MAY13 Maths HL TZ2 Paper 2 Discussion

[Giveme45]

Im assuming we're allowed to discuss, when the moderator allows this thread.

Let's discuss paper 2.

I found the paper quite nice tbh, I did really well even though some questions were difficult. I might even score a full 120 if I'm lucky!

Though, I have two confusions, which bother me quite a lot.

For the question 10, finding the velocity, it was given that acceleration was (60-v)/40, so integrating both sides in respect to t gave the expression:

-ln|60-v|=t/40+c

Now we knew that initially the car was at rest so at t=0, v=0 giving c=-ln60.

We have: -ln|60-v|=t/40-ln60

Now at t=30 we have

-ln|60-v|=3/4-ln60

giving

|60-v|=e^(ln60-3/4)

This gave two results v=31.7 and v=88.3. The two results are there because of absolute value sign. I had no idea which one of the results too chose from the information given, and i assumed that the question was bad and they forgot about the absolute value sign so i went with 31.7 Can someone explain?

For question 12 I think it was, we had something like ydy/dx=cos2x or something like that and when we had to find the general function in form of y=f(x), i was quite baffled, because integrating gave: y²=sin2x+c, so i didnt know how to decide whether the function should be y=sqrt(sin2x+c) or y=-sqrt(sin2x+c). This is quite annoying me, and once again it seemed like not information was given?

I hope I won't lose a lot of marks because of these.

[Christina perera]

In paper 2 section B what was the probability answer? what was the meaning of "1 experiment" was n=1 or 16. Also can someone tell me the 12f, the question on volume on revolution in section B

[Rohan Madhwal]

Im assuming we're allowed to discuss, when the moderator allows this thread.

Let's discuss paper 2.

I found the paper quite nice tbh, I did really well even though some questions were difficult. I might even score a full 120 if I'm lucky!

Though, I have two confusions, which bother me quite a lot.

For the question 10, finding the velocity, it was given that acceleration was (60-v)/40, so integrating both sides in respect to t gave the expression:

-ln|60-v|=t/40+c

Now we knew that initially the car was at rest so at t=0, v=0 giving c=-ln60.

We have: -ln|60-v|=t/40-ln60

Now at t=30 we have

-ln|60-v|=3/4-ln60

giving

|60-v|=e^(ln60-3/4)

This gave two results v=31.7 and v=88.3. The two results are there because of absolute value sign. I had no idea which one of the results too chose from the information given, and i assumed that the question was bad and they forgot about the absolute value sign so i went with 31.7 Can someone explain?

For question 12 I think it was, we had something like ydy/dx=cos2x or something like that and when we had to find the general function in form of y=f(x), i was quite baffled, because integrating gave: y²=sin2x+c, so i didnt know how to decide whether the function should be y=sqrt(sin2x+c) or y=-sqrt(sin2x+c). This is quite annoying me, and once again it seemed like not information was given?

I hope I won't lose a lot of marks because of these.

Shouldn't the integration of cos2x give (1/2)sin(2x) + c ? And I think IB wanted us to just leave it in the square root form, because when you think about it, y = underroot x has both the positive and negative values in it's range. So when you write sqrt(1/2 sin2x + C), you're not limiting your range, unless you went and specifically wrote (y>=0) which I'm assuming you didn't. But that's simply my opinion, I could be wrong.

[Giveme45]

Im assuming we're allowed to discuss, when the moderator allows this thread.

Let's discuss paper 2.

I found the paper quite nice tbh, I did really well even though some questions were difficult. I might even score a full 120 if I'm lucky!

Though, I have two confusions, which bother me quite a lot.

For the question 10, finding the velocity, it was given that acceleration was (60-v)/40, so integrating both sides in respect to t gave the expression:

-ln|60-v|=t/40+c

Now we knew that initially the car was at rest so at t=0, v=0 giving c=-ln60.

We have: -ln|60-v|=t/40-ln60

Now at t=30 we have

-ln|60-v|=3/4-ln60

giving

|60-v|=e^(ln60-3/4)

This gave two results v=31.7 and v=88.3. The two results are there because of absolute value sign. I had no idea which one of the results too chose from the information given, and i assumed that the question was bad and they forgot about the absolute value sign so i went with 31.7 Can someone explain?

