[PREP Q and A] AP Calculus BC

<p>
[quote]
1. If c is a positive integer, the lim(x -> +inf) of (x^c)/(e^x) = ?
A) 0 B) 1 C) e D) c! E) DNE</p>

<ol>
<li>Suppose f and g are functions that are everywhere differentiable. Given that g is the inverse function of f and g(-2) = 5 and f '(5) = -1/2, then g '(-2) = ?</li>
</ol>

<p>A) 2 B) 1/2 C) 1/5 D) -1/5 E) -2</p>

<ol>
<li><p>If integral(1 to 4)[f(x)<em>dx] = 6, compute integral(1 to 4)[f(5-x)</em>dx]
A) 6 B) 3 C) 0 D) -1 E) -6

[/quote]
</p></li>
<li><p>A</p></li>
<li><p>Can you check this question for any mistakes? (Sorry, I just don't get it. I jump to the obvious answer: 0, which isn't a choice.</p></li>
<li><p>A</p></li>
</ol>

<p>sorry guys, i thought this thread died, should bump more often. anyways...</p>

<p>vader, that is correct, nice.</p>

<p>RESMonkey, yes A) -1 is the right answer. 3/2 is also correct, FTC FTW. I solved the x^2003 and x^(1/2003) one awhile ago but I remember using just one integral I think. Don't worry about this one the source is not from the AP test but it is noncalc.</p>

<p>You also got 1 and 3 correct. 2 is correctly stated. I'll give you a hint however. An inverse function f is one that returns the inputs of a different function g, so we can write f(g(x)) = x. It can be proved that if f is an inverse function of g, then g is also an inverse function of f. Hope this points you in the right direction.</p>

<p>It can be proved that if f is an inverse function of g, then g is also an inverse function of f.</p>

<p>This is not true unless f and g are both one-to-one functions.</p>

<ol>
<li>E </li>
</ol>

<p>ok, g = 1/f
g'(-2) = [f^-1(-2)]' = 1/f'(f^-1(x)), where f^-1(x) = -2; so since f^-1(x) = -2; and since its an inverse, the x and y are reversed, and g = f^-1, when y-variable = -2, its the x-variable of g, so y of f^-1 = x-of g, and you know g(-2) = 5, so f^-1(5) = -2. so it becomes: [f^-1(-2)]' = 1/f'(5), which is 1 / -1/2 = -2...so E</p>

<p>wow this stuff is mad hard...any tips for reviewing you guys?</p>

<p>
[quote]
I. Sigma[n = 1, +inf] [(-1)^(n+1)] * 1/(2n+1)</p>

<p>II. Sigma<a href="1/n">n = 1, +inf</a>*(3/2)^n</p>

<p>III. Sigma[n = 2, +inf] 1/(n*ln(n))

[/quote]
</p>

<p>Only I converges, and you can use the alternating series test, right? Because II is always bigger than or equal to (1/n) which diverges since it's the harmonic series, and the integral of III is ln(ln(n)). But I'm probably wrong about this.</p>

<p>@siddharthdhami</p>

<p>(a) a = f ' (x) > 0 @ t=2, so velocity is increasing. </p>

<p>(b) t = 12. when v is 55 again the Integration(0 to x) a(t) dt = 0; so the are under curve from 0 to 6 is 60; and are are from 6-10 is -30, and from 10-12 is -30; so 30+30-30-30 = 0; this is made possible at t=12..I'M SURE THERE IS ANOTHER MORE EFFICIENT WAY TO DO THIS THAN JUST GUESSAnd test: but it still still works!!! :)</p>

<p>(c)maximum occurs when f ' = 0; this happens at t=6, 16. BUT in order for it t0 be a max, the graph must switch from positive to 0 to negative; this happens at 6; but not at 16--so the answer is 6. Now to find the value you do Integration(0 to 6) a(t) dt; find the area under the curve its: 115 ft/s </p>

<p>(d) there is no such point becuase a (f ') is defined everywhere in the interval; if such a point existed, it would be at a point where a was undefined!</p>

