This paper is based on FP1 May 2015, but with the parametrics questions replaced by others.
Comments on marking, January 2017
Comments by question, January 2017
1. Solving a cubic
Many of you forgot to standardise the equation first (dividing through by 9) before using the rules for sum-of-roots and product-of-roots.
Some of you remembered to do that, but then ignored your own work and used 25 rather than 25/9 for product of roots.
The sort of mistakes you’ll learn to avoid with more practice.
2. Interval bisection and linear interpolation
Almost everyone was ok with the basic methods; most lost marks were down to not reading the question carefully.
3. Series formulas
Most of you ok with part (a).
Quite a lot didn’t even try part (b), which asks you to calculate (sum from n+1 to 2n).
It’s (sum from 1 to 2n) − plug in “2n” instead of n, in the formula you got in part (a)
minus (sum from 1 to n) − the exact same formula as you got in part (a)
Or (2n/3)(2n+4)(2n+5) − (n/3)(n+4)(n+5)
Then take out common factor (n/3), expand, collect terms, factorise.
4. Complex numbers
Almost everyone had the right basic idea here. But (and this was a problem in Q.8b too), lots of you still often foul up when calculating zz* for complex numbers, in this case (1+√3i)(1−√3i).
It’s the sum of squares. (a+bi)(a−bi) = a2 + b2
Or in other words, zz* = |z|2. Equals modulus squared, equals magnitude squared, equals enlargement-factor squared, equals determinant of the corresponding matrix.
Marks lost by not reading part (b) carefully. It says: give the argument in radians in terms of π. If you write argument = −60 degrees, or argument = −1.047, you lose marks, even though everything you have done has been correct. After Pol(,), use the Alpha key on the calculator, followed by X=, to get from −1.047 to −π/3.
5a. Induction with sequences
Oddly, a lot of you “lost” the 4 in the working here, which obviously messed you up.
5b. Matrices and geometrical transformations
A significant minority got the geometrical transformation represented by RS wrong after calculating the matrix correctly.
In some cases, that’s down to not doing a diagram.
In some, to doing a tiny and almost illegible diagram. Make your diagrams big, and use a ruler.
One of you did a nice, big, clear diagram, and then just copied the information from the columns of the matrix into the diagram wrong.
For such mishaps, it’s good to have a check. |RS|=1. So RS can’t be a reflection. (Reflections always have negative determinants). Since Edexcel never asks about shearings, or scalings by different factors in different directions, it must be a rotation.
6(i). Induction with matrices
Some of you got mixed up with the working here. Write clearly and give yourself plenty of space: that’s the way to avoid such mistakes.
In these questions you always have a choice whether to work out Ak+1 from Ak.A or from A.Ak. Both will give the same result, but one may give easier working than the other. In this case, Ak.A gave easier working.
6(ii). Induction with series
Problems here with taking out the common factor when you come to simplify
(sum from 1 to k)+([k+1]’th term)
It can’t be because I have failed to bang on about common factors, and insist you practise them, enough.
My guess is that a lot of you just didn’t see that (2k+1) is a factor of 4k2−1. Difference of squares! To do further maths, you need to know difference of squares as well as you know your own birthday.
Knowing difference of squares well will also help you with (a+bi)(a−bi). You will see straight away that it is a difference of squares, a2 minus (bi)2. Since i2 is −1, and minus times minus is plus, this difference of squares becomes a sum of squares.
Almost everyone was pretty good with 7(i). Most of you flunked 7(ii). That’s my fault, not giving enough time to that sort of problem. We discussed it on Friday, and you have similar examples to do for homework.
8a. Induction with divisibility
I think some of you were getting tired by this point, because in the divisibility reasoning there was a lot of multiplying different bits of the right hand side of divisibility statements by different factors, which you can’t do.
