FORM GR1268 Question 32

Question:
\(\displaystyle\frac{d}{dx}\int_{x^3}^{x^4}d^{t^2}dt=\)

(A) \(\displaystyle e^{x^6}\left(e^{x^8-x^6}-1\right)\)

(B) \(\displaystyle 4x^3e^{x^8}\)

(C) \(\displaystyle \frac{1}{\sqrt{1-e^{x^2}}}\)

(D) \(\displaystyle \frac{e^{x^2}}{x^2}-1\)

(E) \(\displaystyle x^2e^{x^6}\left(4xe^{x^8-x^6}-3\right)\)

Answer:
(E)

Answer Key:
Let \(f(t)=e^{t^2}\) and \(F'(t)=f(t)\).
\[\begin{align}
\frac{d}{dx}\int_{x^3}^{x^4}f(t)dt
&= \frac{d}{dx}\left.F(t)\right|_{x^3}^{x^4}\\
&=\frac{d}{dx}\left(F(x^4)-F(x^3)\right)\\
&=\frac{d}{dx}F(x^4)-\frac{d}{dx}F(x^3)\\
&=\frac{d}{dx^4}F(x^4)\cdot\frac{d}{dx}x^4-\frac{d}{dx^3}F(x^3)\cdot\frac{d}{dx}x^3\\
&=f(x^4)4x^3-f(x^3)3x^2\\
&=4x^3e^{(x^4)^2}-3x^2e^{(x^3)^2}\\
&=4x^3e^{x^8}-3x^2e^{x^6}\\
&=x^2e^{x^6}\left(4xe^{x^8-x^6}-3\right).
\end{align}\]

FORM GR1268 Question 31

Question:
Of the numbers \(2,3,\) and \(5\), which are eigenvalues of the matrix \(\displaystyle\begin{pmatrix}3 & 5 & 3\\ 1 & 7 & 3\\ 1 & 2 & 8\end{pmatrix}\)?

(A) None     (B) \(2\) and \(3\) only     (C) \(2\) and \(5\) only     (D) \(3\) and \(5\) only     (E) \(2,3,\) and \(5\)

Answer:
(C)

Answer Key:
If \(\lambda\) is an eigenvalue of a square matrix \(\mathbf A\), then \(\mathbf{Av}=\lambda \mathbf v\) for some nonzero vector \(\mathbf v\).
Then \(\mathbf{Av}=\lambda \mathbf{Iv}\), so \((\mathbf A-\lambda \mathbf I)\mathbf v=\mathbf 0\).
If \(\mathbf A-\lambda \mathbf I\) is invertible, then \(\mathbf v=(\mathbf A-\lambda \mathbf I)^{-1}\mathbf 0=\mathbf 0\).
But we want \(\mathbf v\) to be a non-zero vector, so \(\mathbf A-\lambda \mathbf I\) needs to be singular.

\[\begin{align}
\begin{pmatrix}3 & 5 & 3\\ 1 & 7 & 3\\ 1 & 2 & 8\end{pmatrix}-2I&=\begin{pmatrix}1 & 5 & 3\\ 1 & 5 & 3\\ 1 & 2 & 6\end{pmatrix}\quad \text{(singular)}\\
\\
\begin{pmatrix}3 & 5 & 3\\ 1 & 7 & 3\\ 1 & 2 & 8\end{pmatrix}-3I&=\begin{pmatrix}0 & 5 & 3\\ 1 & 4 & 3\\ 1 & 2 & 5\end{pmatrix}\quad \text{(non-singular)}\\
\\
\begin{pmatrix}3 & 5 & 3\\ 1 & 7 & 3\\ 1 & 2 & 8\end{pmatrix}-5I&=\begin{pmatrix}-2 & 5 & 3\\ 1 & 2 & 3\\ 1 & 2 & 3\end{pmatrix}\quad \text{(singular)}\\
\end{align}\]

FORM GR1268 Question 30

Question:
For what positive value of \(c\) does the equation \(\log x=cx^4\) have exactly one real solution for \(x\)?

(A) \(\displaystyle\frac{1}{4e}\)     (B) \(\displaystyle\frac{1}{4e^4}\)     (C) \(\displaystyle\frac{e^4}{4}\)     (D) \(\displaystyle\frac{4}{e^{1/4}}\)     (E) \(\displaystyle 4e^{1/4}\)

Answer:
(A)

Answer Key:
Depicted below are graphs of \(f(x)=\log x\) and \(g(x)=cx^4\) with \(c=1\).

We want to adjust \(c\) so that two curves touch.

Suppose the two curves touch at \(x=x_0\).

Then \(f'(x_0)=g'(x_0)\), so
\[\begin{align}
\frac{1}{x_0}&=4cx_0^3\\
1&=4cx_0^4\\
x_0^4&=\frac{1}{4c}\\
x_0&=\left(\frac{1}{4c}\right)^{\frac{1}{4}}.
\end{align}\]
We also have \(f(x_0)=g(x_0)\). Thus,
\[\begin{align}
\log x_0&=cx_0^4\\
\log\left(\frac{1}{4c}\right)^{\frac{1}{4}}&=c\cdot\frac{1}{4c}\\
\frac{1}{4}\log\frac{1}{4c}&=\frac{1}{4}\\
\log\frac{1}{4c}&=1\\
\frac{1}{4c}&=e\\
4c&=\frac{1}{e}\\
c&=\frac{1}{4e}.
\end{align}\]

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