FORM GR1268 Question 29

Question:
A tree is a connected graph with no cycles. How many nonisomorphic trees with \(5\) vertices exist?

(A) \(1\)     (B) \(2\)     (C) \(3\)     (D) \(4\)     (E) \(5\)

Answer:
(C)

Answer Key:
A tree with \(5\) vertices have \(5-1=4\) edges.
To see this, starting with a node, attach edge + node one at a time to construct a tree.

The degree of a node is the number of edges attached ("incident") to the node.
Every edge is incident to two nodes.
So, the sum of degrees have to be \(4\times2=8\).

Isomorphic trees have the same degree sequence.
A degree sequence lists the degree of nodes in an descending order.
In other words, if \((d_1,d_2,d_3,d_4,d_5)\) is a degree sequence, \(d_i\ge d_j\) when \(i\lt j\) and \(d_1+d_2+d_3+d_4+d_5=8\).

All possible degree sequences are \((4,1,1,1,1)\), \((3,2,1,1,1)\), and \((2,2,2,1,1)\).
Corresponding trees are shown below.

FORM GR1268 Question 28

Question:
Let \(f\) be a one-to-one (injective), positive-valued function defined on \(\mathbb R\). Assume that \(f\) is differentiable at \(x=1\) and that in the \(xy\)-plane the line \(y-4=3(x-1)\) is tangent to the graph of \(f\) at \(x=1\). Let \(g\) be the function defined by \(g(x)=\sqrt x\) for \(x\ge0\). which of the following is FALSE?

(A) \(f'(1)=3\)

(B) \(\displaystyle\left(f^{-1}\right)^{\prime}(4)=\frac{1}{3}\)

(C) \((fg)'(1)=5\)

(D) \((g\circ f)'(1)=\frac{1}{2}\)

(E) \((g\circ f)(1)=2\)

Answer:
(D)

Answer Key:
The line \(y-4=3(x-1)\) has a constant derivative, \(3\).
Because \(f\) is tangent to the line at \(x=1\), we have \(f'(1)=3\).

The line goes through the point \((x,y)=(1,4)\).
The slope of \(f^{-1}\) is the reciprocal of the slope of \(f\) at that point.
Thus, (B) is true.

For (C), since \(\displaystyle g'(x)=\frac{1}{2\sqrt x}\), we have
\[\begin{align}
(fg)'(1)&=f'(1)g(1)+f(1)g'(1)\\
&=3\cdot\sqrt1+4\cdot\frac{1}{2\sqrt1}\\
&=5.
\end{align}\]

For (D),
\[\begin{align}
(g\circ f)'(1)&=(g(f(x))'(1)\\
&=\frac{dg}{df}(f(1))\frac{df}{dx}(1)\\
&=\frac{1}{2\sqrt{f(1)}}\cdot3\\
&=\frac{1}{2\sqrt4}\cdot 3\\
&=\frac{3}{4}.
\end{align}\]

For (E), \((g\circ f)(1)=g(f(1))=g(4)=\sqrt{4}=2\).

FORM GR1268 Question 27

Question:
\((1+i)^{10}=\)

(A) \(1\)     (B) \(i\)     (C) \(32\)     (D) \(32i\)     (E) \(32(i+1)\)

Answer:
(D)

Answer Key:
\[\begin{align}
(1+i)^{10}&=\left(\sqrt2 e^{\frac{\pi}{4}i}\right)^{10}\\
&=2^{\frac{10}{2}}e^{\frac{10\pi}{4}i}\\
&=32e^{\frac{5\pi}{2}i}\\
&=32\left(\cos\frac{5\pi}{2}+i\sin\frac{5\pi}{2}\right)\\
&=32\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\\
&=32(0+i)\\
&=32i.
\end{align}\]

FORM GR1268 Question 26

Question:
\[\begin{align}
3x&\equiv 5\pmod{11}\\
2y&\equiv 7\pmod{11}
\end{align}\]
If \(x\) and \(y\) are integers that satisfy the congruences above, then \(x+y\) is congruent module \(11\) to which of the following?

(A) \(1\)     (B) \(3\)     (C) \(5\)     (D) \(7\)     (E) \(9\)

Answer:
(D)

Answer Key:
Multiply the first equation by \(2\) to obtain \(6x\equiv10\pmod{11}\).
Multiply the second equation by \(3\) to obtain \(6y\equiv21\equiv10\pmod{11}\).
Add the two equations to obtain \(6(x+y)\equiv20\equiv9\pmod{11}\).

