True, but it felt super-awkward.
On many graded FRQs on the CB website, many perfect-scoring answers had the answers circled. So that shouldn’t be a problem.
Wow, this thread really exploded.
I think I’m getting a 5 @Mathophile26
Yeah, I think I’ll probably get a 5 too. A 4 at the very least.
Ah nice @mundanewarrior! I took BC, not AB (since I took the classes for AB and BC this year).
I thought I did okay. I was exceptional in class and thought I’d do great on the test. I felt like the FRQ’s were much harder than what we did in class. I just need a 4. What do you think you’d need to score on the frq assuming you got 15mc wrong, in order to achieve a 4.
@GemmyJ http://appass.com/calculators/calculusab
Pretty accurate score calculator
Thanks! I just tested it out and it gave me a 4 based on how well I thought I did. Although I was only 3 composite points away from a 3. Should I assume I got a 4 or 3?
@GemmyJ if u were exceptional in class, i think you got that 4 just fine 
also pray the curve will be in our favor this year too lol
I was, you still think so, I got all the frqs, but blanked out on 4(potato) and the last one. I hope it’s really good as well, its the first time with 4 letters for multiple choice
@GemmyJ The curve should hopefully be a bit more generous this year. It’s a new version of the test so usually they’re a little more lenient.
The FRQs have been released. My attempt at #1:
A) Volume ≈ 250.3 + 314.4 + 5*6.5 = 176.3 cu. ft.
B) overestimate because A(h) is decreasing
C) integral from 0 to 10 of f(h) dh ≈ 101.325
D) V(h) = integral from 0 to h of f(x) dx
dV/dt = f(h) (dh/dt)
at h = 5,
dV/dt = (50.3)/(e + 5) * 0.26 = 1.694 cu. ft. per min.
2:
A) integral from 0 to 2 of f(t) dt = 20.051 lbs
B) f’(7) = -8.119 or -8.120
After 7 hours, the rate at which the bananas are removed is decreasing at a rate of 8.119 or 8.120 lbs/hr^2.
C) f(5) = 13.796, g(5) = 11.532
The number of pounds of bananas is decreasing since f(5) > g(5).
D) 50 - integral from 0 to 8 of f(t) dt + integral from 3 to 8 of g(t) dt = 23.347 lbs.
3:
A) f(-6) = 7 - 0.524 = 3
f(5) = 7 - 0.5pi2^2 + 0.5 32 = 10 - 2pi
B) f is increasing on [-6, -2] and [2, 5] because f’ is positive
C) Check critical points (x = -2, 2) and endpoints (x = -6, 5)
f(-6) = 3
f(-2) = 7
f(2) = 7 - 0.5pi2^2 = 7 - 2pi
f(5) = 10 - 2pi
absolute min value is 7 - 2pi
D) f"(-5) = -1/2
f"(3) DNE because the one-sided limits of f"(x) do not agree
(as x -> 3-, lim f"(x) = 2, but as x -> 3+, lim f"(x) = -1)
Stupid me… I got mixed up and thought of 7 - 2pi as 7 - pi, and thought that 3 was the minimum 
Oh, well, should only be one point. I’m pretty sure I got the other 8 on that question.
Hey, where are the FRQs anyway? I don’t see them on the College Board website.
@RandomGuyy17 You can find them here (scroll to the bottom):
Calc AB:
http://apcentral.collegeboard.com/apc/members/exam/exam_information/1997.html
Calc BC:
http://apcentral.collegeboard.com/apc/members/exam/exam_information/8031.html
4:
A) @ t = 0 dH/dt = -0.25(91-27) - 16
Tangent line: y = -16t + 91
H(3) ≈ -48 + 91 = 43°C
B) d2H/dt2 = -0.25 dH/dt = -0.25(-0.25(H-27)) = (1/16)(H - 27)
Since d2H/dt2 > 0, H is concave upward. Therefore, answer in (A) is an underestimate
C) integral of (G - 27)^(-2/3) dG = integral of -dt => G = (-1/3 t + C)^3 + 27,
G(0) = 91 => C = 4
G(3) = (-1/3*3 + 4)^3 + 27 = 54°C
5:
A) Vp(t) = (2t - 2)/(t^2 - 2t + 10), crit. # at t = 1
For [0, 1), Vp(t) < 0, and for (1, 8], Vp(t) > 0. So P is moving left on [0, 1)
B) Vq(t) = (t - 3)(t - 5), crit. # at t = 3, 5
For [0, 3) & (5, 8], Vq(t) > 0, and for (3, 5), Vq(t) < 0. So P and Q are moving in the same direction at (1, 3) and (5, 8]
C) Aq(t) = 2t - 8, Aq(2) = -4
Speed is decreasing since Vq(2) and Aq(2) have opposite signs
D) Xq(3) = 5 + integral from 0 to 3 of Vq(t) dt
= 5 + (1/3 t^3 - 4t^2 + 15t)|from t = 0 to t = 3
= 5 + 18 = 23