Question

(WILL DO BRAINLIST)

Answer 1

You have a function $y= \dfrac{3}{2} \left(1+\sin \left(\dfrac{2t+1}{2}\cdot \pi\right) \right)$.
Since the range of the function $y=\sin x$ is $[-1,1]$ you have that

$-1\le \sin \left(\dfrac{2t+1}{2}\cdot \pi\right)\le 1 \\ 0\le 1+\sin \left(\dfrac{2t+1}{2}\cdot \pi\right) \le 2 \\ 0\le \dfrac{3}{2} \left(1+\sin \left(\dfrac{2t+1}{2}\cdot \pi\right) \right)\le 3$.
Answer: The range of given function is [0,3] and the correct choice is C.