For question 12 I think it was, we had something like ydy/dx=cos2x or something like that and when we had to find the general function in form of y=f(x), i was quite baffled, because integrating gave: y²=sin2x+c, so i didnt know how to decide whether the function should be y=sqrt(sin2x+c) or y=-sqrt(sin2x+c). This is quite annoying me, and once again it seemed like not information was given?

I hope I won't lose a lot of marks because of these.

Shouldn't the integration of cos2x give (1/2)sin(2x) + c ? And I think IB wanted us to just leave it in the square root form, because when you think about it, y = underroot x has both the positive and negative values in it's range. So when you write sqrt(1/2 sin2x + C), you're not limiting your range, unless you went and specifically wrote (y>=0) which I'm assuming you didn't. But that's simply my opinion, I could be wrong.

Yes thats the integration cos2x, and that ultimateley gives y²=sin2x+c since the integration of y is also y²/2. What do u mean leave it just in the square root form?

y=sqrt(sin2x+c) and y=-sqrt(sin2x+c) are two different equations. But IB stated they only wanted one of them since they wanted us to give the general form y=f(x).

And I'm confused about how we were supposed to know which one of these they wanted.

[Garmr]

In paper 2 section B what was the probability answer? what was the meaning of "1 experiment" was n=1 or 16. Also can someone tell me the 12f, the question on volume on revolution in section B

I can't remember the question but I think it was binomial and I added the two probabilities. I wasn't really sure what it was asking for too, and generally I'm not too confident on the probability questions in this paper.

Edited May 11, 2013 by Garmr

[Rohan Madhwal]

Im assuming we're allowed to discuss, when the moderator allows this thread.

Let's discuss paper 2.

I found the paper quite nice tbh, I did really well even though some questions were difficult. I might even score a full 120 if I'm lucky!

Though, I have two confusions, which bother me quite a lot.

For the question 10, finding the velocity, it was given that acceleration was (60-v)/40, so integrating both sides in respect to t gave the expression:

-ln|60-v|=t/40+c

Now we knew that initially the car was at rest so at t=0, v=0 giving c=-ln60.

We have: -ln|60-v|=t/40-ln60

Now at t=30 we have

-ln|60-v|=3/4-ln60

giving

|60-v|=e^(ln60-3/4)

This gave two results v=31.7 and v=88.3. The two results are there because of absolute value sign. I had no idea which one of the results too chose from the information given, and i assumed that the question was bad and they forgot about the absolute value sign so i went with 31.7 Can someone explain?

For question 12 I think it was, we had something like ydy/dx=cos2x or something like that and when we had to find the general function in form of y=f(x), i was quite baffled, because integrating gave: y²=sin2x+c, so i didnt know how to decide whether the function should be y=sqrt(sin2x+c) or y=-sqrt(sin2x+c). This is quite annoying me, and once again it seemed like not information was given?

I hope I won't lose a lot of marks because of these.

Shouldn't the integration of cos2x give (1/2)sin(2x) + c ? And I think IB wanted us to just leave it in the square root form, because when you think about it, y = underroot x has both the positive and negative values in it's range. So when you write sqrt(1/2 sin2x + C), you're not limiting your range, unless you went and specifically wrote (y>=0) which I'm assuming you didn't. But that's simply my opinion, I could be wrong.

Yes thats the integration cos2x, and that ultimateley gives y²=sin2x+c since the integration of y is also y²/2. What do u mean leave it just in the square root form?

y=sqrt(sin2x+c) and y=-sqrt(sin2x+c) are two different equations. But IB stated they only wanted one of them since they wanted us to give the general form y=f(x).

And I'm confused about how we were supposed to know which one of these they wanted.

What im trying to say is that both the negative and positive values of the function since when you write y = underroot x, both the positive values of negative values of y are included in the range. So in y = underroot 4, you will have both 2 and -2 as the answers

[Christina perera]

In paper 2 section B what was the probability answer? what was the meaning of "1 experiment" was n=1 or 16. Also can someone tell me the 12f, the question on volume on revolution in section B

I can't remember the question but I think it was binomial and I just did P(X=1) + P(X=16). I wasn't really sure what it was asking for too, and generally I'm not too confident on the probability questions in this paper

I thought it was quite weird. I think 1 experiment should be n=1. So p(x=3) is not possible. What did others put?