If m | A+B
then m | k(A+B) for any k. You can multiply the whole right hand side by any number you like
but you can’t say m | kA + nB, multiplying different bits of the right hand side by different factors
2 | 3+1 implies 2 | (3+1)×57, or 2| (3+1)×19, or 2 | (3+1)×(any number you like)
2 | 4×3 + 3×1
8b. Complex numbers
Tired again. A surprising number of you wrote
|z1 + z2| = |z1| + |z2|
but that’s saying that the as-the-crow-flies journey from the origin to z1, and then from z1 on to z2, equals the as-the-crow-flies journey from the origin straight to z1+z2.
Only true if z1 and z2 are exactly in line with each other and the origin, i.e. have the same argument (direction).
From school to Tesco and then to PC World is not the same distance as from school straight to PC World.
|z1z2| = |z1||z2| (enlarge by a factor A, then enlarge by a factor B, and that’s the same as enlarging by a factor AB); but that’s a different matter.
Then in the second part there was the same problem as in Q.4, about calculating (a+bi)(a−bi)
Worked answers for January 2016 paper (based on FP1 May 2015) at:
My comments from marking that paper: read below or click here to download as pdf.
For each question, make sure you now know how to do it; where you went wrong earlier this month; what idea you must consolidate in order to be confident with similar questions in future. When you find a question where you can’t puzzle that out, ask me.
Re-sit paper, 2016
Organise yourself to spend 90 minutes under exam conditions doing this paper:
Then I’ll mark the paper for you.
Comments from marking early-January 2016 FP1 papers
Apart from Preslava and Sharif, and Taija with the first six questions, and Wei-Kong with the middle part of the paper, you all did very badly in the FP1 test.
In a way, it was so bad that I have some hope. We still have time to turn this round. Just. Only just.
Many, many marks were lost by not reading the questions carefully, or by writing such things as 6 × 5 = 35, or similar slips. Meanwhile, through the fog I could see that you know many of the basic new ideas.
I know that with practice you can get to the level of seeing that 6 × 5 is 30; or that 4 × 5k is not 20k; or that when a question says “given that 5 is a root” it’s not sensible to answer it by plugging in 5 to check that 5 is a root…
Marking it was like watching a football team play after they’ve got really unfit and done no practice for weeks. They remember the offside rule, or some tricksy gambit they recently learned; but they trip over their bootlaces, they kick wide when facing open goals, they get breathless when they run a few dozen yards, or they stand around looking baffled while an opposing player races through their defence.
The coach watching the football team is dismayed, but he knows that if he makes the team practise hard they’ll get back to form.
Some of you will say that you did practise over the Christmas holidays. I answer: yes, but not enough. You came back with your mathematical knowledge not sharpened by practice, but buried under a million other thoughts.
The test was not specially hard. A couple of questions involved a bit more working than average, but you did just as badly on the completely plain-vanilla and straightforward questions.
My plan: in January, and through to mid-February we will focus on M2, and get that work completed and nailed down. Then we will have maybe two weeks before the March mock exams in which we will focus on really intensive practice on FP1. If you work hard enough, including over half-term, that will get you good grades in the March mocks.
In the meantime, please use my catch-up sessions.
Question 1: cubic equation, given that x=5 is a root, find the other two roots.
This was an absolutely standard, plain-vanilla question, and should have been an easy 5 marks to start the paper with. In fact only Taija and Preslava got 5/5, and no-one else got more than 2/5.
Some of you plugged 5 into the equation to check it was a root. That was a waste of your time, since the question said: “Given that x=5 is a solution…” Some of you started doing the question by the (unnecessarily long, but not difficult) method in the mark scheme and then stopped part way through. Why? Solving the quadratic you get by dividing (x−5) into the cubic is no harder than solving any other quadratic by completing the square or the quadratic formula.
Some of you tried to use the easier sum-of-roots and product-of-roots method. Problem: you forgot to divide through by 9 to make the coefficient of x3 equal to 1 before using the sum-of-roots and product-of-roots rules.
Think about it a minute, and it’s obvious you have to do something like that.
