**Official June 2013 SAT II Math 2 Thread**

<p>Hmm. True.</p>

<p>@RMIBstudent. Yes, please report ambiguous questions to <a href="mailto:satquestion@info.collegeboard.org">satquestion@info.collegeboard.org</a>. I already reported this one, but there’s no harm in reporting it again.</p>

<p>There’s still the fact, though, that leading coefficients can’t be zero.</p>

<p>johnstucky: oh, good idea. Thanks.</p>

<p>Also, there was a very weird question about F(x) = (x-2)F(x-2)…how did you do that? I skipped it, thought it was a typo.</p>

<p>They said in the question that it was 5th degree, making the polynomial odd. An odd function MUST have at least one real root.</p>

<p>I think you’re missing the domain question on your list. What’s in the domain of (x-2)/(x+1)(x-3) or something along those lines. All real numbers except -1 and 3.</p>

<p>Another ambiguous one was the cone one. If anyone reports anything, report that question</p>

<p>Yes, leading coefficients can’t be 0, but the problem said that the variables could be any rational number, contradicting that statement. That’s why I think that that question is unfair.</p>

<p>reported even if you wrote zero it can only help everyone please report the ambiguity</p>

<p>Okay can anyone confirm the (x^2)/9 + (y^2)/25 question for me? I’m freakish out, I was almost 99% sure that answer D was a plus sign</p>

<p>@RMIBstudent</p>

<p>i think the question was F(x) = (x-1)F(x-1) = (1.4-1)F(1.4-1) = (0.4)(2.7)</p>

<p>zachattack: wait, are you absolutely certain that the question specified a 5th degree polynomial?</p>

<p>BurnOut: darn. lol</p>

<p>Hey Niceboat, is that a typo on #30? There were no exponents involved in that question…?</p>

<p>Well. I don’t think anyone can be sure, up to this point.</p>

<p>And, if someone say they are certain, half of us would deny it. lol. xD.</p>

<p>You’re right burnout. Since F is already a function. So there’s no squared.</p>

<p>I’ll change it.</p>

<p>How was the ice cube one 25?</p>

<p>Sent from my PH44100 using CC</p>

<p>for 37, how’d you reach to the conclusion of pi/b?
I totally yolo’d that question and chose 2pi/a</p>

<p>Pretty simple.</p>

<p>You just take the original period (1 for tan) and divided it by the coefficient of the variable which is b. so pi.b</p>

<p>So for example period of this graph sin(2x+5)
Sin’s normal period is 2pi.
2pi/2= pi.</p>

<p>It’s a giver if you know how to do it. </p>

<p>Ice cube is 25. Just plug it back in.</p>

<p>Pretty sure it did</p>

<p>Niceboat, can you delete 33? I think we’re all sure that question wasn’t on there.</p>

<p>The amplitude never affects the period/ frequency. It’s all whatever’s inside the tan x term. I guessed as well (kinda). I figured the natural period of tan x is pi, and i graphed tan5x to see that there were a TON of periods squished together, so pi / b seemed right.</p>

<p>That would require me to shift all the numbers. And tbh, I’m fairly sure it existed. It was one of the beginning question. It was a simple exponential problem.</p>