HL Maths 2018 discussion

Kbvdpo

Quick question...
Seeing as p2 came up on p1 English... Is there any possibility that any paper 2 maths such as stats, theorems or trigonometry could come up on p1 tomorrow or are they definitely 2 separate papers?
Already terrified of failing HL never mind having to study both papers tonight...??
Is this right for p1 or am I forgetting anything;
Algebra
Functions
Logs
Complex Numbers
Induction
Differentiation
Calculus
Integration
Sequences & Series
Financial Maths



Also, just in case I really freak out tomorrow, is the ordinary course p1 the same as hl???

UnknownEntity

Have you even looked at the 2017 paper? Trig came up in Q9 Paper 1

Kbvdpo

UnknownEntity wrote: »

Have you even looked at the 2017 paper? Trig came up in Q9 Paper 1

So can anything come up on either paper or is trig a part of P1?

Celtron

UnknownEntity wrote: »

Have you even looked at the 2017 paper? Trig came up in Q9 Paper 1

That was a functions question with trig, they're allowed do that considering it is a function. Other than that no, like stats and probability can't come up on paper .

Kbvdpo

Celtron wrote: »

That was a functions question with trig, they're allowed do that considering it is a function. Other than that no, like stats and probability can't come up on paper .

Hopefully not as I’m stressed enough as is 😬 the thought of failing terrifies me but hopefully all will be fine and they don’t for some reason put P2 on 1 like in English

Peregrine

AFAIK, there's nothing on the syllabus restricting what they can ask on each paper. It's been several years since I last read the syllabus though so don't quote me on that.

I'd check the syllabus.

Celtron

This is an extract of a letter from the SEC to the Management authorities of Secondary Schools about the leaving cert maths course:
Paper 1 will generally test the material in Strand 3, Strand 4, and Strand 5 of the new syllabus. There will
be two sections on the paper, as follows:
o Section A: Concepts and Skills 150 marks
o Section B: Contexts and Applications 150 marks

Paper 2 will generally test the material in Strand 1 and Strand 2 of the new syllabus, along with material
from Section 3.4 in Strand 3: Length, Area and Volume. There will be two sections on the paper, as follows:
o Section A: Concepts and Skills 150 marks
o Section B: Contexts and Applications 150 marks
Note: Any synthetic geometry examined on the paper will be based on the geometry set out in the
syllabus. There will no longer be a question with internal choice on this topic.
Note: Some additional material (deferred since 2011) will be examinable in 2015 and subsequent years.
Sample questions based on this material, at both Higher Level and Ordinary Level, have been prepared, and
ten copies are enclosed.
Both Paper 1 and Paper 2 will be presented as a combined question-and-answer booklet.

Hope that is of use, and anyways if you look at the Active Maths Books by Oliver Murphy who is heavily involved in making the papers, book one covers all topics in paper 1, book 2 covers all topics in paper 2, and it even says this on the cover of the book

Kbvdpo

Turns out it was only calculus & sequences/series anyway...

eclipsechaser

On the A2019 question, the sequence repeats itself every 6 terms. So A7=A1, A8=A2 etc.

Therefore, you can subtract as many 6's from the term number to find the correct answer. A2019 = A2013 = A2007 = ... = A3 = -2. They'll want that answer rather than just A2019 = A2018 - A2017 (which is true and will likely get partial credit).

So A2019 = -2.

It involves modular arithmetic which isn't really on the course. Some amount of sequences and series!

spurious

Celtron wrote: »

Oliver Murphy who is heavily involved in making the papers,

Are you sure? Did he say that, or do people just think it?
There is a HUGE confidentiality clause in SEC contracts, for good reason.

**edit...just realised it may have been a typo of 'marking' the papers.

Drumm...

eclipsechaser wrote: »

On the A2019 question, the sequence repeats itself every 6 terms. So A7=A1, A8=A2 etc.

Therefore, you can subtract as many 6's from the term number to find the correct answer. A2019 = A2013 = A2007 = ... = A3 = -2. They'll want that answer rather than just A2019 = A2018 - A2017 (which is true and will likely get partial credit).

So A2019 = -2.

It involves modular arithmetic which isn't really on the course. Some amount of sequences and series!

I got it as a trig function with
TN= 4cos((n-1) π/3)
Answer still works to be -2

What did anyone get for the last question or max of h(x)?

Celtron

Ya sorry it was. My applied maths teacher told us he met him few years ago and that he's involved with setting that exam and has some input on the regular maths exam as well

Celtron

Max value via differentiation

lundrum

Drumm... wrote: »

I got it as a trig function with
TN= 4cos((n-1) π/3)
Answer still works to be -2

What did anyone get for the last question or max of h(x)?

