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2013 Student Delegate Invitational
Precalculus Individual – Solutions
tion of the aforementioned formula yields:
p
√
q
√
1
2− 3
[ABC] = (15)(12)
= 45 2 − 3
2
2
1. Answer: (C) (x = 3) Because f (x) is monotonic the only way for some number of iterations
of f (x) upon itself to equal x is for f (x) to equal
x. If f (x) > x then f (f (x)) > x and each further iteration can only increase f (x). The same
is true if f (x) < x. In other words, iterations
of f (x) are a one way function; subsequent iterations cannot both increase and decrease the output. Once this fact is accepted,
√ solving the problem is simple; just solve x = 6 + x. This yields
that x = {−2, 3} but plugging each back into
the
√ function to verify reveals that only 3 works.
( 4 6= −2; -2 is an extraneous solution resulting
from the squaring of x in the first step.)
6. Answer: (E)p(18π)The graph is a semicircle is
disguise. y = 36 − (x + 2)2 thus y 2 + (x + 2)2 =
36 for y ≥ 0. The area of the semicircle is then
2
simply r 2π = 18π.
7. Answer: (E) The domain of the function is simply any value of x such that f (x) is defined. In
this case it means that the value under the radical
must be greater than or equal to 0, that the value
inside the log must be greater than 0, and that
the denominator must not be not equal to 0. Put
into equations: 8 − x ≥ 0, x2 + 3x + 2 > 0, and
x2 + 3x + 2 6= 1 (in order to avoid having the denominator equal 0.) Solving these
(−∞,
8],
n yields
√
√ o
−3+ 5 −3− 5
(−∞, −2) ∪ (−1, ∞) and x 6=
, 2
.
2
The
intersection
three
of these
requirements is
√ √
−3− 5
−3− 5
−∞, 2
∪
, −2 ∪ (−1, 8].
2
2. Answer: (B) The requested value sin2 (θ) cos2 (θ)
may be expressed as sin2 (2θ)/4 by the double
angle formula for sine. Using the double angle formula for the known value cos(4θ) yields
1 − 2sin2 (2θ) = 43 Rearranging this equation
yields sin2 (2θ) = 81 the value of the objective
function is 14 of this value.
3. Answer: (C) By the definition of odd f (x) =
−f (−x). By the definition of even f (x) = f (−x).
Combining these equations yields 2f (x) = 0.
Thus f (x) = 0 for all values of x.
4. Answer: (D) The question is really asking for
20112011 mod 100. This is simply 112011 mod100.
Simply writing out the first few terms shows that
the mods of the powers of 11 base 100 follow the
pattern of 11, 21, . . . 91, 01 . . .. meaning that
it cycles every ten powers. Dividing 2011 by 10
yields remainder 1 meaning that the total remainder is the first term in the pattern of 10 which is
11.
5. Answer:
(C) The simplest way to solve
this problem is to use the formula [ABC] =
1
2 AB sin(C) where A and B are two sides of the
triangle and C is the included angle. This then
π
requires the sine of 12
which is easily obtained via
the sine half angle formula:
r
θ
1 − cos(θ)
sin
=
2
2
π
so that: sin
=
12
p
2−
2
√
3
. A quick applica-
8. Answer: (C) Each root of x6 = 64 will be of
the form 2cis(θ) the squares of the roots will then
be 4cis(2θ). The absolute value of each of the
squares is 4 and there are 6 so (4)(6) = 24
9. Answer:
(D) The range of the function is
√
[ 37, 211]. Since the function is continuous, we
know that y achieves each value between these
end points. The number of integers in the interval is simply 211 − 6, since 211 is the greatest
integer in √
the interval and 6 is the greatest integer
less than 37.
10. Answer: (D) Adding the logs and exponentiating both sides yields y = ex. this would be a line,
but it is undefined for all x, y ≤ 0 due to the original logarithmic form. Thus it is a ray of slope
e beginning with an open circle at the origin and
extending upward into the first quadrant.
11. Answer: (C) This problem can be broken into
three parts: x + 1 + x1 , x2 + x12 , and x3 + x13 . Solving for the first part is easy; it’s just simply 5 + 1.
The next may be obtained by squaring the original equation. This yields x2 + 2 + x12 = 25 thus
x2 + x12 = 23. The last part is obtained by cubing
the first equation. x3 + 3x + x3 + x13 = 125 simply
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2013 Student Delegate Invitational
subtract (3)(x + x1 ) = 15 to get x3 +
Summing these three parts gives 139.
Precalculus Individual – Solutions
1
x3
= 110.
12. Answer: (E) The determinant of any valid product of matrices is simply the product of the individual determinants. Since each of the matrices
is a two-by-two, we know that the product is defined. The determinant of C is (7)(13)−(91)(1) =
0. Thus the product is zero.
13. Answer: (A) This problem is analogous to placing 7 balls in 5 bins. One way to approach this
problem is to partition the 7 dumplings into 5
separate regions each representing a unique individual. In order to do this we must arrange 7
dumplings and 4 partitions in some order. Thus
we have 11 things to arrange
and four must be
11
partitions. This is simply
= 330.
4
14. Answer:
(A) By our cosine addition
formula, we have that cos(8x + 3x) =
cos(8x) cos(3x) − sin(8x) sin(3x) and cos(8x −
3x) = cos(8x) cos(3x)+sin(8x) sin(3x). Summing
these equations yields that cos(8x+3x)+cos(8x−
3x) = 2 cos(8x) cos(3x). So cos(8x) cos(3x) =
cos(11x) + cos(5x)
by dividing and combining
2
the terms inside the cosines. Thus since m < n
we have m = 5 and n = 11. We are looking for
115 mod 5 which equals 15 mod 5 since 11 ≡ 1
mod 5 =⇒ 15 = 1.
