tag:blogger.com,1999:blog-52444586741173631592024-03-06T01:20:04.422-08:00Daily GRETwo GRE questions a day. One Math. One Physics.Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.comBlogger35125tag:blogger.com,1999:blog-5244458674117363159.post-29840178447972522252011-09-05T21:08:00.000-07:002011-09-05T21:11:32.578-07:00Zai geen, sayonara, goodbye (for now at least).<br />
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.<br />
<br />
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).<br />
<br />
- Paul<br />
<br />
<span class="Apple-style-span" style="color: white;">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.</span><br />
<span class="Apple-style-span" style="color: white;">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.</span><br />
<span class="Apple-style-span" style="color: white;"><br /></span>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com5tag:blogger.com,1999:blog-5244458674117363159.post-38266474886492211062011-09-05T20:53:00.001-07:002011-09-05T20:53:55.743-07:00Math 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?
<br />
<br />
<ol>
<li>None</li>
<li>One</li>
<li>Two</li>
<li>Three</li>
<li>Four</li>
</ol>
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Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div>
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<b>Choice 4 is the answer.</b><br />
<br />
$f$ has a horizontal tangent line only when the derivative of $x$ is 0. The derivative of $f$ is:<br />
\[f^\prime(x) = e^{-x^2-x^{-2}}+(-2x+2x^{-3})xe^{-x^2-x^{-2}}=(1-2x^2+2x^{-2})e^{-x^2-x^{-2}}=0.\] This is only zero in two cases: in the limit $\displaystyle\lim_{x\rightarrow 0} f^\prime(x)$ and when $1-2x^2+2x^{-2}=0$.<br />
<br />
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.<br />
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Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2tag:blogger.com,1999:blog-5244458674117363159.post-27449339529413540062011-09-04T21:43:00.001-07:002011-09-04T21:43:58.700-07:00Math GRE - #33How many integers from 1 to 1000 are divisible by 30 by not by 16?<br />
<br />
<ol>
<li>29</li>
<li>31</li>
<li>32</li>
<li>33</li>
<li>38</li>
</ol>
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Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div>
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<b>Choice 1 is the answer.</b><br />
<br />
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.<br />
<br />
Note that the factorization of 30 is: $2\cdot 3\cdot 5.$<br />
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).<br />
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.<br />
<br />
There are four such numbers: 8, 16, 24, 32. These numbers correspond to the multiples: 240, 480, 720, 960.<br />
<br />
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.</div>
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Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-8564323702119914402011-09-03T19:57:00.001-07:002011-09-03T19:57:48.640-07:00Math GRE - #32Let $\mathbb{R}$ 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:<br />
<br />
<div style="text-align: center;">
"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."</div>
<div style="text-align: center;">
<br /></div>
<div style="text-align: center;">
</div>
<ol>
<li style="text-align: left;">For each $s$ in $\mathbb{R}$, there does not exist an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$.</li>
<li style="text-align: left;">For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.</li>
<li style="text-align: left;">There exists an $s$ in $\mathbb{R}$ such that for each $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.</li>
<li style="text-align: left;">There exists an $s$ in $\mathbb{R}$ and there exists an $r$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.</li>
<li style="text-align: left;">For each $r$ in $\mathbb{R}$, there exists an $s$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.</li>
</ol>
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Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div>
