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Worked Problems

Worked problem 1 () 2000 Paper I


f(x) = ax x3 1 + x2,

where a is a constant.

Show that, if a 98, then

f(x) 0

for all x.

First thoughts

When I play tennis and I see a ball that I think I can hit, I rush up to it and smack it into the net. This is a tendency I try to overcome when I am doing mathematics. In this question, for example, even though I’m pretty sure that I can find f(x), I’m going to pause for a moment before I do so. I am going to use the pause to think about two things.

  • I’m going to think about what to do when I have found f(x).
  • I’m going to try to decide what the question is really about.

Of course, I may not be able to decide what to do with f(x) until I actually see what it looks like; and I may not be able to see what the question is really about until I have finished it; maybe not even then — some questions are not really about anything in particular. I am also going to use the pause to think a bit about the best way of performing the differentiation. Should I simplify first? Should I make some sort of substitution? Clearly, knowing how to tackle the rest of the question might guide me in deciding the best way to do the differentiation. Or it may turn out that the differentiation is fairly straightforward, so that it doesn’t matter how I do it.

Two more points occur to me as I re-read the question.

  • I notice that the inequalities are not strict (they are rather than >). Am I going to have to worry about the difference?
  • I also notice that there is an ‘If ... then’, and I wonder if this is going to cause me trouble. I will need to be careful to get the implication the right way round. I mustn’t try to prove that if f(x) 0 then a 98.

    It will be interesting to see why the implication is only one way — why it is not an ‘if and only if’ question. It may be just ‘if’ because ‘only if’ isn’t true; or it may be just ‘if’ because the examiners thought that the question was long enough without the ‘only if’.

Don’t turn over until you have spent a little time thinking along these lines.

Doing the question

Looking ahead, it is clear that the real hurdle is going to be showing that f(x) 0. How am I going to do that? Two ways suggest themselves. First, if f(x) turns out to be a quadratic function, or an obviously positive multiple of a quadratic function, I should be able to use some standard method: looking at the discriminant (‘b2 4ac’), etc; or, better, completing the square. But if f(x) is not of this form, I will have to think of something else: maybe I will have to sketch a graph. I’m hoping that I won’t have to sketch a graph, because then I could be here all night.

I’ve just noticed that f is an odd function, i.e. f(x) = f(x). (Did you notice that?) That is helpful, because it means that f(x) is an even function.12 If it had been odd, there could have been a difficulty, because all odd functions (at least, those with no vertical asymptotes) cross the horizontal axis y = 0 at least once and cannot therefore be positive for all x.

Now, how should I do the differentiation? I could divide out the fraction, giving

x3 1 + x2 = x x 1 + x2


f(x) = (a 1)x + x 1 + x2 = bx + x 1 + x2, ()

where I have set b = a 1 to save writing. (Is this right? I’ll just check that it works when x = 2. Yes, it does: f(2) = 2a 8 5 = 2(a 1) + 2 5.) This might save a bit of writing, but I don’t at the moment see it helping me towards a positive function. On balance, I think I’ll stick with the original form.

Another thought: should I differentiate the fraction using the quotient formula or should I write it as x3(1 + x2)1 and use the product rule? I doubt if there is much in it. I never normally use the quotient rule — it’s just extra baggage to carry round. But on this occasion, since the final form I am looking for is a single fraction and the denominator using the quotient rule is a square and therefore non-negative, I will.

One more thought: since I am trying to obtain an inequality, I must be careful throughout not to cancel any quantity which might be negative; or at least if I do cancel a negative quantity, I must remember to reverse the inequality.

Here goes (at last):

f(x) = a 3x2(1 + x2) 2x(x3) (1 + x2)2 = a + (2a 3)x2 + (a 1)x4 (1 + x2)2 .

I had to do a bit of algebra to obtain the second equation.

This is working out as I had hoped: the denominator is certainly positive and the numerator is a quadratic function of x2. I can finish this off most elegantly by completing the square. There are two ways of doing this. Just to be clear in my mind, I’m going to write the numerator as

A + Bx2 + Cx4,

where A = a, B = 2a 3 and C = a 1. Then I can complete the square in two ways:

A1 + (B2A)x2)2 + C B24Ax4


Cx2 + B2C)2 + A B24C. ()

It doesn’t seem to matter which I use. The second expression () looks a bit simpler

If a 9 8, then certainly C > 0. That means that the first of the two terms of () is positive:
C(x2 + B2C)2 > 0.

The second term of () can be written in terms of a as follows:

A B24C = a (2a 3)2 4(a 1) = 4a(a 1) (2a 3)2 4(a 1) = 8a 9 4(a 1). ()

It is very pleasing to see the numerator 8a 9; it is exactly what I want, because it is non-negative if a 9 8. That is what I needed to complete the question.


Now I can look back and analyse what I have done.

On the technical side, it seems I was right not to use the simplified form of x3(1 + x2) given in (). This would have lead to the quadratic (b + 1) + (2b 1)x2 + bx4, which isn’t any easier to handle than the quadratic involving a and just gives an extra opportunity to make an algebraic error.

Although introducing new variables A, B and C seemed at first to complicate the problem, it was useful to have them when it came to completing the square, which would have been a bit of a mess had I worked directly from the quartic expression with a in it.

Actually, I see on re-reading my solution that I have made a bit of a meal out of the ending. I needn’t have completed the square at all; I could used the inequality a 98 immediately after finding f(x), since a appears in f(x) with a plus sign always:

f(x) = a + (2a 3)x2 + (a 1)x4 (1 + x2)2 9 8 + (9 4 3)x2 + (9 8 1)x4 (1 + x2)2 = (x2 3)2 8(1 + x2)2 0

as required.

However, my unnecessarily elaborate proof, involving completing the square, makes the role of a a bit clearer than it is in the shorter alternative. In fact, I can see how to answer my question about ‘if and only if’, namely is it the case that f(x) 0 for all x if and only if a 9 8? The answer is Yes. Looking at (), I see that if f(x) 0 for all x, then certainly C 0, otherwise for large enough x the first (squared) term would be dominant and negative. But if 0 C < 1 8, then B < 0 and there is a value of x2 for which the squared term in () vanishes, leaving only the second term which is negative, as can be seen from (). That means f(x) < 0 for this value of x contradicting our assumption. Therefore, f(x) 0 for all x implies that C 1 8 and a 9 8.

I wonder why the examiner wanted me to investigate the sign of f(x). The obvious reason is to see what the graph looks like. We can now see what this question is about. It is clear that the examiners really wanted to set the question: ‘Sketch the graphs of the function ax x3(1 + x2) in the different cases that arise according to the value of a’ but it was thought too long or difficult. It is worth looking back over my working to see what can be said about the shape of the graph of f(x) when a < 98.

(I leave that to you to think about!)

12 You can see this easily by sketching a ‘typical’ odd function, or by differentiating the Maclaurin expansion of an odd function.