I think I got the max to be 17, what did you get?

Drumm...

lundrum wrote: »

I think I got the max to be 17, what did you get?

Yeah I got 17 too but wasn't too sure. Thanks 👌

UnknownEntity

17 here too

lundrum

Drumm... wrote: »

Yeah I got 17 too but wasn't too sure. Thanks ��

UnknownEntity wrote: »

17 here too

Yeet

What did you get for the angle of the tangent of cos2x? I got 60 degrees

Drumm...

lundrum wrote: »

Yeet

What did you get for the angle of the tangent of cos2x? I got 60 degrees

It was -60 I think

UnknownEntity

I said 60° but the thing that confused me is what they meant by the "positive sense of the x-axis". Does that mean that the answer is actually 120°??

lundrum

UnknownEntity wrote: »

I said 60° but the thing that confused me is what they meant by the "positive sense of the x-axis". Does that mean that the answer is actually 120°??

I drew it out and so I think the fact that tan of the angle was negative is actually irrelevant. The tangent makes like a right angled triangle with the x and y axis and the angle is the one between the x axis and the line

greenwhale

Anyone know what were the right answers for the very last part of 9, the bit about commenting on the area and perimeter as n tends to infinity?

lundrum

greenwhale wrote: »

Anyone know what were the right answers for the very last part of 9, the bit about commenting on the area and perimeter as n tends to infinity?

I'm like 80% sure the area was 0 and the perimeter was infinite

UnknownEntity

greenwhale wrote:

Anyone know what were the right answers for the very last part of 9, the bit about commenting on the area and perimeter as n tends to infinity?

I think for the area you had to say it's tending towards 0 and for the perimeter, it's increasing forever.

UnknownEntity

How many marks would the last part of 9 be worth? I only got one correct out of the 2

lundrum

UnknownEntity wrote: »

How many marks would the last part of 9 be worth? I only got one correct out of the 2

Me too. I'd guess if you got 1 of the 2 you'd probably get like 3/5 or so. Very few people I know got that far at all

greenwhale

lundrum wrote: »

I'm like 80% sure the area was 0 and the perimeter was infinite

So then would the answer for the earlier part where you had to find what fraction of original triangle would remain be equal to 0 aswell? Or am I confusing two different things sorry

UnknownEntity

greenwhale wrote:

so then would the answer for the earlier part where you had to find what fraction of original triangle would remain be equal to 0 aswell? Or am i confusing two different things sorry

Yep that's right. Did you get it?

greenwhale

UnknownEntity wrote: »

Yep that's right. Did you get it?

Yeah ended up with that but i though I made a mistake somewhere because they mentioned a fraction

UnknownEntity

What did you guys get for Q9ci?

UnknownEntity

And 9dii

lundrum

UnknownEntity wrote: »

And 9dii

I got h = 17 for c (i) and for d(ii) I got like 430000 or something like that

UnknownEntity

lundrum wrote:

I got h = 17 for c (i) and for d(ii) I got like 430000 or something like that

Beautiful

greenwhale

lundrum wrote: »

I got h = 17 for c (i) and for d(ii) I got like 430000 or something like that

Can't remember exactly but I think those numbers look familiar

Drumm...

For d I got the formula as 3^n/2^(n-1)
Stick in 35 and you get a crazy big answer

Drumm...

For c I got 17 too

lundrum

Did either of you get something like 2115 for Q 4 a (i) [the one about the difference of the sums of the first two rows]

UnknownEntity

lundrum wrote:

Did either of you get something like 2115 for Q 4 a (i) [the one about the difference of the sums of the first two rows]

I think so

Patricklouth

She would you reckon any type or marking scheme be available?Also what did yous make of it what she yous thinking you got grade wise

Celtron

Anyone else think Q9 was a nice question? I'm 99.9% sure I got it all out right, ran out of time for the very last bit, not too great for the wordy type questions so left the area one blank and said that the perimeter is increasing according to the formula indefinitely as n approaches infinity? And just on a side note to anyone reading this, don't get bogged down if you didn't get parts of a question out , they mark the exam on the bell curve and don't finalise the marking scheme until they see where people found it tough so for an example in a 25 mark question if the majority found it tough and not many got it out then they would only give it 5 marks out of the 25 ( tongue in cheek example)

kanadams123

For question 2 part C which value of x was supposed to be used to find sum to infinity? Was it x=1 or the answer from part b (i got x=8)?

