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This was the last question on the May 2016 Calculus option paper for IB HL.  It’s worth nearly a quarter of the entire marks – and is well off the syllabus in its difficulty.  You could make a case for this being the most difficult IB HL question ever.  As such it was a terrible exam question – but would make a very interesting exploration topic.  So let’s try and understand it!

Part (a)

First I’m going to go through a solution to the question – this was provided by another HL maths teacher, Daniel – who worked through a very nice answer.  For the first part of the question we need to try and understand what is actually happening – we have the sum of an integral – where we are summing a sequence of definite integrals.  So when n = 0 we have the single integral from 0 to pi of sint/t.  When n = 1 we have the single integral from pi to 2pi of sint/t.  The summation of the first n terms will add the answers to the first n integrals together.

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This is the plot of y = sinx/x from 0 to 6pi.  Using the GDC we can find that the roots of this function are n(pi).  This gives us the first mark in the question – as when we are integrating from 0 to pi the graph is above the x axis and so the integral is positive. When we integrate from pi to 2pi the graph is below the x axis and so the integral is negative.  Since our sum consists of alternating positive and negative terms, then we have an alternating series.

Part (b i)

This is where it starts to get difficult!  You might be tempted to try and integrate sint/t – which is what I presume a lot of students will have done.  It looks like integration by parts might work on this.  However this was  a nasty trap laid by the examiners – integrating by parts is a complete waste of time as this function is non-integrable.  This means that there is no elementary function or standard basic integration method that will integrate it.  (We will look later at how it can be integrated – it gives something called the Si(x) function).  Instead this is how Daniel’s method progresses:

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Hopefully the first 2 equalities make sense – we replace n with n+1 and then replace t with T + pi.  dt becomes dT when we differentiate t = T + pi.  In the second integral we have also replaced the limits (n+1)pi and (n+2)pi with n(pi) and (n+1)pi as we are now integrating with respect to T and so need to change the limits as follows:

t = (n+1)(pi)

T+ pi = (n+1)(pi)

T = n(pi).  This is now the lower integral value.

The third integral uses the fact that sin(T + pi) = – sin(T).

The fourth integral then uses graphical logic.  y = -sinx/x looks like this:

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This is the same as y = sinx/x but reflected in the x axis.  Therefore the absolute value of the integral of  y = -sinx/x  will be the same as the absolute integral of y = sinx/x.  The fourth integral has also noted that we can simply replace T with t to produce an equivalent integral.  The last integral then notes that the integral of sint/(t+pi) will be less than the integral of sint/t.  This then gives us the inequality we desire.

Don’t worry if that didn’t make complete sense – I doubt if more than a handful of IB students in the whole world got that in exam conditions.  Makes you wonder what the point of that question was, but let’s move on.

Part (b ii)

OK, by now most students will have probably given up in despair – and the next part doesn’t get much easier.  First we should note that we have been led to show that we have an alternating series where the absolute value of u_n+1 is less than the absolute value of u_n.  Let’s check the requirements for proving an alternating series converges:

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We already have shown it’s an absolute decreasing sequence, so we just now need to show the limit of the sequence is 0.

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OK – here we start by trying to get a lower and upper bound for u_n.  We want to show that as n gets large, the limit of u_n = 0.  In the second integral we have used the fact that the absolute value of an integral of a function is always less than or equal to the integral of an absolute value of a function.  That might not make any sense, so let’s look graphically:

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This graph above is y = sinx/x.  If we integrate this function then the parts under the x axis will contribute a negative amount.

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But this graph is y = absolute (sinx/x).  Here we have no parts under the x axis – and so the integral of absolute (sinx/x) will always be greater than or equal to the integral of y = sinx/x.

To get the third integral we note that absolute (sinx) is bounded between 0 and 1 and so the   integral of 1/x will always be greater than or equal to the integral of absolute (sinx)/x.

We next can ignore the absolute value because 1/x is always positive for positive x, and so we integrate 1/x to get ln(x). Substituting the values of the definite integral gives us a function of ln which as n approaches infinity approaches 0.  Therefore as this limit approaches 0, and this function was always greater than or equal to absolute u_n, then the limit of absolute u_n must also be 0.

Therefore we have satisfied the requirements for the Alternating Series test and so the series is convergent.

