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,
\[\begin{align}
\log x_0&=cx_0^4\\
\log\left(\frac{1}{4c}\right)^{\frac{1}{4}}&=c\cdot\frac{1}{4c}\\
\frac{1}{4}\log\frac{1}{4c}&=\frac{1}{4}\\
\log\frac{1}{4c}&=1\\
\frac{1}{4c}&=e\\
4c&=\frac{1}{e}\\
c&=\frac{1}{4e}.
\end{align}\]