You were given the equation as 9x3−33x2−55x−25=0 The same equation, with the same roots, could be written as: 18x3−66x2−110x−50=0 or 27x3−99x2−165x−75=0 The sum of the roots can’t be simultaneously equal to 33, 66, and 99; and the product of the roots can’t be simultaneously 25, 50, and 75.
There has to be some rule about standardising the equation before you use the sum-of-roots and product-of-roots rules.
The rule is the simplest it could possibly be: make the coefficient of x3 (for a cubic; or x2 for a quadratic; or x4 for a quartic) equal to one. So with this equation you had to divide through by 9.
Question 2: You’re given that an equation has one root in the interval (13,14), and you’re asked to find approximations to the root:
a) by interval bisection
b) by linear interpolation It’s a completely standard, plain-vanilla question.
Problems: 1. Some of you wasted effort on proving that the equation did have a root in the interval. But this question had already told you that.
2. Some of you didn’t attempt the interval bisection at all. Some of you did the correct calculations for interval bisection, then threw away your marks by failing to write at the end: the root is in the interval (13.25,13.5)
3. Some got the linear interpolation calculation all correct, then threw away marks by failing to write at the end: next approximate value for the root is 13+0.455=13.455. (The question asks you for an approximate value between 13 and 14. 0.455 obviously can’t be it, can it?) Some of you just wrote 13.455 at the end of your working. I’ve given you a mark for that, and probably Edexcel would too, but you should write in sentences. The end of your working should be: Next approximation [or next guess] for α = 13.455
4. Some did the linear interpolation basically correctly, but ignored the fact that the question asked for an answer to three decimal places. To get an answer good to three decimal places, you have to calculate f(13) and f(14) to at least 4 d.p., don’t you? 13.46 is a correct answer to 2 d.p., but it can’t be a correct answer to 3 d.p.!
5. Some got the right diagram for linear interpolation, but then wrote: horizontal side of big triangle/ horizontal side of small triangle = vertical side of small triangle/ vertical side of big triangle Wrong way round. It should be small/big = small/big, not big/small = small/big Everyone makes slips. But when you see an answer which is obviously nonsense – like, the smaller triangle has a bigger horizontal side than the bigger triangle – then don’t just write it down and carry on.
Check where you’ve made the slip.
Question 3. (a) Using standard formulas to get a given formula for a sum, and then (b) using that given formula (for the sum from r=1 to n) to find the sum of the same series from r=n+1 to r=2n
Problems: 1. Some of you got the right standard formulas, but then made blunders taking out the common factor and got into a mess. At least one of you got the right standard formulas, took out the common factor correctly… and then just stopped two lines short of finishing the question. Why?
2. Quite a few didn’t even try part (b)
3. The question said “hence” in part (b), which means: use the result of part (a) to do part (b). Some you tried to do part (b) without using part (a), and got into a mess that way.
4. Some of you wrote that the sum from r=n+1 to 2n was the sum from r=1 to n plus the sum from r=1 to n. This could be true only if 5+6+7+8 = 1+2+3+4+1+2+3+4 which it obviously doesn’t.
The correct answer: sum from r=n+1 to 2n is the sum from r=1 to 2n minus the sum from r=1 to n. Remember the trolleys with weights? The sum of the contents of all the trolleys numbered n+1 to 2n is equal to the sum of the contents of all the trolleys numbered 1 to 2n, minus what’s on the trolleys numbered 1 to n. Or: the total marks you get on questions 5 to 8 equal the total marks you got on questions 1 to 8, minus the total marks you got on question 1 to 4.
5. Some of you had sort-of-forgotten the difference between proof-by-induction and using standard formulas, though it is after all rather like the difference between carpentry and Lego. So you produced stuff which was a weird mixture of bits of working from proofs-by-induction and bits of standard formulas.
You were asked to do a complex-number division; then to find modulus and argument of the answer; then to plot three complex numbers on an Argand diagram.
1. Some of you didn’t even try to do the Argand diagram. Or did it but plotted only two numbers (correctly, but only two).