[Garmr]

In paper 2 section B what was the probability answer? what was the meaning of "1 experiment" was n=1 or 16. Also can someone tell me the 12f, the question on volume on revolution in section B

I can't remember the question but I think it was binomial and I just did P(X=1) + P(X=16). I wasn't really sure what it was asking for too, and generally I'm not too confident on the probability questions in this paper

I thought it was quite weird. I think 1 experiment should be n=1. So p(x=3) is not possible. What did others put?

I thought the same thing as you, for n=1, x=3 gave 0, and x=1 was 0.25, so i just put 0.25, but I had honestly no clue

Oh damn, for that I thought the 16 trials constituted "1 experiment" so I got something like 0.261. You're probably correct.

It didn't say "experiment" anywhere else on the question so I got confused =.=

[Rohan Madhwal]

In paper 2 section B what was the probability answer? what was the meaning of "1 experiment" was n=1 or 16. Also can someone tell me the 12f, the question on volume on revolution in section B

I can't remember the question but I think it was binomial and I just did P(X=1) + P(X=16). I wasn't really sure what it was asking for too, and generally I'm not too confident on the probability questions in this paper

I thought it was quite weird. I think 1 experiment should be n=1. So p(x=3) is not possible. What did others put?

I thought the same thing as you, for n=1, x=3 gave 0, and x=1 was 0.25, so i just put 0.25, but I had honestly no clue

Oh damn, for that I thought the 16 trials constituted "1 experiment" so I got something like 0.261. You're probably correct.

It didn't say "experiment" anywhere else on the question so I got confused =.=

The language on that paper was quite ambiguous if I was very honest. I'm sure that a lot of people will slip up on that question and hence IB might alter the curve accordingly or something. Or at least I hope so haha.

Also for the last question, I managed to get the expression, but was it just me or was the differentiation on that quite tough ? And it was 6 marks, i looked and it tried to fiddle around with it and went "meh" cause I had a few questions left behind which I wanted to try again (Which I managed to do by the end, so it was a good sacrifice) If anyone got it, could you show me how you did it ?

[Christina perera]

In paper 2 section B what was the probability answer? what was the meaning of "1 experiment" was n=1 or 16. Also can someone tell me the 12f, the question on volume on revolution in section B

I can't remember the question but I think it was binomial and I just did P(X=1) + P(X=16). I wasn't really sure what it was asking for too, and generally I'm not too confident on the probability questions in this paper

I thought it was quite weird. I think 1 experiment should be n=1. So p(x=3) is not possible. What did others put?

I thought the same thing as you, for n=1, x=3 gave 0, and x=1 was 0.25, so i just put 0.25, but I had honestly no clue

Oh damn, for that I thought the 16 trials constituted "1 experiment" so I got something like 0.261. You're probably correct.

It didn't say "experiment" anywhere else on the question so I got confused =.=

The language on that paper was quite ambiguous if I was very honest. I'm sure that a lot of people will slip up on that question and hence IB might alter the curve accordingly or something. Or at least I hope so haha.

Also for the last question, I managed to get the expression, but was it just me or was the differentiation on that quite tough ? And it was 6 marks, i looked and it tried to fiddle around with it and went "meh" cause I had a few questions left behind which I wanted to try again (Which I managed to do by the end, so it was a good sacrifice) If anyone got it, could you show me how you did it ?

Even I put it as 0.25. Any idea in 12f the question on volume of revolution was it rotated around y=1?

[Zuler]

For the binomial one, I simply did P (X=1) + P(X=3) with n=16 and p=0.25, why would it be anything else?

Also for the volume, wasn't it simply pi* S f(x)^2 - g(x)^2 dx, where g(x) = 1? So my integral was:

pi * 0->pi/2 S (sin2x + 3) - 1^2 dx (the square over f(x) cancelled the root)

= pi * 0-> pi/2 S (sin2x +2) dx = 13.0 units^3

[Christina perera]

For the binomial one, I simply did P (X=1) + P(X=3) with n=16 and p=0.25, why would it be anything else?

