Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
Mu Alpha Theta National Convention 2004
Mu Individual
NOTA = none of these answers
1)
Find the average value of f(x)  sin
cos2
1
2
cos4  1
(D)
4
A)
2)
(B) 1
1
e2
(D) 7
(E) NOTA
1
e2
(E) NOTA
(C) 8
(D) 16
(E) NOTA
(C) 0
(D)
(C) e – 1
(B) 4
(D)
3x  x3
x  4x2  5
Evaluate: lim
If i 
(B) 
1 , evaluate: (i + 2)
(A)
38 315i
(D) 63
7)
(C) 12
Solve for x: log5 (log 4(log2 x)) log3 27  10log 3
(A) 
6)
cos2 1

2
2
Find the area of the region bounded by the graphs of y  e x , the x-axis,
and the lines x = 0 and x = 2.
(A) 2
5)
(C)
(E) NOTA
(B) 15
(A) 1
4)
(B) 1 cos2
The six-digit number 3730N5, where N is the tens digit, is divisible by 21.
What is the sum of the positive integral factors of N?
(A) 4
3)
x
on the interval 4  x  0 .
2
3
4
(E) NOTA
6
(B) 117  44i
(E) NOTA
(C) 278  29i
Which of the following statements is/are true? Assume that f is a function
whose domain and range are subsets of the real numbers.
1)
2)
3)
If f is continuous everywhere, then f is differentiable everywhere.
If f is differentiable everywhere, then f is continuous everywhere.
8
If f is continuous and f(x)  3 for every x in [4, 8], then 4 f(x) dx  9.
Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
(A) 2 only
8)
6
2
(B) 2 2
(C) 3 6
(B) 8
(C) 6
Simplify the following:
 /12
Evaluate: 0
(A)
12)
1
ln 3
2
(E) NOTA
Evaluate:
(D) 4
(E) NOTA
(C) 2sin(2x)
(B) 0
(E) NOTA
dx
cot(3x)
(B)
ln 2
6
(C) ln
3
2
(D) 1
(E) NOTA
1 2 1 1 1 1
     
2 3 4 3 8 6
(A) Diverges
13)
2
2
(D)
sin(3x) cos(3x)

. Assume nonzero denominators.
sin x
cosx
(A) sin(2x) cos(2x)
(D) 2
11)
(E) NOTA
When an infinitely thin circular plate of metal is heated in an oven, its
radius starts increasing at a rate of 1 centimeter per second while still
maintaining a circular shape. At what rate is the plate’s area increasing
the moment its radius is 5 centimeters? Express your answer in square
centimeters per second.
(A) 10
10)
(C) 1 and 3 (D) 1, 2, 3
What is the eccentricity of the ellipse with equation 4x2  8x  8y2 16y  52?
(A)
9)
(B) 1 and 2
(B)
15
2
Given the differential equation
(C)
4
3
(D)
 1 1
(B) ln  
2e 2
(E) NOTA
(E) NOTA
2
dy
 xe y  x , find the value of y when x = 1,
dx
given that y = 0 when x = 0.
e1/3 
(A) ln
 1
 2

1
(D) 
2
7
3
3 1 
(C) ln  
2 2e
Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
14)
Seven coins are flipped behind a screen, and a reliable source tells you
that at least two of them are heads. What is the probability that at most
five of them are heads?
(A)
15)
29
30
Find
16)
7
8
(C)
13
15
(D)
23
24
(E) NOTA
dy
for y  xln x if x > 0.
dx
2xln x ln x
x
(A)
(D)
(B)
(C) (ln x)xlnx1
(B) xlnx2
xlnx
ln x
(E) NOTA
Find the value of x2  y2  z2 , where x, y, and z satisfy the following system.
x+y–z=8
2x – y = 3
y+z=6
(A) 38
17)
(D) 107
(E) NOTA
(B) 3
(C) 1
(D) 0
(E) NOTA
A rectangular sheet of paper measures 48 decimeters by 55 decimeters.
One corner is folded onto the diagonally opposite corner and then the
paper is creased. The length of the crease is m / n decimeters, where m
and n are relatively prime natural numbers. What is the value of m n ?
(A) 7519
19)
(C) 69
How many extrema (maximum or minimum) does the function
3
2
f(x)  (x  3) (x  4) have on the interval 3  x  6 ?
(A) 2
18)
(B) 42
(B) 5320
(C) 4063
(D) 3559
(E) NOTA
What is the volume of the solid generated when the region bounded by
the graphs of y  x3 , y = 0, y = 8, and the y-axis is rotated about the line x = 4?
(A)
64
3
(B) 20
(C)
384 
5
(D)
96
7
(E) NOTA
Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
20)
Four hundred students answer a multiple-choice question with five possible
choices. The students are in one of three sets A, B, or C. Set A knows the
correct answer, Set B guesses incorrectly and Set C guesses correctly. The
probability of a student guessing correctly is 1/5. If 180 of the 400 students
give a wrong answer, then how many students knew the correct answer?
(A) 175
21)
(B) 145
(B)
5
26
(D)
13
4
(E) NOTA
(B) 621.4
(C) 711.6
(D) 716.8
(E) NOTA
(B)
8

