AP Calculus AB / Mr. Hansen
3/10/2004

 

           Name: ____________________________

Annotated Key for Test on Chapter 8

 

Dear Students:

I apologize for the large number of errors I made in the questions as originally posed. There were errors in problems 1, 3, 6, 7, and 9, and the wording of problem 8 should have clarified that the given region was the base of a solid. The most serious errors were in problems 3, 7, and 9, and I understand how exasperating you must have found these problems during the stressful testing situation.

My only comfort is that nobody is perfect, and even the AP exams may have errors despite the enormous time and expense that goes into their creation. No AP exam would be likely to have 5 errors, I grant you, but errors do occur from time to time.

As for the test results, you know that they were disappointing, with a curved class mean of only 49% (F). In fact, there were 7 students whose raw scores were at or below 4 points out of 40. Any time a third of the class nets no more than one question out of 10, something is seriously wrong.

I briefly considered throwing out problems 3, 7, and 9 altogether. However, this would have made the situation even worse. Nearly everyone’s score would have gone down, and the class mean (curved) would have dropped from 49% to 46%.

Because I try (though I do not always succeed) to be a reasonable person, I felt I must offer a Happy Re-Test on Thursday, 3/11/2004. I will forgive the generally dismal class performance on the March 10 test if you will forgive me at least one or two of the errors I made in posing the questions.

Sincerely,

Mr. Hansen

 

 

 

 

Part I: Multiple choice, no calculator.

 

 

1. B

Like many of the problems on the test, this one was riddled with typos in the original version. The parabola for f ¢ was depicted as having a global minimum (i.e., vertex) when x = –1, as well as zeros at –2 and 1. This, of course, is impossible, since parabolas have their vertex midway between their zeros. However, most students managed to obtain choice B despite the errors.

 

 

2. B

Use R = 9, r = x = y2. Then, by the method of washers,
dV = (pR2 – pr2)dy = p(81 – y4)dy.
Finally, the algebra is not too bad to find V = ò dV = pò03 (81 – y4)dy.

 

 

3. A

Choice A was given as 1/6 (typographical error) in the original version of the test. That choice was closest to the true answer of 5/24. Note that the integral runs from x = 0 to x = 0.5 since the functions intersect at (0.5, 1.5).

 

 

4. C

Factor y ¢ as 4x(x – 1)(x – 3). By inspection, critical points are at x values of 0, 1, and 3.

 

 

5. B

Most people got this. Note that you need only determine the sign of y ¢, which can easily be found from the factored form. Since y ¢ = 4x(x – 1)(x – 3) = (pos.)(pos.)(neg.) when x = 2, y ¢(2) < 0.

 

 

6. A

As originally posed (with typographical error in choice D), this problem had two correct answers. However, nobody chose either of the correct answers. Here are two approaches to the problem:

 

 

 

“Precal” method (recommended):

Since y = (something)2, y ³ 0 for all x.
Is y ever equal to 0?
Yes, when ln x + 1 = 0 Û ln x = –1 Û x = 1/e.
Since there is a unique x for which y = 0, and since y can never be less than 0, there is a unique global minimum.

“The Calculus” method (requires more work, but less thinking):

y ¢ = 2(ln x + 1)/x; set equal to 0 to see that x = 1/e is the only possible critical pt.
y ¢¢ = (–2 ln x)/x2 by Quotient Rule [steps omitted], which is > 0 when x = 1/e
Note that there are no endpoints to check, since the domain is (0, +¥).
Conclusion: Since concavity is upward (y ¢¢ > 0) at the only possible critical pt., there is a global minimum there.

 

 

7. E

Your imperfect instructor messed this one up. As a result, the correct answer (.616) did not appear among the choices given, and you were supposed to choose the closest answer. A situation like this would be extremely rare in a real AP exam. However, the rule remains that you should always choose the best answer if no answer is precisely correct.

 

 

8. B

Use ; then compute

 

 

9. E

This was Mr. Hansen’s last and most embarrassing goof-up of the test. However, I looked at your scratch work, and I wasn’t the only person who should be embarrassed here. Quite a few people tried to use FTC1. THAT APPROACH DOES NOT WORK HERE, since there is no way to find an antiderivative (in closed form) for the given function. The antiderivatives you wrote were creative, yes, but entirely incorrect. The correct solution is to use your calculator to evaluate the integral, wrt x, from 1 to 2. Answer: 17.341.

 

 

10. D

Take r = 1, R = x1/3 + 1. You need an inner and outer radius since the axis of revolution is y = –1, not the x-axis. That gives dV = (pR2 – pr2)dx = p((x1/3 + 1)2 – 12)dx. Integrate as x goes from 0 to 3 to get 32.152. Note that this is one place where adding a fifth parameter (e) to your fnInt command can really save you a lot of time. By specifying e as .1, you get your answer in 1 second instead of the 15 or 16 seconds that the default accuracy of your calculator would require.