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ficha formativa2

Authored by Sílvia Machado

Mathematics

11th Grade

Used 2+ times

ficha formativa2
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24 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 2 pts

Dada a função 𝑓 (𝑥) = 2 𝑠𝑒𝑛 𝑥/2, podemos afirmar que esta função possui:

período 𝜋.

período 2 𝜋.

período 4 𝝅.

período 2 𝜋.

2.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

O valor mínimo da função seno ocorre quando:

  x=π+2kπ, k∈Zx=\pi+2k\pi,\ k\in Z  

  x=π2+2kπ, k∈Zx=\frac{\pi}{2}+2k\pi,\ k\in Z  

  x=3π2+2kπ, k∈Zx=\frac{3\pi}{2}+2k\pi,\ k\in Z  

  x=2kπ, k∈Zx=2k\pi,\ k\in Z  

3.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Qual é o Contra-Domínio da função f(x)=sin⁡x−6f\left(x\right)=\sin x-6  ?

[−7,−5][-7,-5]  

[5, 7][5,\ 7]  

[0, 6][0,\ 6]  

[−1,1][-1,1]  

4.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Qual é o Contra Domínio da função f(x)=cos⁡(x+4)f(x)=\cos(x+4)  

[3, 5][3,\ 5]  

[−4, 4][-4,\ 4]  

[−1,1][-1,1]  

[−5,−3][-5,-3]  

5.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Quais são as soluções da equação cos⁡(2x)=1\cos\left(2x\right)=1  

  x=kπ, k∈Zx=k\pi,\ k\in Z  

  x=π6+2kπ ∨x=\frac{\pi}{6}+2k\pi\ \vee    x=−π6+2kπ, k∈Zx=-\frac{\pi}{6}+2k\pi,\ k\in Z  

x=π3+2kπ ∨x=\frac{\pi}{3}+2k\pi\ \vee   x=−π3+2kπ, k∈Zx=-\frac{\pi}{3}+2k\pi,\ k\in Z  

  x=2kπ, k∈Zx=2k\pi,\ k\in Z  

6.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Considerando o intervalo x∈[0, 2π]x\in\left[0,\ 2\pi\right]  , quais as soluções para sin⁡(x)=−12\sin\left(x\right)=-\frac{1}{2}  ?

S={π6, 5π6}S=\left\{\frac{\pi}{6},\ \frac{5\pi}{6}\right\}  

S={7π6,11π6}S=\left\{\frac{7\pi}{6},\frac{11\pi}{6}\right\}  

S={4π3, 5π3}S=\left\{\frac{4\pi}{3},\ \frac{5\pi}{3}\right\}  

S={−π6, −5π6}S=\left\{-\frac{\pi}{6},\ -\frac{5\pi}{6}\right\}  

7.

MULTIPLE CHOICE QUESTION

5 mins • 10 pts

Se levarmos em consideração todos os números reais, qual é o conjunto solução da equação tan⁡(x−π3)−tan⁡(π4)=0\tan\left(x-\frac{\pi}{3}\right)-\tan\left(\frac{\pi}{4}\right)=0  ?

S={  }S=\left\{\ \ \right\}  

  x=7π12+kπ, k∈Zx=\frac{7\pi}{12}+k\pi,\ k\in Z  

   x=7π12+2kπ, k∈Zx=\frac{7\pi}{12}+2k\pi,\ k\in Z  

  x=π4+kπ, k∈Zx=\frac{\pi}{4}+k\pi,\ k\in Z  

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