MAY13 Maths HL P1 TZ2 Discussion

[Giveme45]

Hahaha uhh cause when you put y=2 into the equation, the x will cancel out That's the rule for horizontal asymptote's isnt it ? try graphing it with your ti-84.

Oh wow I'm such a retard. I'm sorry I thought you ment the range of the function since we also had to calculate that. And I thought you said the upper bound would have been an asymptote.

Anyway, it is an asmyptote of the function for the domain of real numbers, but it was not required on the graph at all, and if you represented the asymptote as an upper limit, then you could actually lose marks, since the domain was restricted.

The reason for this is that the equation couldve been written 2-5/(x+2). So at infinite x, y=2. So that's why y would theoretically be a horizontal asymptote,

if the domain was all real numbers. Though the domain was restricted to x between -1 and 8. So on this domain, there is no horizontal asymptote.

Edited May 11, 2013 by Giveme45

[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

[bomaha]

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

Ask in the paper 2 Topic. It is now opened. But to answer your question, P(X=1)+P(X=3). For the volume question, i think i got pi^2 + pi.

[kinguncaged1]

There was one with a poisson distributions with cars getting rented out on the weekend or something. What was the answer to that one?!

[Giveme45]

Post paper 2 questions in the paper 2 thread. And the answer was 137 dollars.

[Christina perera]

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40

(60-v)dv/dt=1/40

Integrating both sides would give

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

so at t=0, v=0 gives c=-ln60.

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

so at t=30

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

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

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that

there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because

I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

which is the k=1 and k=3 you are talking about?

[MylakIB]

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40

(60-v)dv/dt=1/40

Integrating both sides would give

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

so at t=0, v=0 gives c=-ln60.

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

so at t=30

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

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

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that

there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because

I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

What do u mean by cats=28?

[Giveme45]

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40

(60-v)dv/dt=1/40

Integrating both sides would give

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

so at t=0, v=0 gives c=-ln60.

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

so at t=30

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

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

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that

there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because

I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

which is the k=1 and k=3 you are talking about?

The constants of integration for question 12. Im sure about the k=1, dont remember if i actually put k=3 for the second part though, perhaps i put something else.

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40

(60-v)dv/dt=1/40

Integrating both sides would give

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

so at t=0, v=0 gives c=-ln60.

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

so at t=30

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

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

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that

there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because

I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

What do u mean by cats=28?

There was a normal distribution question about asking how many cats out of 10000 have a lifespan of smaller than 10. The answer was 28.

[MylakIB]

Hah

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40
(60-v)dv/dt=1/40
Integrating both sides would give
-ln|60-v|=t/40+c
so at t=0, v=0 gives c=-ln60.
We have ln|60-v|=ln60-t/40
so at t=30
ln|60-v|=ln60-3/4
|60-v|=e^(ln60-3/4)

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that
there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because
I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

which is the k=1 and k=3 you are talking about?

The constants of integration for question 12. Im sure about the k=1, dont remember if i actually put k=3 for the second part though, perhaps i put something else.

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40
(60-v)dv/dt=1/40
Integrating both sides would give
-ln|60-v|=t/40+c
so at t=0, v=0 gives c=-ln60.
We have ln|60-v|=ln60-t/40
so at t=30
ln|60-v|=ln60-3/4
|60-v|=e^(ln60-3/4)

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that
there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because
I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

What do u mean by cats=28?

There was a normal distribution question about asking how many cats out of 10000 have a lifespan of smaller than 10. The answer was 28.

You mean by that, that the probability that the cat had a lifespan smaller than 10 was 0.0028? Well, having a normally distributed data with variance of 9.5 an mean of 13.5 the probability is about 0.1281

[Giveme45]

Hah

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40

(60-v)dv/dt=1/40

Integrating both sides would give

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

so at t=0, v=0 gives c=-ln60.

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

so at t=30

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

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

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that

there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because

I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

which is the k=1 and k=3 you are talking about?

The constants of integration for question 12. Im sure about the k=1, dont remember if i actually put k=3 for the second part though, perhaps i put something else.

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40

(60-v)dv/dt=1/40

Integrating both sides would give

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

so at t=0, v=0 gives c=-ln60.

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

so at t=30

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

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

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that

there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because

I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

What do u mean by cats=28?

There was a normal distribution question about asking how many cats out of 10000 have a lifespan of smaller than 10. The answer was 28.

