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may 2015 Math HL paper 1

Math HL Paper 1  

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  1. 1. How hard did you find the Math HL Paper 1?

    • Super easy
    • Easier than expected, but challenging
    • Pretty hard!
    • I'm going to fail :(


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It was a disaster! Holy broccoli it was crazy.

1. Time was Extremely limited. I needed a solid hour to attempt all questions to the extent of my knowledge. 
2. The questions weren't easy either! :( 

I probably left around 50-55 points untouched out of the 120.. assuming I get everything else correct (which I probably wont) it was still extremely difficult! 

Tell me how you guys did! Hopefully It didn't go as badly for you! 

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Personally, I found it pretty challenging.. Some questions had me stumped but I'm hoping I'll get a 6 but I'm sure to lose a lot of marks in Section B   :(

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I don't know what happened to me, God must have taken over my brain for the paper, because for the first time ever, I was able to do all of Section B!! So I think I got about 97 marks in total... 

 

I also did about 6 full practice exams beforehand.

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It was a disaster! Holy broccoli it was crazy.

1. Time was Extremely limited. I needed a solid hour to attempt all questions to the extent of my knowledge. 

2. The questions weren't easy either! :(

I probably left around 50-55 points untouched out of the 120.. assuming I get everything else correct (which I probably wont) it was still extremely difficult! 

Tell me how you guys did! Hopefully It didn't go as badly for you! 

 

I feel the same exact way! I hope I got everything I did answer right, because I had at least 40-50 points off already from not getting to the problems...... Way harder (and definitely did worse) than any past paper 1 test I took for practice! Here's to hoping the other papers make up for it...  :truce:

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I thought it was much easier than the level I had expected based on the mock exams we gave. I knew how to attempt the questions and got a lot of the answers. I'm pretty confident that I'll get 70+ which would be a dream come true anyway :) definitely not what I expected haha (obviously, there were challenging questions including some I couldn't complete but the day I completely understand the whole paper is never coming so I was happy). I'm just glad it's over :) 

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Question to those of you in TZ2.. How did you manage to solve the polynomial with solutions in arithmetic sequences in Section B? I just kept getting a contradiction, Anyone else encounter this problem? I can post the solution and show that it is an invalid condition.

 

 

 

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That freaking tanx integral man! I spent a good 10 minutes trying to solve it, then left it with just finding dx in terms of du lol. I also had problems with the complex number which I know is supposed to not be hard since its similar to cube roots of unity.. But man.. It completely went out of my head haha..

Other than that, I thought the exam was doable and the first 5 or six questions were pretty nice. Section B was cool, since they didn't throw in any terrible vector questions lol. 

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I did TZ2 and  I left about 10 points untouched, overall I think I lost about 30. If I had a little more time, I could easily get 8 more marks. Generally, this paper could be considered friendly, if it wasn't for the 2 hours. At one point I totally panicked and in such condition it is difficult to arrive at a correct answer or even at an idea on how to solve a problem. However, I still don't lose my hopes for a 7, if the grade boundaries do not go up and provided that I did fine in other questions.

Edited by ordinaryguy

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Question to those of you in TZ2.. How did you manage to solve the polynomial with solutions in arithmetic sequences in Section B? I just kept getting a contradiction, Anyone else encounter this problem? I can post the solution and show that it is an invalid condition.

OMG Really!? I wasted like so much time on that question! And after that i just gave up and moved on to the last part...

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You're not alone.

Here is my solution:

 

Solution:

 
As i remember:
solutions a,b,c form an arithmetic sequence and
a + b + c = 6
ab + ac + bc = 18
abc = -6
 
so if we let b = u:
a = u-d
b = u
c = u+d
,where d is the common difference.
 
