How hard is Calc BC exam?

<p>I forgot to mention that I am a sophomore that is selfstudying BC. I take precalc at school. Does this change the dynamics of things?</p>

<p>For the Barrons, I can get about 80% of the stuff right, although the ones I miss are just super clever manipulations.</p>

<p>learn some identities...relax...learn some clever math tricks...relax.</p>

<p>you'll do fine, just study. at the same time, don't burn yourself out before the exam!</p>

<p>So would a 5 on physics C (both) be impressive if no one in your school got it?</p>

<p>Yes, it would be.</p>

<p>That sucks. I am gonna go to a special math/science boarding school, where 4's are shamed upon, and everyone gets a 5...</p>

<p>So wait... How would you guys compare Barrons to the actual AP Exam? And just how much do you have to get right to get a 5? And for BC, are the AB type questions chosen from the AB Exam itself?</p>

<p>i did about 200 problems out of the barrons book the 2 days before the BC exam and i got a 5 on BC. its a decent book, but make sure to use multiple sources cuz they usually focus well on certain subjects but are weaker on others. use the barrons with the princeton review book, for example.</p>

<p>you need roughly 62.5% to get a 5. which is relatively difficult because you need to know what you're doing in case there are questions that are impossible. Last year, for example, there was a free reponse question with an inseperable differential equation. i was able to leave that blank and do the rest of the test because i knew that i would have a better chance acing the other stuff. if you know everything really well, you can be safe leaving a free response question blank like i did and still get a 5.</p>

<p>So do they really give super impossible problems? Cuz I can only get about 70%-90% on barrons for the practice problems!!! I am screwed!!!</p>

<p>If you are referring to dy/dx=y+6 from last year's test, that isn't impossible....</p>

<p>Yeah, I can't remember there being an inseparable differential equation on the test. In fact, I'm reasonably sure that I got all the free response questions. Of course, I have a notoriously bad memory when it comes to remembering test questions, but yeah. AP tests also do have multiple versions, right? So it's possible that one version had something. Dunno.</p>

<p>Honestly, the test is really pretty easy. I did BC a while ago, through correspondence school. When I decided to switch to taking math at the local university, I had to take the AP test. I dunno. The math's all really pretty straightforward. I don't know exactly how well self studying works, as I've got absolutely no experience, but yeah. Honestly, if you're getting 70-90% on practice tests, you should prolly be fine.</p>

<p>i meant impossible for me =) also, there were different forms of the test. i believe that it was not dy/dx = y + 6, that'd be easy.</p>

<p>yeah im self studying it right now too. its not really hard, just some cool math tricks. i like calc bc, its fun stuff.</p>

<p>Well, the only thing that hits me is the insane tricks barrons pulls out. </p>

<p>Honestly, if I get a 4, will colleges like MIT and Caltech look down on me?</p>

<p>I am thinking about getting more prep books. I have barrons, and Arco's from 2000. What else?</p>

<p>Also, some of the trouble I have is with memorizing those damn inverse trig derivatives and integrals. Any tips?</p>

<p>a tip? arctan and arcsin should be the only ones you should really know for the APs ^^</p>

<p>HELP plz~!
Let f and g be functions that are differentiable for all real numbers x and that have the following properties.
i) f '(x) = f (x) - g (x)
ii) g '(x) = g (x) - f (x)
iii) f (0) = 5
iv) g (0) = 1
Find f (x) and g (x). Show your work. </p>

<p>I know I have to use Maclaurin and Taylor (maybe power?) series on this one but... can anyone help me out here? (Btw this is 94-BC6)</p>

<p>Yeah... I am screwed...</p>

<p>If I get a 4 on this exam, will it hurt me in college admissions? I am only a sophomore, but most ppl here took this stuff in like 7th grade or w/e.</p>

<p>For the question posted:</p>

<p>g(x) = 3 - 2e^(2x)
f(x) = 3 + 2e^(2x)</p>

<p>"
I know I have to use Maclaurin and Taylor (maybe power?) series on this one but... can anyone help me out here? (Btw this is 94-BC6)"</p>

<p>This is my solution, it has absolutely nothing to do with power series:</p>

<p>f'(x) + g'(x) = 0
f'(x) = -g'(x)
=> f(x) = -g(x) + c
Let x = 0 => f(0) = -g(0) + c
Hence c = 6
f(x) = 6 - g(x)
g'(x) = g(x) - (6 - g(x))
g'(x) = 2 g(x) - 6
Let y = g(x)
dy/dx = 2y-6
dy/dx - 2y = -6
I.F=> e^(-2x)
e^(-2x)(dy/dx - 2y) = -6e^(-2x)
d/dx(e^-2x * y) = -6e(-2x)
Integrating both sides,
e^(-2x)*y = 3e^(-2x) + c
y = g(x) = 3 +ce^(2x)
g(0) = 3 + c = 1
=> c = -2
f(x) = 6 - g(x)
f(x) = 3 + 2e^(2x)</p>