MATH GRE

UNOFFICIAL TESTS
SOLUTIONS
Answers

1. B 23. A 45. A
2. A 24. D 46. E
3. A 25. D 47. E
4. E 26. C 48. A
5. D 27. C 49. D
6. C 28. E 50. A
7. C 29. C 51. E
8. B 30. D 52. D
9. D 31. D 53. E
10. D 32. D 54. C
11. A 33. E 55. A
12. B 34. A 56. C
13. B 35. A 57. C
14. B 36. C 58. B
15. A 37. E 59. B
16. B 38. A 60. D
17. E 39. E 61. C
18. D 40. C 62. A
19. B 41. B 63. E
20. C 42. D 64. A
21. C 43. A 65. C
22. A 44. D 66. B
 FORM MSDC01

MATHEMATICS TEST
SOLUTIONS

Last typeset: June 29, 2020

This test is NOT an official GRE Mathematics Subject Test.

R

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 Z 2 p
1. 16 − 4x2 dx =
−2

(A) 2 (B) 4 (C) 2π (D) 4π (E) 8π

Solution:
We could work this out with trig substitution (make the substitution x = 2 sin u), but the
easiest thing to do is to think geometrically:
p
y = 16 − 4x2
y 2 = 16 − 4x2
4x2 + y 2 = 16
x2 y2
+ =1
4 16
So the integrand is the upper half of an ellipse with semimajor axis a = 4 and semiminor axis
b = 2. The area of this half-ellipse is 12 πab = 4π.

1
2. Four semicircular arcs are inscribed in a square as shown in the figure above. Find the ratio
of the shaded area to the area of the square.

(A) 21 (π − 2) (B) 14 (π − 2) (C) 14 (π − 1) (D) 12 (4 − π) (E) 14 (4 − π)

Solution:
Suppose the square has side length 2x, so that its area is 4x2 . Now we can look at half of one
of the “petals” as in the following figure.

The shaded area is the difference of a quarter-circle with area 14 πx2 and a right triangle with
area 21 x2 , giving us ( 14 π − 21 )x2 . There are eight of these figures in the overall picture, so the
shaded area is (2π − 4)x2 . Thus the ratio of the shaded area to the area of the square comes
(2π − 4)x2 1
out to 2
= (π − 2).
4x 2

2
3. The line y = x + 1 is tangent to which of the following curves at x = 1?

√
(A) y = x
√
(B) y = x+1
√
(C) y = x−1
√
(D) y =2 x
√
(E) y =2 x−1
Solution:
When x = 1, we have y = 1 + 1 = 2, so the curve must pass through (1, 2); eliminate A, C,
√ dy 1
and E. Then, the derivative of y = x + 1 is = √ , which evaluated at x = 1 gives us 21 .
dx 2 x
√ dy 1
On the other hand, the derivative of y = 2 x is = √ , which evaluated at x = 1 gives us
dx x
1. Since the line y = x + 1 has a slope of 1, the answer must be D.

3


4. What are the most specific conditions under which the statement P ∧ (P → Q) → Q is
true?

(A) If and only if P is true

(B) If and only if Q is true
(C) If and only if P and Q have the same truth value
(D) For all truth values of P and Q
(E) For no truth values of P and Q
Solution:
We can check this with a truth table:


P Q P →Q P ∧ (P → Q) P ∧ (P → Q) → Q
T T T T T
T F F F T
F T T F T
F F T F T
Thus the statement is true regardless of the truth values of P and Q. If you think about what
the statement means, this makes sense: if P is true, and P implies Q, then Q must be true as
well.

4
5. Suppose f and g are continuously differentiable functions with the following properties:
f (x) > 0 and g(x) > 0 for all x ∈ R.
f 0 (x) > 0 for all x ∈ R.
g 0 (x) < 0 for all x ∈ R.
Which of the following functions is NOT necessarily monotonic?

2 f (x)

(A) g(x) (B) f (x) − g(x) (C) f (x)g(x) (D) (E) g f (x)
g(x)
Solution:
Let’s work out each in turn by differentiation, looking at the signs we get:
d
2
A: g(x) = 2g(x)g 0 (x) = 2(+)(−) < 0
dx
d
f (x) − g(x) = f 0 (x) − g 0 (x) = (+) − (−) > 0


B:
dx
d
C: f (x)g(x) = f 0 (x)g(x) + f (x)g 0 (x) = (+)(+) + (+)(−)
dx
d f (x) g(x)f 0 (x) − f (x)g 0 (x) (+)(+) − (+)(−)
D: =
2 = >0
dx g(x) g(x) (+)2
d
g f (x) = g 0 f (x) f 0 (x) = g 0 (+) · (+) = (−)(+) < 0



E:
dx
Notice that with option C, f 0 (x)g(x) is positive while f (x)g 0 (x) is negative; however, we have
no idea which is greater magnitude, so we cannot conclude the sign for any given x. It’s
entirely possible for this result to be positive for some x and negative for others, so f (x)g(x)
does not have to be monotonic.

5
6. Let f : R → R be a function defined on the real numbers. Which of the following ensures
that lim f (x) = ∞?
x→−∞

(A) For all ε < 0, there exists a δ < 0 such that x < δ implies f (x) < ε.
(B) For all ε > 0, there exists a δ < 0 such that x > δ implies f (x) > ε.
(C) For all ε > 0, there exists a δ < 0 such that x < δ implies f (x) > ε.
(D) For all δ > 0, there exists an ε < 0 such that x > δ implies f (x) > ε.
(E) For all δ < 0, there exists an ε > 0 such that x < δ implies f (x) > ε.
Solution:
This one is really about just remembering how limits are defined when involving infinity.
Since x needs to approach −∞, it needs to be able to be less than any negative real number
δ, since f (x) needs to approach ∞, it needs to be able to be greater than any positive real
number ε. Past that, just remember every analysis student’s favorite phrase: “For every ε
there exists a δ...”

6
 z
7. Suppose z is a complex number such that |z| = 8. Which of the following is equal to ?
4
(Here z stands for the complex conjugate of z.)

z z 1 16 16
(A) (B) (C) (D) (E)
16 16 16z z z
Solution:
√
Recall that |z| = z · z, so we can calculate hence:
√
z·z =8
z · z = 64
64
z=
z
z 1 64 16
= · =
4 4 z z
If you didn’t remember the above equation for |z|, though, you could always try letting z = 8i
and eliminating the other choices! (You wouldn’t want to try z = 8 because then you can’t
tell the difference between z and z.)

7
8. For which of the following shapes does the set of rotation and reflection symmetries form
an abelian group? (Assume that sides and angles that appear to be congruent are in fact
congruent.)

(A) (B) (C) (D) (E)

Solution:
Let Dn stand for the dihedral group of order 2n, which consists of n reflection symmetries
and n rotation symmetries 360/n degrees apart, including the identity transformation.
For n = 3 or greater, Dn is NOT abelian, so this eliminates B (D3 ), C (D5 ), and D (D4 ).
However, E only has two lines of reflection symmetry (one horizontal and one vertical) and
two rotations (0◦ and 180◦ ), so its group is D2 . This is isomorphic to the Klein 4-group V4 ,
which is abelian.
The circle, which has infinitely many reflection and rotation symmetries, has as its symmetry
group the orthogonal group O(n, R), but you don’t really need to know that — all you need
to know is that reflections and rotations (other than 0◦ or 180◦ ) don’t commute in general.

8
 Z x2 √
9. Let g(x) = cos( t) dt. What is the value of g 0 (π)?
3

(A) −2π (B) −π (C) −2 (D) −1 (E) 0

Solution:
We can use Leibniz’s Rule to evaluate this:
Z b(x)
d
f (t) dt = f b(x) b0 (x) − f a(x) a0 (x)



dx a(x)
√
Here we have a(x) = 3, b(x) = x2 , and f (t) = cos( t). Thus we compute:
√  √
g 0 (x) = cos x2 · 2x − cos( 3) · 0
= 2x cos x

Substituting x = π, we have g 0 (π) = 2π · −1 = −2π.

9
10. Let P = (3, −1, 2, 5) and Q = (1, 2, 2, 4) be two points in R4 . Which of the following points
lies on the line in R4 connecting P and Q?

(A) (5, 5, 2, 3)
(B) (1, 5, 3, −1)
(C) (−1, 3, 2, 0)
(D) (−2, 6, 3, 1)
(E) (−3, 8, 2, 2)
Solution:
The line connecting P and Q can be parametrized as

r(t) = P + (Q − P )t = (3 − 2t, −1 + 3t, 2, 5 − t).

This instantly eliminates B and D since their third coordinates are not 2. From here, we look
at each remaining option in turn:
• If 3 − 2t = 5, then t = −1, but this would imply that −1 + 3(−1) = 5. Eliminate A.
• If 3 − 2t = −1, then t = 2, but this would imply that −1 + 3(2) = 3. Eliminate C.
• If 3 − 2t = −3, then t = 3, which also implies that −1 + 3(3) = 8 and 5 − 3 = 2.

