SAT II: Math Level 2 - Help with a few questions!

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1

<p>Hey,</p>

<p>I am asking these questions from the CollegeBoard book that has all the practice tests.</p>

<p>Q6. The intersection of a cube with a plane could be which of the following?</p>

<p>I. Square
II. Parallelogram
III. Triangle</p>

<p>(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II and III</p>

<p>The correct answer is E... but how? I guess I just couldn't visualize this problem. Can you help?</p>

<p>Q40. If f(x) = x^3 - 4x^2 - 3x + 2, which of the following statements are true?</p>

<p>I. The function f is increasing for all x >= 3.
II. The equation f (x) = 0 has two nonreal solutions.
III. f (x) >= (-16) for all x>=0</p>

<p>(A) I only
(B) II only
(C) I and II
(D) I and III
(E) II and III</p>

<p>The correct answer is D. I factorized the equation on my graphing calculator, and the roots confirmed that II was out, so it was down to I and/or III. But, I got stuck after that. How do I proceed? A wild idea I had was to differentiate the equation and find the maxima/minima and increasing/decreasing nature from that. But I got this idea after I finished the test. Any help with this?</p>

<p>Q41. This question has a graph, so I can't copy it down here. If any of you have the Subject Tests Practict Tests book, then please look into Q41 of the Math Level 2 paper, and help me out with this...</p>

<p>Q42. The set of all real numbers x such that [square root of x^2] = -x consists of</p>

<p>(A) zero only
(B) nonpositive real numbers only
(C) positive real numbers only
(D) all real numbers
(E) no real numbers</p>

<p>The correct answer is B. I went back-n-forth between B and D for a long time, before selecting D. The reasoning I had used was... suppose I put in a positive real integer (say 9) in the LHS, then squaring 9, I get 81. Now 81 has two square roots, right? A positive and a negative one? So, the RHS holds true! But, seems like that is not the case. Where did I go wrong?</p>

<p>That's all. But it would be highly appreciated if you all could get back ASAP. The exam's only tomorrow :D</p>

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<p>For six, think of the corner of a cube. A square is a parallelogram.</p>

3

<p>For #42, let’s say you have a negative real number -n. sqrt((-n)^2) = n, which is equivalent to -(-n). I think you’re a bit confused on when the square root of a number results in both a positive and a negative root. You only need to do that when you’re solving a quadratic equation (e.g., x^2 = 25).</p>

4

<h1>40 Just graph it and look at the graph, that’s what I would do. There is a minimum at x=3, so everything after that has to be increasing (since it’s the “lowest” point). The y-value at x=3 is -16 (from looking at the table/trace/value), so since this is the minimum, every value after x=0 IS going to be greater than or equal to -16.</h1>

<p>PS: You won’t need calc to solve any of the problems (so you won’t need to differentiate anything).</p>

5

<p>For Q42: simply set y1 to be sqrt(x^2) and y2 to be -x.
graph, and see what points they have in common in what locations.</p>

<p>dont make it complicated. to follow up on what the above poster said, graphing functions out and simply observing is a good strategy!</p>

6

<p>Well, you’re either in the exam right now, or done with it, but here are some explanations in case you wanted to know anyway!</p>

<p>For (I) in Q6, imagine a cube lying on a flat surface with a plane intersecting through the cube parallel to the flat surface. This intersection results in a square. For (II), imagine the same cube in the same position, except now the plane going through it is at some tilt (any tilt, except one that is parallel or perpendicular to the flat surface). This would produce a parallelogram. For (III), which I surmise is where your problem lies with this question, imagine the same cube, except now the plane intersects only a corner of the cube (it would look like the plane has been pierced with a corner). This would produce a triangle.</p>

<p>For Q40, did you try graphing the function? I and III would become clear from the visualization, but I’m assuming you’re looking for an analytic interpretation. First off, you could definitely differentiate the function and find the x-values at which the derivative is 0, the relative (local) max’s and min’s of the function. Undoubtedly one of the solutions would be (3, -16), which is a local min of f. By the Extreme Value Theorem, this means that all f’(c) to the near left of x=3 is negative, and all f’(c) to the right of x=3 is positive, which means that I is true. Now you can also infer III, because if (3, -16) is a relative min for all x>=0, then all f(x) for all x>=0 is greater than or equal to three f(3)=(-16).</p>

<p>Q42 implies that if you put in an x (any x, which for now can be positive or negative), then the outcome is the opposite of x. Also remember that a radical that’s “bare” (without a + or - designation in front of it) is automatically assumed to imply the positive square root only. If you put in 9, square it, and take the <em>positive</em> square root of 81 (again, what the radical on the left hand side of the equation implies), you do not get -x, or -9 on the right side. If you put in (-9), square it, and take the positive square root of 81, you get 9/-(-9)/-x. Therefore, this equation applies only to all real numbers that are not positive (because it can apply to 0 also).</p>

<p>Hopefully this helps!!</p>

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<p>Thanks for all the suggestions people! Surely helped :D</p>