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<b>Choice 3 is the answer.</b></div>
<div style="text-align: left;">
<b><br /></b></div>
<div style="text-align: left;">
If you've got an innate talent for logic, you could probably tell that choice 3 is the answer after a bit of thinking. </div>
<div style="text-align: left;">
<br /></div>
<div style="text-align: left;">
Another way is to use quantifiers to translate the statement into formal logic.</div>
<div style="text-align: left;">
The statement:</div>
<div style="text-align: left;">
<br /></div>
<div style="text-align: left;">
<div style="text-align: center;">
"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."</div>
<div>
<br /></div>
<div>
can be translated as: \[\forall s\in\mathbb{R},\;\exists r\in\mathbb{R},\; f(r)>0\rightarrow g(s)>0.\]</div>
<div>
<br /></div>
<div>
We can negate this statement (imagine the negation as an arrow, flipping each quantifier as it flies past it) to be: \[\exists s\in\mathbb{R},\;\forall r\in\mathbb{R},\; f(r)>0\wedge g(s)\leq0.\]</div>
<div>
<br /></div>
<div>
This is choice 3 precisely.</div>
</div>
</div>
</div>
</div>
Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-70684938886748513802011-09-01T22:34:00.000-07:002011-09-01T22:34:19.193-07:00Math GRE - #31Let $\text{F}$ 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 $t=1$?<br />
<br />
<ol>
<li>$-\dfrac{1}{4}$</li>
<br />
<li>$-\dfrac{1}{4\sqrt{2}}$</li>
<br />
<li>$0$</li>
<br />
<li>$\sqrt{2}$</li>
<br />
<li>$3\sqrt{2}$</li>
</ol>
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Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div>
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<b>Choice 3 is the answer.</b><br />
<br />
Recall that <a href="http://en.wikipedia.org/wiki/Work_(physics)#Force_and_displacement">work</a> 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.</div>
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Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com1tag:blogger.com,1999:blog-5244458674117363159.post-45671905592405892432011-08-31T23:03:00.000-07:002011-08-31T23:05:39.588-07:00Math GRE - #30What is \[\int_{-3}^3{|x+1|\,dx}?\]<br />
<ol><li>0</li>
<li>5</li>
<li>10</li>
<li>15</li>
<li>20</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 3 is the answer.</b><br />
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This is a very basic question. The usual method of dealing with absolute values in integrals is to split it into a sum of two integrals and remove the absolute values. Since $|x+1|$ is negative on $[-3, 1)$ and positive on $[-1, 3]\,\,\,$ , we can split the integral into: \[\int_{-3}^3{|x+1|\,dx}=-\int_{-3}^{-1}{(x+1)\,dx}+\int_{-1}^3{(x+1)\,dx}=10.\] Another way of doing this problem is to imagine the graph of $|x+1|$ and realize that the integral is the sum of two 45-45-90 triangles (one with base 2 and height 2, the other with base 4 and height 4). Simply add the area of triangles up and we're done.<br />
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<b>Bonus question:</b> Given that $c$ is a constant, what is \[\int_{-\infty}^{\infty}{e^{-|x+c|}\,dx}?\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-61436527727876844902011-08-30T21:52:00.001-07:002011-08-30T21:54:58.323-07:00Math GRE - #29If $z=e^{2\pi i/5},$ then \[1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=\] <br />
<ol><li>$0$</li>
<li>$4e^{3\pi i/5}$</li>
<li>$5e^{4\pi i/5}$</li>
<li>$-4e^{2\pi i/5}$</li>
<li>$-5e^{3\pi i/5}$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 5 is the answer.</b><br />
<br />
Recall that the <a href="http://en.wikipedia.org/wiki/Root_of_unity#Summation">sum of all the $n-th$ roots of unity</a> for $n>1$ ($n=5$ in this case) is 0. For this particular question, this means that \[1+z+z^2+z^3+z^4 = 0.\] This motivates us to write the sum in the question as: \[\begin{eqnarray*}<br />
(1+z+z^2+z^3+z^4)+4z^4(1+z+z^2+z^3+z^4)+5z^9 & = & 0+4z^4\cdot0+5z^9 \\<br />
& = & 5z^9.<br />
\end{eqnarray*}\] Now we can easily evaluate the sum: \[\begin{eqnarray*}<br />