15. Answer:
(B) Expanding yields sin2 (x) +
2 sin(x) cos(x) + cos2 (x). This simplifies to 1 +
sin(2x) the period of this graph is simply 2π
2 = π.
16. Answer: (C) The sum of the first 47 odd numbers is the sum of the finite arithmetic series
(1 + 93)(47)
1, 3, 5, . . . 93. The sum is just
=
2
2209, since 1 is the first term, 93 is the final term,
and there are 47 terms.
17. Answer: (A) cos(2θ) = 2 cos2 (θ)−1
√ by the dou2 35
ble angle formula. Plugging in
for cos(θ)
17
yields −9/289 this becomes simply 9 when multiplied by −289.
18. Answer: (E) Call the roots a, b, c, d, and e.
The sum 1/a + 1/b + 1/c + 1/d + 1/e simplifies to:
abcd + abce + abde + acde + bcde
abcde
This is simply the sum of the roots taken 4 at a
time divided by the sum of the roots taken 5 at a
time. This is just
56
1
− 14
1
= −4
19. Answer: (A) We have that: −684◦ is coterminal with 36◦ and 846◦ is coterminal with 126◦ .
By the cosine addition formula cos(90◦ + 36◦ ) =
cos(90◦ ) cos(36◦ ) − sin(90◦ ) sin(36◦ ) this becomes
− sin(36◦ ) which cancels with sin(36◦ ) to yield 0.
20. Answer: (A) The surface formed is an ellipse
parallel to the horizontal. The major axis is the
distance from the highest point up one side to
the lowest point on the other. This distance
is calculated by a right triangle whose legs are
the diameter of the base and the relative difference in heights. The diameter is 12 and the difference is 5 meaning the major axis has length
13. The minor axis is simply the diameter of
length 12. The area of the ellipse is calculated
inor) π(13)(12)
by Area = π(M ajor)(M
.
= 39π.
4
4
21. Answer: (E) The total horizontal distance Hansol must travel is 20ft (the length of the hall.)
She must also tranverse it 7 times from side to
side meaning she must move a total of 21ft perpendicular to the first 20. The most efficient way
to achieve this is through a hypotenuse and the
lengths conveniently form a 20, 21, 29 right triangle.
22. Answer:
(B) The expression is simply
91π
91
13 cis( 13 ). 91π/13 = 7π and cis(7π) is −1 thus
the imaginary component is 0.
23. Answer: (A) We have f (0) = E = 18 so E = 18.
f (3) = 81A + 27B + 9C + 3B + E = 354 and
f (−3) = 81A − 27B + 9C − 3B + E = −138.
Summing these two equations yields 162A+18C +
2E = 216. Substituting 18 for E and dividing by
6 yields 27A + 3C = 30 adding E 2 (324) to this
yields 354.
24. Answer: (B) Group factoring the top yields
that 3x6 + 4x4 + 7x4 − 48x2 − 64x − 102 =
(3x2 + 4x + 7)(x4 − 16). After canceling out the
x4 − 16 terms from the top and bottom, we have
that f (x) = 3x2 +4x+7 for x ∈
/ {−2, 2}. Clearly I
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2013 Student Delegate Invitational
Precalculus Individual – Solutions
is false since f (x) is undefined at x = 2. II is true
since 3(22 )+4(2)+7 = 27. (It is valid to divide out
(x4 −16) as long as we approach, but do not reach,
2.) III is true since 3(361) + 4(19) + 7 = 1166. IV
is false since 3(361)+4(−19)+7 = 1014 6= −1166.
Thus the true statements are II and III.
Subtracting our equations yields:
2
1 1
1
1
S= + +
+
+ ···
3
3 9 27 81
The right side is simply and infinite geometric
1
1
3
3
whose sum is
1 = 2 . Multiplying by 2 to
1− 3
isolate S gives that S = 43 . Multiplying by the
16 factored out at the beginning of the question
yields 12.
25. Answer: (D) To find the inverse simply replace
x with y and y with x and re-solve for y. This
2x
yields y =
.
1−x
26. Answer: (C) Set up the system of equations
t + 8 = 3n and t − 2 = 2n where t is the number of tall giraffes and n is the number of notso-tall giraffes. Solving yields that t = 22 and
n = 10. This gives 32 giraffes with 4 legs each
and 2 humans (Steve and Jeremy) with two legs
each. (32)(4) + (2)(2) = 132.
∞
X
27. Answer: (C) Factoring out 16 gives 16
i=0
i
.
3i
Let
1 2 3 4
1
1 2 3
S = + + + +· · · =⇒ S = + + +· · ·
3 9 27 81
3
9 27 81
28. Answer: (D) The smallest natural number that
can be expressed as the sum of two sets of cubes
of distinct natural numbers is 1729. The two sets
of natural numbers are 12 and 1 and 10 and 9.
This number is commonly known as the HardyRamanujan number. 1 + 7 + 2 + 9 = 19.
29. Answer: (A) The area of a triangle is base times
height over 2. (3)(4)
= 6. (3)(4) sin( π6 ) is also 6.
2
√ 30. Answer: (D) We have that arcsin 22 = π/4,
√ arccos −2 3 = 5π/6, and arctan(−1) = −π/4.
These sum to 5π/6.
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