10
11. If f (x) = x1/x , what is f 0 (2)?
√ √ √ √ √
2 2 2 2 2
(A) + log 2 (B) log 2 (C) (1 − log 2) (D) (1 + log 2) (E) (−1 + log 2)
4 4 4 4 4
Solution:
The standard way to work this is to use logarithmic differentiation:

y = x1/x
log y = log x1/x
1
= log x
x  
d d 1
log y = log x
dx dx x
y0 1 1 −1
= · + 2 log x
y x x x
1
y 0 = y · 2 (1 − log x)
x
√
1/2 1 2
Substituting x = 2, we have x = 2 · 2 (1 − log 2) = (1 − log 2).
2 4
There’s a neat trick that can make this quicker though. It turns out that if f and g are both
functions of x, then while f g is neither a power function nor an exponential function, we can
treat it first like a power function and then like an exponential function, and add the two
results together:
d g
f = gf g−1 · f 0 + f g ln f · g 0
dx

11
 1
12. Brian is playing a crane game where the chance of winning a plush toy is each time. What
3
is the probability that it takes him exactly 5 tries to win 3 plush toys?

2 8 4 10 40
(A) (B) (C) (D) (E)
81 81 243 243 243
Solution:
 2  3
2 1 4
The probability of losing two games and winning three, in that order, is · = 5.
3 3 3
However, there are multiple orders in which these games could have happened; the fifth
game is always a win, but the first four games consist of two wins and two losses. Thus we
need the number of arrangements of the “word” WWLL, which is

4! 24
= = 6.
2!2! 2·2
4 8
Multiplying the above probability by this number of arrangements, we have 6 · 5
= .
3 81

12
13. A supplier is manufacturing disposable paper cups in the shape of a right circular cone. The
slant height of each cone is to be 4 inches long. What should be the diameter of the open
circular base of the cup, so that it holds the maximum possible volume of water?

√ √ √ √ √
4 3 8 3 2 6 4 6 8 6
(A) (B) (C) (D) (E)
3 3 3 3 3
Solution:
1 2
First let’s sketch the situation: The volume of the cone is V = πr h. Since the slant height
3
is 4, we know that r2 + h2 = 42 .

At this point, we could either eliminate r or h; however, eliminating r makes the algebra
much nicer, so that we don’t have to deal with square roots! Let r2 = 16 − h2 , which gives us
1 16 1
V = π(16 − h2 )h = πh − πh3 .
3 3 3
r
dV 16 2 16
Differentiating, we have = π − πh ; solving for h, we have h = . This means that
r dh
√ 3 3
16 32 4 6
r = 16 − = = . Therefore the diameter that maximizes the volume of the paper
3
√ 3 3
8 6
cup is d = inches.
3

13
 Z x p
14. For how many values of x does t 9 + t2 dt = 0?
−3

(A) None (B) One (C) Two (D) Three (E) Four
Solution:
First of all, if we let x = −3, then our integral covers no area and we get 0.
√
Next, suppose x = 3. Since t is an odd function and 9 + t2 is an even function, their product
is odd. This means we’re integrating an odd function over a symmetric interval, so we again
get 0.
To see
Z xthat these are the only solutions, notice that by the Fundamental Theorem of Calculus,
d p p
t 9 + t2 = x 9 + x2 , which only is zero at t = 0. This means that the function
dx −3
defined by our above integral only changes direction once, so the two zeros we found are the
only ones.

14
15. Which of the following most closely resembles the graph of the curve defined by the polar
equation r = 1 − 2 cos θ?

(A) (B) (C)

(D) (E)

Solution:
First of all, notice that since r is a function of cos θ, we have

r(−θ) = 1 − 2 cos(−θ) = 1 − 2 cos θ = r(θ).

Therefore the curve should be symmetric about θ = 0, which is the horizontal axis. Eliminate
A and D.
Now notice that the curve should touch the origin whenever r = 0. Setting 1 − 2 cos θ = 0, we
1 π π
get cos θ = , which occurs at θ = and θ = − . The only remaining curve that intersects
2 3 3
the origin twice is E.

15
16. For a hexadecimal (base 16) number, let the digits a, · · · , f represent the numbers 10, · · · , 15
in base 10. Which of the following is a divisor of 14dhex ?

(A) 1chex (B) 25hex (C) 28hex (D) 2dhex (E) 33hex
Solution:
As stated in the directions, the hexadecimal number system has the digits 0-9, as well as
the extra digits a, b, c, d, e, and f, which correspond to 10, 11, 12, 13, 14, and 15 in base 10
respectively.
First let’s convert 14d to base 10:

14dhex = 1 · 162 + 4 · 161 + 13 · 160 = 256 + 64 + 13 = 333

Now let’s convert each of the answer choices to base 10:

A: 1chex = 1 · 161 + 12 · 160 = 28
B: 25hex = 2 · 161 + 5 · 160 = 37
C: 28hex = 2 · 161 + 8 · 160 = 40
D: 2dhex = 2 · 161 + 13 · 160 = 45
E: 33hex = 3 · 161 + 3 · 160 = 51
Since 333 = 32 · 37, the correct divisor is 25hex .
A smart way to go about this would be to factor 333 first and then, seeing that 37 is a factor,
convert 37 into base 16 and see that it’s 25hex .

16
  
1 −2 0 2
17. Let A = 3 2 −1 4. Which of the following is a basis for the null space of A?
1 6 −1 0

   
 1 −1 
(A) 3 ,  1 
1 3
 
 
 0 
 
0
(B)  
0

 
0
 
 

 −4 
 
2

(C)
 8 
 

4
 
   
 2
 −6 
1  1 
(D)  , 
8  0 

 
0 4
 
     

 −4 2 −6 
     
 , , 1 
2 1

(E)  8  8  0 

 
4 0 4
 

Solution:
The null space of a matrix A is the set of all vectors v such that Av = 0. Normally to compute
this, we’d augment by a column of zeros and row-reduce, but since this is multiple-choice,
there’s no reason to do that!
A: This answer is bogus — a 3 × 4 matrix can’t be multiplied by a 3 × 1 column vector. (In
fact, this is the basis of the column space.)
B: Multiplying by the zero vector trivially gives us 0, so no surprises here. This could be
the answer if the matrix has a full rank of 3, so let’s keep going.
C: Multiplying A by this vector does in fact give us 0, so we can eliminate B.
D: Multiplying A by either of these vectors also gives us 0, and neither is a multiple of the
other. Therefore our null space has dimension at least 2; eliminate C.
E: Multiplying A by each of these three vectors gives us 0; however, notice that the first
vector is the sum of the other two. This means the set is not linearly independent, and
therefore can’t be a basis. Eliminate E, leaving D as the correct answer.

17
18. Which quadrants are contained in the preimage of quadrant III of the complex plane under
the mapping z 7→ z 3 ?

(A) Quadrant I only

(B) Quadrant III only
(C) Quadrants I and III only
(D) Quadrants I, II, and III only
(E) Quadrants I, III, and IV only
Solution:
Let a complex number z be represented in polar form as z = reiθ ; then by de Moivre’s
theorem, we have z 3 = r3 e3iθ . In other words, each complex number with argument θ will
be sent to a new complex number with argument 3θ.
Now let’s look at where each quadrant goes in turn:
π
• Quadrant I contains complex numbers with argument 0 < θ < ; these will be sent to
2
3π
complex numbers with argument 0 < 3θ < , encompassing quadrants I, II, and III.
2
π
• Quadrant II contains complex numbers with argument < θ < π; these will be sent to
2
3π
complex numbers with argument < 3θ < 3π, encompassing quadrants IV, I, and II.
2
3π
• Quadrant III contains complex numbers with argument π < θ < ; these will be sent
2
9π
to complex numbers with argument 3π < 3θ < , encompassing quadrants III, IV, and
2
I.
3π
• Quadrant IV contains complex numbers with argument < θ < 2π; these will be sent
2
9π
to complex numbers with argument < 3θ < 6π, encompassing quadrants II, III, and
2
IV.
Thus the preimage of quadrant III under the mapping z → z 3 includes quadrants I, III, and
IV.

18
 x 0 2 4 6
f 0 (x) 4 1 7 −1
f 00 (x) −1 3 0 −3
19. The function f is twice differentiable for all real x. Values of f 0 (x) and f 00 (x) are given for
selected values of x in the table above. Which of the following statements must be true?
I. f 0 has a local maximum at x = 4.
II. f has a point of inflection somewhere in the interval (0, 2).
III. There exists a c ∈ [0, 6] for which f 00 (c) = −4.

(A) I only (B) II only (C) III only (D) I and II only (E) II and III only
Solution:
I. f 0 has a local maximum when f 00 changes sign from positive to negative.
It’s tempting to pick this one, since for x = 2 → 4 → 6, f 00 (x) goes positive → zero →
negative. However, it’s possible that the graph of f 00 merely touches the x-axis at x = 4,
goes back up, and then crosses through it some time afterward, say at x = 5. All we
know is that there has to be a sign change somewhere between 4 and 6.
II. f has a point of inflection when f 00 changes sign.
At first we might think to use the Intermediate Value Theorem to say that there has to
be a sign change between −1 and 3. Note that we did not say that f is twice continuously
differentiable. Nevertheless, Darboux’s theorem says that the derivative of a function
still satisfies the intermediate value property, so our sign change is still guaranteed.
III. Again the Intermediate Value Theorem won’t really help us here, since the table only
shows values of f 00 between −3 and 3. However, notice that on the interval [4, 6], the
−1 − 7
average rate of change of f 0 (x) is = −4. Since f 00 is twice differentiable, f 0 is
6−4
differentiable (as well as continuous), so by the Mean Value Theorem, there must exist
a c ∈ [4, 6] such that f 00 (c) = −4.

