October 2009 SAT II - Math Level 2

Test Preparation SAT Subject Test Preparation
81

<p>For the # of distinct roots of h(x) = f(x)g(x), I put 3 or 4.</p>

<p>Logically, if f(x)=/=g(x) then they cannot have the same two distinct roots. However, they can both share 1 distinct root and thus have 3 distinct roots between the two.</p>

<p>However, now that I think of it, if we’re saying that there are indistinct roots as in other repeated roots too, they can technically have 2, 3, or 4 distinct roots because f(x)=/=g(x) if one of them has additional, indistinct roots (IE, one has roots 2, 2, 4, and the other has roots 2 and 4, that means that f(x)=/=g(x) but they have the same 2 distinct roots to the equation).</p>

<p>:(</p>

82

<p>@steelersngators</p>

<p>I got 8 as well. I did 3 * 3 * 3 * 5 and then just bruteforced all of the factors and ended up with 8. </p>

<p>Which question was this? Was the answer choice E?</p>

83

<p>Hold on, for the h(x) question, I think the answer was that it could be 2,3, or 4.</p>

<p>Let f(x) = (x-2)(x-3), g(x) = (x-2)(x-4)
That’s 3 unique solutions for h(x) then, no?
Similarly you could say let f(x) = (x-2)(x-4) and let g(x) = (x-2)(x-4) and get two.
And finally you could say let f(x) = (x-2)(x-3) and let g(x) = (x-4)(x-5) and get four.</p>

<p>Correct me if I’m doing something wrong here.</p>

<p>EDIT: Oh for heaven’s sake, did the question say that f(x) can’t equal g(x)? :frowning:
EDIT2: But let f(x) = (x-2)(x-4) and let g(x) = 3(x-2)(x-4) and you still get two roots! :D</p>

84

<p>@RedCatharsis I agree with you and I was also thinking that (x-2)(x-3) is different than 2(x-2)(x-3) but I put 3 or 4 as well and now I’m thinking that its 2,3, or 4</p>

85

<p>@smarts1</p>

<p>I don’t remember the exact wording of the question nor which answer choice I chose. I remember picking the one that said 8 though.</p>

86

<p>Monster read my explanation. The reason why we can’t set the two functions equal is that the question itself says that they are NOT equal.</p>

<p>However, the question doesn’t explain whether or not to take into account INDISTINCT roots. I ended up choosing 3 and 4 anyways.</p>

87

<p>I’m quite sure it was asking for real roots.</p>

88

<p>Aaaaargh. Stupid test. I was getting 780 on the official practice test, 740-780 on Sparknotes and 650-700 on Barron’s. Today, I totally sucked! This was so so so much harder than those practice tests!
Omit 5, guess 5! and probably whoknowshowmany wrong!</p>

89

<p>Has anyone taken this test before? Was today’s test harder? And does anyone know much about the curve?</p>

<p>I’ve heard -6 has been 800 a lot in the past. Maybe we’ll get -7 or possibly even -8.</p>

90

<p>Ya’ll got that one that was like 1 + a^2?</p>

<p>And there was the question w/ the graphs of -sinx, abs(sinx), cos x, the answer was I and II, right?</p>

<p>@Abatis: It doesn’t matter if it’s asking for real roots or not, you can still get 2,3,4 roots…well, 2 is debatable, as ESSENTIAL and I have elucidated to.</p>

91

<p>

What exactly do you mean by that? And if you let f(x) = (x-2)(x-4) and let g(x) = 3(x-2)(x-4), wouldn’t you get two distinct roots?</p>

92

<p>@RedCatharsis</p>

<p>Y was 0 when x was pi. Y was 1 when x was 3pi/2 and Y was 0 when x was 2pi. I believe that’s what the curve looked like.</p>

<p>I got I and II as well.</p>

93

<p>@RedCatharsis yes because 1+tan ^2 x=sec^2 x (trig identity), and it was I and II</p>

94

<p>RedCatharsis-i got 1 +a squared as well. and for the 3 graphs, thats what i got too. cos x was the only that didn’t match the graph they showed</p>

95

<p>monstor344, I believe that that is debatable in the sense that the functions are arguably identical. Idk, I think there’s a technicality.</p>

<p>Either way, it’s starting to look like 2,3,4 roots…Unless somebody has a good explanation?</p>

96

<p>I have a good explanation… for why I left it blank, lol.</p>

97

<p>Did anyone here bother with the last question about the cone? I had like 1 minute left when I got to it and was like, “screw this…”</p>

98

<p>I got 4.8 but I’m not sure</p>

99

<p>what was the answer to the -sinx abs(sinx) and cosx question?</p>

100

<p>It’s 4.8 I checked every single answer. i moved pretty fast, but I forgot to check the h(x) roots thing. :&lt;/p>

<p>I remember that there wasn’t really anything tricky on this test…I mean I didn’t spend a lot of time on any problem until the New York Population one.</p>