From plugging in to the first equation we get that
 
(u-d) + u + (u+d) = 6
3u = 6
u = 2 (this is confirmed in the condition, which asked to show that u=2)
 
For the common difference we plug into the second equation:
 
(2-d)*2 + (2-d)(2+d) + 2(2+d) = 18
4 - 2d + 4 - d^2 + 4 + 2d = 18
-d^2 = 18 - 12
d^2 = -6
 
which means the common difference is complex.
However, if we plug it in into the third equation:
 
(2-d)*2*(2+d) = -6
 
2*(4 - d^2) = -6
4 - d^2 = -3
d^2 = 7
then d = sqrt(7)
Edited by Whateverthis
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You're not alone.

Here is my solution:

 

Solution:

 
As i remember:
solutions a,b,c form an arithmetic sequence and
a + b + c = 6
ab + ac + bc = 18
abc = -6
 
so if we let b = u:
a = u-d
b = u
c = u+d
,where d is the common difference.
 
From plugging in to the first equation we get that
 
(u-d) + u + (u+d) = 6
3u = 6
u = 2 (this is confirmed in the condition, which asked to show that u=2)
 
For the common difference we plug into the second equation:
 
(2-d)*2 + (2-d)(2+d) + 2(2+d) = 18
4 - 2d + 4 - d^2 + 4 + 2d = 18
-d^2 = 18 - 12
d^2 = -6
 
which means the common difference is complex.
However, if we plug it in into the third equation:
 
(2-d)*2*(2+d) = -6
 
2*(4 - d^2) = -6
4 - d^2 = -3
d^2 = 7
then d = sqrt(7)

 

 

Um, they asked you to find abc in the next question, it wasn't given. Only p and q were (6 and 18). I got an answer for that with no contradictions.

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a lot of students who take physics would find it very difficult like my colleague because they didn't manage to study maths with physics, but I did, before physics I studied maths

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You're not alone.

Here is my solution:

 

Solution:

 
As i remember:
solutions a,b,c form an arithmetic sequence and
a + b + c = 6
ab + ac + bc = 18
abc = -6
 
so if we let b = u:
a = u-d
b = u
c = u+d
,where d is the common difference.
 
From plugging in to the first equation we get that
 
(u-d) + u + (u+d) = 6
3u = 6
u = 2 (this is confirmed in the condition, which asked to show that u=2)
 
For the common difference we plug into the second equation:
 
(2-d)*2 + (2-d)(2+d) + 2(2+d) = 18
4 - 2d + 4 - d^2 + 4 + 2d = 18
-d^2 = 18 - 12
d^2 = -6
 
which means the common difference is complex.
However, if we plug it in into the third equation:
 
(2-d)*2*(2+d) = -6
 
2*(4 - d^2) = -6
4 - d^2 = -3
d^2 = 7
then d = sqrt(7)

 

 

Um, they asked you to find abc in the next question, it wasn't given. Only p and q were (6 and 18). I got an answer for that with no contradictions.

 

 

Well if you use the third equation, then you can get the answer. but try using the second one, with p = 18

Edited by Whateverthis

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Okay, I think this was a P1 question. The last one on Section A about logarithms - part a) asked about the range of base values. Was the question for real numbers or complex ones?

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There was also the integral of: (sin x + cos x)/ (sin x- cos x) i got ln(sin x- cos x) is that right? used substitution. 

and for the last log question in section a what was the condition for the logarithmic function? 

Edited by mikigalli97

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There was also the integral of: (sin x + cos x)/ (sin x- cos x) i got ln(sin x- cos x) is that right? used substitution. 

and for the last log question in section a what was the condition for the logarithmic function?

A not 0, not 1, not negative

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TZ1
What was that probability question. 5 games, if lose 3 consecutively the coach is out. prob (win for a specific game) is 1/2 prob (draw) is 1/6, prob (lose) is 1/3. Results of one game independent of the others Find P (coach says bye bye)

I didn't get get it in the exam but after paper 2 i tried to solve it i got 21/243 (or 7/81)

There were like 20 answers among my grade

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