<p>@vader1990</p>

<p>For part B, that is actually the most efficient way. If you actually look at the area of the a(t) curve from t=0 to 6, and then look at the area from t=6 to 12, you will realize that it's the exact same quadrilateral, and because t=0 to 6 is positive and t=6 to 12 is negative, they cancel each other out, which then gives you the conclusion that v(12) is also 55 ft/s.</p>

<p>@shravas</p>

<p>I do believe that 1 converges due to the alternating series test, but I also believe that 2 converges due to the ratio test.
let An = (1/n)<em>(3/2)^n
let A(n+1) = (1/(n+1))</em>(3/2)^(n+1)</p>

<p>lim |(1/(n+1))<em>(3/2)^(n+1)</em>1/((1/n)<em>(3/2)^n)| = lim|(2/3)</em>n/(n+1)|
using L'Hopital's rule, it then equals lim|2/3| as n approaches infinity, which is 2/3 < 1. Not sure about 3 though, I'm too tired to think :]</p>

<p>^^ @vader1990</p>

<p>Nice job almost everything is right...except...in part (c) where you have to find the car's absolute maximum velocity...I think the area should be 60 right??? (Area from 0 to 6)</p>

<p>^siddharthdhami </p>

<p>115 IS is the CORRECT ANSWER, I'll tell you why: You are correct, the area is 60, but you overlooked that initially at t=0, velocity was 55; so the total velocity is 55 + Integration(0 to 6) a(t) dt = 60 +55 = 115 :)</p>

<p>@bfeng:</p>

<p>I think that 2 diverges, not converges, because the if you put it in the form of (a) * (r)^(n-1), then you get the ratio as 3/2, which is greater than 1, so it converges.</p>

<p>a. it's increasing because a(t) > 0.</p>

<p>b. at t= 12 because the net change in velocity over t:(0,12) is zero</p>

<p>c. 55 +30 + 30 = 115 ft/sec at t = 6. The net change after only decreases this high value.</p>

<p>d. never, because there is not nuff change in v(t) caused by a(t) (using area, of course) to knock 55 down to zero.</p>

<p>@diamondbacker. you are right. 1 converges, 2 diverges. Can you explain 3 though? I can't get that one.</p>

<p>Math brains, a question for you guys (I really need help, since I suck at calc)</p>

<p>If dy/dt=ky and k is a nonzero constant, then y could be...
(A)2e^kty (B)2e^kt (C)e^kt+3 (D)kty=5 (E)0.5ky^2 + 0.5</p>

<p>^your answer is (b)</p>

<p>seperate the variables: dy / y = kt-----> integrate both sides ----->
ln[y] = kt + C ------>so y= Ce^(kt), where c is a constant, (b) is the only choice where this is the case :)</p>

<p>^but then, when you get rid of the ln[y], wouldnt the constant also go INTO the e^(kt+C)? please enlighten me</p>

<p>Yes that constant would. In essence you have to deal with two constants. </p>

<p>ln[y] = kt + C => y = e^(kt+C) = e^(kt) * e^C</p>

<p>The last step follows from properties of exponents. But e^C is another constant, so we can just call it K. Therefore y = Ke^(kt). You could also use C1 for the first constant and then C.</p>

<p>^my answer was correct right? </p>

<p>anyway new question:
what is the derivative of 2 apples? (seriously, no joke?, its an uber implicit diff question, a, p, l, e and s are all variables :))</p>

<p>vader1990: differentiate with respect to what variable?</p>

<p>lol...I was just kidding (even though I used the word seriously); ok, here's a "real" question:</p>

<p>NO CALCULATOR</p>

<p>Integration (0 and 2pi) [sin(x)^4] dx</p>

<p>d/da [2 apples] = 2pples
d/dp [2 apples] = 4aples
d/dl [2 apples] = 2appes
d/de [2 apples] = 2appls
d/ds [2 apples] = 2apple</p>

<p>:)</p>