Compare this equation to
\(6\cdot1=6\equiv6\pmod{11}\)
\(6\cdot3=18\equiv7\pmod{11}\)
\(6\cdot5=30\equiv8\pmod{11}\)
\(6\cdot7=42\equiv9\pmod{11}\)
\(6\cdot9=54\equiv10\pmod{11}\)

FORM GR1268 Question 25

Question:

The graph of the derivative \(h'\) is shown above, where \(h\) is a real-valued function. Which of the following open intervals contains a value \(c\) for which the point \((c,h(c))\) is an inflection point of \(h\)?

(A) \((-2,-1)\)     (B) \((-1,0)\)     (C) \((0,1)\)     (D) \((1,2)\)     (E) \((2,3)\)

Answer:
(A)

Answer Key:
An inflection point is where a curve switches from concave to convex, or vice versa.
A curve is concave if \(h^{\prime\prime}\lt0\) and convex if \(h^{\prime\prime}\gt0\).
At an inflection point, \(h^{\prime\prime}=0\).
Thus, we are looking for a point where \(h'\) is flat (the slope is zero).

There is such a point in the interval \((-2,-1)\).
Since \(h^{\prime\prime}\) changes from \(\lt0\) to \(\gt0\), \(h\) changes from being concave to being convex.

Alternatively, you can use integration.
The area between \(h'\) and \(x\)-axis is initially decreasing.
From \(x=-2\) to around \(x=-1.5\), the area decreases at an increasing speed.
From around \(x=-1.5\) to around \(x=-0.5\), the area decreases at a decreasing speed.
From around \(x=-0.5\), the area increases at an increasing speed.

FORM GR1268 Question 24

Question:
Consider the system of linear equations
\[\begin{align}
w&+&3x&+&2y&+&2z&=0\\
w&+&4x&+&y&&&=0\\
3w&+&5x&+&10y&+&14z&=0\\
2w&+&5x&+&5y&+&6z&=0\\
\end{align}\]
with solutions of the form \((w,x,y,z)\), where \(w,x,y,\) and \(z\) are real. Which of the following statements is FALSE?

(A) The system is consistent.
(B) The system has infinitely many solutions.
(C) The sum of any two solutions is a solution.
(D) \((-5,1,1,0)\) is a solution.
(E) Every solution is a scalar multiple of \((-5,1,1,0)\).

Answer:
(E)

Answer Key:
Plug in \((-5,1,1,0)\) and confirm that it is a solution.
Thus, (D) is true.

Since the system has a solution, it is consistent.
Thus, (A) is true.

Let \(\displaystyle \mathbf A= \begin{bmatrix}1 & 3 & 2 & 2\\ 1 & 4 & 1 & 0\\ 3 & 5 & 10 & 14\\ 2 & 5 & 5& 6\end{bmatrix}\).
If \(\mathbf v=\begin{bmatrix}w\\x\\y\\z\end{bmatrix}\) is a solution,then \(\mathbf{Av}=\mathbf 0\).
Then for any scalar \(k\), we have \(k\mathbf{Av}=\mathbf{A}(k\mathbf v)=\mathbf 0\), so \(k\mathbf v\) is also a solution.
Thus, (B) is true.

If \(\mathbf v\) and \(\mathbf u\) are solutions, \(\mathbf{Av}=\mathbf 0\) and \(\mathbf{Au}=\mathbf 0\),
so \(\mathbf{Av}+\mathbf{Au}=\mathbf{A(v+u)}=\mathbf 0\).
In other words, \(\mathbf v+u\) is a solution.
Thus, (C) is true.

Since answer choices (A) through (D) are true, it must be that (E) is false.

FORM GR1268 Question 23

Question:
Let \((\mathbb Z_{10}, +,\cdot)\) be the ring of integers modulo \(10\), and let \(S\) be the subset of \(\mathbb Z_{10}\) represented by \(\{0,2,4,6,8\}\). Which of the following statement is FALSE?

(A) \((S,+,\cdot)\) is closed under addition modulo \(10\).
(B) \((S,+,\cdot)\) is closed under multiplication modulo \(10\).
(C) \((S,+,\cdot)\) has an identity under addition modulo \(10\).
(D) \((S,+,\cdot)\) has no identity under multiplication modulo \(10\).
(E) \((S,+,\cdot)\) is commutative under addition modulo \(10\).

Answer:
(D)

Answer Key:
\(S\) contains all even remainders.

(A) is true because even + even = even, and even number has an even remainder when divided by \(10\).

(B) is true because even \(\cdot\) even = even, and even number has an even remainder when divided by \(10\).

(C) is true because \(0\) is additive identity.

(D) is not true because \(6\) is multiplicative identity.
That is,
\(0\cdot6=0\equiv0\pmod{10}\)
\(2\cdot6=12\equiv2\pmod{10}\)
\(4\cdot6=24\equiv4\pmod{10}\)
\(6\cdot6=36\equiv6\pmod{10}\)
\(8\cdot6=48\equiv8\pmod{10}\)

(E) is true addition of integers are commutative.