Victory Quiz #1 Answers and Grading Rubric
Correct answers without work receive full credit.
Question #1 (14 Points)
: What is the domain and range of $f(x) = x \ln(x)$?
Answer:
The domain is all positive real numbers. To find the range, first find the relative maximum and minimum of $f$ by using the derivative:
$$f'(x) = \frac{x}{x} + \ln(x)=1+\ln(x).$$
To find the critical points, set the first derivative equal to 0:
$$1+\ln(x)=0\implies \ln(x) = -1\implies x=e^{-1}.$$
To determine if this is a maximum or minimum, use the second derivative test (the first derivative test also works):
$$f''(x) = \frac{1}{x}, f''(e^{-1})>0.$$
Hence, $x=e^{-1}$ is a relative minimum. Finally, determine the end behavior of the function; as $x$ approaches $0$, $f(x)$ approaches 0, and as $x$ approaches $\infty$, $f(x)$ approaches infinity. Hence, the range is $[-e^{-1}, \infty)$.
Scoring:
1 Point for knowing the difference between the domain and range.
1 Point for the correct domain.
3 Points for finding the correct derivative.
3 Points for finding the critical points of the function.
2 Points for determining the relative maximums and minimums of the function.
2 Points for determining the absolute maximum of the function.
2 Points for determining the absolute minimum of the function.
Question #2 (8 Points)
: What is the derivative of $g(f) = f^2 + e^{\pi} \ln(\sqrt{17})$?
Answer:
Notice that $e^{\pi}\ln(\sqrt{17})$ is just a constant, so its derivative is zero. Thus, $g'(f) = 2f$.
Scoring:
Approach 1: Use the limit definition of the derivative.
1 Point for writing the limit definition of the derivative.
1 Point for properly substituting the function into the limit definition.
4 Points for performing the algebra correctly.
2 Points for properly evaluating the limit.
Approach 2: Use derivative rules.
4 Points for recognizing that $e^{\pi}\ln(\sqrt{17})$ is a constant.
1 Point for properly differentiating $e^{\pi}\ln(\sqrt{17})$.
3 Points for properly differentiating $f^2$.
-1 Point if the product, chain, or quotient rules are attempted incorrectly.
Question #3 (23 Points)
: A group is a set $G$ together with a binary operation $\cdot$ typically called multiplication that satisfies the following properties: 1) G is closed under multiplication 2) multiplication is associative; that is, $a(bc) = (ab)c$ 3) there is an identity $e$ in $G$ such that $ae = ea = a$ for all $a$ in $G$ and 4) for each element $a$ in $G$, there is an element $a^{-1}$ in $G$ called the inverse of $a$ such that $aa^{-1} = a^{-1}a = e$. We define the order of a group as the number of elements in the group, and the order of a group element is the number of times the element must be multiplied by itself to reach the identity. If no such number exists, we say that the order of the group element is infinite. Which of the following statements about groups and their orders are true?
(Please note: you may NOT directly search any of these statements.)
I) The set of integers under addition modulo $n$ form a group.
II) For all elements $a$ and $b$ in a group, $ab = ba$.
III) The identity element in a group is unique.
IV) For elements $a$ and $b$ in a group, $(ab)^{-1} = a^{-1} b^{-1}$.
V) The set of real numbers under multiplication form a group.
VI) A group of order 31 can have an element of order 6.
VII) The order of the identity is 1.
VIII) Let $a$ be a group element. If $a^k=e$ (that is, multiplying $a$ by itself $k$ times) for some natural number $k$, then the order of $a$ is $k$.
IX) Every group has an element of order 1.
Answer:
I, III, VII, and IX are true. II and IV are only true in Abelian (or commutative) groups. The correct version of IV is $(ab)^{-1} = b^{-1} a^{-1}$. V is false because $0$ does not have a multiplicative inverse. VI is false by Lagrange's Theorem; the order of an element in a group must divide the order of the group. A counterexample to VIII is $e^2=e$. The order of the identity is 1, but the value of $k$ here is 2. It is however true that if $a^k=e$ for some natural number $k$, then $k$ divides the order of $a$.
