HL Maths 2018 discussion

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?

Thr000w

PatsyJ wrote: »

Anyone?

Multiply both sides by (the denominator (bottom part of the fraction) that contains x). In this case, you multiply both sides by (x + 2) squared. Because it's squared, it has to be positive, so the inequality signs don't flip like they would if you multiplied by a negative number. You can't cross multiply, because you don't know if the denominator containing x is negative or not, and if it is then you'd have to flip the inequality sign.

Once you solve the ensuing quadratic, the answer is -9 ≤ x ≤ -2 - but the question tells us that x can't equal 2, as the denominator would then be zero and make the fraction undefined, so it's -9 ≤ x < 2.

japester

PatsyJ wrote: »

Anyone?

I would have looked at it this way:


(2x-3)/(x+2) >= 3


so either



1. 2x-3 >= 3(x+2), but this is possible only provided both (2x-3) and (x+2) are simultaneously >=0


or


2. 2x-3 <= 3(x+2), but this is possible only provided both (2x-3) and (x+2) are simultaneously <=0


Examining possibility 1 further and solving the inequality will lead to x <= -9
However (2x-3) and (x+2) must also be simultaneously >= 0 in order for this possibility to work out. If (2x-3) >=0 then x >= 3/2 and if (x+2) >=0 then x >= -2. In order for x to be both >= 3/2 and x >= -2 simultaneously, the only possibility is that x >= 3/2. However we already found that the inequality solution is that x<= -9 and as there is no overlap between the sets x >= 3/2 and x<= -9 then there are no possible solutions for x in this case.


Examining possibility 2 further and solving the inequality will lead to x >= -9
However (2x-3) and (x+2) must also be simultaneously <= 0 in order for this possibility to work out. If (2x-3) <=0 then x <= 3/2 and if (x+2) <=0 then x <= -2. In order for x to be both <= 3/2 and x <= -2 simultaneously, the only possibility is that x <= -2. However we already found that the inequality solution is that x>= -9 and as there is overlap between these sets then there we have possible solutions for x in this case. So the possible solution set is -9 <= x <= -2 but the question tells us x cannot be -2 (as division by zero would ensue in this case) and finally we have the possible solution set as -9 <= x < -2


hope this helps

jeonahr

Thr000w wrote: »

Multiply both sides by (the denominator (bottom part of the fraction) that contains x). In this case, you multiply both sides by (x + 2) squared. Because it's squared, it has to be positive, so the inequality signs don't flip like they would if you multiplied by a negative number. You can't cross multiply, because you don't know if the denominator containing x is negative or not, and if it is then you'd have to flip the inequality sign.

Once you solve the ensuing quadratic, the answer is -9 ≤ x ≤ -2 - but the question tells us that x can't equal 2, as the denominator would then be zero and make the fraction undefined, so it's -9 ≤ x < 2.

I was just doing the papers there (it's been assigned to my 5th year class, now 6th year :rolleyes: ) and I totally missed the fact that x can't be equal to 2.

Boomer18

Can anybody please tell me what constitutes a pass in Foundation Maths with the new marking scheme?

ger664

Q5 Paper 2
(a)
sub in (-2,1) into 2x+3y+1=0 and verify its true
(b)
Line AB is perpendicular. Slope m = -2/3 slope AB = 3/2
Equ of AB 3x-2y+8 = 0
Point B got from sim equ with n 2x+3y-51 = 0 B = (6,13)
(c)
Radius ratio is 1:3 so is the Diameter thus the |AB| = Radius of small cicle * 8
Length AB = 4sqrt13 radius = sqrt13/2
Center lies on the point thats splits the line AB into a ratio of 1:7
center = (-1,5/2)
Equation of circle (x+1)^2 + (y-5/2)^2 = 13/4
Expanded out to x^2 + y^2 +2x -5y + 4 =0

Graph of Question

japester

ger664 wrote: »

Q5 Paper 2
(a)
sub in (-2,1) into 2x+3y+1=0 and verify its true
(b)
Line AB is perpendicular. Slope m = -2/3 slope AB = 3/2
Equ of AB 3x-2y+8 = 0
Point B got from sim equ with n 2x+3y-51 = 0 B = (6,13)
(c)
Radius ratio is 1:3 so is the Diameter thus the |AB| = Radius of small cicle * 8
Length AB = 4sqrt13 radius = sqrt13/2
Center lies on the point thats splits the line AB into a ratio of 1:7
center = (-1,5/2)
Equation of circle (x+1)^2 + (y-5/2)^2 = 13/4
Expanded out to x^2 + y^2 +2x -5y + 4 =0

Graph of Question

Many thanks for that Ger, I can confirm I got the same results as you for it