Download Solutions - Mu Alpha Theta

Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Mu Division
1. We have lim(2 x2  x3  16)  8  8  16  32 .
x2
2. Since 189  33  7 and 588  22  3  7 2 , their greatest common factor is (3)(7) = 21.

3. We have n = 2, so
4
0
x 2 dx 
40 2
 0  2  22  42   24 .
2(2)
4. Converting to common bases, we have 2 x  23 y3 and 32 y  3x9 ; thus x  3 y  3 and
2 y  x  9 . Solve this system to obtain (x, y) = (21, 6), so x + y = 27.
5. Differentiate implicitly to obtain 2 x  y  x
and get 18  6  9
dy
dy
 2y
 0 and immediately plug in (9, –6)
dx
dx
dy
dy
dy
 12
 0 , so
 4.
dx
dx
dx
6. Since 25  1mod13 , 252005  (1)2005  1  12 mod13 , so the desired remainder is 12.
7. Expanding by the first row, we get
5
9
0
2
3
6  5
5 6
7
3
6
6
7
9
2
6
5
7
 0(something)  5(21  36)  9(14  30)  219 .
8. If lim f ( x)  L and lim g ( x)  K , then we’re simply interested in the value of LK. Since
x c
x c
L  K  11 and L  K 2  61 , we have, using some algebra,
2
LK 
( L  K )2  ( L2  K 2 ) 112  61

 30 .
2
2
9. Using the standard relations between the roots of a polynomial and its coefficients, we get
that the sum of the squares is (2)2  2(6)  -8 .
10. We have y   8 x  4 x3 and y   8  12 x 2 . Setting the second derivative equal to 0 yields the
potential inflection points x   2 / 3 . Testing numbers around these two values, we find
that the second derivative is positive when  2 / 3  x  2 / 3 , and hence concave up on
that interval. The desired width is 2 2 / 3  2 6/3 .
Page 1
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Mu Division
11. If the radius of the circle is r, then we’re interested in the moment when 2 r  2 r 2 , or r =
dA
dr
dA
 2 r
 2 (1)(4)  8π .
1. Since A   r 2 , we have
, so
dt
dt
dt
12. If bn is a root-mean sequence, then, by inspection, (bn ) 2  3n  2 , so bn  3n  2 . Since
b6  4 and every previous entry after b2 is not an integer, the answer is 6.
13. By Power-of-a-Point, (AE)(EB) = (CE)(ED), so ED = 6. Call the center of the circle O and
draw perpendiculars OF and OG from O to AB and CD, respectively. Since the
perpendiculars bisect their respective chords, FB = 7 and so EF = 5. Also, GD = 5 so
EG
= 1. Notice that triangle EOG (or alternatively, triangle EOF) is a right triangle, the length of
EO the distance from E to the center of the circle. By the Pythagorean Theorem,
( EG)2  (OG)2  ( EO) 2 , making EO  12  52  26 .
14. In 24 seconds, the jogger has covered 24/56 = 3/7ths of the track, and so the other jogger has
ran 4/7ths of the track. If x is the time it takes for the other jogger to run around the track,
4 / 7 24

then, using proportions,
, making x = 42.
1
x
1
15. In order for f to be continuous, lim f ( x)  f (2) , or 2k  10  23  4k , and so k  - .
x2
3
d
F (G ( x))  F (G ( x))G ( x) , so evaluating at x = 1 produces
dx
F (G (1))G (1)  F (4)G (1)  (7)(8)  56 .
16. By the Chain Rule,
17. Notice that y  (2 x  3)4  16 x 4   81 is a 4th degree polynomial and thus, taking
successive derivatives makes everything vanish except the fourth derivative of the x 4 th term.
The answer is 16  4!  384 .
18. After two cards with two numbers are taken out, there are 4 pairs of cards with the same
number that remain. There are 8 C2 ways to pick the cards without restriction, so the
4
1
 .
probability is
7
8 C2
19. Since the given graph is nonnegative on 0  x   , the area is just


0

2sin x  3cos( 2x ) dx  2 cos x  6sin( 2x ) 0  10 .
Page 2
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Mu Division
20. Since A has four columns, B must have four rows, or else the product AB would not be
defined. Since C has three rows, B must also have three columns, or else the product BC
would not be defined. Thus, B has (4)(3) = 12 elements.
21. By the Pythagorean Theorem, AB  62  122  6 5 . From the Angle Bisector Theorem,
6
12
, making AD  2 5 . Now A  arctan 2 and using the Law of Sines on

AD 6 5  AD
triangle ACD, we obtain
CD
2 5

 2 10 . Thus,
sin arctan 2 sin 45
 2 
CD  2 10 sin arctan 2  2 10 
 4 2.
 5
22. The function and the tangent line must have equal value and derivative at the point of
tangency. Hence, f (a)  a 2 and f (a )  16a . The limit is a 0/0 indeterminate form, so by
L’Hopital’s Rule,
lim
x a
f ( x)  a 2
f ( x) f (a) 16a
 lim


