<p>1) A monkey sits atop a tree and sees two birds. He can look at one of the birds by tilting his head up at an angle of 1 above the horizontal. He can look at the second bird by tilting his head down at an angle of 2 below the horizontal. If both birds are a horizontal distance of d meters from the tree (i.e., one is flying directly above the other), how far apart vertically are the two birds, in meters?
(A) d/3
(B) d(sin 1 + sin 2)
(C) d(cos 1 + cos 2)
(D) d tan(1 + 2)
(E) d(tan 1 + tan 2)</p>
<p>2) f(x) = (x^2)-1 if x > 2 and x + 7 if x less than or greater to 2. What is the range of f?
(A) y less than or equal to 1
(B) y > 1
(C) y is greater than or equal to 0
(D) y is equal to
(E) All real numbers</p>
<p>3) A and B are points on circle O, with the center at the origin. In rectangular coordinates A is the point (0, 5), and B is the point ((5 rt 3)/2, 5/2). What is the length of minor arc AB?
(A) 5pi/6
(B) 2pi/3
(C) 5pi/3
(D) 5pi/2
(E) 10pi/3</p>
<p>4) The region of the xy plane that satisfies y > x2 or y > 0 is
(A) a circle
(B) a square
(C) the region of the plane bounded by a parabola
(D) a crescent-shaped region in the plane
(E) the half of the plane above the x-axis</p>
<p>If we take the horizontal line from the position of monkey’s eye as the base of the triangle and his line of sight to be the two hypotenuses, then we will have two triangles; one of of angle 1 and the other (upside down) of angle 2. </p>
<p>Letting h1 be the height of the top bird above the horizontal and h2 the height of the second bird below the horizontal, we find that </p>
<p>tan 1 = h1 / d and tan 2 = h2 / d</p>
<p>So, h1 + h2 = d(tan 1 + tan 2), which is the height.</p>
<ol>
<li>(E)</li>
</ol>
<p>We plug in x = 2 into f(x). The range of the first piecewise part is (3, infty) and the range of the second part is (-infty, 9]. Since they overlap, the range of f(x) is all real numbers.</p>
<ol>
<li>(A)</li>
</ol>
<p>We can find the angle that the origin to the second point make with respect to the origin. We can calculate: </p>
<p>so, the complementary angle is pi/2 - pi/3 = pi/6.</p>
<p>This is the angle between the two points on the circle. Now we have s = r(theta), so s = 5(pi/6).</p>
<ol>
<li>(E)</li>
</ol>
<p>We have an OR condition here, so if y > 0, it is implied that y > x^2, since x^2 greater than or equal to 0. So, we have the whole region above y = 0, which is a half-plane.</p>
<p>For #3, you can use the unit circle values to find out that one point is at 90 degrees and the other at 30 degrees, this gives you pi/2 and pi/6.</p>
<p>So the arc created is 60 degrees. Then I took the circumference of the circle and got pi*2(5) = 10pi. Since 60 degrees is 1/6 of the total circumference, multiple 10pi by 1/6. This gives you 5/3pi or (C).</p>
<p>EDIT - Are the real questions on the actual Math 2 test like this?</p>
<p>Yeah, I got C for 3 as well. And I’m wondering the same thing in terms of the similarity of these questions to the actual ones on the Math II test…</p>
<p>Okay, we can use vectors to solve this problem. For the first point, (0,5), the vector from the origin to the point would be straight up, so it would have an angle of pi/2.</p>
<p>The vector from the origin to the second point would make some angle with respect to the x-axis. We can set up the triangle and solve this using tangent (refer to my original solution). Given this angle, we can find the angle between the first and second vector by taking the complement of this angle. This gives us pi/3.</p>
<p>So essentially, we are trying to find the arc of a circle of radius 5 when the arc is subtended by an angle of pi/3. We can use s = r(theta) here to arrive at the answer.</p>