Also for the volume, wasn't it simply pi* S f(x)^2 - g(x)^2 dx, where g(x) = 1? So my integral was:

pi * 0->pi/2 S (sin2x + 3) - 1^2 dx (the square over f(x) cancelled the root)

= pi * 0-> pi/2 S (sin2x +2) dx = 13.0 units^3

For the binomial one, n should be 1 because the question said 1 experiment and 1 experiment wasn't defined as 16 trials.

For the volume one, it shouldn't be what you said as it was rotated around y=1 not y=0 (the x axis)

[Garmr]

For the binomial one, I simply did P (X=1) + P(X=3) with n=16 and p=0.25, why would it be anything else?

That's what I did as well, but others interpreted "one experiment" as n = 1. I guess we'll just have to wait until markschemes come out or something >_>

For the last question, I kind of left the derivation for d0/dx mid-working because I didn't know how to get the numerator didn't know how to differentiate arctan (8/x) and arctan (13/20-x) properly. Also what did you guys get for the very last part? I think my method for d0/dt was correct but I got a weird answer, which suggests an arithmetic error on my part.

[Rohan Madhwal]

For the binomial one, I simply did P (X=1) + P(X=3) with n=16 and p=0.25, why would it be anything else?

Also for the volume, wasn't it simply pi* S f(x)^2 - g(x)^2 dx, where g(x) = 1? So my integral was:

pi * 0->pi/2 S (sin2x + 3) - 1^2 dx (the square over f(x) cancelled the root)

= pi * 0-> pi/2 S (sin2x +2) dx = 13.0 units^3

For the binomial one, in the question, it said one experiment was carried out, hence n=1 ? Or at least that's what I thought. 1 experiment = 1 trial ? And that's what the other person who's commenting thought as well apparently, so I'm not alone, the English was quite ambiguous

[Zuler]

@Christina: I get what you mean about the binomial, I guess we'll have to see Also for the volume, I remember the question said something along the lines of "The region R under the x-axis and above the line y=1", so isn't that simply two separate functions?

@Garmr: For the last one, I used chain rule so do/dt = do/dx * dx/dt = (the big fraction function) * 0.5?

@Rohan: Ahh right right I didn't remember that part, I hope they realise people found it ambiguous and award credit for both!

Also, I did laugh a bit when I got 47^2 - 14^2 = 2013 as the answer in the first part of that section B question, I bet the examiners did that on purpose

Also, for the weekly income, I got $165 and not $137, and here's what I did:

They said it was given that 3 requests happened in 2 days (mean =2.4), which had a certain probability (can't remember the exact value). Let's call this x.

Then, this could have happened in 4 ways: either 3 requests on one day then 0 the next (which would earn an income of 120), 0 on one day then 3 ($120), 1 on first day then 2 the next ($180) and 2 then 1 ($180)

Using P(A|B) = P(AUB)/P(B)

P (3 requests then 0| 3 requests) = the value we found in a) / x. I got this value as 0.125
P(0 then 3 | 3 total requests) = the value from a) / x
P(1 then 2| 3 total) = P(X=1) * P(X=2) (mean of 1.4)/x I got this value as 0.375
P(2 then 1| 3 total) = the same as above

Then, I did 2*0.125*$120 + 2*0.375*180 = $165. Idk if I'm right, but that's how I went about it

[race2013]

For the binomial one, I simply did P (X=1) + P(X=3) with n=16 and p=0.25, why would it be anything else?

Also for the volume, wasn't it simply pi* S f(x)^2 - g(x)^2 dx, where g(x) = 1? So my integral was:

pi * 0->pi/2 S (sin2x + 3) - 1^2 dx (the square over f(x) cancelled the root)

= pi * 0-> pi/2 S (sin2x +2) dx = 13.0 units^3

I'm kinda confused. For the binomial one, I did what u did but I also subtracted P(X=1)* P(X=3) after adding the two.

[Zuler]

For the binomial one, I simply did P (X=1) + P(X=3) with n=16 and p=0.25, why would it be anything else?

Also for the volume, wasn't it simply pi* S f(x)^2 - g(x)^2 dx, where g(x) = 1? So my integral was:

pi * 0->pi/2 S (sin2x + 3) - 1^2 dx (the square over f(x) cancelled the root)

= pi * 0-> pi/2 S (sin2x +2) dx = 13.0 units^3

I'm kinda confused. For the binomial one, I did what u did but I also subtracted P(X=1)* P(X=3) after adding the two.

