Theta Functions Solutions
FAMAT State Convention 2014
1. A f2=2, f3=3, f4=5, f5=8, f6=13, f7=21, f8=34, f9=55, f10=89.
2. B g(1) = 8 and g(-3) = 4 so g(1) + g(-3) = 12.
3. D We want q(2((u+1)/2)), so we must substitute (u+1)/2 in for x in the q(2x)
equation from the problem. Hence, we get q(x+1) = x2/2 + 5x/2 + 3.
4. A
(𝒙!𝒉)𝟐 !𝟒 𝒙!𝒉 !𝟒!(𝒙𝟐 !𝟒𝒙!𝟒)
𝒉
= 2𝑥 + ℎ + 4.
5. E Substituting, we have xlogx = 100x  logxlogx = log(100x)  (logx)2 = logx + 2
 (logx)2 – logx – 2 = 0  (logx – 2)(logx + 1) = 0  logx = 2  x = 100 and
logx=-1x=1/10.
6. C f2(x) = 1 – 1/(1-1/x) = 1/(1-x). f3(x) = 1/(1-(1-1/x)) = x. f4(x) = 1-1/x. Hence,
the function is cyclic in groups of three. Thus, f2014(x) = f2(x) = 1/(1-x).
7. E x = log7f-1(x) + 12  log7f-1(x) = x – 12  f-1(x) = 7x -12.
8. A f(log3a) = 3log3a -2 =3log3a-log39=3log3a/9=a/9
9. D f(1/27) = 1/3 = loga1/27  a1/3 = 1/27  a = 1/273 = 1/39
10. B f(x+1)+f(x+2)+f(x-1)= 3x+1+3x+2+3x-1 = 3*3x+9*3x+(1/3)*3x = (37/3)*3x=37*3x-1
11. C The remainder is just f(-3) = -43.
12. C The sum of the roots is –b/a = -5.
13. A Factoring, we get y = (x+2)2 + 5. Hence, the minimum value is 5.
14. A f(x) = (x+1)/((x+2)(x+1)). Hence, x cannot be -2 or -1. Thus, the domain is
(-∞,-2)U(-2,-1)U(-1,+∞).
15. A f(x) has a hole at x=-1 and a vertical asymptote at x=-2. See factoring in
above solution. Hence, there is one asymptote.
16. B The sum that is asked is ½ + ¼ + 1/8 + 1/16 + …. which is a geometric series
with first term ½ and common ratio ½. Thus, the sum is (1/2)/(1-1/2) = 1.
17. E Similar to determine the number of zeroes at the end of x! in base 10,
determining the number of zeroes at the end of x! in base 6, we need to
determine the number of factors of 6 in 542!. However, we will first convert
542 to base 10, which is 206. Hence, we must check for the number of factors
of 3, since there will clearly be more factors of 2 than 3. Thus, we have 206/3
= 68.7, 206/9 = 22.9, 206/27 = 7.6, 206/81 = 2.5. Hence, we have 68+22+7+2 =
99 zeroes.
18. B Taking the log of both sides, f(x) = log(1-1/(x+1)) = log(x/(x+1)) = logx –
log(x+1). Thus, the sum is just (log2 – log3) + (log3 – log4) +…+(log2013 –
log2014) = log(1/1007) = -log1007.
19. C f(42) = f(42+5) = f(42+5+5) = …=f(42+5*12) =f(102) = 105.
20. D 0=|x+1| - |x-3|  |x+1| = |x-3|  x+1 = x-3 or x+1 = -(x-3)  x =1.
21. C Since the polynomial has real coefficients, the non-real roots must come in
pairs, so the max is 8.
22. D f(x) = (x2-5x+6)Q(x) + R(x) where R(x) has form ax+b. Thus, f(2) = (0)Q(x) +
2a+b and f(3) = (0)Q(x) + 3a+b. Thus, we have 3 = 2a+b and 11 = 3a+b.
Subtracting the two equations, a=8. Hence, b=-13. Thus, the remainder is 8x-13.
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Theta Functions Solutions
FAMAT State Convention 2014
2x+2
x+1
x+1
x+1
23. A 2*e
– 13e +15=0  (2e -3)(e – 5) = 0  ex+1=3/2 or ex+1=5 
x=ln(3/2) -1, ln5 – 1.
24. B Factoring, we get (x+1)(2x+1)(3x+2). Checking the regions between the
zeroes, we find that the expression is positive for (-1,-2/3) U (-1/2,∞).
25. A Since f(-x) = f(x), f(x) is even, so the axis of symmetry is the y-axis.
26. C f(g(100))+g(f(100)) = log(1002+3)+((log100)2+3) = 7 + log10003.
27. A We have a2-bc = b2-ac  a2-b2 = bc-ac  (a+b)(a-b)=c(b-a)-(a+b)=c
a+b+c=0.
28. A x = 1+f(x)/x  x2 = x + f(x)x2-x=f(x). Hence, f(x) = 0 x=0,1. Testing the
regions in between, see that f(x)<0 for (0,1).
29. A Since it’s x+4, being substituted, it is a shift of 4 units to the left.
30. E Combining logs and anti-logging, we have (x+3)/(x-3) = x-7  x2 – x – 2 = 0 
(x-2)(x+1)=0 x=2,-1. Checking both solutions, we see that both are
extraneous.
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