You mean by that, that the probability that the cat had a lifespan smaller than 10 was 0.0028? Well, having a normally distributed data with variance of 9.5 an mean of 13.5 the probability is about 0.1281

Variance was 9.5². -> GDC input 10000*normalcdf(0,10,13.5,9.5)=2.79*10^3 cats. Sry about the 28.

Edited May 12, 2013 by Giveme45

[MylakIB]

It was for sure 9.5 ( so that the range was 13.5 plus minus sqrt 9.5) (it was 9.5 years^2)

[Giveme45]

It was for sure 9.5 ( so that the range was 13.5 plus minus sqrt 9.5) (it was 9.5 years^2)

How sure are you? I could swear it was 9.5², and someone in the other topic also said the range was about 16 years, which would go with the 9.5². Often in normal distribution questions they give Variance to be a², so that the students can directly see the standart deviation.

Also: do you remember the phrasing of the second part of the question? Did they ask for the number of cats out of 10000 cats that have a lifespan of less than 10?

Edited May 12, 2013 by Giveme45

[MylakIB]

I'm sure that it was 9.5, because they wanted to see whether we can figure out that we need to take square root of this in order to calculate the range...

Edit: they asked about number of cats (out of 10000) that have lifespan smaller than 10 yrs.

Edited May 12, 2013 by MylakIB

[Giveme45]

I'm sure that it was 9.5, because they wanted to see whether we can figure out that we need to take square root of this in order to calculate the range...

I dunno, that does not really reflect the questions of the pastpapers. Also a range of 16 seems to be more realistic to be the answer than 6.16 imo. Well who knows, i feel like i messed up on the second part because i answered 28 instead of 2790, maybe i just forgot.

[Garmr]

I'm sure that it was 9.5, because they wanted to see whether we can figure out that we need to take square root of this in order to calculate the range...

I dunno, that does not really reflect the questions of the pastpapers. Also a range of 16 seems to be more realistic to be the answer than 6.16 imo. Well who knows, i feel like i messed up on the second part because i answered 28 instead of 2790, maybe i just forgot.

I think this is another example of how IB wrote the paper so confusingly this year, because I got puzzled at that part too. I went with what MylakIB did since they said "variance 9.5 years^2", so I thought the SD would be sqrt(9.5) years.

EDIT: Actually, I just thought about it and maybe by that they meant variance = (9.5 years)^2. If so, then darn you IB why couldn't you have written 9.5^2 years^2?

Edited May 12, 2013 by Garmr

[Giveme45]

I'm sure that it was 9.5, because they wanted to see whether we can figure out that we need to take square root of this in order to calculate the range...

I dunno, that does not really reflect the questions of the pastpapers. Also a range of 16 seems to be more realistic to be the answer than 6.16 imo. Well who knows, i feel like i messed up on the second part because i answered 28 instead of 2790, maybe i just forgot.

I think this is another example of how IB wrote the paper so confusingly this year, because I got puzzled at that part too. I went with what MylakIB did since they said "variance 9.5 years^2", so I thought the SD would be sqrt(9.5) years.

EDIT: Actually, I just thought about it and maybe by that they meant variance = (9.5 years)^2. If so, then darn you IB why couldn't you have written 9.5^2 years^2?

THIS. So confusing IB this year. Seriously...

[demure]

Can somebody please specify which countries belong to the Time Zone 2 set up by the IBO?

Edited May 15, 2013 by demure

[Mokk]

Can somebody please specify which countries belong to the Time Zone 2 set up by the IBO?

As far as I know, Time Zone 2 includes Europe, the Middle East and Asia. I'll assume that limits TZ1 to North America, since Australia and South America usually take the November exams.

[JohnSmith]

I'm sure that it was 9.5, because they wanted to see whether we can figure out that we need to take square root of this in order to calculate the range...

I dunno, that does not really reflect the questions of the pastpapers. Also a range of 16 seems to be more realistic to be the answer than 6.16 imo. Well who knows, i feel like i messed up on the second part because i answered 28 instead of 2790, maybe i just forgot.

I think this is another example of how IB wrote the paper so confusingly this year, because I got puzzled at that part too. I went with what MylakIB did since they said "variance 9.5 years^2", so I thought the SD would be sqrt(9.5) years.

EDIT: Actually, I just thought about it and maybe by that they meant variance = (9.5 years)^2. If so, then darn you IB why couldn't you have written 9.5^2 years^2?

Guys there is no way they meant 9.5^2, the SD was definitely sort(9.5)...

[rinik]

I am currently trying the paper 1 complex number question. I Can't answer the proof and and the 1+w argument. Can somebody walk me trough them?