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3"},"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":70,"ordre_questions":"Fixe","questions":[{"id":"cmlk9a4ti0002l104pfto9sn7","texte_question":"

Let $$f(x)$$ have an inverse function $$h(x)$$. Then $$h(f(x))=$$?

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

If $$h(x)$$ is the inverse of $$f(x)$$, then composing $$h$$ with $$f$$ returns the original input:

$$h(f(x))=x.$$

This is the defining property of inverse functions.

Final answer: $$x$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$x$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$f(x)\\cdot h(x)$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj0007l104j65k3mwm","texte_question":"

The function $$f(x)=x^3+2x-4$$ has exactly one real zero. It is between:

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

To find where the real zero is, evaluate $$f(x)$$ at the endpoints of the intervals and look for a sign change.

Compute $$f(0)$$:

$$f(0)=0^3+2(0)-4=-4$$

Compute $$f(1)$$:

$$f(1)=1^3+2(1)-4=1+2-4=-1$$

There is no sign change between $$0$$ and $$1$$ because both values are negative.

Compute $$f(2)$$:

$$f(2)=2^3+2(2)-4=8+4-4=8$$

Now compare $$f(1)$$ and $$f(2)$$:

$$f(1)=-1$$

$$f(2)=8$$

There is a sign change from negative to positive, so the zero must be between $$1$$ and $$2$$.

Final answer: $$1$$ and $$2$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$0$$ and $$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$ and $$2$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$2$$ and $$3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$3$$ and $$4$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj000cl1047h71kwqf","texte_question":"

The range of $$y=\\ln(2+\\sin x)$$ is:

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

The sine function satisfies:

$$-1\\le\\sin x\\le1$$

Add $$2$$ to all parts:

$$2-1\\le2+\\sin x\\le2+1$$

So:

$$1\\le2+\\sin x\\le3$$

Now apply the natural logarithm.

Since $$\\ln(x)$$ is increasing for positive $$x$$, the inequality directions stay the same:

$$\\ln(1)\\le\\ln(2+\\sin x)\\le\\ln(3)$$

Simplify:

$$\\ln(1)=0$$

So, the range is:

$$0\\le y\\le\\ln(3)$$

Final answer: $$[0,\\ln(3)]$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$(-\\Infty,\\ln(3)]$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$[0,\\ln(3)]$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$[-1,1]$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(-1,1)$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj000hl104rgaiel7m","texte_question":"

Solve for $$b$$: $$\\log_b(b^4)=8$$

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

Use the definition of logarithms:

$$\\log_b(b^4)=8$$

means

$$b^8=b^4.$$

Divide both sides by $$b^4$$:

$$b^4=1$$

Solve:

$$b=\\pm1$$

But a logarithm base must satisfy:

$$b>0$$ and $$b\\ne1$$

The value $$b=1$$ is not allowed, and $$b=-1$$ is not allowed because the base must be positive.

So, there is no real value of $$b$$ that satisfies the equation.

Final answer: No real solution

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$0$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$1$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"No real solution","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj000ml104vo1zifuf","texte_question":"

Let $$f^{-1}$$ be the inverse function of $$f(x)=x^3-5$$. Then $$f^{-1}(x)=$$?

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

Start with:

$$y=x^3-5$$

Swap $$x$$ and $$y$$ to find the inverse:

$$x=y^3-5$$

Solve for $$y$$:

Add $$5$$ to both sides:

$$x+5=y^3$$

Take the cube root of both sides:

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

So:

$$f^{-1}(x)=\\sqrt[3]{x+5}$$

Final answer: $$\\sqrt[3]{x+5}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{1}{x^3+5}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$(x+5)^3$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\sqrt[3]{x+5}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{(x+5)^3}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj000rl1041u54v4h2","texte_question":"

The set of $$x$$-intercepts of the graph of $$f(x)=x^3-x^2-4x+4$$ is:

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

The $$x$$-intercepts occur where:

$$f(x)=0$$

Set the function equal to zero:

$$x^3-x^2-4x+4=0$$

Factor by grouping:

Group the first two terms and the last two terms:

$$(x^3-x^2)+(-4x+4)=0$$

Factor out common factors:

$$x^2(x-1)-4(x-1)=0$$

Factor out $$(x-1)$$:

$$(x-1)(x^2-4)=0$$

Factor the difference of squares:

$$x^2-4=(x-2)(x+2)$$

So:

$$(x-1)(x-2)(x+2)=0$$

Set each factor equal to $$0$$:

$$x-1=0$$ gives $$x=1$$

$$x-2=0$$ gives $$x=2$$

$$x+2=0$$ gives $$x=-2$$

So, the set of $$x$$-intercepts is: $$\\{-2,1,2\\}$$

Final answer: $$\\{-2,1,2\\}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\{-2,2\\}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\{-1,1,4\\}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\{-2,1,2\\}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\{-4,-1,1\\}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj000wl104539dyy1g","texte_question":"

If the domain of $$f$$ is restricted to the open interval $$\\left(-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right)$$, then the range of $$f(x)=e^{2\\tan x}$$ is:

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

On the interval $$\\left(-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right)$$, the tangent function is continuous and its range is:

$$-\\infty<\\tan x<\\infty$$

Multiply by $$2$$:

$$-\\infty<2\\tan x<\\infty$$

Now apply the exponential function:

$$y=e^{2\\tan x}$$

For any real number $$t$$, $$e^t>0$$.

Also, as $$t\\to-\\infty$$, $$e^t\\to0$$ but never equals $$0$$.

As $$t\\to\\infty$$, $$e^t\\to\\infty$$

So, the range of $$e^{2\\tan x}$$ is:

$$0<y<\\infty$$

This is the set of all positive real numbers.

Final answer: the set of all positive real numbers.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"the set of all real numbers","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"the set of positive real numbers","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"the set of nonnegative real numbers","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"the set of real numbers in the interval (0,1]","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj0011l104tyec5lb9","texte_question":"

Which of the following represents a reflection of the graph of $$y=f(x)$$ in the $$y$$-axis?

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

A reflection in the $$y$$-axis changes every point $$(x,y)$$ on the graph to $$(-x,y)$$.

This means we replace $$x$$ with $$-x$$ in the function.

So, the reflected graph is: $$y=f(-x)$$

Final answer: $$y=f(-x)$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$y=-f(x)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$y=f(-x)$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$y=f(|x|)$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$y=-f(-x)$$","correcte":false,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj0016l104qsq4hr7f","texte_question":"

The smallest positive $$x$$ for which the function $$f(x)=\\cos\\left(\\frac{x}{4}\\right)+2$$ is a maximum is:

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

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

The maximum value of cosine is:

$$\\cos(\\theta)=1$$

So, we need:

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

Cosine equals $$1$$ when its angle is:

$$\\frac{x}{4}=2\\pi n,$$

where $$n$$ is an integer.

Solve for $$x$$:

$$x=8\\pi n$$

The smallest positive $$x$$ occurs when $$n=1$$:

$$x=8\\pi$$

Final answer: $$8\\pi$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$2\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$4\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$6\\pi$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$8\\pi$$","correcte":true,"explication":"","imageUrl":"$undefined"}]},{"id":"cmlk9a4tj001bl1047m426dtj","texte_question":"

Evaluate: $$\\sin(\\arccos(\\frac{1}{2}))$$

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

Let:

$$\\theta=\\arccos\\left(\\frac{1}{2}\\right).$$

This means:

$$\\cos(\\theta)=\\frac{1}{2}$$

and $$\\theta$$ is in the range:

$$0\\le\\theta\\le\\pi$$

Since $$\\cos(\\theta)=\\frac{1}{2}$$ and $$\\theta$$ is in $$[0,\\pi]$$, the angle is:

$$\\theta=\\frac{\\pi}{3}$$

Now compute:

$$\\sin(\\theta)=\\sin\\left(\\frac{\\pi}{3}\\right)=\\frac{\\sqrt{3}}{2}$$

Final answer: $$\\frac{\\sqrt{3}}{2}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$-\\frac{\\sqrt{3}}{2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{\\sqrt{3}}{2}$$","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"$$$-\\frac{1}{2}$$","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"$$$\\frac{1}{2}$$","correcte":false,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Functions "],"date":"2026-02-13T02:15:52.035Z","modified":"2026-02-20T11:29:30.438Z"}}]]}]}]]