19
20. Find the volume of the solid formed by revolving about the y-axis the region in the first
2
quadrant bounded by the curves y = e−x and x = 2.

(A) π(1 − e−4 ) (B) 2π(1 − e−4 ) (C) π(1 + e−2 ) (D) 2π(1 − e−2 ) (E) 2π(1 + e−2 )
Solution:
We can use the Shell Method to find the volume:
Z 2 Z 2
2
V = 2πxf (x) dx = 2πxe−x dx
0 0

Substitute u = −x2 and du = −2x dx:

Z −4
V = πeu (−du)
0
Z 0
=π eu du
−4
0
u
= πe 
−4
= π(1 − e−4 )

20
21. Suppose a curve C is parametrized by the equations x = f (t) and y = g(t), where f and g
are twice-differentiable functions. If f 0 (t) 6= 0, which of the following expressions gives the
d2 y
value of 2 when it exists?
dx

f 0 (t)g 00 (t) − g 0 (t)f 00 (t)

(A)
3
f 0 (t)
f 0 (t)g 00 (t) − g 0 (t)f 00 (t)
(B)
2
f 0 (t)
f 0 (t)g 00 (t) + g 0 (t)f 00 (t)
(C)
3
f 0 (t)
f 0 (t)g 00 (t) + g 0 (t)f 00 (t)
(D)
2
g 0 (t)
f 0 (t)g 00 (t) − g 0 (t)f 00 (t)
(E)
3
g 0 (t)

Solution:
dy g 0 (t)
First, the first derivative is = 0 .
dx f (t)
d2 y dy 0
The easiest way to remember the second derivative for parametric equations is = :
dx2 dx

d

g 0 (t)
 f 0 (t)g 00 (t) − g 0 (t)f 00 (t)

2
dy 0 dt f 0 (t) f 0 (t) f 0 (t)g 00 (t) − g 0 (t)f 00 (t)
= 0
= =
f 0 (t)

3
dx f (t) f 0 (t)

21
 ZZ
22. Evaluate cos(x2 ) dx dy, where Ω is the triangular region pictured above.
Ω

1 1 1
(A) sin 1 (B) cos 1 (C) sin 1 (D) (1 − sin 1) (E) (1 − cos 1)
2 2 2
Solution:
In order to compute this integral, we need to switch the order of integration:
ZZ Z 1Z x
2
cos(x ) dx dy = cos(x2 ) dy dx
Ω 0 0
Z 1 x
2
= y cos(x ) dx

0 0
Z 1
= x cos(x2 ) dx
0
1 1
= sin(x2 )

2 0
1
= sin 1
2

22
 ex + e−x
23. If f (x) = , then f −1 (x) =
ex − e−x
r r r r r
x+1 x−1 x+1 x−1 x+1
(A) log (B) log (C) log (D) log (E) log
2 2 x−1 x+1 1−x
Solution:
We find f −1 (x) by writing y = f (x), swapping x and y, and then solving for y again.

ex + e−x
y=
ex − e−x
ey + e−y
x= y
e − e−y
x(e − e ) = ey + e−y
y −y

xey − xe−y = ey + e−y

At this point, we can multiply both sides of the equation by ey , which allows us to solve for
y:

xe2y − x = e2y + 1
xe2y − e2y = x + 1
(x − 1)e2y = x + 1
x+1
e2y =
x−1
x+1
2y = log
x−1
1 x+1
y = log
2 x−1
r
x+1
= log
x−1

(This problem was adapted from GCTM’s 1999 Written Test.)

23
24. Which of these rings has the largest number of units?

(A) Z6 (B) Z7 (C) Z8 (D) Z (E) Z × Z

Solution:
A unit is an element of a ring that has a multiplicative inverse.
For Zn , the units are those values of n that are coprime to n.
• For Z6 , the units are 1 and 5.
• For Z7 , the units are 1, 2, 3, 4, 5, and 6. Note that this means Z7 is a field (which is always
true of Zp for a prime p).
• For Z8 , the units are 1, 3, 5, and 7.
For Z, the only units are 1 and −1 — all the other multiplicative inverses are fractions. Then,
for Z × Z, the multiplicative identity is (1, 1), so the only units are (1, 1), (1, −1), (−1, 1), and
(−1, −1).

24
25. Let y = f (x) be a nonzero solution to the differential equation

y 00 + 4y 0 + 7y = 0.

Which of the following statements must be true?

I. The equation f (x) = 0 has infinitely many solutions.
II. lim f (x) = 0.
x→∞
 
III. lim f (x) = ∞.
x→−∞

(A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III
Solution:
The characteristic equation of this differential equation is λ2 + 4λ + 7 = 0, which we can solve
with the quadratic formula:
√
√
p
−4 ± (−4)2 − 4(1)(7) −4 ± −12
λ= = = −2 ± i 3
2(1) 2

Since our solutions are nonreal, the general solution to our system of equations is
√ √
y = c1 e−2t cos( 3t) + c2 e−2t sin( 3t).

I. Since our solution oscillates around 0, there are indeed infinitely many solutions.
II. The e−2t factor means that this is a decaying oscillator, so it indeed tends toward zero.
We could prove this if we wanted using the Squeeze Theorem.
III. While the magnitude of f (x) does grow greater as x → −∞, it also returns back to 0
infinitely often, which means that this limit does not exist.

25
26. In a room with ten people, some pairs of people shake hands, while some don’t. Nobody
shakes the same person’s hand more than once. Each person in the room is then asked
whether they shook hands with an even or odd number of people. Which of the following
statements is true?

(A) The number of people who answered “even” must be even.

(B) The number of people who answered “even” must be odd.
(C) The number of people who answered “odd” must be even.
(D) The number of people who answered “odd” must be odd.
(E) The number of people who answered “odd” could be even or odd.
Solution:
Each handshake that occurs is between exactly two people, so if we were to add up each
person’s individual handshake count to make a total, we would count each handshake twice.
That is, if Alice and Bob shake hands, then the Alice-Bob handshake will be counted once
when Alice gives her count and once when Bob gives his count.
This means the sum of all the handshake counts is double the number of handshakes —
in particular, that sum is an even number. However, if an odd number of people answered
“odd” to the question, then the sum of all the handshake counts would be odd, so the number
of people who answered “odd” must be even.
You might be able to better visualize this if you think of it as a graph theory problem — each
person is a vertex, and each handshake is an edge. Then this property says that the number
of vertices of odd degree must be even. (This is even sometimes called the “Handshake
Theorem” in graph theory.)

26
27. Let g be the function defined by g(x, y, z) = z 2 exy for all real x, y, and z. The maximum
possible value M of the directional derivative of g at the point (2, 0, −1) in the direction of
some vector u ∈ R3 falls within which of the following ranges?

(A) 1 < M < 2 (B) 2 < M < 3 (C) 3 < M < 4 (D) 4 < M < 5 (E) M > 5
Solution:
Recall that the directional derivative in the direction of a unit vector u is given by ∇g · u. The
maximum possible value is then the gradient ∇g:
 
∂g ∂g ∂g
= yz 2 exy , xz 2 exy , 2zexy


∇g(x, y, z) = , ,
∂x ∂y ∂z
√
Thus ∇g(2, 0, −1) = (0, 2, −2), the magnitude of which is 2 2 ≈ 2.828.

27
 dn
 
kπ
28. cos x = sin x + if and only if which of the following congruences holds?
dxn 2

(A) k−n≡0 (mod 2)

(B) k−n≡1 (mod 2)
(C) k−n≡1 (mod 4)
(D) k−n≡2 (mod 4)
(E) k−n≡3 (mod 4)
Solution:
One way to solve this would be to use the sum of angles formula:
 
kπ kπ kπ
sin x + = sin x cos + cos x sin
2 2 2

d kπ kπ
From here, if n = 1, then we have cos x = − sin x = sin x cos + cos x sin . This means
dx 2 2
kπ kπ
we need cos = −1 and sin = 0, which is true if k = 2. Thus k − n = 1 ≡ 1 (mod 4).
2 2
d3
 
kπ
Another possibility would be to let n = 3, giving us 3 cos x = sin x = sin x + . Since
dx 2
sin x has period 2π, k can be any multiple of 4, and the result follows.

28
29. Find the remainder when 6293 is divided by 11.

