## Monday, 5 September 2011

### Zai geen, sayonara, goodbye (for now at least).

With the start of school looming so close, I fear that I won't have enough time to update this blog daily. Thus, I have decided to take a break on posting questions 'til winter vacation comes. I'll still be in the background gathering questions, but they won't be released to the public until December.

Thanks for all the comments, you guys really taught me a lot. I hope I was able to help all of you too (even if it was just the slightest bit).

- Paul

PS: The title of this post is from a song my friend wrote. His alias is LittleColumbus. This text is in white so that I can properly attribute him while not turning this post into a bad advertisement.

PPS: It would be awesome if you guys spread the word about this blog. I'd love to get more people commenting so we can have more discussions.

Labels:
Math GRE,
Physics GRE

### Math GRE - #34

$f$ is the function defined by: \[
f(x)=\begin{cases}
xe^{-x^{2}-x^{-2}} & \text{if } x\neq0\\
0 & \text{otherwise.}\end{cases}\]
At how many values of $x$ does $f$ have a horizontal tangent line?

$f$ has a horizontal tangent line only when the derivative of $x$ is 0. The derivative of $f$ is:

\[f^\prime(x) = e^{-x^2-x^{-2}}+(-2x+2x^{-3})xe^{-x^2-x^{-2}}=(1-2x^2+2x^{-2})e^{-x^2-x^{-2}}=0.\] This is only zero in two cases: in the limit $\displaystyle\lim_{x\rightarrow 0} f^\prime(x)$ and when $1-2x^2+2x^{-2}=0$.

The quadratic-like equation above can be solved by solving $x^2-2x^4+2=0.$ This equation has two non-zero solutions. These solutions, along with the with the solution obtained in the limit above, gives us three horizontal tangent lines in total.

- None
- One
- Two
- Three
- Four

Solution :

**Choice 4 is the answer.**

$f$ has a horizontal tangent line only when the derivative of $x$ is 0. The derivative of $f$ is:

\[f^\prime(x) = e^{-x^2-x^{-2}}+(-2x+2x^{-3})xe^{-x^2-x^{-2}}=(1-2x^2+2x^{-2})e^{-x^2-x^{-2}}=0.\] This is only zero in two cases: in the limit $\displaystyle\lim_{x\rightarrow 0} f^\prime(x)$ and when $1-2x^2+2x^{-2}=0$.

The quadratic-like equation above can be solved by solving $x^2-2x^4+2=0.$ This equation has two non-zero solutions. These solutions, along with the with the solution obtained in the limit above, gives us three horizontal tangent lines in total.

Labels:
Math GRE

### Physics GRE - # 34

Eigenfunctions for a rigid dumbbell rotating about its centre have a $\phi$ dependence of the form $\psi(\phi)=Ae^{im\phi}$, where $m$ is a quantum number and $A$ is a constant. Which of the following values of $A$ will properly normalize the eigenfunction?

For $\psi(\phi)$ to be normalized on $a$ to $b$, recall the normalization condition: \[\int_a^b{\left|\psi(\phi)\right|^2}d\phi=1.\] Since $\phi$ ranges from $0$ to $2\pi$, we have: \[\int_0^{2\pi}{\left|\psi(\phi)\right|^2}d\phi=\int_0^{2\pi}{\left|Ae^{im\phi}\right|^2}d\phi=\int_0^{2\pi}{|A|^2}d\phi=1\] where the second equality holds as $|e^{im\phi}|=1$ for any $m$ and any $\phi$.

Thus, we know that: \[\int_0^{2\pi}{|A|^2}d\phi=2\pi|A|^2=1\implies |A|=\dfrac{1}{\sqrt{2\pi}}.\]

- $\sqrt{2\pi}$
- $2\pi$
- $(2\pi)^2$
- $\dfrac{1}{\sqrt{2\pi}}$
- $\dfrac{1}{2\pi}$

Solution :

**Choice 4 is the answer.**

For $\psi(\phi)$ to be normalized on $a$ to $b$, recall the normalization condition: \[\int_a^b{\left|\psi(\phi)\right|^2}d\phi=1.\] Since $\phi$ ranges from $0$ to $2\pi$, we have: \[\int_0^{2\pi}{\left|\psi(\phi)\right|^2}d\phi=\int_0^{2\pi}{\left|Ae^{im\phi}\right|^2}d\phi=\int_0^{2\pi}{|A|^2}d\phi=1\] where the second equality holds as $|e^{im\phi}|=1$ for any $m$ and any $\phi$.

