HL Maths 2018 discussion

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

DaleyBlind

japester wrote: »

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

Would I lose marks if I gave my answer as 2pi/3, without writing "radians" afterwards. Not too worried about it but just wondering, and guessing I wouldn't lose many for leaving it out.

japester

DaleyBlind wrote: »

Would I lose marks if I gave my answer as 2pi/3, without writing "radians" afterwards. Not too worried about it but just wondering, and guessing I wouldn't lose many for leaving it out.

Ah no, your answer is still correct and the question didn't specify whether the angular value was to be in degrees or radians either. You could possibly lose a tiny amount for omission of the units in this case but again, depending on how the marks pan out across the board, there may be no penalty at all - generally if an angular value refers to pi, as it does in your case, it is assumed that it is measured in radians anyway.

japester

My solutions to Q7, more to follow:



(a) t(x) = 35.96 = k.ln(1-x/80) = k.ln(1-35/80) = -0.575364
which means k = -62.4995 = -62.5 to 1 d.p.


(b) t(x) = 100 = -62.5ln(1-x/80)


so -1.6 = ln(1-x/80) so e^-1.6 = 1-(x/80)
so x = 63.848
so number of wpm Jack can learn in 100 days of practice is 64 to nearest whole number


(c) plug in the values to calculator:



x = 0 => t(x) = 0
x = 10 => t(x) = 8.3457
x = 20 => t(x) = 17.9801


etc.


Use values to plot t(x) on exam script. This is a log-based function that cuts through the origin



(d)


can do these calculations in the head:


x= 0, p(x) = 0
x = 10, p(x) = 15
x = 20, p(x) = 30


etc


Use values to plot p(x) on exam script. This is a linear function that cuts through the origin


(e)


(i) h(x) = p(x)-t(x) so where h(x) = 0 then p(x) = t(x) so we are looking for the points of intersection of p(x) and t(x). They clearly intersect anyway where x=0 and, from my own estimation based on the graphs, they also intersect around x = 65.


(ii) h(x) = 1.5x +62.5ln(1-x/80)


so h'(x) = 1.5 + 62.5[1/(1-{x/80}) . (-1/80)] = 1.5 - 0.78125(80/[80-x])


max value of h(x) occurs when h'(x) = 0


so 1.5 - 0.78125(80/[80-x]) = 0 => x = 38.333


so max value of h(x) occurs where x = 38.333


at this point h(x) = 1.5(38.333) +62.5[ln(1-{38.333/80})] = 16.73


so max value of h(x) = 17 to nearest whole number

japester

My solutions to Q8 and Q9:


Q8


(a) where graph intersects y-axis, x=0. At this point f(x) = f(0) = 1/(sqrt[2.pi]) so coordinates of A are (0,1/(sqrt[2.pi]))


(b) area of shaded rectangle = 2/(sqrt[2.pi.e]) = 0.484 to 3 d.p


(c) f(x) = [1/sqrt(2.pi)].e^(-0.5x*x)


so f'(x) = [1/sqrt(2.pi)].e^(-0.5x*x).(-x)


At point C, x=1



so f'(x) at point C = -0.24197


Now as the rate of change is negative at this point we can deduce that the function f(x) must be decreasing at C (which is in correspondance with what we can see visually from the graph shown)


(d) At point B, x = -1


we know f'(x) = [1/sqrt(2.pi)].e^(-0.5x*x).(-x) and so
f''(x) is therefore e^(-0.5x*x)/[sqrt(2.pi)] * (x*x - 1)


therefore, with x = -1 it turns out that at point B, f''(x) is zero and so
point B must be a point of inflection (again this is consistent with our visual observations of the graph where we can see that the max value of tangent to the graph of f(x) would occur at point B i.e. rate of change f'(x) is at its maximum here


Q9


(a) Complete table by observation just looking at the diagrams:



0 1 2 3



1 3 9 27



1 3/4 9/16 27/64


(b)



(i) # black traingles in step n will be 3^n (deduce by observing the numbers produced in the table above)


(ii) here 10^9 < 3^k


so log10^9 < log3^k
so 9 < klog3
so k > 18.863


so # black triangles first exceeds 10^9 when k = 19


(c)


(i) fraction of original triangle remaining at step n = 3^n/[2^(2n)] by observing the numbers in the table earlier


so here 3^n/[2^(2n)] < 1/100
so [2^(2n)]/(3^n) > 100
so 2^(2n) > 100.(3^n)
so log(2^(2n)) > log(100.(3^n))
so 2nlog2 > log100 + log(3^n)
so 0.602n > 2 + 0.477n
so 0.1248n > 2
so n > 16


therefore, when h = 17, the fraction of the original triangle remaining will be < 1/100 of the original triangle


(ii) The fraction remaining is a geometric progression where a = 1 and r = 3/4


so the general term of the progression, tn, will be a.r^(n-1) = 1.(0.75)^(n-1) = 0.75^(n-1)


after an infinite number of steps, n = infinity and so tn = 0 at this point so
we conclude that the fraction of the original triangle remaining after an infinite number of steps of the pattern is zero


(d)


(i) table completed by observation of the diagrams earlier:



0 1 2 3 4



3 9/2 27/4 81/8 243/16



(ii)


total perimeter in step n = 3^(n+1)/[2^n] from observation of the numbers in the table above


so in step 35, the total perimeter must be 4368328.818 = 4368329 to nearest unit


(iii)


As n tends to infinity we can see that the total area of the black triangles tends to zero, this is confirmed by our result in part (c) (ii) and is what we expect by observation since the "white area" is getting larger and larger


As n tends to infinity we can see that the total perimeter of the black triangles is getting larger and larger and tends to infinity. The result we got from part(d) (ii) confirms that the total perimeter gets large very quickly, even for a relatively small step value of 35, so we can deduce that as n tends to infinity, this total perimeter value will also tend to infinity (this is also clear if we consider the expression 3^(n+1)/[2^n]and see that as n increases 3^(n+1) will increase more rapidly that 2^n so that the value of the expression, as n tends to infinity, will be infinity.

lundrum

What did you think of that paper 2? I thought it was much worse than p1

PatsyJ

How do you do Q1 part b) on Paper 1, the question about solving the inequality?

DavidAdam

All the articles in the national newspapers were very positive about HL maths paper 1. The impression given was that teachers found the paper much easier and doable than in previous years. But I have yet to talk to any student from my school or any other schools that found this paper anyway easy. The all found it very difficult and many were in utter despair and this include the potential H1's.
We found paper 2 today somewhat easier.

tuisginideach

The opposite in my house - Paper 1 doable, thrown by paper 2.

Cork Lass

My daughter thought paper 1 was fine but thought paper 2 was “horrific”. Says everyone in her school felt the same.

PatsyJ

In our school some came out of paper 1 in tears and other came out of paper 2 in tears.

Not very nice.

PatsyJ wrote: »

How do you do Q1 part b) on Paper 1, the question about solving the inequality?

Anyone?