(A) 6 (B) 7 (C) 8 (D) 9 (E) 10

Solution:
By Fermat’s Little Theorem, we know that 610 ≡ 1 (mod 11).
This means that 6293 = 610·29+3 = (610 )29 · 63 ≡ 63 (mod 11).
Hence could either find the remainder manually when dividing 63 = 216 by 11, or we could
aggressively reduce mod 11 whenever possible:

62 = 36 ≡ 3 (mod 11)
3 2
6 = 6 · 6 ≡ 6 · 3 = 18 ≡ 7 (mod 11)

29
 sin x − tan x
30. lim =
x→0 x3

1 1
(A) −1 (B) − (C) 0 (D) (E) The limit does not exist.
2 2
Solution:
0
Seeing the indeterminate form , we could use L’Hôpital’s Rule, but if we do so in its current
0
state, the results will be hairy. Instead, noticing sin x in the numerator and at least one x in
the denominator, let’s play with the expression algebraically:
sin x
sin x − tan x sin x − cos x sin x(1 − sec x) sin x 1 − sec x
3
= = = ·
x x3 x3 x x2
sin x 1 − sec x
This means that we can split our limit into lim · lim . The former limit is 1,
x x→0 x→0 x2
which you can check with L’Hôpital’s Rule if you wish, so we just need to evaluate the latter.
Since sec 0 = 1, let’s go ahead and use L’Hôpital’s Rule:

1 − sec x L’H sec x tan x

lim = lim
x→0 x2 x→0 2x
0 tan x sec x
Again we have the indeterminate form , but we can split this into lim · lim . The
0 x→0 x x→0 2
former limit comes out to 1 again by a quick application of L’Hôpital’s Rule, and the latter
1 1
comes out to . Thus the overall limit is .
2 2

30
31. Suppose the power series a0 + a1 (x + 1) + a2 (x + 1)2 + a3 (x + 1)3 + · · · is used to represent
2x
the function f (x) = 2 . What is the radius of convergence of this power series?
x +1
√ √
(A) 1 (B) 2 (C) 3 (D) 2 (E) ∞
Solution:
We could go through the work of finding a Taylor series expansion about x = −1 and then
determining the radius of convergence using the ratio test. However, the easiest way to think
about this is to imagine f as a function of a complex number:

2z
f (z) =
z2 + 1
Note that this function has singularities (simple poles, in fact) at z = i and z = −i. Therefore,
the disk of convergence centered at z = −1 can only extend until it hits either of these
singularities (which it actually hits both of at the same time).

√
Thus the radius of convergence is the distance from z = −1 to z = i, which is 2.

31
32. What is the set of all vectors v that satisfy the equation (2, 3, 4) × v = (−3, 2, 1)?

(A) ∅
n o
(B) (−5, −14, 13), (8, 12, −13)
n o
(C) (5, 14, −13), (−8, −6, −12)
n o
(D) (−8, −12, 13), (8, 6, −12)
n o
(E) (x, y, z) : 3x − 2y − z = 0

Solution:
Remember that for two vectors a and b, a×b will be perpendicular to both a and b. However,
notice that (2, 3, 4) · (−3, 2, 1) = (2)(−3) + (3)(2) + (4)(1) = 4. Since this dot product isn’t
zero, there’s no way that (2, 3, 4) × v could possibly be (−3, 2, 1) no matter what v is.

32
33. Suppose x is the smallest positive integer that satisfies the following congruences:

x≡1 (mod 4)
x≡2 (mod 5)
x≡3 (mod 7)

Which of the following must be true?

(A) x ≡ 2 (mod 9)
(B) x ≡ 4 (mod 9)
(C) x ≡ 6 (mod 9)
(D) x ≡ 8 (mod 9)
(E) There is no such value of x.
Solution:
The Chinese Remainder Theorem says that since the moduli are coprime, we can definitely
find a solution:

x = 1(5 · 7)(5 · 7)−1 −1 −1

4 + 2(4 · 7)(4 · 7)5 + 3(4 · 5)(4 · 5)7

Here, the notation (5 · 7)−1

4 stands for the multiplicative inverse of 5 · 7 (mod 4).

• 5 · 7 = 35 ≡ −1 (mod 4), so its inverse is −1 (mod 4).

• 4 · 7 = 28 ≡ 3 (mod 5), so its inverse is 2 (mod 5).
• 4 · 5 = 20 ≡ −1 (mod 7), so its inverse is −1 (mod 7).
Therefore the solution above gives x = 1(35)(−1) + 2(28)(2) + 3(20)(−1) = −35 + 112 − 60 =
17. This solution is unique mod 4 · 5 · 7 = 140, so it is the smallest positive solution, which
means x ≡ 8 (mod 9).
A faster way to do this would be to notice that if x ≡ 1 (mod 4), then x must be odd, and if
also x ≡ 2 (mod 5), then x must end in a 7. From here, we could let x = 10n+7 and substitute
this into the last congruence to find a suitable value of x, or we could just start looking at 3
more than multiples of 7 and keep going until we find one that ends in a 7 while satisfying
the first congruence.

33
34. Consider the following algorithm, which takes an input integer n>1 and prints a decimal
number.
input(n)
t := 0
k := 1
s := 1
while (k < n) {
a := 1/k
t := t + s*a
k := k + 2
s := s * -1
}
output(t)
If the input integer is 500, which of the following will be the output when truncated after the
hundredths digit?

(A) 0.59 (B) 0.69 (C) 0.72 (D) 0.78 (E) 0.81
Solution:
The effect of this algorithm is to approximate the infinite series:

1 1 1
1− + − + ···
3 5 7
In particular:
• k is the index of the term being computed, which increases by 2 each time.
• n is the maximum index added to the summation.
• s is the sign of the term, which is repeatedly flipped by multiplying by -1.
• a is the absolute value of the term, which is then multiplied by s to produce the alternating
series.
• t is the total of all the terms so far.
Since the value of n given is 500, the last term appended is -1/499. Then the error for the
series is bounded above by the absolute value of the next term, which would have been 1/501,
guaranteeing that our approximation is correct to the nearest hundredth.
The value of this series can be found by remembering the Taylor series for arctan x:

x3 x5 x7
arctan x = x − + − + ···
3 5 7
π
The value of this series evaluated at x = 1 is ≈ 0.78.
4

34
35. Let y = f (x) be a solution to the differential equation y 0 = y 3 − 3y + 2, y(0) = a, where a ∈ Z.
For how many values of a is lim f (x) finite?
x→∞

(A) One (B) Two (C) Three (D) Four (E) More than four
Solution:
What we’re looking for is the equilibrium solutions to the differential equation, which are
found when y 0 = 0. By inspection, y 0 = 0 when y = 1; we can divide y 3 − 3y + 2 by y − 1 to
factor it, for example with synthetic division.

1 0 −3 2
1 1 1 −2
1 1 −2 0

Therefore we have y 0 = (y − 1)(y 2 + y − 2) = (y − 1)(y − 1)(y + 2), so y = 1 and y = 2 are the

equilibrium solutions.
We can then build a sign chart for y 0 :
− + +
−2 1

Now we can classify what happens for initial conditions in these regions.
• If y(0) < −2, then y 0 < 0, and the solution will tend toward −∞.
• If −2 < y(0) < 1, then y 0 > 0, and the solution will tend toward a limit of 2.
• If y(0) > 1, then y 0 > 0, but the solution will tend toward +∞.
Thus lim f (x) is finite for the initial condition y(0) = a if a = −1, 0, 1, or 2.
x→∞

35
36. The above graph of y = fZ(x), defined for −2 ≤ x ≤ 3, consists of a semicircle and two line
x
segments. Define g(x) = f (t) dt, and let A, B, and C be defined as follows:
0

A = The maximum value of g(x) for −2 ≤ x ≤ 3

B = The number of inflection points of g(x) for −2 ≤ x ≤ 3
C = The number of points at which g(x) is not differentiable for −2 ≤ x ≤ 3
Which of the following correctly ranks the values of A, B, and C?

(A) A<B<C
(B) A<C=B
(C) B<A<C
(D) C<A<B
(E) C=B<A
Solution:
Let’s look at each of A, B, and C:
• For A, note that if we integrate backwards from 0, we get a positive result. In particular,
Z −2
π
g(−2) = g(x) dx = ≈ 1.57.
0 2
• For B, g has an inflection point wherever g 00 changes sign. Since g 00 = f 0 , we see from the
graph that f 0 changes sign at x = −1 (where f switches from decreasing to increasing)
and x = 1 (where f switches from increasing to decreasing), so g has two inflection
points.
• For C, g is not differentiable wherever g 0 is not defined. Since g 0 = f , we see from the
graph that f is defined for all x in the interval −2 ≤ x ≤ 3. (Don’t let the sharp corners
fool you — that’s where f , not g, fails to be differentiable!) So, C = 0.
Ordering these, we have C < A < B.

36
37. Suppose (x − λ1 )(x − λ2 )(x − λ3 ) is the characteristic polynomial of the matrix
 
3 1 0
A = 1 2 1  .
0 1 3

1 1 1
Calculate the quantity + + .
λ1 λ2 λ1 λ3 λ2 λ3

1 1 2 4 19
(A) (B) (C) (D) (E)
2 3 3 9 12
Solution:
λ1 + λ2 + λ3
The quantity we want can be rewritten as .
λ1 λ2 λ3
The sum of the eigenvalues is the trace of the matrix, which is 3 + 2 + 3 = 8. We could also
then find the determinant of the matrix, which is the product of the eigenvalues, but we can
actually figure out the eigenvalues directly by looking at the matrix:
• The rows have a constant sum of 4, so 4 is one of the eigenvalues (corresponding to the
vector (1, 1, 1)T ).
• By inspection,

subtracting 3I from the main diagonal will give two identical rows of
0 1 0 , so 3 is one of the eigenvalues.
• Since the trace is 8, the remaining eigenvalue must be 1.
8 2
Therefore the product of the eigenvalues is 3 · 4 · 1 = 12, so or desired quantity is = .
12 3

37
38. If G is a group of order 60, then G does not necessarily have a subgroup of order:

