My mother recently gave me a book of old ‘A’ level exam papers. Here’s a 4 mark (ie low mark) question from the June 1982 University of London “Syllabus B” paper, that I think I’d really struggle with – I’ll try it later:

Given that show that is a factor of . Factorise completely and hence state the set of values for which .

This one (also 4 marks), seems a bit easier though:

The functions and , each with domain , where:

and ,

are defined by

and ,

Write down and simplify and expression for , giving its domain of definition. Sketch the graph of .

5 thoughts on “Maths ‘A’ level dumbing down: the proof”

ah now – I could do the first one but had to google trig functions to do the second one. Mind you better than my finals papers which I discovered at some point and couldn’t understand the questions!

Actually, the first one wasn’t as horrific as I thought, so maybe I’d be on my way to an E grade after all.

So I worked through the problem my way (seldom the right way) and made a few interesting-to-me discoveries. At least the way I did it seems a bit tough for students in a test environment.

“(3-x) is a factor of P(x)” is the same as “P(3) is 0”. Easy to show that by calculation.

It’s even easy to calculate P(3) by hand. You might notice that it is easier to calculate P(3)/3 and gets you the same information (that the result is zero) (but explaining that is more work).

Once you divide the polynomial by the given factor, you have a quadratic which can be solved by the normal formula. In fact, you should be able to solve this particular quadratic by inspection.

After I did that, I realized that the constant term (3) suggests which values to probe as roots. The product of roots is 3, and one is 3, and so the others ought to be 1 or -1 (all the coefficients are integral and the highest term’s coefficient is -1).

Once you know all the roots, you know all the zero crossings and can figure out where the function is negative.

Oh, I just remembered: it is negative exactly where an odd number of the factors is negative and the others are non-zero. Since one root is doubled, it is exactly where the other root is negative and the doubled root is not zero: when x is greater than 3. f is zero when any of the factors is zero: when x is 3 or 1. So the answer is: when x is 1 or greater or equal to 3.

So all that is straightforward if you are comfortable with polynomials. Probably not if you only learned by rote (unless this looks like your rote questions). So this looks like a good question. It may even work well for partial marks. But it would surprise me if a majority of students, even undergrads, could do this under time pressure.

I find it interesting that the two problems define functions using different notations.

The second question is a little trickier because one has to understand and describe that the composition has a reduced domain ([pi/2, pi]). All the more confusing because each function’s defining expression is well defined for all of R. I don’t think students are very good with domains.

Hugh, interesting reply and a few insights that passed me by. But it’s surely not that difficult a question? I just thought must be case that and it turned out to be not so difficult to work out what a, b and c were. When I wrote the question out this afternoon I thought it would be difficult, but it wasn’t really. The second one I could work out in my head even as I typed it out, thought it was really quite trivial. But that’s just me I guess.

ah now – I could do the first one but had to google trig functions to do the second one. Mind you better than my finals papers which I discovered at some point and couldn’t understand the questions!

Actually, the first one wasn’t as horrific as I thought, so maybe I’d be on my way to an E grade after all.

So I worked through the problem my way (seldom the right way) and made a few interesting-to-me discoveries. At least the way I did it seems a bit tough for students in a test environment.

“(3-x) is a factor of P(x)” is the same as “P(3) is 0”. Easy to show that by calculation.

It’s even easy to calculate P(3) by hand. You might notice that it is easier to calculate P(3)/3 and gets you the same information (that the result is zero) (but explaining that is more work).

Once you divide the polynomial by the given factor, you have a quadratic which can be solved by the normal formula. In fact, you should be able to solve this particular quadratic by inspection.

After I did that, I realized that the constant term (3) suggests which values to probe as roots. The product of roots is 3, and one is 3, and so the others ought to be 1 or -1 (all the coefficients are integral and the highest term’s coefficient is -1).

Once you know all the roots, you know all the zero crossings and can figure out where the function is negative.

Oh, I just remembered: it is negative exactly where an odd number of the factors is negative and the others are non-zero. Since one root is doubled, it is exactly where the other root is negative and the doubled root is not zero: when x is greater than 3. f is zero when any of the factors is zero: when x is 3 or 1. So the answer is: when x is 1 or greater or equal to 3.

So all that is straightforward if you are comfortable with polynomials. Probably not if you only learned by rote (unless this looks like your rote questions). So this looks like a good question. It may even work well for partial marks. But it would surprise me if a majority of students, even undergrads, could do this under time pressure.

I find it interesting that the two problems define functions using different notations.

The second question is a little trickier because one has to understand and describe that the composition has a reduced domain ([pi/2, pi]). All the more confusing because each function’s defining expression is well defined for all of R. I don’t think students are very good with domains.

Hugh, interesting reply and a few insights that passed me by. But it’s surely not that difficult a question? I just thought must be case that and it turned out to be not so difficult to work out what a, b and c were. When I wrote the question out this afternoon I thought it would be difficult, but it wasn’t really. The second one I could work out in my head even as I typed it out, thought it was really quite trivial. But that’s just me I guess.