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How was IAL C34?

This poll is closed

How did you find the IAL C34 June 2016 paper?

Very hard 19%
Hard 37%
Okay 30%
Easy 7%
Very easy7%
Total votes: 27
Any boundary predictions or answers?

Here are the few I remember - will add to the OP as people post. :smile:

Area of the triangle = 181118\sqrt{11}

Time: t = 11.63

Inequality: 2(a−b)3<x<2(a+b)\frac{2(a-b)}{3}<x<2(a+b)

Upper limit = 3000, k = 0.614, nearest integer for dP/dt = 9

Spoiler

Volume of revolution = 17π10\frac{17\pi}{10}

Binomial expansion: a = - 1/4, coefficient of x³ = 23/16

Trapezium rule: approximation ≈ 0.875, exact = -3 + √15, magnitude of error = 2 × 10^-3

Model answers posted here.
(edited 7 years ago)

Scroll to see replies

Da heck happened with LaTeX.

I'd say it was a lengthy paper, but it was definitely do-able.

I personally struggled with time, which I usually don't, probably because I was being extra cautious not to trip up on simple algebra mistakes.
(edited 7 years ago)
RIP LaTeX

Not bad but it was longer than usual. I think the first time I've ever appreciated having 2.5 hrs for a paper.
Volume of revolution = 17pi/10
how did u solve the integration by parts question ??
Original post by Mohammad_besani
how did u solve the integration by parts question ??


Solution to Q6:

Spoiler

Here is a sketch of graph oh y=arctan(x).
image063.gif2.3KB
Original post by AKucukuncular
Here is a sketch of graph oh y=arctan(x).


1 mark for Shape and 1 mark for Asymptotes, I reckon.
In the binomial expansion: I don't remember x^2 / x^3 coefficients but, a = -1/4 yeah?

You had to solve a quadratic as far I remember, to get two values of a and you had to reject the value greater then zero.
Reply 9
Avatar for tgygt
tgygt
2
Do you think the grade boundary is gonna be like that of 2015 June or January?
January series grade boundaries are horrible
Reply 10
Avatar for tgygt
tgygt
2
Original post by SaadKaleem
In the binomial expansion: I don't remember x^2 / x^3 coefficients but, a = -1/4 yeah?

You had to solve a quadratic as far I remember, to get two values of a and you had to reject the value greater then zero.


Yes
Original post by SaadKaleem
1 mark for Shape and 1 mark for Asymptotes, I reckon.


Yes :smile:
Reply 12
Avatar for Ayman!
Ayman!
OP
15
Original post by tgygt
Do you think the grade boundary is gonna be like that of 2015 June or January?
January series grade boundaries are horrible


Well, it is a June paper so I see it being similar to June 2015 and 2014.

Original post by SaadKaleem
In the binomial expansion: I don't remember x^2 / x^3 coefficients but, a = -1/4 yeah?

You had to solve a quadratic as far I remember, to get two values of a and you had to reject the value greater then zero.


Correct, a = -1/4. I think the coefficient of x³ = 23/16.
How about the double long division? Oh, I had a blast while solving that question. You first had to divide the x^4....... by (x+4), then divide the resulting x^3......... by (x-3) to obtain the values of A and B.

[ (x+4) and (x-3) from the quadratic simplification of x^2+x-12 ]
Original post by Ayman!

Correct, a = -1/4. I think the coefficient of x³ = 23/16.


Exactly, that fraction for x^3 looks familiar! :tongue:

Original post by tgygt
Do you think the grade boundary is gonna be like that of 2015 June or January?January series grade boundaries are horrible
Probably will be like the previous June papers (I hope so :dontknow:)
(edited 7 years ago)
Original post by Ayman!
Any boundary predictions or answers?

Here are the few I remember - will add to the OP as people post. :smile:

Area of the triangle: 18sqrt(11)

Time: t = 11.63

Inequality: 2/3 x (a - b) < x < 2(a + b)

∫x(2^x) dx = (8ln2 - 3)/(ln2)^2

Upper limit = 3000, k = 0.614, nearest integer for dP/dt = 9

Volume of revolution = 17pi/10

Binomial expansion: a = - 1/4, coefficient of x³ = 23/16

EDIT: RIP LaTeX. :frown:


OH MY GODDDDD that integration by parts one!!!! :frown: i got the answer you got, first then it seemed a bit fishy because there was ln^2 at the bottom idk what clicked in my brain but in a sudden moment of stupidity i re-did it as u=2^x and dv/dx= x :frown:

do you think I'll get a mark for this atleast? God I really feel like killing myself right now!!
Original post by TheMadHatteress
OH MY GODDDDD that integration by parts one!!!! :frown: i got the answer you got, first then it seemed a bit fishy because there was ln^2 at the bottom idk what clicked in my brain but in a sudden moment of stupidity i re-did it as u=2^x and dv/dx= x :frown:

do you think I'll get a mark for this atleast? God I really feel like killing myself right now!!


That would have resulted in both x2 and 2x/ln(2) in the second integration. How did you manage to do the second integral?
Original post by TheMadHatteress
OH MY GODDDDD that integration by parts one!!!! :frown: i got the answer you got, first then it seemed a bit fishy because there was ln^2 at the bottom idk what clicked in my brain but in a sudden moment of stupidity i re-did it as u=2^x and dv/dx= x :frown:

do you think I'll get a mark for this atleast? God I really feel like killing myself right now!!


but if you put u = 2^x, wouldn't it be infinite integration by parts?
Reply 18
Avatar for Yua
Yua
14
anyone have the paper/ms?
Original post by jackien1
but if you put u = 2^x, wouldn't it be infinite integration by parts?


As i said sudden moment of sheer stupidity. Did that in the last minue whats wrong with me :frown: lost a good, easy six marks

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