(A) 2 (B) 3 (C) 4 (D) 5 (E) 6

Solution:
Since G has order 60, Cauchy’s theorem for groups says that any prime p that divides 60, G
has an element of order p. Since 60 = 22 · 3 · 5, G must have elements of order 2, 3, and 5, and
therefore these elements will therefore generate cyclic subgroups of their respective orders.
Then, Sylow’s first theorem guarantees that there exist subgroups for all prime powers that
divide 60, so G must have a subgroup of order 22 = 4. Thus by process of elimination, we
can conclude that G does not necessarily have a subgroup of order 6.
If you’re curious about a concrete example, consider the direct product A4 ×Z5 . Recall that A4
is the alternating group on 4 elements, consisting of all the even permutations of 4 elements,
of which there are 4!/2 = 12; this means that A4 × Z5 will contain 12 · 5 = 60 elements. The
elements of A4 include:
• the identity permutation,
• the 3-cycles (such as (123)), and
• the 2, 2-cycles (such as (12)(34)).
These cycle types have orders 1, 3, and 2 respectively, so they once again generate cyclic
subgroups of the same order. In addition, taking any two distinct 2, 2-cycles will generate
a subgroup of order 4, which is isomorphic to the Klein 4-group V4 . However, if we take any
two 3-cycles, or a 3-cycle and a 2, 2-cycle, we will generate the entirety of A4 ; therefore the
only subgroups of A4 have order 1, 2, 3, 4, or 12. Notably, A4 has no subgroup of order 6.
Now, consider A4 ×Z5 . The identity of Z5 , i.e. 0, has order 1, while the other elements all have
order 5. This means that if our subgroup contains any element of the form (a, b) with a ∈ A4
and b ∈ Z5 \ {0}, then the order of this subgroup will be divisible by 5. Thus if we’re going
to construct a subgroup of order 6, it must only contain elements of the form (a, 0), which
means it must be isomorphic to a subgroup of A4 . However, we’ve just shown A4 doesn’t
contain a subgroup of order 6, so this is impossible.

38
 (−1)n n
 
39. Let {an } be the sequence defined by an = 1+ . Calculate lim sup an − lim inf an .
n n→∞ n→∞

e−1 e−1 e2 − 1
(A) 0 (B) (C) (D) (E) +∞
e e2 e
Solution:
 x n
Recall that lim 1 + = ex . What we have here, then, is an interleaving of two sequences
n→∞ n
1 n −1 n
     
— one is 1+ , which converges to e, and the other is 1+ , which converges
n n
1 1 e2 − 1
to e−1 . Therefore lim sup an = e and lim inf an = , giving a difference of e − = .
n→∞ n→∞ e e e

39
40. Let P3 (R) be the vector space of all polynomials of degree at most 3 with real coefficients.
Consider the following subspaces of P3 (R):
U = {p ∈ P3 (R) : p(0) = 0}
V = {p ∈ P3 (R) : p(−1) = p(1) = 0}
Which of the following statements are true?
I. U ∩ V is a subspace of P3 (R).
II. U ∪ V is a subspace of P3 (R).
III. dim(U ) + dim(V ) = dim (P3 (R)).

(A) I only (B) III only (C) I and II only (D) I and III only (E) I, II, and III
Solution:
Remember that subspaces must be closed under addition and scalar multiplication.
I. An element of U ∩ V is a polynomial for which p(−1) = p(0) = p(1) = 0. If we add two
such polynomials together or multiply one by a real number, the zeros are unchanged,
so U ∩ V is a subspace.
II. This is not true. Let p(x) = x and q(x) = x2 − 1. Then p(x) + q(x) no longer has any of
these roots, so U ∪ V is not closed under addition and therefore is not a subspace.
III. Since cubic polynomials have four coefficients, P3 (R) is isomorphic to R4 , so it has
dimension 4. From here, every time we specify one of the roots, we introduce a linear
dependency among the coefficients of our polynomials. Suppose an element of P3 (R)
is written as p(x) = ax3 + bx2 + cx + d:
• If p(0) = 0, then we know that d = 0.
• If p(1) = 0, then we know that a + b + c + d = 0.
• If p(−1) = 0, then we know that −a + b − c + d = 0.
Each linear dependency reduces the dimension of our space by 1, so dim(U ) = 3 and
dim(V ) = 2. This means that our given equation is false; the correct statement would
be dim(U ) + dim(V ) − dim(U ∩ V ) = dim(P3 (R)), since dim(U ∩ V ) = 1.

40
 Z
41. Let C be the semicircular path from (0, 0) to (2, 0) pictured above. Evaluate F · dr, where
C
F(x, y) = (x2 + y 2 )i + 2xyj.

8 16
(A) 2 (B) (C) 3 (D) 4 (E)
3 3
Solution:
We could parametrize the semicircle and evaluate our integral directly, but notice what happens
when we look at the cross-partials of the components of F:

∂ 2 ∂
(x + y 2 ) = 2y = (2xy)
∂y ∂x

This means that we’re taking the line integral of a gradient of some function, so the integral
result is path-independent by the Fundamental Theorem of Line Integrals. We could instead
just choose to integrate over the straight line path from (0, 0) to (2, 0), or better yet, we could
reconstruct the original function (the potential) from its gradient.
Z
1
(x2 + y 2 ) dx = x3 + xy 2 + g(y)
3
 
∂ 1 3
x + xy 2 = 2xy + g 0 (y)
∂y 3
1
Since g 0 (y) = 0, our potential function is x3 + xy 2 + C. Thus the value of our integral is
 (2,0) 3
1 3  8
x + xy 2  = .
3 (0,0) 3

41
 r
√ dy
q p
42. If y = x+ x + x + x + · · ·, then =
dx

1 1 y 2y + 1 1 − 2y
(A) (B) (C) (D) (E)
2y − 1 2y + 1 2y − 1 y y
Solution:
√
We can rewrite this expression as y = x + y, or y 2 = x + y. From here we can use implicit
differentiation:
d 2 d
(y ) = (x + y)
dx dx
dy dy
2y =1+
dx dx
dy dy
2y − =1
dx dx
dy
(2y − 1) =1
dx
dy 1
=
dx 2y − 1

(This problem was adapted from a Mu Alpha Theta test.)

42
43. Neisha has four index cards, each of which has a different set written on it, as shown above.
First, Neisha chooses a card at random, and lets A be the set written on that card. Then, she
replaces the card and shuffles the cards, chooses a second card at random, and lets B be the
set written on the second card. If F is the set of all functions with domain A and codomain
B, what is the probability that F is a countable set?

1 1 3 3 9
(A) (B) (C) (D) (E)
2 4 8 16 16
Solution:
Recall that the number of functions with domain A and codomain B is |B||A| .
The cardinalities of Z, Q, R, and {0, 1} are ℵ0 , ℵ0 , 2ℵ0 , and 2, respectively. (Also remember
that 2ℵ0 is strictly greater than ℵ0 , which is why we say R is uncountable.)
The only possible exponentiations that will lead to a cardinality of ℵ0 or less are 22 = 4
and ℵ0 2 = ℵ0 ; anything else will give an uncountable cardinality. This means that the only
combinations that give a countable set of functions are:

{0, 1} → {0, 1} {0, 1} → Z {0, 1} → Q

Thus 3 of the 16 possible combinations lead to a countable set of functions.

43
44. Consider the function f defined as
 c

5/2
if x ≥ 1
f (x) = x
0 if x < 1,

where c is a real constant. Suppose f is the probability distribution of a continuous random

variable X. What is the expected value of X?

(A) 1 (B) 3 (C) 6 (D) 9 (E) ∞

Solution:
In order to be a probability distribution, the area under the pdf curve must be 1. Hence we
can integrate to find the value of c:
Z ∞ ∞
2
cx−5/2 dx = − cx−3/2  = 1

1 3 1
2
c=1
3
3
c=
2
Now let’s find E(X):
∞ ∞
3 ∞
Z Z
3 −3/2
E(X) = xf (x) dx = x dx = − √  = 3
1 1 2 x 1

44
45. Suppose S is a set of continuous functions on [3, 5] such that for each f ∈ S, the following
properties hold:
f (3) = 2
f (5) = 4
f 0 (x) > 0 for all x ∈ [3, 5]
Z 4 
−1
Calculate sup f (x) dx .
f ∈S 2

(A) 4 (B) 6 (C) 8 (D) 10 (E) 14

Solution:
We can visualize the integral of f −1 as measuring area between the curve y = f (x) and the
Z 4
y-axis, in which case we’d really be writing it as f −1 (y) dy. (Remember, the actual variable
2
name we use doesn’t matter — it’s just a dummy variable in the end.)
More precisely, if f is an invertible and differentiable function such that f (a) = c and f (b) = d,
Z b Z d
then f (x) dx + f −1 (y) dy = bd − ac.
a c

From here, the supremum will be approached as the graph of f looks as much as possible
like what’s given in the figure above — close to 2 for as long as possible, then shooting up to
4 as x reaches 5. Thus the supremum of all possible areas to the left is 10.

45
 Z x
46. For each integer n ≥ 0, define Pn (x) = tn e−t dt. Which of the following recurrences is
0
satisfied by Pn (x) for all n ≥ 1?

(A) Pn (x) = nPn−1 (x)

(B) Pn (x) = xn e−x + nPn−1 (x)
(C) Pn (x) = xn e−x − nPn−1 (x)
(D) Pn (x) = xn e−x + nPn−1 (x) − Pn−1 (x)
(E) Pn (x) = xn e−x − nPn−1 (x) + Pn−1 (x)
Solution:
Let u = tn and dv = e−t dt, in which case du = ntn−1 dt and v = −e−t . Then, using integration
by parts, we have:
Z x x Z x
u dv = uv  − v du

0 0 0
x Z x
= tn e−t  − −e−t · ntn−1 dt


0
Z0 ∞
n −x
=x e +n tn−1 e−t dt
0
= xn e−x + nPn−1 (x)

46
47. The vertex-edge graph above depicts a relation ∼ on the set S = {a, b, c, d}. For any x ∈ S
and y ∈ S, an arrow drawn from x to y on the graph signifies that x ∼ y.
What is the minimum number of additional arrows that must be drawn so that the relation
represented by the resulting vertex-edge graph is transitive?

