Online Course Discussion Forum

Sprint Camp AIME

 
 
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Sprint Camp AIME
by Ethan Sun - Monday, December 18, 2023, 8:30 PM
 

Arghhh why do I always have questions that I can't answer...

Literally every single homework I have a couple I can't really solve...

I don't really know how to question 7 and 10...

Help would be great, Thanks!


Edit: For Q10 all the results from the square roots are rational

For Q7 i think substituting y for b^x or something

 
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Re: Sprint Camp AIME
by John Lensmire - Tuesday, December 19, 2023, 12:59 PM
 

The problems are meant to be difficult, so there's nothing wrong with asking for help! In fact, it's great that you're looking at the problems early enough to have time to ask and then try to solve with a hint. That's a great way to learn!

For 7: Letting $z = b^x$ is a good start, so we get $z^2 + 3z - 1$. This is a quadratic so it's graph (max/min, where increasing, etc.) should be fairly easy to understand. One annoyance: We need to consider cases where $b > 1$ and $0 < b < 1$. Why? For example, if $b > 1$ then we know that $z=b^x$ is always increasing on the interval $[-1, 1]$, so it's max value is $z=b$ and it's minimum value is $z=b^{-1}$. What values do these give on the quadratic?

For 10: Let me try the following hint to start. Note: $$\begin{aligned} 1 + \frac{1}{1^2} + \frac{1}{2^2} &= \frac{9}{4} = \left( \frac{3}{2}\right)^2 \\ 1 + \frac{1}{2^2} + \frac{1}{3^2} &= \frac{49}{36} = \left( \frac{7}{6}\right)^2 \\ 1 + \frac{1}{3^2} + \frac{1}{4^2} &= \frac{169}{144} = \left( \frac{13}{12}\right)^2 \\ \end{aligned}$$So in general try to write $$1 + \frac{1}{k^2} + \frac{1}{(k+1)^2}$$ as a perfect square. Note this problem is very tricky!