Part (c)

Part (c) is at least accessible for level 6 and 7 students as long as you are still sticking with the question.  Here we note that we have been led through steps to prove we have an alternating and convergent series.  Now we use the fact that the sum to infinity of a convergent alternating series lies between any 2 successive partial sums.  Then we can use the GDC to find the first few partial sums:

Screen Shot 2016-05-21 at 10.30.29 PMAnd there we are!  14 marks in the bag.  Makes you wonder who the IB write their exams for – this was so far beyond sixth form level as to be ridiculous.  More about the Si(x) function in the next post.

 

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IB HL Calculus P3 May 2016:  The Hardest IB Paper Ever?

IB HL Paper 3 Calculus May 2016 was a very poor paper.  It was unduly difficult and missed off huge chunks of the syllabus.  You can see question 5 posted above. (I work through the solution to this in the next post).  This is so far off the syllabus as to be well into undergraduate maths.  Indeed it wouldn’t look out of place in an end of first year or end of second year undergraduate calculus exam.  So what’s it doing on a sixth form paper for 17-18 year olds?   The examiners completely abandoned their remit to produce a test of the syllabus content – and instead decided that a one hour exam was the time to introduce extensions to that syllabus, whilst virtually ignoring all the core content of the course.

A breakdown of the questions

1) Maclaurin- on the syllabus.  This was reasonable.  As was using it to find the limit of a fraction.  Part (c) requires use of Lagrange error – which students find difficult and forms a very small part of the course.  If this was the upper level of the challenge in the paper then fair enough, but it was far from it.

2) Fundamental Theorem of Calculus – barely on the syllabus – and unpredictable in advance as to what is going to be asked on this.  This has never been asked before on any paper, there is no guidance in the syllabus, there was no support in the specimen paper and most textbooks do not cover this in any detail.  This seems like an all or nothing question – students will either get 7 or 0 on this question.  Part (c) for an extra 3 marks seems completely superfluous.

3) Mean Value Theorem – a small part of the syllabus given dispropotionate exam question coverage because the examiners seem to like proof questions.  This seems like an all or nothing question as well – if you get the concept then it’s 7 marks, if not it’ll likely be 0.

4) Differential equations –  This question would have been much better if they had simply been given the integrating factor /separate variables question in part (b), leaving some extra marks to test something else on part (a) – perhaps Euler’s Method?

5) An insane extension to the syllabus which took the question well into undergraduate mathematics – and hid within it a “trap” to make students try to integrate a function that can’t actually be integrated.  This really should have been nowhere near the exam.  At 14 marks this accounted for nearly a quarter of the exam.

Content unassessed

The syllabus is only 48 hours and all schools spend that time ploughing through limits and differentiability of functions, L’Hopital’s rule, Riemann sums, Rolle’s Theorem, standard differential equations, isoclines, slope fields, the squeeze theorem, absolute and conditional convergence, error bounds, indefinite integrals, the ratio test, power series, radius of convergence.  All of these went pretty much unassessed.  I would say that the exam tested around 15% of the syllabus content.  Even the assessment of alternating series convergence was buried inside question 5 – making is effectively inaccessible to all students.

The result of this is that there will be a huge squash in the grade boundaries – perhaps as low as 50-60% for a Level 6 and 25-35% for a level 4.    The last 20 marks on the paper will probably be completely useless – separating no students at all.  This then produces huge unpredictability as dropping 4-5 marks might take from from a level 5 to level 3 or level 6 to level 4.

Teachers no longer have any confidence in the IB HL examiners

One of my fellow HL teachers posted this following the Calculus exam:

At various times throughout the year I joke with my students about how the HL Mathematics examiners must be like a group of comic book villains sitting in a lair, devising new ways to form cruel questions to make students suffer and this exam leads me to believe that this is not too far fetched of a concept.

And I would tend to agree.  Who wants students to be demoralised with low scores and questions they can’t succeed on.  Surely that should not be an aim when creating an exam!

I’ve taught the HL Calculus Option for the last 4 years – I think the course is a good one.  It’s difficult but a rewarding syllabus which introduces some of the tools needed for undergraduate maths.  However I no longer have any confidence in the IB or the IB examiners to produce a fair test to examine this content.  Many other HL teachers feel the same way.  So what choice is left?  Abandon the Calculus option and start again from scratch with another option?  Or continue to put our trust in the IB, when they continue to let teachers (and more importantly the students) down?

 

 

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