I found the both of them but im pretty sure it said only 1?

UnknownEntity

kanadams123 wrote:

For question 2 part C which value of x was supposed to be used to find sum to infinity? Was it x=1 or the answer from part b (i got x=8)?

How did you find a sum to infinity using x=1? Correct me if I'm wrong but I thought that the geometric ratio for x=1 was greater than 1.

It should've only been possible to find a sum to infinity using x=8 because that had a ratio less than 1.

UnknownEntity

And btw I for the ratio I mean -1<r<1, for a sum to infinity

spurious

Patricklouth wrote: »

She would you reckon any type or marking scheme be available?Also what did yous make of it what she yous thinking you got grade wise

The official final marking scheme will be online before the date of viewing of scripts.
It can't go up now as during the examining period it may be tweaked a bit.

kanadams123

UnknownEntity wrote: »

How did you find a sum to infinity using x=1? Correct me if I'm wrong but I thought that the geometric ratio for x=1 was greater than 1.

It should've only been possible to find a sum to infinity using x=8 because that had a ratio less than 1.

Tbh i am after forgettiing what i did now! I remember my final answer was 8 and whatever sum to infinity i got for that! I do remember choosing that answer over something else though!
I didnt have time to finish the last question so couldnt look back over my answers so its hard to remember exactly what i did 😂 lol

Patricklouth

What answer didn't you say get for the some of infinity was it 128 or something like that

UnknownEntity

Patricklouth wrote:

What answer didn't you say get for the some of infinity was it 128 or something like that

Think so

japester

I'm working my way through paper 1 at the moment and I'll try to post up my solutions to it before the night is out for anyone who may be interested

japester

My solutions to the first 3 questions below, more to follow - any errors on my part please let me know:







Q1 (a) x=2, y =-1, z=5
(b) -9<=x<-2


Q2 (a) common ratio r = t2/t1 = t3/t2
so (5x-8)/(x*x) = (x+8)/(5x-8)
keep working on the algebra from there to get the equation desired


(b) Subsitute 1 for x in f(x) to prove f(1) = 0
As f(1) = 0 then x-1 must be a factor of the cubic equation
So we can divide the cubic by x-1 to get a resulting quadratic
The division results in the quadratic x*x - 16x + 64



Now the other roots can be found by observation (or the well known formula -b +- sqrt(b*b - 4ac)/2a )


By observation, x*x - 16x + 64 is (x-8)(x-8) so the other roots are identical, each with the value 8


(c) common ratio = (5x - 8)/(x*x)


If x = 1 then common ratio is -3, if x = 8 common ratio is 0.5


For a finite sum to infinity, the common ratio must always be some
value between -1 and +1 so x must be 8 in this case to guarantee
a finite sum to infinity


The value of the sum to infinity = a/(1-r) = (x*x)/(1-0.5) = 64/0.5

= 128


Q3 (a) h(x) = Cos(2x)
Differentiating this gives h'(x) = -2Sin(2x)
When x = pi/3, h'(x) = -sqrt(3)
Therefore angle made by tangent with positive sense of x-axis is
120 degrees


(b) average value of h(x) over the interval given =



1/[pi/4 - 0] * Integral of Cos(2x) over the limits 0 to pi/4


= 4/pi * [Sin(pi/2)/2 - Sin(0)/2] = 2/pi

lundrum

My answer to 3 (i) was 60 degrees, do you think I'd lose many marks? Seems quite ambiguous as to which angle is with the x axis

japester

My solutions to Q4, Q5 and Q6 follow:


Q4 (a) A proof by mathematical induction:


Show P(1) is true, so in this case



(CosQ + iSinQ)^1 = CosQ + iSinQ


and Cos(1.Q) + iSin(1.Q) = CosQ + iSinQ


so we have proven the relationship holds for P(1)


Next we assume it holds for P(k) i.e. that


(CosQ + iSinQ)^k = Cos(k.Q) + iSin(k.Q)


From there we now try to prove that the relationship holds for P(k+1)


i.e. to show that (CosQ + iSinQ)^(k+1) = Cos([k+1]Q) + iSin([k+1]Q)


Now (CosQ + iSinQ)^(k+1) can be written as


(CosQ + iSinQ)^k . (CosQ + iSinQ)


and we have assumed from P(k) that

(CosQ + iSinQ)^k = Cos(k.Q) + iSin(k.Q)


so (CosQ + iSinQ)^(k+1) can be written as


[Cos(k.Q) + iSin(k.Q)].(CosQ + iSinQ)


