SAT1000 - P329

At $t = 0$ , there is no part of the circle above the $x$ -axis. That means $x(0) = 0$ and $f(0) = \cos (x(0)) = \cos(0) = 1$ , so the graph of $f(t)$ should go through $(0, 1)$ .

At $t = 1$ , $O$ is at the origin, and exactly half the circle is above the $x$ -axis. That means $x(1) = \pi$ and $f(1) = \cos (x(1)) = \cos(\pi) = -1$ , so the graph of $f(t)$ should go through $(1, -1)$ .

When $\angle AOB$ is a right angle, $\triangle AOB$ is an isosceles right triangle with a height of $\frac{\sqrt{2}}{2}$ , and exactly one fourth of the circle is above the $x$ -axis. That means $t = 1 - \frac{\sqrt{2}}{2} \approx 0.293$ , $x(0.293) = \frac{\pi}{2}$ and $f(0.293) = \cos (x(0.293)) = \cos(\frac{\pi}{2}) = 0$ , so the graph of $f(t)$ should go through $(0.293, 0)$ .

The only graph that goes through $(0, 1)$ , $(1, -1)$ , and $(0.293, 0)$ is $\boxed{B}$ .