(A) Two (B) Three (C) Four (D) Five (E) Seven
Solution:
We need to make sure that there is an arrow x → y whenever there is a path from x to y:
• a→b→a
• b→a→b
• c→a→b
• a→b→d
• c→a→b→d
Thus we need five additional arrows, as illustrated below.

47
 1/n
nn

48. lim =
n→∞ n!

(A) 1 (B) e (C) e1/e (D) ee (E) The limit does not exist.
Solution:
n n n n 1/n
We can rewrite our limit as L = lim · · · ... · . Taking the logarithm of both
n→∞ 1 2 3 n
1  n n n n
sides gives us log L = lim log + log + log + . . . + log . This looks remarkably
n→∞ n 1 2 3 n
like a Riemann sum:
n
1X n
log L = lim log
n→∞ n k
k=1
Z 1
1
= log dx
0 x
Z 1
=− log x dx
0
1
= −(x log x − x) = 1

0

Since log L = 1, we have L = e.

48
49. Michael brought 12 identical cookies to work. In how many ways can he distribute those
cookies to his four coworkers so that each coworker gets at least one cookie?
        

11 12 12 13 15
(A) (B) (C) (D) (E)
3 3 4 4 3
Solution:
This kind of problem is usually solved via the so-called “stars-and-bars” method, though we
have to deal with the condition that each coworker gets at least one cookie. We’ll do that by
giving one cookie to each coworker ahead of time, which leaves 8 cookies to distribute. Now
we think of those cookies as 8 “stars”, and we think of our 4 coworkers as being separated by
3 “bars”, all of which we can arrange in a line:

? ? ? ? | ? || ? ? ?

The arrangement above would mean that of the remaining 8 cookies, the four coworkers get
4, 1, 0, and 3, respectively.
 
11! 11
The number of arrangements of 8 “stars” and 3 “bars” is = .
8!3! 3

49
 Z 1
sin t
50. Estimate dt to the nearest thousandth.
0 t

(A) 0.942
(B) 0.943
(C) 0.944
(D) 0.945
(E) 0.946
Solution:
This is what power series are made for! Begin with the Maclaurin series for sin t:

t3 t5 t7
sin t = t − + − + ···
3! 5! 7!
Divide through by t:
sin t t2 t4 t6
= 1 − + − + ···
t 3! 5! 7!
Integrate:
x
t3 t5 t7
Z
sin t
dt = t − + − + ···
0 t 3 · 3! 5 · 5! 7 · 7!
Evaluate at x = 1:
Z 1
sin t 1 1 1
dt = 1 − + − + ···
0 t 3 · 3! 5 · 5! 7 · 7!
1 1 1
=1− + − + ···
18 600 7 · 7!
Since 7! = 5040, 7 · 7! will be well above the error bound of 0.0005 we need to be correct to the
1
nearest thousandth, so we can just sum the first three terms. In particular, = 0.11111 · · · ,
9
1 1 1
so = 0.05555 · · · , and = 0.16666 · · · , so = 0.00166 · · · . Thus we can calculate:
18 6 600

1.00166
−0.05555
0.94611

50
51. Which of the following abelian groups of order 360 is NOT isomorphic to the other four?

(A) Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z5
(B) Z2 ⊕ Z2 ⊕ Z6 ⊕ Z15
(C) Z2 ⊕ Z3 ⊕ Z6 ⊕ Z10
(D) Z3 ⊕ Z4 ⊕ Z5 ⊕ Z6
(E) Z2 ⊕ Z6 ⊕ Z30
Solution:
The direct sum Zm ⊕ Zn is isomorphic to Zmn if and only if m and n are coprime. This means
that, in particular, Z2 ⊕ Z2 ∼
6 Z4 (it’s actually isomorphic to V4 instead). Thus choice D is the
=
odd one out. All of the other choices are produced by only combining coprime orders:

Z2 ⊕ Z2 ⊕ (Z2 ⊕ Z3 ) ⊕ (Z3 ⊕ Z5 ) ∼
= Z2 ⊕ Z2 ⊕ Z6 ⊕ Z15
Z2 ⊕ Z3 ⊕ (Z2 ⊕ Z3 ) ⊕ (Z2 ⊕ Z5 ) ∼
= Z2 ⊕ Z3 ⊕ Z6 ⊕ Z10
Z2 ⊕ (Z2 ⊕ Z3 ) ⊕ (Z2 ⊕ Z3 ⊕ Z5 ) ∼
= Z2 ⊕ Z6 ⊕ Z30

51
52. Let f : (0, 1) → (0, ∞) be a uniformly continuous function. Which of the following statements
are true?
I. If {xn } is a Cauchy sequence in (0, 1), then {f (xn )} is a Cauchy sequence in (0, ∞).
 
II. If lim xn exists, then lim f (xn ) = f lim xn .
n→∞ n→∞ n→∞
 
III. If {xn } and {yn } are two Cauchy sequences in (0, 1), then f (xn ) − f (yn ) is a Cauchy
sequence in (0, ∞).

(A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III
Solution:
I. This is a basic fact from real analysis: uniform continuity preserves Cauchy sequences.
Let ε > 0; since f is uniformly continuous, there exists a δ > 0 such that |x − y| < δ
implies |f (x) − f (y)| < ε. Then, since {xn } is Cauchy, there exists an N > 0 such that
m, n > N implies |xm − xn | < δ, which then implies |f (xm ) − f (xn )| < ε. Thus {f (xn )}
is Cauchy.
II. Let lim xn = x, and let ε > 0. Since f is continuous, there exists a δ > 0 such that
n→∞
|xn − x| < δ implies |f (xn ) − f (x)| < ε. By the definition of limit, for this δ, there exists
an N > 0 such that n > N implies |xn − x| < δ, which then implies |f (xn ) − f (x)| < ε.
Thus lim f (xn ) = f (x). (This is actually just a property of continuous functions, as
n→∞
long as lim f (xn ) exists in the first place, which is what uniform continuity guarantees
n→∞
here.)
III. Let ε > 0. By I above, we know that {f (xn )} and {f (yn )} are also Cauchy sequences.
ε
This means that we can find an N1 such that m, n > N1 implies |f (xm ) − f (xn )| <
2
ε
and an N2 such that m, n > N2 implies |f (ym ) − f (yn )| < ; letting N = max {N1 , N2 },
2
both implications are satisfied. Now we can use the Triangle Inequality:


  
 f (xm ) − f (ym ) − f (xn ) − f (yn )  = f (xm ) − f (ym ) − f (xn ) + f (yn )



=  f (xm ) − f (xn ) − f (ym ) − f (yn ) 
ε ε
< + =ε
2 2
 
Thus f (xn ) − f (yn ) is a Cauchy sequence as well.

52
 I
1
53. Calculate tan z dz, where C is the circle |z − 1| = 1 parametrized counterclockwise.
2πi C

(A) −2 (B) −1 (C) 0 (D) 1 (E) 2

Solution:
π
The given circle contains exactly one of the singularities of tan z at :
2

π
This means that we just need to calculate the residue of tan z at z0 = . Doing so from the
2
p(z)
definition would be annoying, but we can use a shortcut formula: If f (z) = , where
q(z)
p(z0 )
p(z0 ) 6= 0, q(z0 ) = 0, and q 0 (z0 ) 6= 0, then Res f (z) = 0 . In this case we have p(z) = sin z
z=z0 q (z0 )
π
sin 2
and q(z) = cos z, so Resπ tan z = = −1.
z= 2 − sin π2
I I
1
From here, we know that tan z dz = 2πi · Resπ tan z, so tan z dz = Resπ tan z = −1.
C z= 2 2πi C z= 2

53
54. Suppose A is a 3 × 3 matrix with the property that A3 = A. Which of the following must be
true?

(A) The eigenvalues of A are distinct.

(B) If A is invertible, then AT = A.
(C) A2 is the identity matrix or the zero matrix.
(D) The trace of A2 equals the rank of A.
(E) The absolute value of the trace of A3 equals the rank of A.
Solution:
Since A3 = A, we have A3 − A = A(A + 1)(A − 1) = 0; this means that the only eigenvalues of
A can be 0, 1, or −1. We can eliminate the incorrect answer choices by carefully constructing
example.
 
0 0 0
• If A = 0 0 0, then A3 = A, but its only eigenvalue is 0, repeated three times.
0 0 0
Eliminate A.
 
1 0 0
• If A = 0 0 21 , then A3 = A and det A = −1, but AT 6= A. Eliminate B.
0 2 0
 
1 0 0
• If A = 0 0 0, then A3 = A, but A2 = A as well. Eliminate C.
0 0 0
For D and E, there are a few facts that will help us out:
• The trace of a matrix is the sum of its eigenvalues.
• If λ is an eigenvalue of A, then λn is an eigenvalue of An .
• The rank of a matrix is equal to the sum of its nonzero eigenvalues.
From these, we can see that squaring 0, 1, and −1 will give 0, 1, and 1, respectively; this means
that adding up the eigenvalues of A2 is the same as counting the nonzero eigenvalues  of 
A.
1 0 0
This is not so for A3 , which can still have −1 as an eigenvalue. As a result, if A = 0 −1 0,
0 0 0
3
then the trace of A is 0, while the rank of A is 2.