5z^9=5\left(e^{2\pi i/5}\right)^9 & = & 5\left(e^{2\pi i/5}\right)^4\cdot\left(e^{2\pi i/5}\right)^5 \\<br />
& = & 5\left(e^{2\pi i/5}\right)^4 \\<br />
& = &5e^{8\pi i/5}=-5e^{3\pi i/5}.<br />
\end{eqnarray*}\]<br />
</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-26395337141796955592011-08-29T22:07:00.001-07:002011-08-30T18:46:27.080-07:00Math GRE - #28What is the volume of the solid formed by revolving about the x-axis the region in the first quadrant of the xy-plane bounded by the coordinate axes and the graph of the equation: \[y=\frac{1}{\sqrt{1+x^2}}?\]<br />
<ol><li>$\dfrac{\pi}{2}$</li>
<li>$\pi$</li>
<li>$\dfrac{\pi^2}{4}$</li>
<li>$\dfrac{\pi^2}{2}$</li>
<li>$\infty$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 4 is the correct answer.</b><br />
<br />
Recall that the <a href="http://en.wikipedia.org/wiki/Solid_of_revolution#Disc_method">volume of revolution</a> around the x-axis is given by the formula \[V=\int{\pi y(x)^2}\,dx.\] In this case, we can substitute for $y$ and plug in the limits of integration to obtain: \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx.\]<br />
Almost everyone with some calculus experience has had the great displeasure of memorizing: \[\int{\frac{1}{1+x^2}}\,dx = \arctan x + C.\] Now we can put it to good use! We can see that the answer is (if you'll pardon my poor notation): \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx=\pi\left[\arctan\infty-\arctan0\right]=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}.\]<br />
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<b>Interesting tidbit:</b> <a href="http://en.wikipedia.org/wiki/Gabriel's_Horn">Here's</a> a funny little thing you'll often see in first year calculus. It's called Gabriel's Horn. It's got finite volume but INFINITE surface area. Just something interesting I wanted to share.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com4tag:blogger.com,1999:blog-5244458674117363159.post-34965372546103316642011-08-28T23:29:00.000-07:002011-08-28T23:32:56.961-07:00Math GRE - #27Which of the following is the best approximation of $\sqrt{1.5}(266)^{3/2}$?<br />
<br />
<ol><li>1000</li>
<li>2700</li>
<li>3200</li>
<li>4100</li>
<li>5300</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 5 is the answer.</b><br />
<br />
With some simplification, we note that: \[\begin{eqnarray*}<br />
\sqrt{1.5}(266)^{3/2}=\sqrt{\frac{3}{2}}\cdot(266)^{3/2} & = & 266\sqrt{\frac{3\cdot 266}{2}} \\ <br />
& = & 266\sqrt{399}\approx266\cdot20\approx 5300.<br />
\end{eqnarray*}\]<br />
Those of us familiar with Taylor series may try approximations with calculus, however this question teaches us that too much knowledge may be a little bit dangerous if misapplied. And that most of the time, there are simpler solutions than you'd think.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2tag:blogger.com,1999:blog-5244458674117363159.post-80108524208556982062011-08-27T22:24:00.000-07:002011-08-27T22:24:24.135-07:00Math GRE - #26A total of $x$ feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of $x$?<br />
<br />
<ol><li>$\dfrac{x^2}{9}$</li>
<li>$\dfrac{x^2}{8}$</li>
<li>$\dfrac{x^2}{4}$</li>
<li>$x^2$</li>
<li>$2x^2$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 2 is the answer.</b><br />
<br />
Let's call one side of the fence $a$ and the other side $b$. Since we are only looking at 3 sides of a rectangular fence (e.g. we're fencing in our back yard and one side is the wall of the house), let's call the side length that we have to fence twice. So we have: \[x = 2a+b.\] The area of the fence is \[A = ab = a(x-2a)\] where the last equality comes from our equation for $x$. To maximize the area, we take the derivative of $A$ with respect to $a$ and set equal to 0 to obtain: \[0 = x-4a\implies a=\frac{x}{4}.\] We know this represents the maximum since the second derivative of $A$ is always negative. Using our equation for the perimeter, we now know that: \[x=2a + b\implies b=\frac{x}{2}.\] Hence the maximum possible area is: \[A=ab=\frac{x}{4}\cdot\frac{x}{2}=\frac{x^2}{8}.\]<br />