Scoring:
2 Points for each true statement correctly marked as true.
1 Points for each false statement not marked as true.
1 Point for correctly marking all of I, III, VII, and IX true.
1 Point for each answer with a reasonable rationale (even if wrong).
Question #4 (25 Points)
: What words and expressions should go in the blanks in the following proof to correctly prove the vector Thales theorem: if
s
and
v
lie on one line through
0
,
t
and
w
are on another line through
0
, and
w
-
v
is parallel to
t
-
s
, then
v
= a
s
and
w
= a
t
for some scalar a.
(Notes: Boldfaced font indicates a vector. Each blank indicates a single word or a single expression. An expression in this context is any sequence of symbols that includes no equals sign. Answers should be given inline with some kind of separator (e.g.: color, blanks). You may search the phrase "vector Thales theorem", but you may not copy parts of the skeleton proof below.)
Proof:
Since
w
-
v
is ________ to
t
-
s
, there exists a scalar a such that
w
-
v
= ________ = ______. Then, since
v
is on the same line as ____, __ = ___ for some scalar b. Likewise, since
w
is on the _____ _____ as
t
, there exists a scalar c such that
w
= ______. Thus,
w
-
v
= ______ and
w
-
v
= c
t
- c
s
. It follows that (c - a)
t
+ (a - b)
s
= ____. Since ____ and
t
point in ______ ______ from
0
, they are _______ _______. Thus, c - a = 0 and a - b = 0. Therefore,
v
= a
s
and
w
= a
t
. QED.
Answer:
Since
w
-
v
is __parallel__ to
t
-
s
, there exists a scalar a such that
w
-
v
= _a(
t
-
s
)__ = __a
t
- a
s
__. Then, since
v
is on the same line as _
s
_, __
v
__ = __b
s
__ for some scalar b. Likewise, since
w
is on the _same_ _line_ as
t
, there exists a scalar c such that
w
= __c
t
__. Thus,
w
-
v
= __a
t
-a
s
__ and
w
-
v
= c
t
- c
s
. It follows that (c - a)
t
+ (a - b)
s
= _
0
_. Since __
s
_ and
t
point in __different__ __directions__ from
0
, they are __linearly__ __independent__. Thus, c - a = 0 and a - b = 0. Therefore,
v
= a
s
and
w
= a
t
. QED.
Scoring:
3 Points for a diagram/image of the setup.
1 Point for each correct blank.
2 Points for four correct blanks.
2 Points for eight correct blanks.
1 Point for twelve correct blanks.
1 Point for sixteen correct blanks.
Question #5 (30 Points)
: Suppose $Y$ is a normally distributed random variable with mean 40 and variance $\sigma^2$. To the nearest integer, what value of $\sigma$ makes the probability that $Y$ is between 20 and 60 equal to 50%?
Answer:
First, compute the normalized z-score associated with $20$:
$$z_{20}=\frac{20-40}{\sigma}.$$ Since the normal distribution is symmetric and the mean of $Y$ is 40, the probability that $Y$ is between 20 and 40 should be the same as the probability that $Y$ is between 40 and 60. Let $X$ denote a normally distributed random variable with mean 0 and variance 1. Using a calculator, we can compute the value $z$ such that the probability $X$ is less than $z$ is 25%. (The value 25% comes from the fact that the normal distribution is symmetric about its mean.) This comes out to be about -0.674. Hence,
$$\frac{20-40}{\sigma}\approx -0.674.$$
Solving this equation for $\sigma$ yields $\sigma = 30.$
Scoring:
5 Points for computing the normalized z-score.
6 Points for noting the normal distribution is symmetric around the mean.
14 Points for finding the z-score cutoff.
2 Points for setting the z-score equal to -0.674.
2 Points for solving for sigma.
1 Point for correctly rounding to the nearest integer.