54
 x 0 1 2 3 4
f (x) 1 −5 −7 −2 3

55. Selected values of a polynomial f (x) of degree 4 are given in the table above. What is the
value of f (5)?

(A) −9 (B) −4 (C) 2 (D) 7 (E) 11

Solution:
Since f is a polynomial of degree 4, its fourth differences ∆4 f (x) should be constant. First
let’s build a table of finite differences:

x 0 1 2 3 4
f (x) 1 −5 −7 −2 3
∆f (x) −6 −2 5 5
2
∆ f (x) 4 7 0
∆3 f (x) 3 −7
∆4 f (x) −10

From the table we see that ∆4 f (x) = −10, so we build the value of f (5) backwards from here:

x 0 1 2 3 4 5
f (x) 1 −5 −7 −2 3 −9
∆f (x) −6 −2 5 5 −12
∆2 f (x) 4 7 0 −17
3
∆ f (x) 3 −7 −17
∆4 f (x) −10 −10

Thus f (5) = −9.

55
  
56. Let g(z) be an analytic function such that g(z) = 3 for all z in the open disk |z| < 2. If
g(1) = 3i, find all possible values of g(−1).

(A) {3i}
(B) {−3i}
(C) {3i, −3i}
(D) {3, −3, 3i, −3i}
(E) {z ∈ C : |z| = 3}
Solution:
In complex analysis, there are plenty of statements that boil down to “if an analytic function
has [some property], then it must be constant.” One such property is having a constant
modulus in an open disk. There are many ways to show this; this is one of them.
Suppose g(x+iy) = u(x, y)+iv(x, y), where u and v are real-valued functions of real variables
x and y. Then |g(x, y)|2 = u(x, y)2 + v(x, y)2 , which we are claiming is a constant. This means
that their partial derivatives must be zero:

∂ 2 ∂ 2
(u + v 2 ) = 0 (u + v 2 ) = 0
∂x ∂y
∂u ∂v ∂u ∂v
2u + 2v =0 2u + 2v =0
∂x ∂x ∂y ∂y
∂u ∂v ∂u ∂v
u +v =0 u +v =0
∂x ∂x ∂y ∂y
∂u ∂v
Now, the Cauchy-Riemann equations tell us that for an analytic function, = and
∂x ∂y
∂u ∂v
= − . We can use these to rewrite our equations above entirely in terms of v:
∂y ∂x
∂v ∂v ∂v ∂v
u +v =0 −u +v =0
∂y ∂x ∂x ∂y

Multiplying the first equation by u and the second equation by v and then adding them
together, we end up with:
∂v
(u2 + v 2 ) =0
∂y
∂v
If u2 + v 2 = 0, then g(z) = 0, so g is constant. Otherwise, if = 0, which means that v is
∂y
∂v
constant and therefore = 0 as well. Applying the Cauchy-Riemann equations once more,
∂x
we can show that u is constant as well, and that therefore g must be constant.
Thus we must have g(−1) = 3i.

56
 θ
57. Let C1 be the curve defined by the polar equation r = for θ ≥ 0, and let C2 be the circle
θ+1
defined by the equation x2 + y 2 = 1. Let C = C1 ∪ C2 .
Which of the following statements are true with respect to the standard topology on R2 ?
I. For any point P ∈ C, there exists an open disk centered at P containing some point
Q ∈ C such that P 6= Q.
S
II. For any collection F of open disks such that C ⊆ F, there exists a finite subcollection
G ⊆ F such that C ⊆ G.
III. For any pair of points P ∈ C and Q ∈ C, there exists a continuous function f : [0, 1] → C
such that f (0) = P and f (1) = Q.

(A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III
Solution:
The graph of C1 is a spiral that approaches the unit circle C2 , as shown in the figure below.
The resulting hsape is sometimes called the “topologist’s whirlpool”, and it’s very similar to
the well-known “topologist’s sine curve.”

I. This is saying that every point of C is a limit point of C; this is true, since C is the closure
of C1 and is therefore closed.
II. This is saying that C is compact; since C is closed and bounded, the Heine-Borel theorem
tells us that C is compact with respect to the standard topology on R2 .
III. This is saying that C is path-connected; however, much like the topologists’s sine curve,
while C is connected, it is NOT path-connected. Actually proving this rigorously on
the exam isn’t necessary — you just need to see the similarities between the curves and
make a quick judgment. If you want to see a proof, though, there’s one provided in
Introduction to Topology, Applied and Pure by Adams and Franzosa.

57
 Z ∞
1
58. dx =
0 1 + eax

1 1 log 2 a
(A) (B) a log 2 (C) (D) (E)
a a log 2 a log 2
Solution:
Letting u = eax , we have du = aeax dx. We don’t have an extra eax anywhere, but that’s fine;
du
we can make the same substitution again and write du = au dx, or = dx. Hence we have:
au
Z ∞ Z ∞
1 ∞
Z
1 1 du 1
ax
dx = · = du
0 1 + e 1 1 + u au a 1 u(u + 1)
1 1 1
The fraction can be shown via partial fractions to equal − , which we can
u(u + 1) u u+1
use to integrate:

1 ∞ 1
Z  
1 1
∞
− du = log u − log(u + 1) 
a 1 u u+1 a 1
1 u ∞
= log
a u+1 1 

1 1
= log 1 − log
a 2
1
= (0 + log 2)
a
log 2
=
a

A slicker way to do this would be to multiply the top and bottom through by e−ax .

58
59. To the nearest thousand, approximately how many roots does the function f (x) = e−x sin(x2 )
have on the interval [0, 100]?

(A) 1000 (B) 2000 (C) 3000 (D) 4000 (E) 5000
Solution:
√ √ √
The zeros of sin(x2 ) will occur at the square roots of the zeros of sin x: x = 0, π, 2π, 3π,
√
and so on. Therefore we need to estimate the value of n such that nπ ≈ 100.
√
nπ ≈ 100
nπ ≈ 10 000
10 000
n≈
π
10 000
/
3
≈ 3 333

59
60. Circle C is tangent to the graph of y = x2 at the origin and has the same curvature as the
parabola at the point of tangency. What is the radius of circle C?

1 1 2 3
(A) (B) (C) (D) (E) 2
3 2 3 2
Solution:
Every GRE has a problem or two that uses some random formula you’ve likely forgotten. In
this case, it’s the curvature formula.
One common formula for curvature (that doesn’t require that we reparametrize in terms of
arclength) is:
kT0 (t)k
κ=
kr0 (t)k
√
Parametrizing our parabola as r(t) = (t, t2), we have r0 (t) = (1,2t), so kr0 (t)k = 1 + 4t2 .
r0 (t) 1 2t
Then we can calculate T(t) = 0 (t)k
= √ ,√ ; differentiating again, we
 kr 1 +4t 2 1 + 4t2
4t 2
end up with T0 (t) = − 3/2
, .
2
(1 + 4t ) (1 + 4t2 )3/2
kT0 (0)k
Now r0 (0) = (1, 0) and T0 (0) = (0, 2), so κ = = 2. The osculating circle therefore
kr0 (0)k
1 1
has radius r = = .
κ 2
However, since we’re working in 2D, there’s an even faster formula we can use when y is a
function of x:
|y 00 |
κ=
(1 + (y 0 )2 )3/2
Since y 0 = 2x and y 00 = 2, this becomes:

|2|
κ= =2
(1 + (0)2 )3/2
1
Again we have κ = 2, so r = .
2

60
61. Let V be the vector space of continuous functions C → C under pointwise addition and
scalar multiplication by complex numbers, and let A be the set of fourth roots of unity in the
complex plane. For each α ∈ A, define the set Sα = {cos αz, sin αz, eαz }.
[
What is the dimension of the subspace of V spanned by Sα ?
α∈A

(A) 3 (B) 4 (C) 6 (D) 8 (E) 12

Solution:
The fourth roots of unity are A = {1, i, −1, −i}, so the sets Sα are:

S1 = {cos z, sin z, ez }
Si = cos iz, sin iz, eiz = cosh z, i sinh z, eiz
 

S−1 = cos(−z), sin(−z), e−z = cos z, − sin z, e−z

 

S−i = cos(−iz), sin(−iz), e−iz = cosh z, −i sinh z, e−iz

 

From here, we notice immediately that by ignoring complex scalar multiples, our subspace
is spanned by eight functions:

sin z, cos z, sinh z, cosh z, ez , e−z , eiz , e−iz



However, we can remove even more redundancy using Euler’s identity:

eiz = cos z + i sin z

e−iz = cos(−z) + i sin(−z) = cos z − i sin z
ez = cos(−iz) + i sin(−iz) = cosh z + sinh z
e−z = cos(iz) + i sin(iz) = cosh z − sinh z

Thus we only need four elements to span this space.

(If we REALLY wanted to be sure we can’t span the space with even fewer, we could calculate
the Wronskian, but that would be overkill at this point. That being said, if you’re bored one
day, you’re welcome to work that out ... you should get 4.)