</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-80306305433227198022011-08-26T18:20:00.000-07:002011-08-26T18:23:12.421-07:00Math GRE - #25Let $k$ be the number of roots to the equation $f(x)=e^x+x-2$ in the interval $[0, 1]$, and let $n$ be the number of roots that are not in $[0,1]$. Which of the following is true?<br />
<br />
<ul><li>$k=0$ and $n=1$</li>
<li>$k=1$ and $n=0$</li>
<li>$k=n=1$</li>
<li>$k>1$</li>
<li>$n>1$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 2 is the answer.</b><br />
<br />
We can solve this problem by using the <a href="http://en.wikipedia.org/wiki/Intermediate_value_theorem">Intermediate Value Theorem</a>.<br />
By evaluating $f$ at 0 and 1, we see that: \[f(0)=e^0+0-2<0\] \[f(1)=e^1+1-2>0.\]<br />
By the Intermediate Value Theorem, there exists a $c\in[0,1]$ such that $f(c)=0.$ Thus we know for sure that there is at least one root in [0,1].<br />
<br />
Taking the derivative of $f$, we note that: \[f^\prime(x)=e^x+1>0\] for all $x$. <br />
Thus $f$ is increasing everywhere and so it must only have one root. Since this root is in [0,1], $k=1.$ As we cannot have roots anywhere outside of [0,1], we also have $n=0$. This gives choice 2 as the answer.<br />
<br />
<b>Bonus question:</b> The Intermediate Value Theorem may seem obvious, but it can also be used to prove many fun facts. One such fact is that if you draw a circle anywhere in the universe, there will be two points directly opposite to each other on the circle (i.e. 180 deg. apart) that have the exact same temperature. Can you prove this?<br />
<br />
If you get stuck, here is some <a href="http://hippomath.blogspot.com/2011/06/equal-temperatures-at-two-opposite.html">intuition</a> as to why this interesting fact is true.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2tag:blogger.com,1999:blog-5244458674117363159.post-53107970945206138272011-08-25T22:53:00.000-07:002011-08-26T04:02:25.186-07:00Math GRE - #24What is the greatest possible area of a triangular region with one vertex at the centre of a unit circle and the other two vertices on the circle?<br />
<br />
<ol><li>$\frac{1}{2}$</li>
<li>$1$</li>
<li>$\sqrt{2}$</li>
<li>$\pi$</li>
<li>$\frac{1+\sqrt{2}}{4}$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">Choice 1 is the answer.<br />
<br />
The <a href="http://en.wikipedia.org/wiki/Triangle#Using_trigonometry">area of any triangle</a> with sides $a$, $b$, and angle $\theta$ between $a$ and $b$ is: \[A = \frac{1}{2}ab\sin\theta\]<br />
Since the sides have the same length as the radius of the circle, $a=b=1.$ <br />
Therefore, \[A=\frac{1}{2}\sin\theta\] which equals $\frac{1}{2}$ at the maximum of $\theta=\frac{\pi}{2}$.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2tag:blogger.com,1999:blog-5244458674117363159.post-23383047521840337242011-08-24T03:45:00.000-07:002011-08-24T03:49:19.294-07:00Math GRE - #23Suppose B is a basis for a real vector space V of dimension greater than 1. Which of the following statements could be true?<br />
<br />
<ol><li>The zero vector of V is an element of B.</li>
<li>B has a proper subset that spans V.</li>
<li>B is a proper subset of a linearly independent subset of V.</li>
<li>There is a basis for V that is disjoint from B.</li>
<li>One of the vectors in B is a linear combination of the other vectors in B.</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">Let us first recall the definition of a basis:<br />
<div style="text-align: center;">"A basis for a vector space is a <b>linearly independent</b> set of vectors that <b>spans the vector space</b>".</div><div style="text-align: left;"><br />
</div><div style="text-align: left;">B is linearly independent because it is a basis. This removes choice 5.<br />
<br />
Because a basis is linearly independent, the zero vector cannot be part of it. <br />
This removes choice 1.<br />
<br />
For B to have a proper subset that spans V, there must exist vectors in B which are not linearly independent (since we can remove vectors and still span V). However, this is impossible since B is a basis (i.e. we cannot remove any vectors and still hope to span V). This removes choice 2.<br />
<br />