2

3

(C)
26
3
(D)
10 16
 3


(E) NOTA
The sum of the sides of a right triangle is 540. The sides are in an
arithmetic progression. What is the length of the hypotenuse?
(B) 215
(C) 220
(D) 235
(E) NOTA
Find the sum of the coordinates of the point in the graph of xy 2  x3y  6
that has a horizontal tangent line.
(A) 1
26)
(C)
1
(A) 210
25)
28
27
Evaluate: 0 (2x 1)2 sin(x) dx
(A) 14
24)
(E) NOTA
Let N be a four digit number with no two of its digits the same and none of its
digits zero. Let S be the sum of the digits of N. The maximum value possible
for N S is what?
(A) 608.2
23)
(D) 135
Find the value of a + b + c + d so that the graph of y  ax3  bx2  cx  d has a
relative minimum at (0, 0) and a relative maximum at (3, 4).
(A) 1
22)
(C) 185
(B) 2
(C) 3
(D) 4
(E) NOTA
How many solutions are there to the equation cos x  sin(2x)  cos(3x)  0 ,
where x lies on the interval [2 ,2] ?
(A) 15
(B) 13
(C) 11
(D) 9
(E) NOTA
Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
27)
A solid has its base given by the region bounded by the graphs of y  x ,
x = 0, x = 1, and the y-axis. If the cross sections of this solid perpendicular
to the x-axis are in the shape of regular dodecagons, what is the volume of
the solid?
4
(1 3 )
3
(A)
28)
(B)
5
2

8 64
(C)
3 3 3

16 32
(D) 2 
2
4
(E) NOTA
Let f(x)  x2  2mx  m , where x is a real number. Let r1 and r2 be the zeroes
2
2
3
3
of f (not necessarily distinct). Find the sum of all m such that r1  r2  r1 r2 .
(A)
29)
5
4
(B) 1
(D)
1
2
(E) NOTA
Let s be the sum of all distinct integer solutions to the equation
2
(x2  x  5) x  4x3  1. If f(x)  (x 1)(x  2)(x  3) , find the value of f '(s) .
(A) –3
30)
3
4
(C)
(B) –2
(C) –1
(D) 0
(E) NOTA
Which of the following mathematicians is most associated with the
discovery of calculus?
(A) Plato
(B) Pythagoras
(C) Newton (D) Napier
(E) NOTA
Tiebreaker #1 Let f(x) be a continuous and differentiable function. The table
1
below gives the value of f(x) and f’(x) at several values. Let g(x) =
and
f (x)
h(x) = x3 + 2x. Find the value of the derivative of h(f(x)) at x = 1.
x
1
f(x) -3
f’(x) -5
2
-8
-4
3
-9
3
4
0
16

Tiebreaker #2
Evaluate
e

x
dx
 e x
Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
Mu Alpha Theta National Convention 2004
Mu Individual
Answers
#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Answer
C
C
B
D
A
B
E
D
A
D
B
D
E
E
A
B
A
#
18
19
20
21
22
23
24
25
26
27
28
29
30
TB1
TB2
TB3
Answer
D
C
E
B
B
D
E
D
B
Thrown out
A
C
C
-145
/2
Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
1) (C) Average value =
1
0
x
cos2  1
.
sin dx 