61
62. Which of the following must be true of a function f : R2 → R?
∂f ∂f
I. If and exist and are continuous at (0, 0), then f is differentiable there.
∂x ∂y
II. If f has directional derivatives in all directions at (0, 0), then f is differentiable there.
∂2f ∂2f
III. If and exist at (0, 0), then they are equal there.
∂x∂y ∂y∂x

(A) None (B) II only (C) I and II only (D) I and III only (E) II and III only
Solution:
 xy
 if (x, y) 6= (0, 0)
I. Let f (x, y) = x2 + y2 . Then f is identically zero along both the
0 if (x, y) = (0, 0)
∂f ∂f
x-axis and the y-axis, so = = 0. However, the limit of f along the line y = x is
∂x ∂y
1
, so f is not even continuous, much less differentiable.
2
p
II. Let f (x, y) = 3 x2 y. Then if u = (cos θ, sin θ) is a unit vector, the directional derivative
f (h cos θ, h sin θ) − f (0, 0) √
3
of f at (0, 0) is fu0 (0, 0) = lim = cos2 θ sin θ, which exists for
h→0 h
∂f ∂f
any θ. We can get the partial derivatives and in particular by letting θ = 0 and
∂x ∂y
π
θ = , respectively, both of which give 0. However, this would imply that the directional
2
derivative equals ∇f (0, 0) · u = 0, but our expression in terms of θ is not always 0. Thus
f is not differentiable.
To visualize f , see https://www.math.tamu.edu/~tom.vogel/gallery/node17.html.
 3 3
 x y − xy if (x, y) 6= (0, 0)
III. Let f (x, y) = x2 + y 2 . Then, using the quotient rule, the partial
0 if (x, y) = (0, 0)

derivatives are:
∂f x4 y + 4x2 y 3 − y 5 ∂f x5 − 4x3 y 2 − xy 4
= =
∂x (x2 + y 2 )2 ∂y (x2 + y 2 )2

f (x, 0) − f (0, 0)
Notice that setting x = 0 gives us fx (0, y) = −y, so fx (0, 0) = lim = 0.
x→0 x
Similarly, fy (x, 0) = x and fy (0, 0) = 0. Differentiating with respect to the other variable,
∂ 2 f  2 
0 (0, 0) = lim fx (0, y) − fx (0, 0) = −1, but ∂ f 
we have = fxy =
∂y∂x (x,y)=(0,0) y ∂x∂y (x,y)=(0,0)
 
y→0
0 (0, 0) = lim fx (x, 0) − fx (0, 0) = 1. Thus the mixed partials are not equal despite
fyx
x→0 x
both being defined.

62
 1 2
− 23
 
3 3
63. Consider the matrix equation QRx = b, where Q = − 23 2 1
 is an orthogonal matrix,
 
3 3
2 1 2
  3 3 3
3 1 4 11
R = 0 1 5 is an upper triangular matrix, and b = −10.
  
0 0 9 −65
 
x1
If x = x2 , what is the value of x3 ?
x3

(A) −6 (B) −3 (C) 3 (D) 9 (E) 12

Solution:
Since Q is orthogonal, we have Q−1 = QT , so multiplying this on both sides of the equation
we have Rx = QT b:
   1
− 23 23
  
3 1 4 x1 3 11
0 1 5  x2  =   32 2
3
1 
3  −10

0 0 9 x3 −3 2 1 2 −65
3 3

We can get the value of x3 by multiplying the last row of each matrix by the column vector
on each side of the equation:

0x1 + 0x2 + 9x3 = − 32 (11) + 13 (−10) + 23 (−65)

9x3 = −54
x3 = −6

63
64. Let A and B be ideals of a ring R, and define the ideals A + B and AB as follows:

A + B = a + b : a ∈ A, b ∈ B

AB = a1 b1 + · · · + an bn : ai ∈ A, bi ∈ B, i ∈ {1, . . . , n} , n ∈ {1, 2, . . .}

Which of the following correctly orders A + B, AB, and A ∩ B via inclusion?

(A) AB ⊆ A+B ⊆ A∩B

(B) A∩B ⊆ AB ⊆ A+B
(C) A∩B ⊆ A+B ⊆ AB
(D) AB ⊆ A∩B ⊆ A+B
(E) A+B ⊆ A∩B ⊆ AB
Solution:
Rather than thinking about this in the abstract, the easiest way to handle this would be to
look at the ideals of a very familiar ring — Z. Let A = 4Z and B = 6Z. Then:
• A + B is the set of all integers that can be written as 4x + 6y; since gcd(4, 6) = 2, we have
A + B = 2Z.
• AB is the set of all integers that can be written as 4x1 · 6y1 + . . . + 4xn · 6yn . We can
simplify this to 24(x1 y1 + . . . + xn yn ); therefore AB = 24Z.
• A ∩ B is the set of all integers that are divisible by both 4 and 6; since lcm(4, 6) = 12, we
have A ∩ B = 12Z.
Thus the correct order of inclusion is AB ⊆ A ∩ B ⊆ A + B. (Remember, when n is smaller,
nZ is a “larger” ideal because it contains more elements.)

64
65. For each pair of integers a and b, let the set {a + bn : n ∈ Z} be denoted a + bZ. Consider
the topology T on Z whose basis is {a + bZ : a, b ∈ Z and b 6= 0}. Which of the following
statements is false?

(A) (Z, T ) is Hausdorff.

(B) (Z, T ) is totally disconnected.
(C) The set {0} is closed under T .
(D) No nonempty open sets of T are finite.
(E) Exactly two sets of T are both open and closed.
Solution:
If you’d like to look this topology up later, it’s called the “evenly spaced integer topology” or
“arithmetic progression topology.”
A: (Z, T ) is Hausdorff if for any two elements x, y ∈ Z, we can find two disjoint open sets
that “separate” them. In this case, all we need to do is find an integer n that does not
divide y − x, and then the sets x + nZ and y + nZ will separate them.
B: (Z, T ) is totally disconnected if any set with at least two elements is disconnected. Suppose
we have two elements x, y ∈ S ⊆ Z, and again find an integer n that does not divide
y − x. Then let U = x + nZ and V = (x + 1 + nZ, · · · , x + (n − 1) + nZ) — in other
words, V is the union of all arithmetic sequences with difference n other than U . Then
U ∩ V = ∅, while (U ∩ S) ∪ (V ∩ S) = S, so S must be disconnected.
C: {0} is closed if its complement Z \ {0} is open. Consider the following open sets:

1 + 2Z, 2 + 4Z, 4 + 8Z, 8 + 16Z, · · ·

The union of these open sets is Z \ {0}. Since the union of any number of open sets in a
topology is open, Z \ {0} is open, so {0} must be closed.
D: Since the basis elements are all infinite, the only way we could possibly get a finite
open set is by an intersection. It suffices to consider what happens when intersecting
basis elements: if two basis elements a + bZ and c + dZ have at least one element k in
common, then they will again overlap after any multiple of lcm(b, d), so we can explicitly
calculate that (a + bZ) ∩ (c + dZ) = k + lcm(b, d)Z. Since we’re only allowed to take finite
intersections, all open sets must be infinite.
E: This may be true on the standard topology on R, but it’s not true here! For example, the
set 1 + 2Z is open since it’s a basis element, but since its complement 0 + 2Z is also open,
1 + 2Z is closed as well. In fact, it can be shown that any basis element is both open and
closed!

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66. Let an abelian group M form a left module over a ring R. We say that a subset S of M is a
spanning set of M if for every m ∈ M , there exist {r1 , r2 , . . . , rn } ⊆ R and {s1 , s2 , . . . , sn } ⊆ S
such that
Xn
m= ri si .
i=1

We call this spanning set S minimal if no proper subset of S is a spanning set for S. In
addition, S is called a basis for M if m = 0 implies r1 = r2 = · · · = rn = 0.
Which of the following statements must be true?
I. If R is finite, then any basis S of M is finite.
II. If S and S 0 are two distinct minimal spanning sets of M , then |S| = |S 0 |.
III. If R is a field and S is a finite minimal spanning set of M , then S is a basis of M .

(A) II only (B) III only (C) I and II only (D) I and III only (E) II and III only
Solution:
You can think of a module intuitively as being just like a vector space, with the elements of
the group M being like the “vectors,” except instead of the “scalar” coefficients coming from
a field, they come from a ring.
I. One example of a module is R[x], the set of all polynomials with coefficients in a ring R.
Even if R is finite — say, perhaps R = Z5 — the basis 1, x, x2 , x3 , · · · is infinite.


II. Another example of a module is any abelian group, which can be turned into a module
using Z as the ring of coefficients. In fact, we can even make the group be Z as well, so
we’re just multiplying and adding integers.
Now, an easy spanning set for Z would be S = {1}, since every integer can be written as
a · 1 for some a ∈ Z. This spanning set is minimal, since if you take 1 out, there’s nothing
left!
A different spanning set would be S 0 = {2, 3}, since gcd(2, 3) = 1 and therefore every
integer can be written as a · 2 + b · 3 for some a, b ∈ Z. This spanning set is also minimal,
since removing 2 would only span 3Z and removing 3 would only span 2Z.
6 |S 0 |. This is a way in which modules are different from vector spaces —
However, |S| =
they don’t necessarily have the invariant basis number condition.
III. If R is a field, then M is actually a vector space after all! Vector spaces do have the
invariant basis number condition. Since M has a finite minimal spanning set S, that set
S must be a basis. (It’s a good thing we have the finiteness condition given, because
otherwise we’d need to invoke Zorn’s Lemma to claim that every vector space has a
basis!)
Many thanks to the folks at https://discord.sg/math for helping me put together this question!

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