Similarly, we cannot add any more linearly independent vectors to B because it already spans all of V. Thus B cannot be a proper subset of a linearly independent subset of V. This removes choice 3.<br />
<br />
The only choice left is choice 4. Thus choice 4 is the answer (there are in fact an infinite number of bases that are disjoint from B).<br />
<br />
</div></div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-79542067604394469402011-08-23T22:07:00.001-07:002011-08-23T22:07:17.231-07:00Math GRE - #22Consider the function $f$ defined by $f(x)=e^{-x}$ on the interval [0, 10]. Let $n>1$ and let $x_0,x_1,\ldots,x_n$ be numbers such that: \[0= x_0 < x_1 < \cdots < x_n =10.\] Which of the following is greatest? <br />
<br />
<ol><li>$\displaystyle\sum_{j=1}^n{f(x_j)(x_j-x_{j-1})}$</li>
<li>$\displaystyle\sum_{j=1}^n{f(x_{j-1})(x_j-x_{j-1})}$</li>
<li>$\displaystyle\sum_{j=1}^n{f\left(\frac{x_j+x_{j-1}}{2}\right)(x_j-x_{j-1})}$</li>
<li>$\displaystyle\int_0^{10}{f(x)dx}$</li>
<li>$0$</li>
</ol><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 2 is the answer.</b><br />
<br />
The first three choices are Riemann sums with choice 1 the right sum, choice 2 the left sum and choice 3 the midpoint sum. Choice 4 is the actual area of $f$ from 0 to 10.<br />
<br />
Since $e^{-x}$ is a decreasing function, the left Riemann sum is the greatest out of all the choices (see diagram below).<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNy7p0FzPirxujtPbu4pi5Vqr-PDahvflZg8vSvW4dBdgrsvtRSX9NvKpTJrkYXKHIyfBBKCfBra5Ur8Hh8GyEe4Q53rM5oCMQKprRjfgccbaJuLu2Be5e3-98H3L7Fp-FgEfu-uCEpfJN/s1600/LeftRiemann.svg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="256" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNy7p0FzPirxujtPbu4pi5Vqr-PDahvflZg8vSvW4dBdgrsvtRSX9NvKpTJrkYXKHIyfBBKCfBra5Ur8Hh8GyEe4Q53rM5oCMQKprRjfgccbaJuLu2Be5e3-98H3L7Fp-FgEfu-uCEpfJN/s320/LeftRiemann.svg" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br />
</div>Thus, choice 2 (the left Riemann sum) is the answer.</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-86943002998692022622011-08-22T01:40:00.000-07:002011-08-22T01:40:02.279-07:00Math GRE - #21Let $x$ and $y$ be uniformly distributed, independent random variables on [0, 1]. The probability that the distance between $x$ and $y$ is less than $\frac{1}{2}$ is:<br />
<br />
<ul><li>$\dfrac{1}{4}$ </li>
<br \>
<li>$\dfrac{1}{3}$</li>
<br \>
<li>$\dfrac{1}{2}$</li>
<br \>
<li>$\dfrac{2}{3}$</li>
<br \>
<li>$\dfrac{3}{4}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">The answer is $\frac{3}{4}$.<br />
<br />
If $x$ and $y$ are uniformly distributed and independent random variables on [0, 1], this means that they are in the unit square.<br />
<br />
We require $|x-y| < \frac{1}{2}$, i.e. we need \[-\frac{1}{2} < x - y < \frac{1}{2}\].<br />
On the unit square, we can draw two regions which represent the inequality above: \[y < x+\frac{1}{2}\] \[y > x - \frac{1}{2}.\]<br />
If we can find the area of this region (shown below), then we are done the question.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2Ud4N_8amfmfMP3iaoTIqRmbDvT-WdJ2QWGDS7Pg3TkijuRT5FlI2BxCP8Y5GHF6qNB7CK_hg_Df4ap8UbtIf976mNSP00icL1REsCZAvQB5INMgWZ0K96VMlaOcqBjKZ3ikaU-c5doId/s1600/RandVarPlot.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="319" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2Ud4N_8amfmfMP3iaoTIqRmbDvT-WdJ2QWGDS7Pg3TkijuRT5FlI2BxCP8Y5GHF6qNB7CK_hg_Df4ap8UbtIf976mNSP00icL1REsCZAvQB5INMgWZ0K96VMlaOcqBjKZ3ikaU-c5doId/s320/RandVarPlot.png" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">The region we are looking at.</td></tr>
</tbody></table><br />
I'll leave it as an exercise to show that the filled out region above is $\frac{3}{4}$ of the unit square as we wanted.</div></div></div><br />
Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-27779247278940775592011-08-21T00:33:00.000-07:002011-08-21T00:34:27.609-07:00Math GRE - #20A fair coin is tossed 8 times. What is the probability that more of the tosses will result in heads than tails?<br />
<br />
<ul><li>$\dfrac{1}{4}$</li>
<li>$\dfrac{1}{3}$</li>