Thus, we know that: \[\int_0^{2\pi}{|A|^2}d\phi=2\pi|A|^2=1\implies |A|=\dfrac{1}{\sqrt{2\pi}}.\]

Labels:
Physics GRE

## Sunday, 4 September 2011

### Math GRE - #33

How many integers from 1 to 1000 are divisible by 30 by not by 16?

There are 33 multiples of 30 under 1000 (1000/30 = 33.33..). So we only need to focus our attention on choices 1 to 4.

Note that the factorization of 30 is: $2\cdot 3\cdot 5.$

Now lets consider multiples of 30 in the form: $30n$. For $30n$ to be divisible by 16, $n$ must be divisible by 8 (since we will have a factor of 2 from 30 and another factor of 8 from $n$ to make a factor of 16).

Since $n$ can only go from 1 to 33 (as there are only 33 multiples of 30 under 1000), we only need to find numbers under 33 that are divisible by 8.

There are four such numbers: 8, 16, 24, 32. These numbers correspond to the multiples: 240, 480, 720, 960.

All the multiples remaining are not divisible by 16. Thus there are 33 - 4 = 29 multiples of 30 under 1000 that are divisible by 30 but not 16.

- 29
- 31
- 32
- 33
- 38

Solution :

**Choice 1 is the answer.**

There are 33 multiples of 30 under 1000 (1000/30 = 33.33..). So we only need to focus our attention on choices 1 to 4.

Note that the factorization of 30 is: $2\cdot 3\cdot 5.$

Now lets consider multiples of 30 in the form: $30n$. For $30n$ to be divisible by 16, $n$ must be divisible by 8 (since we will have a factor of 2 from 30 and another factor of 8 from $n$ to make a factor of 16).

Since $n$ can only go from 1 to 33 (as there are only 33 multiples of 30 under 1000), we only need to find numbers under 33 that are divisible by 8.

There are four such numbers: 8, 16, 24, 32. These numbers correspond to the multiples: 240, 480, 720, 960.

All the multiples remaining are not divisible by 16. Thus there are 33 - 4 = 29 multiples of 30 under 1000 that are divisible by 30 but not 16.

Labels:
Math GRE

### Physics GRE - #33

A radioactive nucleus decays, with the activity shown in the plot below. The half-life of the nucleus is:

The initial activity is $6\cdot 10^3$ (yes, 6 since the line marked $10^3$ is really $1\cdot 10^3$ and you start counting up from there). At $3\cdot 10^3,$ the time is around 5 to 10 minutes. Thus the only possible answer is choice 2.

- 2 min
- 7 min
- 11 min
- 18 min
- 23 min

Solution :

**Choice 2 is the answer.**

The initial activity is $6\cdot 10^3$ (yes, 6 since the line marked $10^3$ is really $1\cdot 10^3$ and you start counting up from there). At $3\cdot 10^3,$ the time is around 5 to 10 minutes. Thus the only possible answer is choice 2.

Labels:
Physics GRE

## Saturday, 3 September 2011

### Math GRE - #32

Let $\mathbb{R}$ be the set of real numbers and let $f$ and $g$ be functions from $\mathbb{R}$ into $\mathbb{R}$. Which of the following is the negation of the statement:

"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."

- For each $s$ in $\mathbb{R}$, there does not exist an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$.
- For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.
- There exists an $s$ in $\mathbb{R}$ such that for each $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.
- There exists an $s$ in $\mathbb{R}$ and there exists an $r$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.
- For each $r$ in $\mathbb{R}$, there exists an $s$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.

Solution :

**Choice 3 is the answer.**

If you've got an innate talent for logic, you could probably tell that choice 3 is the answer after a bit of thinking.

Another way is to use quantifiers to translate the statement into formal logic.

