We’ll keep things ticking along while working on a couple meatier posts. So, time for another TNDOT. This one is a double-barrelled question from the 1913 Victorian Pass Algebra Matric exam. Have fun.
7. If 
be the roots of 
, express, in terms of 
the value of
and of 
Prove that
can have no real value between 2 and 7, 
being a real quantity.
UPDATE (20/09/24)
OK, this one got short circuited a little, but no matter. We’re about to post another one, in a similar vein. The solutions below will follow J. D.’s, and thanks also to Sentinel (and very much to Red Five for digging up the exams).
These questions can look quite forbidding, but they would have been pretty standard in the day, with standard tools available to help. In particular, students taking the 1913 exam would have been very familiar with Vièta’s formulas, relating the roots of a quadratic (or higher) polynomial to the coefficients:
(These formulas are easily proved by noting that 
and then equating coefficients.) Given the problem above, any student would be expecting Vièta’s formulas to come into the solution at some point.
Now to tackle the first fraction sum, we begin with 
being a solution of the quadratic:
We’re interested in the quantity 
, so we rewrite this equation as,
and then
Using this equation together with the corresponding equation for 
, we find
Now Vièta kicks in. By a very standard manipulation (at least for the time),
(The general point is, any symmetric polynomial of and can alway be written as some combo of the “elementary” symmetric polynomials, 
and 
.)
Plugging back in we obtain,
We’re done.
The second fraction sum can be handled similarly, by first putting everything over a common denominator. Then, noting the symmetry in and , and using the expressions for 
and , everything can be written in terms of 
and 
, and then simplified quite easily. However, J. D. has noted a quicker path.
The trick is that the term 
almost appears in this second fraction. So, we just rewrite the sum to make this explicit. There are various ways to do this and J. D. approached it one way. An alternative approach is,
![\color{Plum}\boldsymbol{\mbox{\bf THING2}=\dfrac{1}{c-a\alpha^2}+ \dfrac{1}{c-a\beta^2}=\dfrac{1}{c}\left[\dfrac{1}{1-\frac{a}{c}\alpha^2}+ \dfrac{1}{1-\frac{a}{c}\beta^2}\right]=\dfrac{1}{c}\left[\dfrac{1}{1-\frac{\alpha}{\beta}}+ \dfrac{1}{1-\frac{\beta}{\alpha}}\right]\,,}](https://mathematicalcrap.com/wp-content/ql-cache/quicklatex.com-de3a35e40fa006e2ddf8712f1ab6463d_l3.svg "Rendered by QuickLaTeX.com")
where we have used the Vièta formula for in the last step. Then, getting rid of the fractions within fractions, it turns out that the thing in square brackets is exactly 1. Therefore,
Finally, J.D. has noted that one can, and arguably should, watch for special cases, where a denominator is zero. This is a useful exercise, but was probably not considered necessary to answer the exam question and it may muddy the waters to get into it here. So we’ll leave it.
For the final problem, Sentinel gives the natural and modern calculus approach. There is also the postmodern approach: graduates of the Leigh-Lancaster-Stacey School of Mathematical Wankery can plug the function into their CAS, zoom up the graphing screen or whatever and declare the problem solved.
Back on Planet Earth, J.D. has given a very nice presentation of the non-calculus approach, which, as Red Five noted, is undoubtedly what would have been expected in 1913. Here is J. D.’s solution, with minor tinkering.
Let’s suppose the rational function spits out the quantity 
. Then,
We then want to show this is impossible for between 2 and 7. This turns out to be easy, by getting rid of the fraction:
Collecting terms, we arrive at the quadratic equation,
It is then a question of for which values of the equation has no solutions. And that’s just a discriminant thing:
Plugging that into our CAS Factorising, we see
It follows that the discriminant is negative exactly if 
, which gives the impossible values for the original rational function.
A nice Year 10 question, perhaps? In another era?
This was the PASS exam… Oh how times have changed for the worse!
Yes, the past is a very foreign country.
To be fair, the IB and NSW HSC exams (and I’m sure the British A-level papers for some examination boards) still examine Vieta’s formulas quite regularly.
These questions are clear, concise and decent mathematics.
And should be compulsory study in pre-service teaching qualifications.
(I worked through this paper a few months ago so will refrain from posting my solutions for a few days to allow others a chance to have some fun).
RF, if you want put your solutions in a comment, I’ll update the post in a day or so.
The final part I actually turned into a question for my Methods 3&4 class this year.
File attached for anyone interested.
So you think you understand domain and range
Is there a website or a book where these problems are collected?
Given what was probably taught back then these problems are fairly standard; today this is serious stuff . . .
Hi Franz. No, there is no real collection that I know of, and it’s not even clear to what extent the exams still exist. Red Five has been collecting for a bit, and I’ve hunted a bit in the past, and it seems very hit or miss what the big libraries (State, UoM) hold. It seems very random which exams from which years have been preserved, and by whom.
The notable books from the era contain many such problems, but it’s not quite the same as a straight collection. Perhaps at some point a Victorian team can work to systematically collect what there is to collect, and possibly VCAA would not object to its publishing. I don’t have time to think about it now, but it’s on my long-list.
These are my solutions. If you’re still working on the problems, don’t read any further.
The first equation can be rewritten (in general) as
, so
First expression =
.
We also have
, so
First expression =
.
An exceptional case occurs when
, but in this case the original expression is undefined. And of course, the question is meaningless if 
.
For the second part,
and similarly if
and 
are exchanged. Thus:
Second expression =
.
For the third part, the equation
can be rewritten as
The discriminant of this quadratic equation in
is 
. Whenever 
, the discriminant is negative, hence the equation has no solutions.
J.D., you know better than that.
Sorry, feel free to delete my comment then.
It’s fine. People can avert their eyes, and I’ll post another soon. But it was kinda thoughtless.
No, really. When I wrote the response, I thought it would be a positive contribution. If it annoys you or you think it will annoy other people, I’d prefer it if it wasn’t there.
Of course it’s a positive contribution, but twelve hours after posting is not optimal timing.
It’s fine, it’s done. I’ll update in the next day or so, and post another one.
Apologies, I didn’t attach the correct file in my first post. Let me try again.
Here is my tentative answer to the last part.
This can form a good SM34 or MM34 exam question for the current students.
TNDOT6-Q1c
Thanks, Sentinel. I’ll look at yours and J.D.’s solutions in the next day or so, and then update.
An interesting approach, to use calculus like that. My own solution was basically the same as JD’s and I think that in 1913 a calculus solution would not have been expected as there was a separate Calculus and Applied Mathematics paper (also a Trigonometry paper, which Marty has sampled from in the past to create these TNDOT posts).
That doesn’t mean the calculus method doesn’t work or isn’t interesting. It is and perhaps makes it a good starting point for a Methods SAC question (which is itself not a good argument, ironically).
OK, I’ve updated the post with a solution (basically J. D.’s solution with a bit of rewording.) A new TNDOT coming very soon.