<li>$\dfrac{87}{256}$</li>
<li>$\dfrac{23}{64}$</li>
<li>$\dfrac{93}{256}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">The answer is $\dfrac{93}{256}$.<br />
<br />
The chance of exactly 4 heads and 4 tails is: \[\left(\frac{1}{2}\right)^8\cdot\binom{8}{4}=\frac{70}{256}.\]<br />
<br />
Thus, the chance of having more tails than heads or having more heads than tails is: \[1-\frac{70}{256}=\frac{186}{256}.\]<br />
<br />
Because of symmetry, the probability of having more heads than tails is: \[\frac{1}{2}\frac{186}{256}=\frac{93}{256}.\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-67567235325913166772011-08-20T03:28:00.001-07:002011-08-20T03:30:38.530-07:00Math GRE - #19Let $A$ be a $2\times 2$ matrix for which there is a constant $k$ such that the sum of the entries in <b>each row </b>and <b>each column</b> is $k$. Which of the following must be an eigenvector of $A$?<br />
<br />
<ol><li>$\left[\begin{array}{c}<br />
1\\ 0\end{array}\right]$</li>
<li>$\left[\begin{array}{c}<br />
0\\ 1\end{array}\right]$</li>
<li>$\left[\begin{array}{c}<br />
1\\ 1\end{array}\right]$</li>
</ol>Choices:<br />
<ul><li>1 only</li>
<li>2 only</li>
<li>3 only</li>
<li>1 and 2</li>
<li>1, 2, and 3</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>Choice 3 is the only solution.</b><br />
<br />
Since the rows add up to $k$, $\left[\begin{array}{c}1\\ 1\end{array}\right]$ is an eigenvector with eigenvalue $k$. To see that choice 1 and 2 do not work, take the matrix: \[\left[\begin{array}{cc} 1 & 1\\ 1 & 1\end{array}\right]\] which satisfies the given conditions but do not have either choice 1 or choice 2 as eigenvectors.<br />
<br />
<b>Bonus problem:</b><br />
Given a matrix with columns adding up to $k$, prove that $k$ is an eigenvalue (in fact, if the entries in the matrix are all greater than zero, then $k$ is the largest eigenvalue).<br />
<br />
</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-83650870898057159982011-08-18T19:58:00.001-07:002011-08-20T02:16:07.012-07:00Math GRE - #18If we have \[F(x)=\int_e^x{\log{t}dt}\] for all positive $x$, then $F'(x)=$<br />
<br />
<ul><li>$x$</li>
<li>$\frac{1}{x}$</li>
<li>$\log{x}$</li>
<li>$x\log{x}$</li>
<li>$x\log{x}-1$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">I decided to use this question because of it's similarity to the question I posted yesterday. Since the derivative of an integral is itself (<a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part">FTC - Part I</a>), we have \[F'(x)=\frac{d}{dx}\int_e^x{\log{t}dt}=\log{x}.\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-91657210442778984632011-08-17T22:49:00.001-07:002011-08-20T02:16:07.076-07:00Math GRE - #17Let $h$ be the function defined by \[h(x)=\int_0^{x^2}{e^{x+t}dt}.\] Then $h'(1)=$<br />
<br />
<ul><li>$e-1$</li>
<li>$e^2$</li>
<li>$e^2-e$</li>
<li>$2e^2$</li>
<li>$3e^2-e$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">The answer is $3e^2-e$.<br />
<br />
This can be solved with applications of the product rule, the chain rule, and the fundamental theorem of calculus.<br />
<br />
First let us recall that the derivative of an integral is itself, that is: \[\frac{d}{dx}\int_a^x{f(x)}=f(x).\]<br />
When the upper limit is a function, we must apply the chain rule: \[\frac{d}{dx}\int_a^{x^2}{f(x)}=2x\cdot f(x^2).\]<br />
For this particular question, we have: \[h(x)=\int_0^{x^2}{e^{x+t}dt}=e^x\cdot\int_0^{x^2}{e^{t}dt}=e^x\cdot g(x)\] for $g(x)=\int_0^{x^2}{e^{t}dt}$. The product rule tells us that \[h'(x)=e^x\cdot g(x)+e^x\cdot g'(x)=e^x\cdot\int_0^{x^2}{e^{t}dt}+e^x\cdot\left(2x\cdot e^{x^2}\right).\]<br />
Upon substituting $1$ for $x$, we obtain the answer: \[h'(1)=e\cdot(e-1)+e\cdot(2\cdot e)=3e^2-e.\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-6264689517121469152011-08-17T00:28:00.001-07:002011-08-17T19:53:45.524-07:00Math GRE - #16For how many positive integers $k$ does the ordinary decimal representation of the integer $k!$ end in exactly 99 zeros?<br />
<br />
<ul><li>None</li>
<li>One</li>
<li>Four</li>
<li>Five</li>