= Cos(k.Q)CosQ + iSin(k.Q)CosQ + iSinQCos(k.Q) + i*iSin(k.Q)SinQ


= Cos(k.Q)CosQ -Sin(k.Q)SinQ + i[Sin(k.Q)CosQ + SinQCos(k.Q)]


Noting that Cos(A+B) = CosACosB - SinASinB and

Sin(A+B) = SinACosB + CosASinB


= Cos(k.Q + Q) + i[Sin(k.Q + Q)]


= Cos([k+1]Q) + iSin([k+1]Q)


which is what we were trying to prove by induction


so P(k+1) has been proven and so the relationship must hold for all
values of k


(b) Here CosQ = -0.5 and SinQ = sqrt(3)/2 and so the angle Q must be 120 degrees


so the value given must be equivalent to (Cos120 + iSin120)^3


We know from (a) that this is equivalent to Cos(3*120) + iSin(3*120) =

Cos360 + iSin360 = 1 + i.0 = 1


Q5
(a)


(i)
In 1st row, first term is 4 and common difference is 3 so

Sum of first 45 terms in 1st row = S45 = (45/2)[2(4) + (45-1)(3)] = 3150


In 2nd row, first term is 7 and common difference is 5 so
Sum of first 45 terms in 2nd row = S45 = (45/2)[2(7) + (45-1)(5)] = 5265


So difference between sums of first 45 terms in 1st and 2nd rows = 2115


(ii) In 60th row, 1st term is T60 of an arithmetic progression where first term is 4 and common difference is 3 so 1st term in row 60 must be 4 + (60-1)(3) = 181


In 60th row, 2nd term is T60 of an AP where first term is 7 and common difference is 5 so 2nd term of row 60 must be 7 + (60-1)(5) = 302


So in row 60 1st term is 181 and common difference is (302-181) = 121 and so 70th term in row 60 = 181 + (70-1)(121) = 8530


(b)


a3 = a2-a1 = -2
a4 = a3-a2 = -4
a5 = a4-a3 = -2
a6 = a5-a4 = 2
a7 = a6-a5 = 4
a8 = a7-a6 = 2


So it is clear that the pattern repeats after every 6 terms of the sequence i.e. a1 = a7 = a13 = a19 etc and note that 7 divided by 6 = 1 remainder 1, 13 divided by 6 = 2 remainder 1 etc



So a2019 can be found by dividing by 2019 by 6 and looking at the remainder upon division. 2019/6 = 336 remainder 3. So a2019 = a3 = -2


Q6


(a)


where the functions intersect h(x) = k(x) so x = x^3


x = 0 satisfies this equation as 0 = 0^3 (graph confirms this also)

x = 1 satisfies this equation as 1 = 1^3
x = -1 satisfies this equation as -1 = (-1)^3


Using h(x) = x, this means the coordinates of the 3 points of intersection are (0,0) , (1,1) and (-1,-1)


(b)



(i) area enclosed by the graphs of the 2 functions =



2 * [area of triangle made by h(x) from (0,0) to (1,1) to (1,0) minus the area under k(x) from (0,0) to (1,1)]


= 2 * [0.5*1*1 - Integral of x^3 over the limits 0 to 1]
= 2 * [0.5 - (x^4)/4 over the limits 0 to 1] = 0.5 square units


(ii)


k(x) = x^3 so function k maps x to x^3


therefore function inv(k) maps x^3 to x i.e. function inv(k) maps x to x^(1/3) i.e. cbrt(x)


plug in values into calculator for x = 0, x = 0.2, x = 0.4 etc to get corresponing values for inv(k) and then draw on graph provided. The graph I get looks like k(x) put through an axial symmetry where h(x) is the axis.

japester

lundrum wrote: »

My answer to 3 (i) was 60 degrees, do you think I'd lose many marks? Seems quite ambiguous as to which angle is with the x axis

The tangent value of -sqrt(3) results in an angle of -60 degrees when you plug the number into the calculator. So you'd say that this makes an angle of 60 degrees with the negative sense of the x-axis (Tracing the line through the origin you'd see that the line cuts through the 2nd quadrant and splits it into 2 parts, one making an angle of 60 degrees with the negative part of the x-axis and the other angle will be the combination of 30 degree angle made by the line with the positive y-axis plus the 90 degrees separating the positive x and y axes. So you'd say that the angle made with the positive sense of the x-axis in this case is 30+90 = 120 degrees but I'm certain the penalty involved for this small mistake will be very little (depending on how that overall marks pan out nationwide, the penalty could end up being even smaller) so don't fret at all