4
0  (4)
2
2
2) (C) To be divisible by 21, the number has to be divisible by 3 and 7; thus
3  7  3  0 N 5  18 N has to be a multiple of 3. That means N has to be 0, 3, 6, or 9. It’s
easy to just check each of these one by one to see if the original number is divisible by 7. It follows
that 6 is the only one that works. The sum of the positive integral factors of 6 is
1+2+3+6=
12.
3) (B) Area = 0 ex dx  e x
2
  e
2
0
2
 1.
4) (D) Since log3 27  3 and 10lo g3  3 , the equation reduces to log5 (log 4(log2 x)))  0 ; thus
x2
45
0
2
41
 16 .
5) (A) The degree of the numerator is bigger than the denominator’s so as x gets larger on the positive
scale, the numerator dominates. Furthermore, since the numerator is negative and the denominator
positive, the limit is  .
6) (B) By the Binomial Theorem,
6
6
6
6
6
6
6
(i  2)6   i6 20   i5 2   i422   i3 23   i2 24   i125   i0 26 . Using the periodic
0
1
2
3
4
5
6
cycle of powers of i, the above simplifies to 117  44i .
7) (E) Statement 1 is false in general; for example, take f(x) | x | , which is continuous everywhere but
is not differentiable at the origin. Statement 2 is true; it’s actually a theorem. Statement 3 is also true
8
8
because f(x)  3 implies 4 f(x) dx  4 3dx  12  9 .
16  8
2
(x 1)2 (y  1)2

8) (D) Complete the square to get
.

 1; the eccentricity is given by
2
16
8
16
9) (A) We have A  r , so
2
dA
 2 (5)(1)  10 .
dt
dA
dr
 2 r . Plug in all the known values to obtain
dt
dt
Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
10) (D) Combine the fractions and use trig. identities to get
sin(3x) cosx  cos(3x) sinx
sin(3x  x)
2sin(2x)


 2.
sinx cos x
(1/ 2)sin(2x)
sin(2x)
11) (B)
 /12
0
 /12

dx
 /12
1
 0 tan(3x)dx  ln | sec(3x) |
cot(3x)
3
0

1 
  1
lnsec   ln(sec 0)
3 
4  3

 ln 2
1 1/2 1 1
ln 2   ln 2 
3
3 2
 6
1
1
and common ratio of ; its sum is
2
2
1/ 2
2
1
 1. The even terms form a geometric series whose first term is and common ratio ;
1 1 / 2
3
2
2/3
4
4 7
 . Altogether the sum is 1  .
the sum of that is
1 1 / 2 3
3 3
12) (D) The odd terms form a geometric series with first term of
1  x2
y
x 2
y
13) (E) Separate variables to obtain e dy  xe dx ; thus e   e  C . Use the fact that the
2
 1 1
1
curve passes through the origin to get C   . Thus, when x  1, y   ln  .
2
2e 2
n!
C(7,i)
, where C(n,i) 
. Thus,
7
2
(n  i)!i!
P(2  X  5)
(C(7,2)  C(7,3)  C(7,4)  C(7,5)) / 27
14
.
P(X  5 | X  2) 


7
P(X  2)
(C(7,2)  C(7,3)  C(7,4)  C(7,5)  C(7,6)  C(7,7)) / 2
15
14) (E) If X is the number of heads, then P(X  i) 
15) (A) Take natural logs of both sides to obtain lny  (ln x)2 . Differentiate implicitly and we have
1 dy 2ln x
dy 2yln x 2xln x ln x

. Thus,
.


y dx
x
dx
x
x
16) (B) Use your favorite technique of solving 3 x 3 systems to get (x,y,z)  (4,5,1) ; thus
2
2
2
x  y  z  16 25 1 42.
17) (A) We’ll get extrema when the derivative is 0 or undefined. The derivative of f is
f '(x)  (x  4)(x  3)2 (5x  6) ; it’s obviously not undefined, but it equals 0 whenever x is –3, 4, or
6/5. Two of these values lie on the interval –3 < x < 6.
Mu Alpha Theta National Convention 2004
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For each question, NOTA means none of the above answers
18) (D) Let ABCD be a rectangle with AB = 48 and BC = 55. Fold the rectangle such that A coincides
with C. Unfold the rectangle. Let EF be the crease, where E and F are on AD and BC, respectively.
By symmetry, BF = ED = x. Now fold the rectangle again. Notice that the points A (which now
coincides with C), E, and D make a right triangle with legs x and 48 with hypotenuse 55 – x. By the
Pythagorean theorem, (55 x)2  x2  482 ; this yields x = 721/110. Now unfold the rectangle again.
Notice the length of the crease L is the hypotenuse of a triangle with legs of 48 and 55 – 2x. Thus,
3504
721 2
L  482  (55  2x)2  482  (55  2(110
)) 
, making m + n = 3559.
55
2
384
3
19) (C) By the Shell Method, the volume is 2 0 (4  x)(8  x )dx 
.
5
20) (E) (CHANGED TO E) Set A knows; Set B guesses wrong; set C guesses right – 180 answer wrong
 P(guessing wrong) = 4/5 4/5 (# guessing) = 180
# guessing = 225 
# guessing + # who knew = 400  # who knew = 400 – 225 = 175
21) (B) First things first, since the graph passes through the origin, d = 0. Now from the information
given, y '(0)  y '(3)  0 . Since y '  3ax 2  2bx  c , this makes c  0 and
2
0  3a(3)  2b(3)  27a  6b . The graph also passes through (3, 4), so
 8 4 
3
2
4  a(3) b(3)  27a  9b . Solving this yields (a,b)   , ; thus
 27 3 
8 4
28
a b  c  d  
 00 
.
27 3
27
22) (B) largest number on top / smallest number on bottom 9321/15 = 621.4
23) (D) The quickest way to do this problem is probably to use tabular integration:
/
u