<li>Twenty-Four</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><b>There are five positive integer $k$ for which this is true.</b><br />
<br />
Note that the number of zeros depend directly on how many factors of 10 are present in $k!$. However, the number of 10s depend on the number of factors of 2s and 5s. Since there are an abundance of 2 in the factorization of $k!$, we just have to count the number of times a factor of 5 appears. Since one (or more) factor of 5 appear every five numbers, every five numbers have the same number of zeros at the end.<br />
<br />
There is a simple flaw with the argument above, do you see it?<br />
We haven't proven the existence of a single integer $k$ such that $k!$ ends in 99 zeros! It turns out that 400! has 99 zeros (and thus 401! to 404! do as well). I believe that we were supposed to assume the existence of such an integer $k$ and that the question is also a bit poorly worded. However, I could be wrong.<br />
<br />
Can anybody see why there must be an integer $k$ such that $k!$ ends in 99 zeros?</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com3tag:blogger.com,1999:blog-5244458674117363159.post-44266339456818755522011-08-16T00:15:00.000-07:002011-08-16T00:15:27.639-07:00Math GRE - #15Suppose $A$ and $B$ are invertible $n\times n$ matrices. If $A$ and $B$ are similar, which of the following statements are true?<br />
<br />
<ol><li>$A-2I$ and $B-2I$ are similar.</li>
<li>$A$ and $B$ have the same trace.</li>
<li>$A^{-1}$ and $B^{-1}$ are similar matrices.</li>
</ol>Choices:<br />
<ul><li>1 only.</li>
<li>2 only.</li>
<li>3 only.</li>
<li>1 and 2 only.</li>
<li>1, 2, and 3 only.</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;"><div><b>1, 2, and 3 are all true.</b></div><div><b><br />
</b></div><div>Recall that $A$ and $B$ are similar if $B=P^{-1}AP$ for some invertible matrix $P$.</div><div><span class="Apple-style-span" style="font-size: xx-small;"> </span></div><div>Since $\text{tr}(AB)=\text{tr}(BA)$, we know that 2 is true due to: \[\text{tr}(B)=\text{tr}(P^{-1}AP)=\text{tr}(APP^{-1})=\text{tr}(A).\]</div><br />
Moreover, we have \[B^{-1}=(P^{-1}AP)^{-1}=P^{-1}A^{-1}P.\] But by the definition of being similar, this means that $A^{-1}$ and $B^{-1}$ are similar. So 3 is true. Since only the last choice offers 2 and 3 as true, we know that the answer is the last choice: <b>1, 2, and 3 are all true</b>.<br />
<br />
Can you prove why 1 must be true?</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2tag:blogger.com,1999:blog-5244458674117363159.post-49865469487977608632011-08-15T03:09:00.001-07:002011-08-15T03:11:35.467-07:00Math GRE - #14If $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$ (i.e. the floor of x), then which of the choice below is the value of \[\int_0^\infty{\lfloor x\rfloor e^{-x}dx}?\]<br />
<ul><li>$\dfrac{e}{e^2-1}$</li>
<br \>
<li>$\dfrac{1}{e-1}$</li>
<br \>
<li>$\dfrac{e-1}{e}$</li>
<br \>
<li>$1$</li>
<br \>
<li>$\infty$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">Since the floor function is discrete, we can turn this integral into a discrete sum of integrals by noting that:<br />
\[\begin{eqnarray*}<br />
\int_0^\infty{\lfloor x\rfloor e^{-x}dx} & = & \int_0^1{0\cdot e^{-x}dx}+\int_1^2{1\cdot e^{-x}dx}+\int_2^3{2\cdot e^{-x}dx}+\cdots \\<br />
& = & \sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}}<br />
\end{eqnarray*}\]<br />
The sum above turns out to be telescoping, as we can see that:<br />
\[\begin{eqnarray*}<br />
\sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}} & = & \sum_{n=0}^\infty{n\left(e^{-n}-e^{-(n+1)}\right)} \\<br />
& = & \sum_{n=0}^\infty{n\cdot e^{-n}}-\sum_{n=0}^\infty{n\cdot e^{-(n+1)}} \\<br />
& = & \sum_{n=0}^\infty{n\cdot e^{-n}}-\sum_{n=0}^\infty{(n+1)\cdot e^{-(n+1)}}+\sum_{n=0}^\infty {e^{-(n+1)}} \\<br />
& = & \sum_{n=0}^\infty {e^{-(n+1)}}<br />
\end{eqnarray*}\]<br />
Where the last line is due to adjusting the summation indices on the previous line. We can also see that \[\sum_{n=0}^\infty {e^{-(n+1)}}=\frac{1}{e-1}\] as the above is a geometric series. <br />
<br />
