Question

10) (PUC) O gráfico abaixo é da função f. A lei de f é:
a) f(x)=3cos x/2
b) f(x) = 3cos 2x
c) f(x) = 3sen x/2
d) f(x) = 3sen 2x
e) f(x) = 3+ sen x/2

11) (UFRGS) O grafico abaixo representa uma função real f. Esta função é dada por.
a) f(x) = 1- cos x
b) f(x) = 1+ cos x
c) f(x) = cos(x + 1)
d) f(x) = cos(x - 1)
e) f(x) = cos(x +π)

Answer 1

$\Large\boxed{\begin{array}{l}\rm 10)\\\sf Im=[a-b,a+b]=[-3,3]\\+\underline{\begin{cases}\sf a-\backslash\!\!\!b=-3\\\sf a+\backslash\!\!\!b=3\end{cases}}\\\sf 2a=0\\\sf a=\dfrac{0}{2}=0\\\sf a+b=3\\\sf 0+b=3\\\sf b=3\\\sf P=\dfrac{2\pi}{c}=4\pi\\\sf 4\pi c=2\pi\\\sf c=\dfrac{2\pi\div2\pi}{4\pi\div2\pi}\\\\\sf c=\dfrac{1}{2}\\\sf f(x)=a+bsen(cx)\\\sf f(x)=0+3sen\bigg(\dfrac{1}{2}x\bigg)\\\sf f(x)=3\cdot sen\bigg(\dfrac{x}{2}\bigg)\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~c}}}}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm 11)\\\sf Im=[a-b,a+b]=[0,2]\\+\underline{\begin{cases}\sf a-\diagup\!\!\!b=0\\\sf a+\diagup\!\!\!b=2\end{cases}}\\\sf 2a=2\\\sf a=\dfrac{2}{2}\\\sf a=1\\\sf a-b=0\implies a=b\\\sf b=1\\\sf P=\dfrac{\diagdown\!\!\!\!\!\!2\pi}{c}=\diagdown\!\!\!\!\!\!2\pi\\\sf c=1\\\sf f(x)=a+b\cdot cos(cx)\\\sf f(x)=1+1\cdot cos(1x)\\\sf f(x)=1+cos(x)\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~b}}}}\end{array}}$