Uhh, I see your point, but isn't there no need to do that since P(X=1) and P(X=3) are mutually exclusive events, and hence P(A intersect B) = 0?

[Garmr]

@Zuler: Did that as well, but I think they asked for a value for d0/dt at a certain value of t (which I forgot) and I got a really small answer (like something to the power of 10^-4) so I'm not sure if I did it right

As for the binomial question, yeah I hope examiners will realize the ambiguity and credit both.

[Christina perera]

@Christina: I get what you mean about the binomial, I guess we'll have to see Also for the volume, I remember the question said something along the lines of "The region R under the x-axis and above the line y=1", so isn't that simply two separate functions?

@Garmr: For the last one, I used chain rule so do/dt = do/dx * dx/dt = (the big fraction function) * 0.5?

@Rohan: Ahh right right I didn't remember that part, I hope they realise people found it ambiguous and award credit for both!

Also, I did laugh a bit when I got 47^2 - 14^2 = 2013 as the answer in the first part of that section B question, I bet the examiners did that on purpose

Also, for the weekly income, I got $165 and not $137, and here's what I did:

They said it was given that 3 requests happened in 2 days (mean =2.4), which had a certain probability (can't remember the exact value). Let's call this x.

Then, this could have happened in 4 ways: either 3 requests on one day then 0 the next (which would earn an income of 120), 0 on one day then 3 ($120), 1 on first day then 2 the next ($180) and 2 then 1 ($180)

Using P(A|B) = P(AUB)/P(B)

P (3 requests then 0| 3 requests) = the value we found in a) / x. I got this value as 0.125

P(0 then 3 | 3 total requests) = the value from a) / x

P(1 then 2| 3 total) = P(X=1) * P(X=2) (mean of 1.4)/x I got this value as 0.375

P(2 then 1| 3 total) = the same as above

Then, I did 2*0.125*$120 + 2*0.375*180 = $165. Idk if I'm right, but that's how I went about it

I am pretty sure you dint have to complicate it so much. It was straightforward. Total income from 3 requests = 180. This should be multiplied by the probability which was around 0.2 giving the answer in 30s.

[Zuler]

@Christina: I get what you mean about the binomial, I guess we'll have to see Also for the volume, I remember the question said something along the lines of "The region R under the x-axis and above the line y=1", so isn't that simply two separate functions?

@Garmr: For the last one, I used chain rule so do/dt = do/dx * dx/dt = (the big fraction function) * 0.5?

@Rohan: Ahh right right I didn't remember that part, I hope they realise people found it ambiguous and award credit for both!

Also, I did laugh a bit when I got 47^2 - 14^2 = 2013 as the answer in the first part of that section B question, I bet the examiners did that on purpose

Also, for the weekly income, I got $165 and not $137, and here's what I did:

They said it was given that 3 requests happened in 2 days (mean =2.4), which had a certain probability (can't remember the exact value). Let's call this x.

Then, this could have happened in 4 ways: either 3 requests on one day then 0 the next (which would earn an income of 120), 0 on one day then 3 ($120), 1 on first day then 2 the next ($180) and 2 then 1 ($180)

Using P(A|B) = P(AUB)/P(B)

P (3 requests then 0| 3 requests) = the value we found in a) / x. I got this value as 0.125

P(0 then 3 | 3 total requests) = the value from a) / x

P(1 then 2| 3 total) = P(X=1) * P(X=2) (mean of 1.4)/x I got this value as 0.375

P(2 then 1| 3 total) = the same as above

Then, I did 2*0.125*$120 + 2*0.375*180 = $165. Idk if I'm right, but that's how I went about it

I am pretty sure you dint have to complicate it so much. It was straightforward. Total income from 3 requests = 180. This should be multiplied by the probability which was around 0.2 giving the answer in 30s.

Ah, but the problem is, sometimes they can't service all their requests as stated in the question, so if it turned out that they got 3 requests one day and none the other, they could only service 2 and only earn $120. Or of course, if it happened that they got 1 request one day then 2 the next, then they could service all 3 and earn $180.

Also, they said it was given that they did receive 3 requests, so logically shouldn't the answer should be between $120 and $180, not in the low 30s?

Honestly I'm not feeling so good about the paper anymore, and I'm far from an authority on any of this