Therefore, the answer is: \[\int_0^\infty{\lfloor x\rfloor e^{-x}dx}=\sum_{n=0}^\infty{\int_n^{n+1}{n\cdot e^{-x}dx}}=\frac{1}{e-1}\]</div></div></div><br />
Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2tag:blogger.com,1999:blog-5244458674117363159.post-69897671521965030782011-08-14T01:08:00.000-07:002011-08-14T01:08:00.063-07:00Math GRE - #13Determine the set of real numbers x for which the series below is convergent: \[\sum_{n=1}^{\infty}\frac{n!x^{2n}}{n^{n}\left(1+x^{2n}\right)}\]<br />
<ul><li>$\{0\}$</li>
<li>$\{x: -1 \leq x \leq 1\}$</li>
<li>$\{x: -1 < x < 1\}$</li>
<li>$\{x: -\sqrt{e} \leq x \leq \sqrt{e}\}$</li>
<li>$\mathbb{R}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">The answer is $\mathbb{R}$.<br />
<br />
Note that for all numbers $x\in\mathbb{R}$, \[x^{2n}\geq 0\implies \frac{x^{2n}}{1+x^{2n}}\leq \frac{x^{2n}}{x^{2n}}=1.\]<br />
This means that \[\sum_{n = 1}^\infty \frac{n! x^{2n}}{n^n(1 + x^{2n})} < \sum_{n = 1}^\infty \frac{n!}{n^n}.\] Thus, if we can prove that \[\sum_{n = 1}^\infty \frac{n!}{n^n}\] is convergent, then by the comparison test our original series is convergent for all $x$.<br />
<br />
By the ratio test, $\sum_{n = 1}^\infty \frac{n!}{n^n}$ converges as:<br />
\[\begin{eqnarray*}<br />
\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}} & = & \lim_{n\rightarrow\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!} \\<br />
& = & \lim_{n\rightarrow\infty}\frac{n+1}{n}\cdot\left(\frac{n}{n+1}\right)^{n+1} \\<br />
& = & \lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{n} \\<br />
& = & \lim_{n\rightarrow\infty}\left(1-\frac{1}{n+1}\right)^{n} \\<br />
& = & \frac{1}{e}<br />
\end{eqnarray*}\]<br />
Where the last line is essentially a result of the <a href="http://en.wikipedia.org/wiki/E_(mathematical_constant)#Representations">limit definition of e</a>.<br />
<br />
Note that if we hadn't guessed that the answer was $\mathbb{R}$, it would have been safer to do the ratio test on the original sequence directly and obtain the same answer (the method above just saves a bit of work).</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-85535185824028053252011-08-13T02:39:00.000-07:002011-08-13T02:39:05.170-07:00Math GRE - #12A simple one today:<br />
What is the value of the integral: \[\int_0^1{\frac{x}{1+x^2}}dx?\]<br />
<ul><li>$1$</li>
<li>$\frac{\pi}{4}$</li>
<li>$\arctan\frac{\sqrt{2}}{2}$</li>
<li>$\log 2$</li>
<li>$\log\sqrt{2}$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">We can use the substitution $u=1+x^2\implies du = 2x\,dx$. With this substitution, the original integral becomes: \[\int_0^1{\frac{x}{1+x^2}}dx=\frac{1}{2}\int_1^2{\frac{1}{u}}du\]<br />
Upon computing this integral, we obtain: \[\frac{1}{2}\int_1^2{\frac{1}{u}}du=\frac{1}{2}\left. \log{u}\right|_1^2=\frac{1}{2}\log{2}=\log{\sqrt{2}}.\]</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com0tag:blogger.com,1999:blog-5244458674117363159.post-38824261425081765282011-08-12T01:29:00.000-07:002011-08-12T01:29:44.902-07:00Math GRE - #11For all real numbers $x$ and $y$, which of the answers below is the expression \[\frac{x+y+|x-y|}{2}\] equal to?<br />
<br />
<ul><li>The maximum of $x$ and $y$</li>
<li>The minimum of $x$ and $y$</li>
<li>$|x+y|$</li>
<li>The average of $|x|$ and $|y|$</li>
<li>The average of $|x+y|$ and $x-y$</li>
</ul><br />
<div style="margin: 5px 20px 20px;"><div class="smallfont" style="margin-bottom: 2px;">Solution : <input onclick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }" style="font-size: 10px; margin: 0px; padding: 0px; width: 60px;" type="button" value="Show" /> </div><br />
<div class="alt2" style="-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;"><div style="display: none;">The expression is equal to the maximum of $x$ and $y$.<br />
Suppose $x>y$. Then $|x-y|=x-y$ and so \[\frac{x+y+|x-y|}{2}=\frac{x+y+x-y}{2}=x.\]<br />
Now suppose $x<y$. Then $|x-y|=-(x-y)$ and so \[\frac{x+y+|x-y|}{2}=\frac{x+y-(x-y)}{2}=y.\]<br />
\[\therefore\quad\frac{x+y+|x-y|}{2}=\max\{x,y\}.\]<br />
<br />
Can you find a representation for $\min\{x,y\}$?</div></div></div>Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.com2