# 2008 MC Exam Key by ajizai

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```									AP Calculus
2008 MC Key

1. This is asking you for the end behavior, or the horizontal asymptote. Foil numerator and denominator
and see where the bigger exponent is. In this case, they are the same, so the limit is the coefficient of

lim
 2 x  1 3  x   lim 2 x2  7 x  3  2 x2    2 , Answer: B
x   x  1 x  3     x  x 2  2 x  3     x2

2. Rewrite it as x 2 , add one the exponent, multiply by the reciprocal, don’t forget C.

1
x   2
dx   x 2 dx   x 1  C , Answer: D

3. Product Rule: f   x    x 2  2   3  x  1  x 2  2  2 x , but this isn’t a choice. So, you need to factor
3                          2

something common from each term. f   x    x 2  2                     x                             
 2   6 x  x  1   x 2  2   7 x 2  6 x  2 
2          2                                     2

4. Substitution, let u  2 x, du  2dx, dx  du ,   sin  u   cos  u   du    cos u  sin u   C
1    1                                1
2    2                                2

x2  5x2  8            5x  8  8
2

5. Try to plug 0 in first, doesn’t work, so factor. lim                                   lim
x 0   x 2  3x 2     16         3x  16  16
x 0    2

6. If you factor the top function, it turns into x + 2. So plugging a 2 into it, you get 4 which means this is
a line that has a hole in it at x = 2. So it is not continuous, and b/c of that, it can’t be differentiable
either. It has a limit at 2, b/c as you approach 2 from the left and right, the values approach 4.

1

 3t        6t dt  t 3  3t 2 |  4 . This means
1
2
7. You need to integrate velocity to find change in position.
0
0
the position changed by 4. Since its original position is 2, it is now at 6. Answer: B

                    3
8. Chain Rule. f   x   3sin  3x  , f     3sin  3    3 
 2  , Answer: E

9          9            

9. Area accumulator, g  x  is area under the graph given starting at -2. The areas for each choice are as
follows: A (-1.5), B (0), C (3), D (4), E (3), Answer: D

10. Chain Rule: f  x   e 2/ x  e2 x , f   x   e2 x  2 x2 , Answer: D
1                 1
11. Since the function is decreasing, the right Riemann sum is always the least, Answer: C

12. The derivative of the graph given goes positive, negative, positive, leaving only A or B as
possibilities. The graph give does not have a max/min at x = 0, Answer: B

13. Remember on these, find the derivative first, then plug in the value. A common mistake is to find the
derivative of f  x   x2  2x , which equals 2x  2 and then plugging ln x into x, but this is wrong.
1
You should find the derivative of  f  ln x  first, which is f   ln x     . Now plug ln x into 2x  2 ,
x

 2  ln x   2    1  , Answer: A
 
x

14. I would save this one until the end b/c you are doing 5 problems. The first 3 choices are talking about
the first derivative, which we can’t tell enough about from the table of the second derivative. Choice
D is close, b/c the second derivative is 0 at the point, but we don’t know that it changes signs there.
Choice D COULD happen, but doesn’t have to be true. Answer: E

1
15. The only way to do this is substitution. u  x 2  4, du  2 xdx, dx          du , so
2x
x du    1 1         1
 u 2 x  2  u du  2 ln u  C , Answer:    C

1
y
 dy            dy            1         dy cos  xy 
16. Implicit differentiation, cos  xy    x  y   1, x  y                ,                      , from here,
 dx            dx        cos  xy  dx              x
1           1  y cos  xy 
y
dy cos  xy          cos  xy        1  y cos  xy 
be careful w/ you fraction skills.                                                          , Answer: D
dx        x               x              x cos  xy 

17. The equation to the line give is y   x  k , the derivative of the function given is y  2 x  3 . Since
the slope of the line is -1, set that equal to y  , so 1  2x  3 , so x  2 . Plug that back into the
original function, y  1 , so x  y  k  3 , Answer: A

18. Area accumulator, points of inflection come from the 2nd derivative. The 2nd derivative of g can be
found by looking at the first derivative of the graph of f. The graph has a max and min at 2 and 5.