2. Pretty much everyone got the procedure for dividing complex numbers right, but then at least three of you wrote that 1 – 3i2 was -2, not 4. But when you multiply a non-zero complex number by its conjugate, you always get a positive real answer. One of you calculated the answer 4 correctly, but then multiplied the top of the fraction by the bottom number, i.e. 4, instead of dividing. Mistakes like 6/4 = 24 are not due to the paper being “too hard”.
(a) Finding the normal to a hyperbola at a point (it means the line at right angles to the curve); (b) finding where that normal meets the curve again.
We did race through parabolas and hyperbolas, though I think if more of you had done the homework on them with more care you would have been able to do this very straightforward and plain-vanilla question.
The odd thing is that quite a few of you got right the bit which includes new ideas, part (a); and then failed or didn’t attempt the bit which includes no new ideas, part (b). All you need to do for part (b) is plug in y=(8x−45)/2 into the equation xy=9, and solve the resulting quadratic in x. Year 11 stuff. It’s useful to do a diagram so that you can check whether your answer looks right.
Two induction proofs: (a) with matrices, (b) with series
Both parts involved more working than this sort of question usually does (which I guess is why the question was given 12 marks).
Some of you got lost in the working, but often with mistakes that you really shouldn’t make now, like writing 4×5k = 20k or failing to take out common factors.
Some of you failed to pick up the marks which you can always get in a proof-by-induction question, even on the worst of days, by proving Step 1.
As with question 3, a few of you had sort-of-forgotten the difference between proof-by-induction and using standard formulas. So you produced stuff which was a weird mixture of bits of working from proofs-by-induction and bits of standard formulas. This made even less sense here, because all of you knew you were attempting a proof-by-induction in part (a). The question states clearly that part (b) asks for proof by induction too.
Some of you had forgotten that with series, the sum from 1 to (k+1) (which is what you want to work with in Step 2) equals the sum from 1 to k plus the (k+1)th term.
If you’re working, your total earnings for days 1 to (k+1) will equal your earnings from days 1 to k, plus your earnings for the (k+1)th day, no? Or, in an exam, your total marks for questions 1 to (k+1) equal your marks for questions 1 to k, plus your marks for the last question, the (k+1)th?
Question 7. (i) To find for what values of k a matrix is singular (i.e. has determinant = 0) (ii) You’re given a matrix B which transforms T into T’. You’re asked to find the inverse (which transforms T’ into T) and calculate the vertices of T from the vertices of T’, which include an unknown factor c. Then you’re told the area of T’ and asked to calculate the value of c. You do that by calculating the area of T in terms of c, from the vertices, and then equating that to area of T’ divided by det(B).
Problems: 1. A lot of you correctly worked out the equation for k. In fact, everyone apart from those who had got so tired and emotional by Question 7 that they made no attempt at it knew what a singular matrix was and how to work out the determinant. They also knew how to work out the inverse. But then lots of you made slips in (not very heavy) algebra… or just wrote down the equation for k, a perfectly ordinary quadratic equation… and didn’t solve it.
2. Oddly, a lot of you seem to have forgotten how to calculate where a matrix moves a point, the very first thing that we nailed down when studying matrices. Write the point as a single-column matrix, and multiply the transformation matrix onto that single-column matrix.
Question 8. You’re given a parabola.
(a) Find the distance from a typical point to the focus.
(b) Find the equations of tangents at two points, and where they meet.
(c) Prove a result connecting the distances from the two points to the focus and the distance from the meeting point to the focus.
This last part includes algebra a bit more complicated than average with the questions, but on the whole the question is completely plain-vanilla. Pretty much every question on parabolas and hyperbolas asks you to find the equations of tangents (or normals) to the parabola (or hyperbola).
The impression I get from your scripts is that almost everyone was tired and emotional by the time you got to question 8. Most of you didn’t even try to do it. Still, part (a) only asks you to find the distance from the point (3p2, 6p) to the point (3,0). You could still do that even if you’d never heard the words “parabola” or “tangent”.