 8.
2
2
x a 2 x
2a
2a
x a
23. Let a, b, and c be the sides of the triangle, with c the hypotenuse. We have a  b  c  56
and a 2  b 2  c 2  c 2  c 2  2c 2  1250 , so c = 25. So a  b  31 . Now the area of the
triangle is 12 ab ; we derive this value without actually finding out what a and b are:
1
(a  b)2  (a 2  b 2 ) (a  b)2  c 2 (a  b  c)(a  b  c) (31  25)(31  25)
ab 



 84 .
2
4
4
4
4
Of course, letting a = 7 and b = 24 will also do the trick…
24. By the linearity of integrals,

8
4

8
6
8
6
2
2
f ( x) dx   f ( x) dx   f ( x) dx  14  5  9 . Also, we have
6
8
6
4
6
4
f ( x) dx   f ( x) dx   f ( x ) dx . Thus, 2   f ( x) dx  9 , so
25. Write the equation as

( x  1)  10 x  1  25  9 , which factors as
x  1  5  9 , and we quickly arrive at x = 17.
Page 3
6
4
f ( x) dx  -7 .

x 1  5

2
 9 , so
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Mu Division
26. Let N  M 2 . Since the last two digits of N are 25, the units digit of M must be 5. In
particular, if M is 5, then x = 0. Now suppose that M has at least two digits. If the tens digit
of M is 1, 3, 6, or 8, then x = 2. If the tens digit of M is 2 or 7, then x = 6. If the tens digit is
4, 5, or 9, then x = 0. The sum of the possible values of x is 0 + 2 + 6 = 8.
27. By the Intermediate-Value Theorem, for any c  (2,3) we can find an x  (4, 2) such that
f ( x)  c . In the given set, there are 4 elements that are in the interval (–2, 3): –1, 0, e, and
2.
28. To maximize the velocity, we need to set the acceleration function equal to 0. We have
2  2 ln t
a (t )  x (t ) 
, so a (t )  0 at t = e. The first derivative test (with respect to the
t2
velocity function) confirms that this value yields a maximum.
29. The primes with the given property are 11, 13, 31, 17, 71, 37, 73, 79, and 97, for a total of 9
primes.
30. Note that differentiation of a function reverses its “parity,” that is, the derivative of an even
function is odd, and vice versa. Thus, f  is odd and f  is even. From this, the sum of the
missing entries is 6  0  0  18  84  -60 .
31. If (7, 2) is a point on g, then the corresponding point on f is (2, 7). Since f ( x)  4 x3  6 ,
f (2)  26 and g (7)  1/ f (2)  1/ 26 . The slope of the normal line is –26.
32. The resulting solid is an octahedron, which consists of two congruent square pyramids glued
together at the base. The base of the pyramid is a square with diagonal 6, and thus, a side
length of 6 / 2  3 2 . The height of the pyramid is 3. Thus, the volume of the octahedron
is 23 (3 2)2 (3)  36 .
1
dy
1 x3
2
2 2
 x y , then 2 dy  x dx or, after integrating both sides,  
33. If
 C . Since the
dx
y
y 3
1
1 03
solution passes through (2, 3), we have C = –3 , so when x = 0,    3 and y  .
3
y 3
Page 4
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Mu Division
A
34. Consider this figure:
B
C
D
G
E
F
H
H
There are 5 acute triangles in this figure: AEG, DHE, HBC, ACI, and BDF. There is a more
theoretical (and satisfying) way to prove that 5 is the maximum number of acute triangles, but
it’s a tad bit more time consuming and impractical for the purposes of this test.
2
1
 12 4 
35. By the Disc Method, the volume is     dx       8 .
 /12 x
 
 
 /4
36. Note that 28 is a perfect number (a number whose positive integral factors add up to twice the
number). In a perfect number, the sum of the reciprocals of its positive integral factors is
equal to 2. Since 28  22  7 , it has (2 + 1)(1 + 1) = 6 positive integral factors. The harmonic
mean is 6 / 2  3 .
37. The area of the sample space is 1, so the probability will just be the area of the correct graph.
If we let x and y be the numbers, then we want the region on 0 < x, y < 1 such that
x  y  1 , or y  (1  x ) 2 . Integration produces the area of the region and consequently,
the probability:

1
0
1
(1  x ) 2 dx   1  2 x  x dx 
0
Page 5
1
.
6
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – School Bowl – Mu Division
38. We minimize the square of the distance from the center to one focus, which is
(t 2  30)  (8t  5)  t 2  8t  25  (t  4) 2  9 . Since (t  4)2  0 for all real numbers, the
minimum squared distance is 9, making the minimum distance 3.
39. There are 7 choices for the first digit (since the number is in-between 1000 and 8000), 9
choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit.
The total number of possibilities is (7)(9)(8)(7) = 3528.
40. We have
sin(2 )  2sin  cos  2(
3
10
)(
1
10
)  53 and cos(2 )  cos2   sin 2   (
so sin(4 )  2sin(2 ) cos(2 )  2( 53 )( 54 )  -
24
.
25
Page 6
) (
1 2
10
)   54 ,
3 2
10