$$

Swap the variables:

$$x=\\sqrt[3]{y+5}$$

Cube both sides:

$$x^3=y+5$$

Solve for $$y$$:

$$y=x^3-5$$

Replace $$y$$ with the inverse function notation.

Final answer: $$g^{-1}(x)=x^3-5$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$g^{-1}(x)=\\frac{1}{3}x^3+5$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g^{-1}(x)=x+5$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g^{-1}(x)=x^3-5$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$g^{-1}(x)=(x+5)^3$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9lp07k5jgllwo22jcef","texte_question":"

Find the zeros of $$g(x)=x^4-5x^2$$.

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

Problem: Find the zeros of $$g(x)=x^4-5x^2$$.

Solution:

Set the function equal to zero:

$$x^4-5x^2=0$$

Factor the expression:

$$x^2(x^2-5)=0$$

Set each factor equal to zero.

First factor:

$$x^2=0$$

$$x=0$$

Second factor:

$$x^2-5=0$$

$$x^2=5$$

$$x=\\pm\\sqrt{5}$$

Final answer: $$0,\\ -\\sqrt{5},\\ \\sqrt{5}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-5$$, $$5$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$, $$-\\sqrt{5}$$, $$\\sqrt{5}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-\\sqrt{5}$$, $$\\sqrt{5}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$, $$5$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9lt07kbjglll0l2h6g2","texte_question":"

Consider the function:

$$g(x)=\\frac{2}{a}\\sin(ax)$$

For what value of $$a$$ does $$g$$ have a period of $$\\pi$$?

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

Problem: Consider the function: $$g(x)=\\frac{2}{a}\\sin(ax)$$

For what value of $$a$$ does $$g$$ have a period of $$\\pi$$

Solution:

The period of the sine function $$\\sin(ax)$$ is given by $$\\frac{2\\pi}{|a|}$$

The coefficient $$\\frac{2}{a}$$ affects only the amplitude and does not affect the period.

Set the period equal to $$\\pi$$:

$$\\frac{2\\pi}{|a|}=\\pi$$

Solve for $$|a|$$:

$$2\\pi=\\pi|a|$$

$$|a|=2$$

Thus, the possible values of $$a$$ are:

$$a=2$$

$$a=-2$$

Final answer: $$-2$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-2$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9lw07khjgllec4rpfjh","texte_question":"

Consider the function $$g(x)=\\frac{2}{a}\\sin(ax).$$ What is the amplitude of $$g$$?

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

Problem: Consider the function $$g(x)=\\frac{2}{a}\\sin(ax).$$ What is the amplitude of $$g$$.

Solution:

The amplitude of a sine function of the form $$g(x)=A\\sin(ax)$$ is the absolute value of the coefficient $$A$$.

Here, the coefficient of the sine function is $$\\frac{2}{a}$$.

Therefore, the amplitude is $$\\left|\\frac{2}{a}\\right|$$.

Final answer: $$\\frac{2}{|a|}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$a$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\pi.a$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2\\pi.a$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{2}{|a|}$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9ly07knjgllj3a6v54g","texte_question":"

Consider the function $$g(x)=2+\\cos\\left(\\frac{\\pi x}{4}\\right).$$ What is the smallest positive value of $$x$$ for which $$g$$ is a minimum?

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

Problem: Consider the function $$g(x)=2+\\cos\\left(\\frac{\\pi x}{4}\\right).$$ What is the smallest positive value of $$x$$ for which $$g$$ is a minimum?

Solution:

The function $$g(x)=2+\\cos\\left(\\frac{\\pi x}{4}\\right)$$ is minimized when the cosine term is minimized.

The minimum value of cosine is $$-1$$, so we need

$$\\cos\\left(\\frac{\\pi x}{4}\\right)=-1.$$

Cosine equals $$-1$$ when its angle is an odd multiple of $$\\pi$$:

$$\\frac{\\pi x}{4}=\\pi+2\\pi k,$$

where $$k$$ is an integer.

Divide both sides by $$\\pi$$:

$$\\frac{x}{4}=1+2k.$$

Multiply both sides by $$4$$:

$$x=4(1+2k)=4+8k.$$

The smallest positive value occurs when $$k=0$$:

$$x=4.$$

Final answer: $$4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$4$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9m107ktjgll1s3bgxt3","texte_question":"

The domain of $$h(x)=\\frac{\\sqrt{x-5}}{x^2-9}$$ is:

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

Problem: The domain of $$h(x)=\\frac{\\sqrt{x-5}}{x^2-9}$$ is:

Solution:

For the square root to be defined, $$x-5\\ge0.$$

This gives: $$x\\ge5$$

For the denominator to be nonzero,

$$x^2-9\\neq0$$

Factor the denominator:

$$x^2-9=(x-3)(x+3)$$

So,

$$x\\neq3$$ and $$x\\neq-3$$

Combine the conditions:

$$x\\ge5$$ and $$x\\neq3$$

Since $$-3<5$$, it is already excluded by $$x\\ge5.$$

Thus the domain is:

$$x\\ge5,$$ $$x\\neq3$$

Final answer: $$x\\ge5$$ and $$x\\neq\\pm3$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$x\\neq\\pm3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$x\\le5$$ and $$x\\neq\\pm3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$x\\ge5$$ and $$x\\neq\\pm3$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$x>5$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmpz9r9m407kzjgllmgcs7lr3","texte_question":"

If $$f(x)=2x^3-5x+7$$ and $$g(x)=4,$$ then $$g(f(x))=$$?

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

Problem: If $$f(x)=2x^3-5x+7$$ and $$g(x)=4,$$ then $$g(f(x))=$$?

Solution:

The function $$g(x)$$ is a constant function.

This means that for any input value,

$$g(x)=4.$$

Substitute $$f(x)$$ into $$g$$:

$$g(f(x))=4.$$

The value of $$f(x)$$ does not change the output because $$g$$ always equals $$4.$$

Final answer: $$4$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$8x^3-20x+28$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$4x^3-5x+7$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2x^3-5x+7$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$4$$","correcte":true,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Functions "],"date":"2026-06-04T09:04:33.357Z","modified":"2026-06-04T09:04:33.357Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"Functions-Quiz 1"}]}]]}]}]]

Quiz 1

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