(2x  1) ]
sin( x)

8x  4 ]


8 ]


0
dv
2
1

cos( x)
1
2
1
3
sin( x)
cos( x)
Multiply along the arrows to obtain an antiderivative:
(2x  1)2 cos(x) (8x  4) sin(x) 8cos(x)



. Use the Fundamental Theorem of Calculus to
2
3


obtain the answer in choice D.

Mu Alpha Theta National Convention 2004
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For each question, NOTA means none of the above answers
24) (E) 3x + 4x + 5x = 540 x = 45 135 + 180 + 225 = 540 1352 + 1802 = 2252
dy 3x2y  y2
; the curve will have a horizontal tangent

dx 2xy  x3
line when this quantity is 0. This occurs when 3x2y  y2  y(3x2  y)  0 . If y  0 , then we have
2
3
2
x(0)  x (0)  0  6 , a contradiction. Thus, 3x  y ; plug this into the original equation to get
9x5  3x5  6 , or x  1, making y  3 . The sum of the coordinates is 4.
25) (D) Differentiating implicitly, we get that
26) (B) Regroup the equation as cos x  cos(3x)  sin(2x)  0 . Now use the sum-to-product identities
on the first two terms to obtain 2sin(2x) sin(x)  sin(2x)  0 , which simplifies to
sin(2x)(2sinx 1)  0 . On the interval [2 ,2] , sin(2x)  0 has the solution set
 

 5 11 7 
3
,  2 . The equation 2sinx 1  0 has solutions  , ,
,
0, ,  ,
. There
2
6
6 
 2

6 6
are 13 solutions in all.
27) (THROWN OUT) The formula for the area of a regular dodecagon with side length s is
3(2  3 )s2 ; one can derive this by breaking up the dodecagon into 12 isosceles triangles with vertex
angle of 360/12 = 30 degrees and using some trigonometry. At any rate, since s  y  x , the
1
3
2
volume is equal to 0 3(2  3)( x) dx  (2  3) .
2
28) (A) Recall the factorizations a2 b2  (a b)2  2ab and a3 b3  (a b)3  3ab(a b) . Since f is
(2m)
m
 2m  a  b and
 m  ab ,
a quadratic, the sum and product of the roots is given by 
1
1
2
2
respectively. Thus, r1  r2  (2m)2  2(m)  4m2  2m and
3
3
3
3
2
r1  r2  (2m)  3(m)(2m)  8m  6m . Set these two quantities equal to each other to get
(10) 5
 .
8m3 10m2  2m  0 . The sum of the solutions to this equation is 
8
4
29) (C) Let A  x2  x 5 and B  x2  4x  3 . Now for AB  1, we must have either A = 1 (and B can
be anything), B = 0 (in this case A  0 ), or A  1 and B is even. The first case yields x being equal
to –3 or 2. The second case yields x being equal to –3 or –1. The third case only produces noninteger values for B. Thus, the sum of all distinct solutions is s  3 1 2  2 . We have
f '(x)  (x  2)(x  3)  (x  1)(x  3)  (x  1)(x  2) , making f '(s)  0  (1)(1)  0  1 .
30) (C)
Mu Alpha Theta National Convention 2004
Mu Individual
For each question, NOTA means none of the above answers
d
h(x)  h '(f (x))f '(x)  (3(f (x))2  2)f '(x)
dx
Tiebreaker #1 -145
d
h(1)  (3(3)2  2)(5)  145
dx

2

b
0
dx
e xdx
e xdx
lim  2x
 lim

 x  x  b
 1 a a e 2x  1
 e  e
0 e
b
lim arctan e x  lim arctan e x
Tiebreaker #2
b+
0
lim (arctan e b 
a 


0
a
)  lim (  arctan e a ) 
b 
4 a  4
  
 



0
 2 4   4
  2