The statement:

"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."

can be translated as: \[\forall s\in\mathbb{R},\;\exists r\in\mathbb{R},\; f(r)>0\rightarrow g(s)>0.\]

We can negate this statement (imagine the negation as an arrow, flipping each quantifier as it flies past it) to be: \[\exists s\in\mathbb{R},\;\forall r\in\mathbb{R},\; f(r)>0\wedge g(s)\leq0.\]

This is choice 3 precisely.

Labels:
Math GRE

### Physics GRE - #32

As shown below, a coaxial cable having radii $a$, $b$, and $c$ carries equal and opposite currents of magnitude $i$ on the inner and outer conductors. What is the magnitude of the magnetic induction at a point $P$ outside the cable at a distance $r$ from the axis?

The currents cancel out the magnetic fields created by each other, resulting in a net field of zero. More formally, this is an application of Ampere's Law.

- $0$
- $\dfrac{\mu_0 ir}{2\pi a^2}$
- $\dfrac{\mu_0 i}{2\pi r}$
- $\dfrac{\mu_0 i}{2\pi r}\dfrac{c^2-r^2}{c^2-b^2}$
- $\dfrac{\mu_0 i}{2\pi r}\dfrac{r^2-b^2}{c^2-b^2}$

Solution :

**Choice 1 is the answer.**

Labels:
Physics GRE

## Thursday, 1 September 2011

### Math GRE - #31

Let $\text{F}$ be a constant force that is given by the vector $\left(\begin{array}{c}-1\\0\\1\end{array}\right)$. What is the work done by $\text{F}$ on a particle that moves along the path given by $\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)$ between time $t=0$ and $t=1$?

Recall that work can be calculated according to the line integral: \[W=\int_C{\text{F}\cdot d\text{r}}\] along a curve $C$ with force $\text{F}$ and position vector $\text{r}$. In this case, we have \[\text{r}=\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)\implies d\text{r}=\left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)dt.\] The curve goes from 0 to 1, so our integral is: \[W=\int_0^1{\left(\begin{array}{c}-1\\0\\1\end{array}\right)\cdot \left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)}dt=\int_0^1{(-1+0+3t^2)}dt=0\] which gives zero work as we wanted.

- $-\dfrac{1}{4}$
- $-\dfrac{1}{4\sqrt{2}}$
- $0$
- $\sqrt{2}$
- $3\sqrt{2}$

Solution :

**Choice 3 is the answer.**

Recall that work can be calculated according to the line integral: \[W=\int_C{\text{F}\cdot d\text{r}}\] along a curve $C$ with force $\text{F}$ and position vector $\text{r}$. In this case, we have \[\text{r}=\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)\implies d\text{r}=\left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)dt.\] The curve goes from 0 to 1, so our integral is: \[W=\int_0^1{\left(\begin{array}{c}-1\\0\\1\end{array}\right)\cdot \left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)}dt=\int_0^1{(-1+0+3t^2)}dt=0\] which gives zero work as we wanted.

Labels:
Math GRE

### Physics GRE - #31

Which of the following is nearly the mass of the Earth? The radius of the Earth is about $6.4\cdot10^6\,\text{m}$.

If you don't remember the mass of the earth off the bat, recall that at ground level, \[mg=\frac{GmM_E}{R_E^2}\implies M=\frac{gR_E^2}{G}\] where $g$, $G$, and $R_E$ are given constants (a table of constants is given during the exam). We can plug in the numbers and compute \[M_E\approx6\cdot10^{24}\,\text{kg}.\]

- $6\cdot10^{24}\,\text{kg}$
- $6\cdot10^{27}\,\text{kg}$
- $6\cdot10^{30}\,\text{kg}$
- $6\cdot10^{33}\,\text{kg}$
- $6\cdot10^{36}\,\text{kg}$

Solution :

**Choice 1 is the answer.**

If you don't remember the mass of the earth off the bat, recall that at ground level, \[mg=\frac{GmM_E}{R_E^2}\implies M=\frac{gR_E^2}{G}\] where $g$, $G$, and $R_E$ are given constants (a table of constants is given during the exam). We can plug in the numbers and compute \[M_E\approx6\cdot10^{24}\,\text{kg}.\]

Labels:
Physics GRE

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