19. Without a calculator, plug “large” numbers into x, and “small” numbers and see what happens. I’ll try
5  210    210                    5  210 5  0
10 and -10,          10  1 , now try -10,                  5 , Answer: E
1  210 2                        1  210 1  0

20. Make a sign chart w/ zeros at 0, 3, and 6. The chart should go + + - +, which means it changes twice,
at 3 and 6, Answer: D

21. Position is given, velocity increases when the 2nd derivative is positive, or concave up. Answer: A
22. The 2nd derivative isn’t important, they give you a point, (2, 1) and its slope, 4. The equation is
y 1  4  x  2 , plug 1.9 into x, Answer: B

23. If p people have heard the rumor, then N – p people have not. When the problem states “proportional
to the product”, that means multiplied w/ the constant of proportionality ‘k’ in the front, Answer: B

1 2 1 3
24. Separate the variables. ydy  x 2 dx,    ydy  x dx,     y  x  C , plug (3, -2) into the equation,
2

2    3
2 x3
C  7 . Then solve for y, y             14 , so it could be D or E. The original y value given was -2,
3

25. To be differentiable, that means a function is continuous and has the same derivative at a point. This
piecewise function changes at 2, so that’s the x value we need to use. You need to write 2 equations,
one setting each piece equal to each other, and another setting their derivatives equal to each other.
cx  d  x2  cx and c  2 x  c , then plug 2 into x, 2c  d  4  2c and c  4  c . Using the 2nd
equation, c = 2, plug that into the first equation, 4  d  4  4 , d = -4, Answer: B

26. This is one of those I doubt anyone will remember.
dy                  u      4              4
arctan  u                                  2 , Answer: A
dx                1  u 1  16 x
2         2
1
2

1  16  
4

27. If g is the inverse of f, then the x’s and y’s switch. So, g   3 3 is the x, which means they want to
know about when 3 is the y on f  x  . We will use the point f  6  3 , so f   6  2 , and inverse
functions have derivatives that are reciprocals. Answer: A

dy
28. Plug points in, it looks like when y is zero,      is zero, eliminating D and E. Answer: C
dx

29. When the derivative is positive, the function is increasing, Answer: B

30. The left and right sided limits exist, but are not the same, Answer: C

31. Use your calculator and find where the graph is above the x-axis. Answer: B

 f  x  dx  4 into  f  x  dx  4 , -17 + 4, Answer:
2                      5
32. Change                                                                B
5                     2

33. Graph the derivative (the function given) and see how many max/mins it has, Answer: E

34. Antiderivative is the area, and the starting point is -7, Answer: E

35. Put the velocity into your calculator and find the derivative at 3, Answer: B
2

x        8x 2  18x  5   x  5 dx and
3
36. There are two regions you must find,
1
5

  x  5   x        8x 2  18x  5 dx , Answer: B
3

2

37. Integrate (math9) the velocity, that will give you change in position. Initial position is 2, so add 2 to

38. Relative max happens when the derivative goes from positive to negative, Answer: C

39. The antiderivative of f   x  is f  x  , f  1  f  4 , Answer: B

dS       dr dS
40. Find the derivative, plug in values.              8 r ,     8  3 2  , Answer: C
dt       dt dt

41. They are telling you that between 2 and -2, there are no max/mins, but f  2  f  2  0 . The only
way a function can start and end at the same point w/out having a max/min is if it is not continuous.
That would mean that the derivative doesn’t exist at some point, Answer: E

42. Find the area under the function (math9) and divide by 4, Answer: C

43. At the beginning of the table, the velocity is negative, which means the position graph should be going
down, leaving only E and C left. When t = 1, the velocity is 2, eliminating choice E, Answer: C

44. This one could be the most difficult on the exam. They tell us f   x   0 , which means that f   x  is
getting smaller. If you look at table D, the f  x  values change by 2, 2, which means f   x  is 0 and
f   x  is zero, so that one is out. Table E, the f  x  values change by 1.5, 2.5, which means f   x  is
positive. So we have 3 tables left. Table B, the f  x  values change by 2.5, 2, which means f   x  is
negative, but, since the change is 2 in between 3 and 4, it doesn’t work b/c the problem says
f  3  2 . This reason eliminates table C also because it changes by 2 and 1.5. Answer: A

45. The x runs from left to right, so it has to be dx, from 0 to 4. To find the volume, you must multiply by
7, f  x  , and dx, Answer: B

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